Therefore $(\mathfrak{p},I)\models(r,a,\gamma,\delta)$ if and only if
$(\mathfrak{p},I)\models(r,rv,\gamma\alpha,\delta\alpha)$ or
$(\mathfrak{p},I)\models(a,a,\gamma(\alpha-1)v,\delta(\alpha-1)v)$. The same
argument shows that $(\mathfrak{p},I)\models(r,a,\delta,\gamma)$ if and only
if $(\mathfrak{p},I)\models(r,rv,\delta\alpha,\gamma\alpha)$ or
$(\mathfrak{p},I)\models(a,a,\delta(\alpha-1)v,\gamma(\alpha-1)v)$. ∎

######  Proposition 6.10.

Given an auxiliary sentence $\Xi$ and $\lambda\in R\backslash\\{0\\}$, we can
compute $n\in\mathbb{N}$ and for $1\leq j\leq n$,
$(r_{j},r_{j}a_{j},\gamma_{j},\delta_{j})\in R^{4}$ such that, for all prime
ideals $\mathfrak{p}\lhd R$ and ideals $I\lhd R_{\mathfrak{p}}$,

  * •

$(\mathfrak{p},I)\models(r_{j},r_{j}a_{j},\gamma_{j},\delta_{j})$ for some
$1\leq j\leq n$ if and only if $R_{\mathfrak{p}}/\lambda I\models\Xi$, and

  * •

$(\mathfrak{p},I)\models(r_{j},r_{j}a_{j},\delta_{j},\gamma_{j})$ for some
$1\leq j\leq n$ if and only if $R_{\mathfrak{p}}/\lambda I\models D\Xi$

###### Proof.

Let $\Xi$ be the sentence

$\bigwedge_{i=1}^{l^{\prime}}\left|\nicefrac{{d_{i}|x}}{{x=0}}\right|=1\wedge\bigwedge_{i=l^{\prime}}^{l}\left|\nicefrac{{xb_{i}=0}}{{c_{i}|x}}\right|=1.$

Using 2.15 and 6.6, we can compute $n_{i}\in\mathbb{N}$ for $1\leq i\leq l$
and $s_{ij},b_{ij},g_{ij},h_{ij}\in R$ for $1\leq i\leq l$ and $1\leq j\leq
n_{i}$ such that for all prime ideals $\mathfrak{p}\lhd R$ and ideals $I\lhd
R_{\mathfrak{p}}$,

  * •

$R_{\mathfrak{p}}/I\models\Xi$ if and only if
$(\mathfrak{p},I)\models\bigwedge_{i=1}^{l}\bigvee_{j=1}^{n_{i}}(s_{ij},b_{ij},g_{ij},h_{ij})$,
and

  * •

$R_{\mathfrak{p}}/I\models D\Xi$ if and only if
$(\mathfrak{p},I)\models\bigwedge_{i=1}^{l}\bigvee_{j=1}^{n_{i}}(s_{ij},b_{ij},h_{ij},g_{ij})$.

Therefore, for all prime ideals $\mathfrak{p}\lhd R$ and ideals $I\lhd
R_{\mathfrak{p}}$,

  * •

$R_{\mathfrak{p}}/I\models\Xi$ if and only if

$(\mathfrak{p},I)\models\bigvee_{\begin{subarray}{c}\sigma:\\{1,\ldots,l\\}\rightarrow\mathbb{N}\\\
\sigma(i)\leq n_{i}\end{subarray}}\
\bigwedge_{i=1}^{l}(s_{i\sigma(i)},b_{i\sigma(i)},g_{i\sigma(i)},h_{i\sigma(i)}),\text{
and}$

  * •

$R_{\mathfrak{p}}/I\models D\Xi$ if and only if

$(\mathfrak{p},I)\models\bigvee_{\begin{subarray}{c}\sigma:\\{1,\ldots,l\\}\rightarrow\mathbb{N}\\\
\sigma(i)\leq n_{i}\end{subarray}}\
\bigwedge_{i=1}^{l}(s_{i\sigma(i)},b_{i\sigma(i)},h_{i\sigma(i)},g_{i\sigma(i)}).$

We can use 6.7, to replace the conjunction
$\bigwedge_{i=1}^{l}(s_{i\sigma(i)},b_{i\sigma(i)},g_{i\sigma(i)},h_{i\sigma(i)})$,
for each $\sigma$, by a disjunction to produce and $(s^{\prime}_{\sigma
k},b^{\prime}_{\sigma k},g^{\prime}_{\sigma k},\delta^{\prime}_{\sigma k})$
for $1\leq k\leq m_{\sigma}$ such that

  * •

$R_{\mathfrak{p}}/I\models\Xi$ if and only if
$(\mathfrak{p},I)\models(s^{\prime}_{\sigma k},b^{\prime}_{\sigma
k},g^{\prime}_{\sigma k},h^{\prime}_{\sigma k})$ for some $1\leq k\leq
m_{\sigma}$, and

  * •

$R_{\mathfrak{p}}/I\models D\Xi$ if and only if
$(\mathfrak{p},I)\models(s^{\prime}_{\sigma k},b^{\prime}_{\sigma
k},h^{\prime}_{\sigma k},g^{\prime}_{\sigma k})$ for some $1\leq k\leq
m_{\sigma}$.

Applying 6.8 to each $(s^{\prime}_{\sigma k},b^{\prime}_{\sigma
k},h^{\prime}_{\sigma k},g^{\prime}_{\sigma k})$, we compute
$(r^{\prime}_{j},a^{\prime}_{j},\gamma^{\prime}_{j},\delta^{\prime}_{j})$ for
$1\leq j\leq m$ such that

  * •

$R_{\mathfrak{p}}/\lambda I\models\Xi$ if and only if
$(\mathfrak{p},I)\models(r^{\prime}_{j},a^{\prime}_{j},\gamma^{\prime}_{j},\delta^{\prime}_{j})$
for some $1\leq j\leq m$, and

  * •

$R_{\mathfrak{p}}/\lambda I\models D\Xi$ if and only if
$(\mathfrak{p},I)\models(r^{\prime}_{j},a^{\prime}_{j},\delta^{\prime}_{j},\gamma^{\prime}_{j})$
for some $1\leq j\leq m$.

Finally, applying 6.9 to each
$(r^{\prime}_{j},a^{\prime}_{j},\gamma^{\prime}_{j},\delta^{\prime}_{j})$, we
can compute $(r_{i},r_{i}a_{i},\gamma_{i},\delta_{i})$ for $1\leq i\leq n$
such that

  * •

$R_{\mathfrak{p}}/\lambda I\models\Xi$ if and only if
$(\mathfrak{p},I)\models(r_{i},r_{i}a_{i},\gamma_{i},\delta_{i})$ for some
$1\leq i\leq n$, and

  * •

$R_{\mathfrak{p}}/\lambda I\models D\Xi$ if and only if
$(\mathfrak{p},I)\models(r_{i},r_{i}a_{i},\delta_{i},\gamma_{i})$ for some
$1\leq i\leq n$. ∎

###  6.2. Simplification of
$\nicefrac{{\varphi}}{{\psi}}(R_{\mathfrak{p}}/\lambda I)$ and
$\nicefrac{{\varphi}}{{\psi}}(I/\lambda R_{\mathfrak{p}})$

The results of this subsection will be used in sections 7 and 8. In this
subsection, we no longer need to worry about stability under duality.

######  Remark 6.11.

Let $a,b\in R$ and $\alpha,u,v\in R$ be such that $a\alpha=bu$ and
$b(\alpha-1)=av$. For prime ideals $\mathfrak{p}\lhd R$ and ideals $I\lhd
R_{\mathfrak{p}}$,

  * •

$a\in bI$ if and only if $(\mathfrak{p},I)\models(1,u,\alpha,1)$ or
$(\mathfrak{p},I)\models(1,1,v(\alpha-1),1)$, and,

  * •

$aR_{\mathfrak{p}}\supseteq bI$ if and only if
$(\mathfrak{p},I)\models(u,0,\alpha,1)$ or
$(\mathfrak{p},I)\models(1,0,\alpha-1,1)$.

######  Lemma 6.12.

Let $R$ be a recursive Prüfer domain and $\lambda\in R\backslash\\{0\\}$.

  1. (a)

Given $d\in R$, we can compute finite sets $S_{1},S_{2},S_{3}\subseteq R^{4}$,
$\rho:\bigcup_{i=1}^{3}S_{i}\rightarrow\\{1,2,3\\}$ and
$s:\bigcup_{i=1}^{3}S_{i}\rightarrow R$ such that for all
$q\in\bigcup_{i=1}^{3}S_{i}$, $q\in S_{\rho(q)}$, and, for all prime ideals
$\mathfrak{p}\lhd R$ and ideals $I\lhd R_{\mathfrak{p}}$, there exists
$q\in\bigcup_{i=1}^{3}S_{i}$ such that $(\mathfrak{p},I)\models q$, and

$\left|\nicefrac{{d|x}}{{x=0}}(R_{\mathfrak{p}}/\lambda
I)\right|:=\left\\{\begin{array}[]{ll}\left|R_{\mathfrak{p}}/s(q)I\right|,&\hbox{if
$(\mathfrak{p},I)\models q$ and $\rho(q)=1$;}\\\
\left|s(q)R_{\mathfrak{p}}/I\right|,&\hbox{if $(\mathfrak{p},I)\models q$ and
$\rho(q)=2$;}\\\ 1,&\hbox{if $(\mathfrak{p},I)\models q$ and
$\rho(q)=3$.}\end{array}\right.$

Furthermore, if $(\mathfrak{p},I)\models q$ for some
$q\in\bigcup_{i=1}^{3}S_{i}$ and $\rho(q)=2$ then
$s(q)R_{\mathfrak{p}}\supseteq I$.

  2. (b)

Given $b,c\in R$, we can compute finite sets $S_{1},\ldots,S_{5}\subseteq
R^{4}$, $\rho:\bigcup_{i=1}^{5}S_{i}\rightarrow\\{1,\ldots,5\\}$ and
$s:\bigcup_{i=1}^{5}S_{i}\rightarrow R$ such that for all
$q\in\bigcup_{i=1}^{5}S_{i}$, $q\in S_{\rho(q)}$, and, for all prime ideals
$\mathfrak{p}\lhd R$ and ideals $I\lhd R_{\mathfrak{p}}$, there exists
$q\in\bigcup_{i=1}^{5}S_{i}$ such that $(\mathfrak{p},I)\models q$, and

$\left|\nicefrac{{xb=0}}{{c|x}}(R_{\mathfrak{p}}/\lambda
I)\right|:=\left\\{\begin{array}[]{ll}\left|R_{\mathfrak{p}}/\lambda
I\right|,&\hbox{if $(\mathfrak{p},I)\models q$ and $\rho(q)=1$;}\\\
\left|R_{\mathfrak{p}}/cR_{\mathfrak{p}}\right|,&\hbox{if
$(\mathfrak{p},I)\models q$ and $\rho(q)=2$;}\\\ \left|I/bI\right|,&\hbox{if
$(\mathfrak{p},I)\models q$ and $\rho(q)=3$;}\\\ 1,&\hbox{if
$(\mathfrak{p},I)\models q$ and $\rho(q)=4$;}\\\
\left|I/s(q)R_{\mathfrak{p}}\right|,&\hbox{if $(\mathfrak{p},I)\models q$ and
$\rho(q)=5$.}\end{array}\right.$

Furthermore, if $(\mathfrak{p},I)\models q$ for some
$q\in\bigcup_{i=1}^{5}S_{i}$ and $\rho(q)=5$ then $s(q)\in I$. Moreover, if
$b=0$ then we may assume that $S_{3}=S_{4}=S_{5}=\emptyset$.

###### Proof.

(a) If $d=0$ then $\nicefrac{{d|x}}{{x=0}}(N)=0$ for all $R$-modules $N$. So
set $S_{1}=S_{2}=\emptyset$ and $S_{3}:=\\{(1,0,1,1)\\}$.

Suppose $d\neq 0$. Let $\alpha,u,v\in R$ be such that $d\alpha=\lambda u$ and
$\lambda(\alpha-1)=dv$.

  * •

If $(\mathfrak{p},I)\models q_{1}:=(1,0,\alpha-1,1)$ then

$\left|\nicefrac{{d|x}}{{x=0}}(R_{\mathfrak{p}}/\lambda
I)\right|=\left|dR_{\mathfrak{p}}/dvI\right|=\left|R_{\mathfrak{p}}/vI\right|.$

  * •

If $(\mathfrak{p},I)\models q_{2}:=(u,0,\alpha,1)$ then by definition
$uR_{\mathfrak{p}}\supseteq I$, and,

$\left|\nicefrac{{d|x}}{{x=0}}(R_{\mathfrak{p}}/\lambda
I)\right|=\left|\lambda uR_{\mathfrak{p}}/\lambda
I\right|=\left|uR_{\mathfrak{p}}/I\right|.$

  * •

If $(\mathfrak{p},I)\models q_{3}:=(1,u,\alpha,1)$ then $d\in\lambda I$ and
hence $\left|\nicefrac{{d|x}}{{x=0}}(R_{\mathfrak{p}}/\lambda I)\right|=1$.

For all prime ideals $\mathfrak{p}\lhd R$, either $\alpha\notin\mathfrak{p}$
or $\alpha-1\notin\mathfrak{p}$, and, for all ideal $I\lhd R_{\mathfrak{p}}$,
either $uR_{\mathfrak{p}}\supseteq I$ or $u\in I$. Therefore for all prime
ideals $\mathfrak{p}\lhd R$ and ideals $I\lhd R_{\mathfrak{p}}$, either
$(\mathfrak{p},I)\models q_{1}$, $(\mathfrak{p},I)\models q_{2}$ or
$(\mathfrak{p},I)\models q_{3}$. So, set $S_{i}:=\\{q_{i}\\}$ for $1\leq i\leq
3$, and, set $s(q_{1}):=v$, $s(q_{2}):=u$ and $s(q_{3}):=1$.

(b) For all prime ideals $\mathfrak{p}\lhd R$ and ideals $I\lhd
R_{\mathfrak{p}}$,

$\left|\nicefrac{{xb=0}}{{c|x}}(R_{\mathfrak{p}}/\lambda
I)\right|=\left|\frac{(\lambda
I:b)+cR_{\mathfrak{p}}}{cR_{\mathfrak{p}}+\lambda I}\right|.$

First suppose $b=0$. If $c\in\lambda I$ then
$\left|\nicefrac{{xb=0}}{{c|x}}(R_{\mathfrak{p}}/\lambda
I)\right|=\left|R_{\mathfrak{p}}/\lambda I\right|$, and, if
$cR_{\mathfrak{p}}\supseteq\lambda I$ then
$\left|\nicefrac{{xb=0}}{{c|x}}(R_{\mathfrak{p}}/\lambda
I)\right|=\left|R_{\mathfrak{p}}/cR_{\mathfrak{p}}\right|$. We can use 6.11 to
compute finite sets $S_{1},S_{2}\subseteq R^{4}$ such that for all prime
ideals $\mathfrak{p}\lhd R$ and ideals $I\lhd R_{\mathfrak{p}}$, there exists
$q\in S_{1}$ such that $(\mathfrak{p},I)\models q$ if and only if $c\in\lambda
I$, and, there exists $q\in S_{2}$ such that $(\mathfrak{p},I)\models q$ if
and only if $cR_{\mathfrak{p}}\supseteq\lambda I$. Set
$S_{3}=S_{4}=S_{5}=\emptyset$.

Now suppose $b\neq 0$. If $bR_{\mathfrak{p}}\supseteq\lambda I$ then

$\left|\frac{(\lambda I:b)+cR_{\mathfrak{p}}}{cR_{\mathfrak{p}}+\lambda
I}\right|=\left|\frac{\lambda
I+bcR_{\mathfrak{p}}}{bcR_{\mathfrak{p}}+b\lambda I}\right|.$

So

$\left|\nicefrac{{xb=0}}{{c|x}}(R_{\mathfrak{p}}/\lambda
I)\right|:=\left\\{\begin{array}[]{ll}\left|R_{\mathfrak{p}}/\lambda
I\right|,&\hbox{if $b,c\in\lambda I$;}\\\
\left|R_{\mathfrak{p}}/cR_{\mathfrak{p}}\right|,&\hbox{if $b\in\lambda I$ and
$cR_{\mathfrak{p}}\supseteq\lambda I$;}\\\ \left|I/bI\right|,&\hbox{if
$bR_{\mathfrak{p}}\supseteq\lambda I$ and $c\in\lambda I$;}\\\ 1,&\hbox{if
$bcR_{\mathfrak{p}}\supseteq\lambda I$;}\\\ \left|\lambda
I/bcR_{\mathfrak{p}}\right|,&\hbox{if $bR_{\mathfrak{p}}\supseteq\lambda I$,
$cR_{\mathfrak{p}}\supseteq\lambda I$ and $bc\in\lambda
I$.}\end{array}\right.$

Therefore it is enough to compute:

  * •

$S_{1}$ such that $b,c\in\lambda I$ if and only if there exists $q\in S_{1}$
such that $(\mathfrak{p},I)\models q$,

  * •

$S_{2}$ such that $b\in\lambda I$ and $cR_{\mathfrak{p}}\supseteq\lambda I$ if
and only if exists $q\in S_{2}$ such that $(\mathfrak{p},I)\models q$,

  * •

$S_{3}$ such that $bR_{\mathfrak{p}}\supseteq\lambda I$ and $c\in\lambda I$
and if and only if exists $q\in S_{3}$ such that $(\mathfrak{p},I)\models q$,

  * •

$S_{4}$ such that $bcR_{\mathfrak{p}}\supseteq\lambda I$ if and only if there
exists $q\in S_{4}$ such that $(\mathfrak{p},I)\models q$, and,

  * •

$S_{5}$ and for each $q\in S_{5}$, $s_{q}\in R$ such that
$bR_{\mathfrak{p}}\supseteq\lambda I$, $cR_{\mathfrak{p}}\supseteq\lambda I$
and $bc\in\lambda I$ if and only if there exists $q\in S_{5}$ such that
$(\mathfrak{p},I)\models q$ and such that, in this situation
$\left|\nicefrac{{xb=0}}{{c|x}}(R_{\mathfrak{p}}/\lambda
I)\right|=\left|I/s_{q}R_{\mathfrak{p}}\right|$.

It is easy to compute $S_{1},\ldots,S_{4}$ using 6.11 and 6.7.

Let $\epsilon,r,s$ be such that $bc\epsilon=\lambda s$ and
$\lambda(\epsilon-1)=bcr$. By 6.11, $bc\in\lambda I$ if and only if
$(\mathfrak{p},I)\models(1,s,\epsilon,1)$ or
$(\mathfrak{p},I)\models(1,1,r(\epsilon-1),1)$. If
$bR_{\mathfrak{p}}\supseteq\lambda I$, $cR_{\mathfrak{p}}\supseteq\lambda I$
and $(\mathfrak{p},I)\models(1,s,\epsilon,1)$ then

$\left|\nicefrac{{xb=0}}{{c|x}}(R_{\mathfrak{p}}/\lambda
I)\right|=\left|\lambda
I/bcR_{\mathfrak{p}}\right|=\left|I/sR_{\mathfrak{p}}\right|.$

Use 6.11 and 6.7 to compute $S_{5}^{\prime}$ such that
$(\mathfrak{p},I)\models q$ for some $q\in S_{5}^{\prime}$ if and only if
$bR_{\mathfrak{p}}\supseteq\lambda I$, $cR_{\mathfrak{p}}\supseteq\lambda I$
and $(\mathfrak{p},I)\models(1,s,\epsilon,1)$.

If $(\mathfrak{p},I)\models(1,1,r(\epsilon-1),1)$ then $I=R_{\mathfrak{p}}$
and

$\left|\nicefrac{{xb=0}}{{c|x}}(R_{\mathfrak{p}}/\lambda
I)\right|=\left|\lambda
I/bcR_{\mathfrak{p}}\right|=\left|I/1R_{\mathfrak{p}}\right|.$

Let $S_{5}^{\prime\prime}:=\\{(1,1,r(\epsilon-1),1)\\}$ and let
$S_{5}:=S_{5}^{\prime}\cup S_{5}^{\prime\prime}$. Set $s(q):=s$ if $s\in
S_{5}^{\prime}$ and $s(q):=1$ otherwise. Note that in both cases, by
definition, $s(q)\in I$. ∎

######  Proposition 6.13.

Let $R$ be a recursive Prüfer domain, $\lambda\in R\backslash\\{0\\}$ and $Z$
a finite subset of pp-pairs of the form $\nicefrac{{xb=0}}{{c|x}}$ and
$\nicefrac{{d|x}}{{x=0}}$. Let

$T_{Z}:=\\{\mu:Z\rightarrow\\{1,\ldots,5\\}\ |\ \text{ for all
}\nicefrac{{d|x}}{{x=0}}\in Z\ \ \mu(\nicefrac{{d|x}}{{x=0}})\leq 3\\}.$

We can compute $S_{Z}$ a finite subset of $R^{4}$, $\rho_{Z}:S_{Z}\rightarrow
T_{Z}$ and $s_{Z}:S_{Z}\times Z\rightarrow R$ such that for all prime ideals
$\mathfrak{p}\lhd R$ and ideals $I\lhd R_{\mathfrak{p}}$,

  1. (a)

there exists $q\in S_{Z}$ such that $(\mathfrak{p},I)\models q$,

  2. (b)

    1. (1)

if $\rho_{Z}(q)(\nicefrac{{d|x}}{{x=0}})=1$ then
$\left|\nicefrac{{d|x}}{{x=0}}(R_{\mathfrak{p}}/\lambda
I)\right|=\left|R_{\mathfrak{p}}/s_{Z}(q,\nicefrac{{d|x}}{{x=0}})I\right|$

    2. (2)

if $\rho_{Z}(q)(\nicefrac{{d|x}}{{x=0}})=2$ then
$\left|\nicefrac{{d|x}}{{x=0}}(R_{\mathfrak{p}}/\lambda
I)\right|=\left|s_{Z}(q,\nicefrac{{d|x}}{{x=0}})R_{\mathfrak{p}}/I\right|$

    3. (3)

if $\rho_{Z}(q)(\nicefrac{{d|x}}{{x=0}})=3$ then
$\left|\nicefrac{{d|x}}{{x=0}}(R_{\mathfrak{p}}/\lambda I)\right|=1$

  3. (c)

    1. (1)

if $\rho_{Z}(q)(\nicefrac{{xb=0}}{{c|x}})=1$ then
$\left|\nicefrac{{xb=0}}{{c|x}}(R_{\mathfrak{p}}/\lambda
I)\right|=\left|R_{\mathfrak{p}}/\lambda I\right|$

    2. (2)

if $\rho_{Z}(q)(\nicefrac{{xb=0}}{{c|x}})=2$ then
$\left|\nicefrac{{xb=0}}{{c|x}}(R_{\mathfrak{p}}/\lambda
I)\right|=\left|R_{\mathfrak{p}}/cR_{\mathfrak{p}}\right|$

    3. (3)

if $\rho_{Z}(q)(\nicefrac{{xb=0}}{{c|x}})=3$ then
$\left|\nicefrac{{xb=0}}{{c|x}}(R_{\mathfrak{p}}/\lambda
I)\right|=\left|I/bI\right|$

    4. (4)

if $\rho_{Z}(q)(\nicefrac{{xb=0}}{{c|x}})=4$ then
$\left|\nicefrac{{xb=0}}{{c|x}}(R_{\mathfrak{p}}/\lambda I)\right|=1$

    5. (5)

if $\rho_{Z}(q)(\nicefrac{{xb=0}}{{c|x}})=5$ then
$\left|\nicefrac{{xb=0}}{{c|x}}(R_{\mathfrak{p}}/\lambda
I)\right|=\left|I/s_{Z}(q,\nicefrac{{xb=0}}{{c|x}})R_{\mathfrak{p}}\right|$

Furthermore, for all prime ideals $\mathfrak{p}\lhd R$ and ideals $I\lhd
R_{\mathfrak{p}}$, if $(\mathfrak{p},I)\models q$ and
$\rho_{Z}(q)(\nicefrac{{d|x}}{{x=0}})=2$ then
$s_{Z}(q,\nicefrac{{d|x}}{{x=0}})R_{\mathfrak{p}}\supseteq I$, and, if
$(\mathfrak{p},I)\models q$ and $\rho_{Z}(q)(\nicefrac{{xb=0}}{{c|x}})=5$ then
$s_{Z}(q,\nicefrac{{xb=0}}{{c|x}})\in I$. Moreover, if $b=0$ then we may
assume $\rho_{Z}(q)(\nicefrac{{xb=0}}{{c|x}})\in\\{1,2\\}$ for all $q\in
S_{Z}$.

###### Proof.

We prove the proposition iteratively. Let $\lambda\in R\backslash\\{0\\}$ and
$Z$ be a finite set of pp-pairs of the form $\nicefrac{{xb=0}}{{c|x}}$ or
$\nicefrac{{d|x}}{{x=0}}$. If $|Z|=1$ then 6.12 gives the required result.
Suppose that $S_{Z},\rho_{Z}$ and $s_{Z}$ are as in the statement. We
construct $S_{Z^{\prime}},\rho_{Z^{\prime}}$ and $s_{Z^{\prime}}$ for
$Z^{\prime}=Z\cup\\{\nicefrac{{\varphi}}{{\psi}}\\}$.

Suppose that $\nicefrac{{\varphi}}{{\psi}}$ is either of the form
$\nicefrac{{d|x}}{{x=0}}$ or $\nicefrac{{xb=0}}{{c|x}}$ and
$\nicefrac{{\varphi}}{{\psi}}\notin Z$. Let $S_{1},\ldots,S_{5}$,
$s:\bigcup_{i=1}^{5}S_{i}\rightarrow R$ and
$\rho:\bigcup_{i=1}^{5}S_{i}\rightarrow\\{1,\ldots,5\\}$ be as in 6.12 (1) or
(2), as appropriate (if $\nicefrac{{\varphi}}{{\psi}}$ is of the form
$\nicefrac{{d|x}}{{x=0}}$ then set $S_{4}=S_{5}=\emptyset$). By 6.7, for each
$q\in S_{Z}$ and $p\in S_{i}$, we can compute a finite set $S_{q,p,i}\subseteq
R^{4}$ such that for all prime ideals $\mathfrak{p}\lhd R$ and ideals $I\lhd
R_{\mathfrak{p}}$, $(\mathfrak{p},I)\models q$ and $(\mathfrak{p},I)\models p$
if and only if there exists $q^{\prime}\in S_{q,p,i}$ such that
$(\mathfrak{p},I)\models q^{\prime}$. Let $S_{q,i}:=\bigcup_{p\in
S_{i}}S_{q,p,i}$, let $S_{q}:=\bigcup_{i=1}^{5}S_{q,i}$ and let
$S_{Z^{\prime}}:=\bigcup_{q\in S_{Z}}S_{q}$.

For all prime ideals $\mathfrak{p}\lhd R$ and ideals $I\lhd R_{\mathfrak{p}}$,
by assumption, there exists $q\in S_{Z}$ such that $(\mathfrak{p},I)\models q$
and, by 6.12, there exists $1\leq i\leq 5$, such that $(\mathfrak{p},I)\models
q^{\prime}$ for some $q^{\prime}\in S_{i}$. Therefore,
$(\mathfrak{p},I)\models q^{\prime\prime}$ for some $q^{\prime\prime}\in
S_{q,i}\subseteq S_{Z^{\prime}}$. So (a) holds for $S_{Z^{\prime}}$.

Define $\rho_{Z^{\prime}}:S_{Z^{\prime}}\rightarrow T_{Z^{\prime}}$ by setting

$\rho_{Z^{\prime}}(q^{\prime})(\nicefrac{{\sigma}}{{\tau}}):=\left\\{\begin{array}[]{ll}\min\\{\rho_{Z}(q)(\nicefrac{{\sigma}}{{\tau}})\
|\ q^{\prime}\in S_{q}\\},&\hbox{if $\nicefrac{{\sigma}}{{\tau}}\in Z$;}\\\
\min\\{1\leq i\leq 5\ |\ q^{\prime}\in S_{q,i}\text{ for some }q\in
S_{Z}\\},&\hbox{ if
$\nicefrac{{\sigma}}{{\tau}}=\nicefrac{{\varphi}}{{\psi}}$.}\end{array}\right.$

For $\sigma/\tau\in Z$, set
$s_{Z^{\prime}}(q^{\prime},\nicefrac{{\sigma}}{{\tau}})$ to be
$s_{Z}(q,\sigma/\tau)$ for some $q\in S_{Z}$ where $q^{\prime}\in S_{q}$ and
$\rho_{Z^{\prime}}(q^{\prime})(\nicefrac{{\sigma}}{{\tau}})=\rho_{Z}(q)(\nicefrac{{\sigma}}{{\tau}})$.
For $q^{\prime}\in S_{Z^{\prime}}$, set
$s_{Z^{\prime}}(q^{\prime},\nicefrac{{\varphi}}{{\psi}})$ to be $s(p)$ for
some $p\in S_{j}$ where
$j=\rho_{Z^{\prime}}(q^{\prime})(\nicefrac{{\varphi}}{{\psi}})$ and
$q^{\prime}\in S_{q,p,j}$.

Now, for $\nicefrac{{\sigma}}{{\tau}}\in Z$, properties (b) or (c), as
appropriate, are inherited from those properties holding for $\rho_{Z}$ and
$s_{Z}$ and for $\nicefrac{{\varphi}}{{\psi}}$, properties (b) or (c), as
appropriate, are inherited from $\rho$ and $s$. ∎

######  Lemma 6.14.

Let $R$ be a recursive Prüfer domain and $\lambda\in R\backslash\\{0\\}$.

  1. (a)

Given $b,c\in R$, we can compute finite sets $S_{1},\ldots,S_{6}\subseteq
R^{4}$, $\rho:\bigcup_{i=1}^{6}S_{i}\rightarrow\\{1,\ldots,6\\}$ and
$s:\bigcup_{i=1}^{5}S_{i}\rightarrow R$ such that for all
$q\in\bigcup_{i=1}^{6}S_{i}$, $q\in S_{\rho(q)}$, for all prime ideals
$\mathfrak{p}\lhd R$ and ideals $I\lhd R_{\mathfrak{p}}$ with $\lambda\in I$,
there exists $q\in\bigcup_{i=1}^{6}S_{i}$ such that $(\mathfrak{p},I)\models
q$, and,

$\left|\nicefrac{{xb=0}}{{c|x}}(I/\lambda
R_{\mathfrak{p}})\right|:=\left\\{\begin{array}[]{ll}\left|I/\lambda
R_{\mathfrak{p}}\right|,&\hbox{if $(\mathfrak{p},I)\models q$ and
$\rho(q)=1$;}\\\ \left|I/cI\right|,&\hbox{if $(\mathfrak{p},I)\models q$ and
$\rho(q)=2$;}\\\ \left|R_{\mathfrak{p}}/bR_{\mathfrak{p}}\right|,&\hbox{if
$(\mathfrak{p},I)\models q$ and $\rho(q)=3$;}\\\ 1,&\hbox{if
$(\mathfrak{p},I)\models q$ and $\rho(q)=4$;}\\\
\left|s(q)R_{\mathfrak{p}}/I\right|,&\hbox{if $(\mathfrak{p},I)\models q$ and
$\rho(q)=5$;}\\\ \left|R_{\mathfrak{p}}/s(q)I\right|,&\hbox{if
$(\mathfrak{p},I)\models q$ and $\rho(q)=6$.}\end{array}\right.$

Furthermore, if $\rho(q)=5$ and $(\mathfrak{p},I)\models q$ then
$s(q)R_{\mathfrak{p}}\supseteq I$.

  2. (b)

Given $d\in R$, we can compute finite sets $S_{1},S_{2}\subseteq R^{4}$,
$\rho:\bigcup_{i=1}^{2}S_{i}\rightarrow\\{1,2\\}$ and
$s:\bigcup_{i=1}^{2}S_{i}\rightarrow R$ such that for all
$q\in\bigcup_{i=1}^{2}S_{i}$, $q\in S_{\rho(q)}$, for all prime ideals
$\mathfrak{p}\lhd R$ and ideals $I\lhd R_{\mathfrak{p}}$ with $\lambda\in I$,
there exists $q\in\bigcup_{i=1}^{2}S_{i}$ such that $(\mathfrak{p},I)\models
q$, and,

$\left|\nicefrac{{d|x}}{{x=0}}(I/\lambda
R_{\mathfrak{p}})\right|:=\left\\{\begin{array}[]{ll}\left|I/s(q)R_{\mathfrak{p}}\right|,&\hbox{if
$(\mathfrak{p},I)\models q$ and $\rho(q)=1$;}\\\ 0,&\hbox{if
$(\mathfrak{p},I)\models q$ and $\rho(q)=2$.}\end{array}\right.$

Furthermore, if $\rho(q)=1$ and $(\mathfrak{p},I)\models q$ then $s(q)\in I$.

###### Proof.

(a)Let $\alpha,r,s\in R$ be such that $\lambda\alpha=bcr$ and
$bc(\alpha-1)=\lambda s$.

Claim: If $\lambda\in I$ then

$\left|\nicefrac{{xb=0}}{{c|x}}(I/\lambda
R_{\mathfrak{p}})\right|:=\left\\{\begin{array}[]{ll}\left|I/\lambda
R_{\mathfrak{p}}\right|,&\hbox{if (1) $bI\subseteq\lambda R_{\mathfrak{p}}$
and $cI\subseteq\lambda R_{\mathfrak{p}}$;}\\\ \left|I/cI\right|,&\hbox{if (2)
$bI\subseteq\lambda R_{\mathfrak{p}}$ and $\lambda\in cI$;}\\\
\left|R_{\mathfrak{p}}/bR_{\mathfrak{p}}\right|,&\hbox{if (3) $b\in\lambda I$
and $cI\subseteq\lambda R_{\mathfrak{p}}$;}\\\ 1,&\hbox{if (4) $\lambda\in
bI$, $\lambda\in cI$, $r\in I$ and $\alpha\notin\mathfrak{p}$;}\\\
\left|rR_{\mathfrak{p}}/I\right|,&\hbox{if (5) $\lambda\in bI$, $\lambda\in
cI$, $rR_{\mathfrak{p}}\supseteq I$ and $\alpha\notin\mathfrak{p}$;}\\\
\left|R_{\mathfrak{p}}/sI\right|,&\hbox{if (6) $\lambda\in bI$, $\lambda\in
cI$ and $\alpha-1\notin\mathfrak{p}$.}\end{array}\right.$

Note that

$\left|\nicefrac{{xb=0}}{{c|x}}(I/\lambda
R_{\mathfrak{p}})\right|=\left|\frac{(\lambda R_{\mathfrak{p}}:b)\cap
I+cI}{cI+\lambda R_{\mathfrak{p}}}\right|.$

For all $a\in R$, $a\in\textnormal{ann}_{R}I/\lambda R_{\mathfrak{p}}$ if and
only if $aI\subseteq\lambda R_{\mathfrak{p}}$. Therefore the equalities for
conditions $(1)$ and $(2)$ hold. If $\lambda\in bI\subseteq bR_{\mathfrak{p}}$
then $b\neq 0$ since $\lambda\neq 0$ and $(\lambda R_{\mathfrak{p}}:b)\cap
I/\lambda R_{\mathfrak{p}}\cong R_{\mathfrak{p}}/bR_{\mathfrak{p}}$. So the
equality for condition $(3)$ holds.

When proving the equalities for $(4)$, $(5)$ and $(6)$, we may assume $b\neq
0$ and $c\neq 0$ since $\lambda\neq 0$, $\lambda\in cI$ and $\lambda\in bI$.
Moreover,

$\left|\nicefrac{{xb=0}}{{c|x}}(I/\lambda
R_{\mathfrak{p}})\right|=\left|(\lambda
R_{\mathfrak{p}}:b)+cI/cI\right|=\left|\lambda
R_{\mathfrak{p}}+bcI/bcI\right|.$

If $\alpha\notin\mathfrak{p}$ and $r\in I$ then $\lambda\in bcI$. So the
equality for condition $(4)$ holds.

Suppose condition (5) holds. Then $\alpha\notin\mathfrak{p}$ implies $\lambda
R_{\mathfrak{p}}=bcrR_{\mathfrak{p}}$. Since $rR_{\mathfrak{p}}\supseteq I$,
$\left|\nicefrac{{xb=0}}{{c|x}}(R_{\mathfrak{p}}/\lambda
I)\right|=\left|rR_{\mathfrak{p}}/I\right|$. So the equality for condition (5)
holds.

Suppose condition (6) holds. Since $\alpha-1\notin\mathfrak{p}$, $bcI=\lambda
sI$. So $\lambda R_{\mathfrak{p}}+bcI/bcI\cong R_{\mathfrak{p}}/sI$. So the
equality for condition (6) holds. So we have proved the claim.

Given a finite set of conditions of the form $\beta\notin\mathfrak{p}$,
$a\in\lambda I$ or $aR_{\mathfrak{p}}\supseteq\lambda I$, using 6.7 and 6.11,
we can compute a finite set $S\subseteq R^{4}$ such that $(\mathfrak{p},I)$
satisfies these conditions if and only if $(\mathfrak{p},I)\models q$ for some
$q\in S$. So for each conditions ($i$) for $1\leq i\leq 6$ in the claim, we
can compute $S_{i}\subseteq R^{4}$ such that $(\mathfrak{p},I)$ satisfies
$(i)$ and $\lambda\in I$ if and only if there exists $q\in S_{i}$ with
$(\mathfrak{p},I)\models q$. Moreover, it is easy to see that, for all prime
ideals $\mathfrak{p}\lhd R$ and ideals $I\lhd R_{\mathfrak{p}}$, there exists
$1\leq i\leq 6$ such that $(\mathfrak{p},I)$ satisfies $(i)$. Let
$\rho:\bigcup_{i=1}^{6}S_{i}\rightarrow\\{1,\ldots,6\\}$ be such that $q\in
S_{\rho(q)}$. Finally, set $s(q)=1$ if $\rho(q)\leq 4$, $s(q)=r$ if
$\rho(q)=5$ and $s(q)=s$ if $\rho(q)=6$.

(b) The case $d=0$ is done as in 6.12. Suppose $d\neq 0$. Let $\alpha,r,s\in
R$ be such that $d\alpha=\lambda r$ and $\lambda(\alpha-1)=ds$. If either
$\alpha\notin\mathfrak{p}$, or, $\alpha-1\notin\mathfrak{p}$ and
$sR_{\mathfrak{p}}\supseteq I$ then $dI\subseteq\lambda R_{\mathfrak{p}}$. If
$\alpha-1\notin\mathfrak{p}$ and $s\in I$ then $dsR_{\mathfrak{p}}=\lambda
R_{\mathfrak{p}}\subseteq dI$. Since $d\neq 0$, $dI/dsR_{\mathfrak{p}}\cong
I/sR_{\mathfrak{p}}$. Therefore

$\left|\nicefrac{{d|x}}{{x=0}}(I/\lambda
R_{\mathfrak{p}})\right|=\left|\frac{dI+\lambda R_{\mathfrak{p}}}{\lambda
R_{\mathfrak{p}}}\right|=\left\\{\begin{array}[]{ll}1,&\hbox{if
$(\mathfrak{p},I)\models(1,0,\alpha,1)$;}\\\ 1,&\hbox{if
$(\mathfrak{p},I)\models(s,0,\alpha-1,1)$;}\\\
\left|I/sR_{\mathfrak{p}}\right|,&\hbox{if
$(\mathfrak{p},I)\models(1,s,\alpha-1,1)$.}\end{array}\right.$

It is now clear how to define, $S_{1},S_{2}$, $\rho$ and $s$. ∎

######  Proposition 6.15.

Let $R$ be a recursive Prüfer domain, $\lambda\in R\backslash\\{0\\}$ and $Z$
a finite subset of pp-pairs of the form $\nicefrac{{xb=0}}{{c|x}}$ and
$\nicefrac{{d|x}}{{x=0}}$. Let

$T_{Z}:=\\{\mu:Z\rightarrow\\{1,\ldots,6\\}\ |\
\mu(\nicefrac{{d|x}}{{x=0}})\leq 2\\}.$

We can compute $S_{Z}$ a finite subset of $R^{4}$, $\rho_{Z}:S_{Z}\rightarrow
T_{Z}$ and $s_{Z}:S_{Z}\times Z\rightarrow R$ such that, for all prime ideals
$\mathfrak{p}\lhd R$ and ideals $I\lhd R_{\mathfrak{p}}$ with $\lambda\in I$,
if $(\mathfrak{p},I)\models q$ for some $q\in S_{Z}$ then $\lambda\in I$, and,
if $\lambda\in I$ then

  1. (a)

there exists $q\in S_{Z}$ such that $(\mathfrak{p},I)\models q$,

  2. (b)

    1. (1)

if $\rho_{Z}(q)(\nicefrac{{d|x}}{{x=0}})=1$ then
$\left|\nicefrac{{d|x}}{{x=0}}(I/\lambda
R_{\mathfrak{p}})\right|=\left|I/s_{Z}(q,\nicefrac{{d|x}}{{x=0}})R_{\mathfrak{m}}\right|$

    2. (2)

if $\rho_{Z}(q)(\nicefrac{{d|x}}{{x=0}})=2$ then
$\left|\nicefrac{{d|x}}{{x=0}}(I/\lambda R_{\mathfrak{p}})\right|=1$

  3. (c)

    1. (1)

if $\rho_{Z}(q)(\nicefrac{{xb=0}}{{c|x}})=1$ then
$\left|\nicefrac{{xb=0}}{{c|x}}(I/\lambda
R_{\mathfrak{p}})\right|=\left|I/\lambda R_{\mathfrak{p}}\right|$

    2. (2)

if $\rho_{Z}(q)(\nicefrac{{xb=0}}{{c|x}})=2$ then
$\left|\nicefrac{{xb=0}}{{c|x}}(I/\lambda
R_{\mathfrak{p}})\right|=\left|I/cI\right|$

    3. (3)

if $\rho_{Z}(q)(\nicefrac{{xb=0}}{{c|x}})=3$ then
$\left|\nicefrac{{xb=0}}{{c|x}}(I/\lambda
R_{\mathfrak{p}})\right|=\left|R_{\mathfrak{p}}/bR_{\mathfrak{p}}\right|$

    4. (4)

if $\rho_{Z}(q)(\nicefrac{{xb=0}}{{c|x}})=4$ then
$\left|\nicefrac{{xb=0}}{{c|x}}(I/\lambda R_{\mathfrak{p}})\right|=1$

    5. (5)

if $\rho_{Z}(q)(\nicefrac{{xb=0}}{{c|x}})=5$ then
$\left|\nicefrac{{xb=0}}{{c|x}}(I/\lambda
R_{\mathfrak{p}})\right|=\left|s(q,\nicefrac{{xb=0}}{{c|x}})R_{\mathfrak{p}}/I\right|$

    6. (6)

if $\rho_{Z}(q)(\nicefrac{{xb=0}}{{c|x}})=6$ then
$\left|\nicefrac{{xb=0}}{{c|x}}(I/\lambda
R_{\mathfrak{p}})\right|=\left|R_{\mathfrak{p}}/s(q,\nicefrac{{xb=0}}{{c|x}})I\right|$

Furthermore, for all prime ideals $\mathfrak{p}\lhd R$ and ideals $I\lhd
R_{\mathfrak{p}}$, if $\rho_{Z}(q)(\nicefrac{{d|x}}{{x=0}})=1$ and
$(\mathfrak{p},I)\models q$ then $s_{Z}(q,\nicefrac{{d|x}}{{x=0}})\in I$, and,
if $\rho_{Z}(q)(\nicefrac{{xb=0}}{{c|x}})=5$ and $(\mathfrak{p},I)\models q$
then $s_{Z}(q,\nicefrac{{d|x}}{{x=0}})R_{\mathfrak{p}}\supseteq I$.

###### Proof.

This can be proved similarly to 6.13 by replacing 6.12 by 6.14. The extra
condition that $(\mathfrak{p},I)\models q$ for some $q\in S_{Z}$ implies
$\lambda\in I$ can be incorporated using 6.7, since
$(\mathfrak{p},I)\models(1,\lambda,1,1)$ if and only if $\lambda\in I$. ∎

##  7\. Finite modules

In this section we investigate the consequences of $\mathrm{EPP}(R)$ being
recursive, and of $\mathrm{EPP}(R)$ and the radical relation being recursive.
In particular, we show that for a recursive Prüfer domain the theory of
$R$-modules of size $n$ is decidable uniformly in $n$ if and only if
$\mathrm{EPP}(R)$ is recursive, 7.6.

Observe that finite modules over a Prüfer domain $R$ are finite direct sums of
modules of the form $R_{\mathfrak{p}}/I$ where $\mathfrak{p}\lhd R$ is a prime
ideal and $I\lhd R_{\mathfrak{p}}$ is an ideal. There are many ways of seeing
this. If $M$ is finite then $M$ is pure-injective. So, by 2.2 and 2.10, there
exist prime ideals $\mathfrak{p}_{i}\lhd R$ and uniserial
$R_{\mathfrak{p}_{i}}$-modules $U_{i}$ such that $M$ is elementary equivalent,
and hence isomorphic, to $\oplus_{i=1}^{n}U_{i}$. The desired result now
follows from 2.12.

Let $W$ be the set of tuples $(f,g,\overline{a},\gamma)$ where

  1. (i)

$f:X\rightarrow\mathbb{N}$ where $X:=X_{0}\cup\\{\star\\}$, $X_{0}$ is a
finite subset of $R$ and $\star\notin R$,

  2. (ii)

$g:Y\rightarrow\mathbb{N}$ where $Y$ is a finite subset of $R$,

  3. (iii)

$\overline{a}:=(a_{1},\ldots,a_{m})$ is a finite tuple of elements of $R$ and
$\gamma\in R$.

Let $V$ be the set of $(f,g,\overline{a},\gamma)\in W$ such that, for some
$h\in\mathbb{N}$ and for $1\leq i\leq h$, there exist a prime ideal
$\mathfrak{p}_{i}$ and an ideal $I_{i}\lhd R_{\mathfrak{p}_{i}}$ such that
$a_{j}\in I_{i}$ for $1\leq j\leq m$ and $1\leq i\leq h$,
$\gamma\notin\mathfrak{p}_{i}$,

$|\oplus_{i=1}^{h}R_{\mathfrak{p}_{i}}/I_{i}|=f(\star),$
$|\oplus_{i=1}^{h}R_{\mathfrak{p}_{i}}/eR_{\mathfrak{p}_{i}}|=f(e)\ \ \text{ \
for \ }e\in X_{0}$

and

$|\oplus_{i=1}^{h}R_{\mathfrak{p}_{i}}/eR_{\mathfrak{p}_{i}}|\geq g(e)\ \
\text{ \ for \ }e\in Y.$

We write $(\mathfrak{p}_{i},I_{i})_{1\leq i\leq
h}\models(f,g,\overline{a},\gamma)$. By convention,
$\emptyset\models(f,g,\overline{a},\gamma)$ if $f(e)=1$ for all $e\in X$ and
$g(e)=1$ for all $e\in Y$ and in this situation, $(f,g,\overline{a},\gamma)\in
V$.

As in 4.2, $\mathbb{W}$ will denote bounded distributive lattice generated by
$W$ and $\mathbb{V}$ will denote the (prime) filter in $\mathbb{W}$ generated
by $V$.

Let $W_{0}$ be the subset of elements of the form
$(f,\emptyset,\overline{a},\gamma)$ with $|X_{0}|\leq 1$ and let $W_{1}$ be
the subset of elements of the form $(f,\emptyset,\overline{a},\gamma)$ where,
in both cases, $\emptyset$ denotes the function from the empty set to
$\mathbb{N}$. Let $\mathbb{W}_{0}$, respectively $\mathbb{W}_{1}$, denote the
lattice generated by $W_{0}\cup\\{\top,\bot\\}$ in $\mathbb{W}$, respectively
the lattice generated by $W_{1}\cup\\{\top,\bot\\}$ in $\mathbb{W}$.

######  Lemma 7.1.

Let $R$ be a recursive Prüfer domain. There is an algorithm which given $w\in
W_{1}$ returns $\underline{u}\in\mathbb{W}_{0}$, such that $w\in V$ if and
only if $\underline{u}\in\mathbb{V}$.

###### Proof.

Define $\text{clx}(f,\emptyset,\overline{a},\gamma):=(|X_{0}|,\prod_{e\in
X_{0}}f(e))$ and order $\mathbb{N}_{0}\times\mathbb{N}$ lexicographically. For
any $w\in W_{1}$ with $|X_{0}|>1$, we show how to compute
$\underline{u}\in\mathbb{W}_{1}$, such that
$\mathrm{clx}\underline{u}<\mathrm{clx}w$ and $w\in V$ if and only if
$\underline{u}\in\mathbb{V}$. Since the lexicographic order on
$\mathbb{N}_{0}\times\mathbb{N}$ is artinian, this is sufficient to prove the
lemma.

Take $e_{1},e_{2}\in X_{0}$ non-equal. Let $\alpha,r,s\in R$ be such that
$e_{1}\alpha=e_{2}r$ and $e_{2}(\alpha-1)=e_{1}s$. Let $\Omega$ be the set of
pairs of functions $(f_{1},f_{2})$ such that
$f_{1}:X\cup\\{r\\}\rightarrow\mathbb{N}$,
$f_{2}:X\cup\\{s\\}\rightarrow\mathbb{N}$ and

  1. (1)

$f_{1}(e)f_{2}(e)=f(e)$ for all $e\in X$,

  2. (2)

$f_{1}(e_{1})=f_{1}(e_{2})f(r)$, and

  3. (3)

$f_{2}(e_{2})=f_{2}(e_{1})f(s)$.

Let $\underline{u}$ be

$\bigsqcup_{\begin{subarray}{c}(f_{1},f_{2})\in\Omega\\\ f_{1}(e_{2})\neq
1,f_{2}(e_{1})\neq
1\end{subarray}}(f_{1}|_{(X\cup\\{r\\})\backslash\\{e_{1}\\}},\emptyset,\overline{a},\gamma\alpha)\sqcap(f_{2}|_{(X\cup\\{s\\})\backslash\\{e_{2}\\}},\emptyset,\overline{a},\gamma(\alpha-1))$
$\sqcup\bigsqcup_{\begin{subarray}{c}(f_{1},f_{2})\in\Omega\\\
f_{1}(e_{2})=1,f_{2}(e_{1})\neq
1\end{subarray}}(f_{1}|_{X\backslash\\{e_{2}\\}},\emptyset,\overline{a},e_{2}\gamma\alpha)\sqcap(f_{2}|_{(X\cup\\{s\\})\backslash\\{e_{2}\\}},\emptyset,\overline{a},\gamma(\alpha-1))$
$\sqcup\bigsqcup_{\begin{subarray}{c}(f_{1},f_{2})\in\Omega\\\
f_{1}(e_{2})\neq
1,f_{2}(e_{1})=1\end{subarray}}(f_{1}|_{(X\cup\\{r\\})\backslash\\{e_{1}\\}},\emptyset,\overline{a},\gamma\alpha)\sqcap(f_{2}|_{X\backslash\\{e_{1}\\}},\emptyset,\overline{a},e_{1}\gamma(\alpha-1))$
$\sqcup\bigsqcup_{\begin{subarray}{c}(f_{1},f_{2})\in\Omega\\\
f_{1}(e_{2})=1,f_{2}(e_{1})=1\end{subarray}}(f_{1}|_{X\backslash\\{e_{2}\\}},\emptyset,\overline{a},e_{2}\gamma\alpha)\sqcap(f_{2}|_{X\backslash\\{e_{1}\\}},\emptyset,\overline{a},e_{1}\gamma(\alpha-1)).$

Claim: $w\in V$ if and only if $\underline{u}\in\mathbb{V}$

For all prime ideals $\mathfrak{p}\lhd R$, $\alpha\notin\mathfrak{p}$ implies
$e_{1}R_{\mathfrak{p}}=e_{1}\alpha R_{\mathfrak{p}}=e_{2}rR_{\mathfrak{p}}$
and $\alpha-1\notin\mathfrak{p}$ implies
$e_{2}R_{\mathfrak{p}}=e_{2}(\alpha-1)R_{\mathfrak{p}}=e_{1}sR_{\mathfrak{p}}$.
So, $\alpha\notin\mathfrak{p}$ implies

$\left|R_{\mathfrak{p}}/e_{1}R_{\mathfrak{p}}\right|=\left|R_{\mathfrak{p}}/e_{2}R_{\mathfrak{p}}\right|\cdot\left|R_{\mathfrak{p}}/rR_{\mathfrak{p}}\right|$

and $\alpha-1\notin\mathfrak{p}$ implies

$\left|R_{\mathfrak{p}}/e_{2}R_{\mathfrak{p}}\right|=\left|R_{\mathfrak{p}}/e_{1}R_{\mathfrak{p}}\right|\cdot\left|R_{\mathfrak{p}}/sR_{\mathfrak{p}}\right|.$

Suppose that $\mathfrak{p}_{i}\lhd R$ is a prime ideal and $I_{i}\lhd
R_{\mathfrak{p}_{i}}$ is an ideal for $1\leq i\leq h$ such that
$(\mathfrak{p}_{i},I_{i})_{1\leq i\leq
h}\models(f,\emptyset,\overline{a},\gamma)$. By reordering, we may assume that
$\alpha\notin\mathfrak{p}_{i}$ for $1\leq i\leq h^{\prime}$ and
$\alpha-1\notin\mathfrak{p}_{i}$ for $h^{\prime}+1\leq i\leq h$. Let
$f_{1}(\star):=|\oplus_{i=1}^{h^{\prime}}R_{\mathfrak{p}_{i}}/I_{i}|$ and
$f_{2}(\star):=|\oplus_{i=h^{\prime}+1}^{h}R_{\mathfrak{p}_{i}}/I_{i}|$. For
each $e\in X_{0}\cup\\{r\\}$, let
$f_{1}(e)=|\oplus_{i=1}^{h^{\prime}}R_{\mathfrak{p}_{i}}/eR_{\mathfrak{p}_{i}}|$
and for each $e\in X_{0}\cup\\{s\\}$, let
$f_{2}(e)=|\oplus_{i=h^{\prime}+1}^{h}R_{\mathfrak{p}_{i}}/eR_{\mathfrak{p}_{i}}|$.
It follows from the first paragraph of the proof of this claim that
$(f_{1},f_{2})\in\Omega$.

If $f_{1}(e_{2})\neq 1$ then

$(\mathfrak{p}_{i},I_{i})_{1\leq i\leq
h^{\prime}}\models(f_{1}|_{(X\cup\\{r\\})\backslash\\{e_{1}\\}},\emptyset,\overline{a},\gamma\alpha)$

and if $f_{2}(e_{1})\neq 1$ then

$(\mathfrak{p}_{i},I_{i})_{h^{\prime}+1\leq i\leq
h}\models(f_{2}|_{(X\cup\\{s\\})\backslash\\{e_{2}\\}},\emptyset,\overline{a},\gamma(\alpha-1)).$

By definition, if $f_{1}(e_{2})=1$, then $e_{2}\notin\mathfrak{p}_{i}$ for
$1\leq i\leq h^{\prime}$ and if $f_{2}(e_{1})=1$ then
$e_{1}\notin\mathfrak{p}_{i}$ for $h^{\prime}+1\leq i\leq h$. So if
$f_{1}(e_{2})=1$ then

$(\mathfrak{p}_{i},I_{i})_{1\leq i\leq
h^{\prime}}\models(f_{1}|_{X\backslash\\{e_{2}\\}},\emptyset,\overline{a},e_{2}\gamma\alpha)$

and if $f_{2}(e_{1})=1$ then

$(\mathfrak{p}_{i},I_{i})_{h^{\prime}+1\leq i\leq
h}\models(f_{2}|_{X\backslash\\{e_{1}\\}},\emptyset,\overline{a},e_{1}\gamma(\alpha-1)).$

So we have shown that if $w\in V$ then one of the components of the join
defining $\underline{u}$ is in $\mathbb{V}$ and hence
$\underline{u}\in\mathbb{V}$.

Conversely, take $(f_{1},f_{2})\in\Omega$. Suppose $f_{1}(e_{2})\neq 1$ and

$(\mathfrak{p}_{i},I_{i})_{1\leq i\leq
h^{\prime}}\models(f_{1}|_{(X\cup\\{r\\})\backslash\\{e_{1}\\}},\emptyset,\overline{a},\gamma\alpha).$

Since $\alpha\notin\mathfrak{p}_{i}$ for $1\leq i\leq h^{\prime}$,

$\left|\oplus_{i=1}^{h^{\prime}}R_{\mathfrak{p}_{i}}/e_{1}R_{\mathfrak{p}_{i}}\right|=\left|\oplus_{i=1}^{h^{\prime}}R_{\mathfrak{p}_{i}}/e_{2}R_{\mathfrak{p}_{i}}\right|\cdot\left|\oplus_{i=1}^{h^{\prime}}R_{\mathfrak{p}_{i}}/rR_{\mathfrak{p}_{i}}\right|=f_{1}(e_{2})\cdot
f_{1}(r)=f_{1}(e_{1})$

by definition of $\Omega$. So $(\mathfrak{p}_{i},I_{i})_{1\leq i\leq
h^{\prime}}\models(f_{1},\emptyset,\overline{a},\gamma)$.

Suppose that $f_{2}(e_{1})=1$ and

$(\mathfrak{q}_{i},J_{i})_{1\leq i\leq
h^{\prime\prime}}\models(f_{2}|_{X\backslash\\{e_{1}\\}},\emptyset,\overline{a},e_{1}\gamma(\alpha-1)).$

Then

$(\mathfrak{q}_{i},J_{i})_{1\leq i\leq
h^{\prime\prime}}\models(f_{2},\emptyset,\overline{a},\gamma)$

because $e_{1}\notin\mathfrak{q}_{i}$ for $1\leq i\leq h^{\prime\prime}$
implies

$\left|\oplus_{i=1}^{h^{\prime\prime}}R_{\mathfrak{q}_{i}}/e_{1}R_{\mathfrak{q}_{i}}\right|=1=f_{1}(e_{1}).$

So, setting $\mathfrak{p}_{i}:=\mathfrak{q}_{i-h^{\prime}}$ and
$I_{i}:=J_{i-h^{\prime}}$ for $h^{\prime}+1\leq i\leq
h^{\prime}+h^{\prime\prime}=h$,

$(\mathfrak{p}_{i},I_{i})_{1\leq i\leq
h}\models(f,\emptyset,\overline{a},\gamma)$

because $f_{1}(e)f_{2}(e)=f(e)$ for all $e\in X$. We leave the case
$f_{1}(e_{2})\neq 1$ and $f_{2}(e_{1})\neq 1$, the case $f_{1}(e_{2})=1$ and
$f_{2}(e_{1})\neq 1$ and the case $f_{1}(e_{2})=f_{2}(e_{2})=1$ to the reader.

Claim: $\mathrm{clx}\underline{u}<\mathrm{clx}w$

We show that each of the components, $u^{\prime}$, of the lattice combination
defining $\underline{u}$ have $\mathrm{clx}u^{\prime}<\mathrm{clx}w$. We only
consider the components involving $f_{1}$; the result for those involving
$f_{2}$ follows similarly.

If $f_{1}(e_{2})=1$ then
$\mathrm{clx}(f_{1}|_{X\backslash\\{e_{2}\\}},\emptyset,\overline{a},e_{2}\gamma\alpha)<\mathrm{clx}w$
since $|X_{0}|>|X_{0}\backslash\\{e_{2}\\}|$. If $f_{1}(e_{2})>1$ then

$\frac{f_{1}(r)}{f(e_{1})}=\frac{f_{1}(e_{2})f_{1}(r)}{f_{1}(e_{2})f(e_{1})}=\frac{f_{1}(e_{1})}{f_{1}(e_{2})f(e_{1})}<1$

since $f_{1}(e_{1})/f(e_{1})\leq 1$. So

$f_{1}(r)\cdot\prod_{x\in X_{0}\backslash\\{e_{1}\\}}f_{1}(x)\leq
f_{1}(r)\cdot\prod_{x\in
X_{0}\backslash\\{e_{1}\\}}f(x)=\frac{f_{1}(r)}{f(e_{1})}\cdot\prod_{x\in
X_{0}}f(x)<\prod_{x\in X_{0}}f(x).$

Therefore
$\text{clx}(f_{1}|_{(X\cup\\{r\\})\backslash\\{e_{1}\\}},\emptyset,\overline{a},\gamma\alpha)<\text{clx}w$.
∎

######  Lemma 7.2.

Let $R$ be a recursive Prüfer domain. If $\mathrm{EPP}(R)$ is recursive then
there is an algorithm which given $w\in W_{1}$ answers whether $w\in V$ or
not.

###### Proof.

Let $w=(f,\emptyset,\overline{a},\gamma)$ and suppose that $X_{0}=\\{e\\}$
i.e. $w\in W_{0}$. Let $P$ be the set of prime divisors of $f(\star)\cdot
f(e)$. If $P$ is empty then $f(\star)=f(e)=1$ and $w\in V$. Otherwise, for
each $p\in P$, let $n_{p}\in\mathbb{N}_{0}$ and $m_{p}\in\mathbb{N}_{0}$ be
such that $f(\star)=\prod_{p\in P}p^{n_{p}}$ and $f(e)=\prod_{p\in
P}p^{m_{p}}$. For any prime ideal $\mathfrak{p}\lhd R$, $e\in R$ and ideal
$I\lhd R_{\mathfrak{p}}$, if $|R_{\mathfrak{p}}/eR_{\mathfrak{p}}|$
(respectively $|R_{\mathfrak{p}}/I|$) is finite then it is a power of
$|R/\mathfrak{p}|$. Thus a prime power. Hence $w\in V$ if and only if
$(p,n_{p};\overline{a};\gamma;e,m_{p})\in\mathrm{EPP}_{*}(R)$ for all $p\in
P$.

If $w\in W_{1}$ then by 7.1, we can compute $\underline{w}\in\mathbb{W}_{0}$
such that $w\in V$ if and only if $\underline{w}\in\mathbb{V}$. Therefore, by
the previous paragraph there is an algorithm which, given $w\in W_{1}$ answers
whether $w\in V$ or not. ∎

######  Corollary 7.3.

Let $R$ be a recursive Prüfer domain with $\mathrm{EPP}(R)$ recursive. There
is an algorithm which given $r,a,\gamma,\delta\in R$, $n\in\mathbb{N}$,
$e_{j}\in R$ for $1\leq j\leq n$, $N\in\mathbb{N}$ and $N_{j}\in\mathbb{N}$
for $1\leq j\leq n$, answers whether there exist $h\in\mathbb{N}_{0}$, prime
ideals $\mathfrak{p}_{i}\lhd R$ and ideals $I_{i}\lhd R_{\mathfrak{p}_{i}}$
for $1\leq i\leq h$ such that

  1. (1)

$(\mathfrak{p}_{i},I_{i})\models(r,a,\gamma,\delta)$ for $1\leq i\leq h$,

  2. (2)

$\left|\oplus_{i=1}^{h}R_{\mathfrak{p}_{i}}/I_{i}\right|=N$, and

  3. (3)

$\left|\oplus_{i=1}^{h}R_{\mathfrak{p}_{i}}/e_{j}R_{\mathfrak{p}_{i}}\right|=N_{j}$
for $1\leq j\leq n$.

###### Proof.

Applying 6.9, we may reduce to the case where $a=ra^{\prime}$. We may also
assume $r\neq 0$ since $(\mathfrak{p},I)\models(0,b^{\prime},\gamma,\delta)$
implies $I=0$ and hence
$\left|R_{\mathfrak{p}}/I\right|=\left|R_{\mathfrak{p}}\right|$ which is
infinite.

For any prime ideal $\mathfrak{p}\lhd R$ and ideal $I\lhd R_{\mathfrak{p}}$,
$(\mathfrak{p},I)\models(r,ra^{\prime},\gamma,\delta)$ if and only if there
exists $J\lhd R_{\mathfrak{p}}$ such that $I=rJ$ and
$(\mathfrak{p},J)\models(1,a^{\prime},\gamma,\delta)$. Note that, because $R$
is a domain,
$\left|R_{\mathfrak{p}}/rJ\right|=\left|R_{\mathfrak{p}}/J\right|\cdot\left|R_{\mathfrak{p}}/rR_{\mathfrak{p}}\right|$.

Therefore, there exist $h\in\mathbb{N}_{0}$, prime ideals
$\mathfrak{p}_{i}\lhd R$ and ideals $I_{i}\lhd R_{\mathfrak{p}_{i}}$ for
$1\leq i\leq h$ satisfying $(1),(2)$ and $(3)$ if and only if there exist
$N^{\prime},N^{\prime\prime}\in\mathbb{N}$ with $N^{\prime}\cdot
N^{\prime\prime}=N$, $h\in\mathbb{N}_{0}$, prime ideals $\mathfrak{p}_{i}\lhd
R$ and ideals $J_{i}\lhd R_{\mathfrak{p}_{i}}$ for $1\leq i\leq h$ such that

  1. (a)

$(\mathfrak{p}_{i},J_{i})\models(1,a^{\prime},\gamma,\delta)$ for $1\leq i\leq
h$,

  2. (b)

$\left|\oplus_{i=1}^{h}R_{\mathfrak{p}_{i}}/J_{i}\right|=N^{\prime}$, and

  3. (c)

$\left|\oplus_{i=1}^{h}R_{\mathfrak{p}_{i}}/rR_{\mathfrak{p}_{i}}\right|=N^{\prime\prime}$,
$\left|\oplus_{i=1}^{h}R_{\mathfrak{p}_{i}}/e_{j}R_{\mathfrak{p}_{i}}\right|=N_{j}$
for $1\leq j\leq n$.

By 7.2, there is an algorithm which answers whether there exist
$h\in\mathbb{N}_{0}$, prime ideals $\mathfrak{p}_{i}\lhd R$ and ideals
$J_{i}\lhd R_{\mathfrak{p}_{i}}$ for $1\leq i\leq h$ satisfying (a),(b) and
(c). ∎

######  Proposition 7.4.

Let $R$ be a recursive Prüfer domain such that $\mathrm{EPP}(R)$ and the
radical relation are recursive. There is an algorithm which, given
$(f,g,\overline{a},\gamma)\in W$, answers whether there exist
$h\in\mathbb{N}_{0}$, prime ideals $\mathfrak{p}_{i}$ and ideals $I_{i}\lhd
R_{\mathfrak{p}_{i}}$ for $1\leq i\leq h$ such that
$(\mathfrak{p}_{i},I_{i})_{1\leq i\leq h}\models(f,g,\overline{a},\gamma)$.

###### Proof.

Suppose $(f,g,\overline{a},\gamma)\in W$.

Case $Y=\emptyset$: This is 7.2.

Case $|Y|=1$: Suppose that $X_{0}:=\\{e_{1},\ldots,e_{n}\\}$ and $Y:=\\{e\\}$.

If $\gamma\prod_{j=1}^{n}e_{j}\notin\textnormal{rad}eR$ then there exists a
prime ideal $\mathfrak{p}$ such that $e_{j}\notin\mathfrak{p}$ for $1\leq
j\leq n$, $\gamma\notin\mathfrak{p}$ and $e\in\mathfrak{p}$. Then
$|R_{\mathfrak{p}}/e_{j}R_{\mathfrak{p}}|=1$ for $1\leq j\leq n$ and
$|R_{\mathfrak{p}}/eR_{\mathfrak{p}}|>1$. So
$(\mathfrak{p}_{i},I_{i})_{i=1}^{h}\models(f,\emptyset,\overline{a},\gamma)$
if and only if
$(\mathfrak{p}_{i},I_{i})_{i=1}^{h+g(e)}\models(f,g,\overline{a},\gamma)$
where $\mathfrak{p}_{i}:=\mathfrak{p}$ and $I_{i}:=R_{\mathfrak{p}_{i}}$ for
$h<i\leq g(e)$. So $(f,g,\overline{a},\gamma)\in V$ if and only if
$(f,\emptyset,\overline{a},\gamma)\in V$. So we are now in the situation of
case $Y=\emptyset$.

If $\gamma\prod_{j=1}^{n}e_{j}\in\textnormal{rad}eR$ then there exist
$l\in\mathbb{N}$ and $r\in R$ such that $(\gamma\prod_{j=1}^{n}e_{j})^{l}=er$.
Thus, for all prime ideals $\mathfrak{p}$ with $\gamma\notin\mathfrak{p}$,
$|R_{\mathfrak{p}}/eR_{\mathfrak{p}}|\leq\prod_{j=1}^{n}|R_{\mathfrak{p}}/e_{j}R_{\mathfrak{p}}|^{l}$.
Therefore $(\mathfrak{p},I)\models(f,g,\overline{a},\gamma)$ if and only if
there exists $f^{\prime}:X\cup\\{e\\}\rightarrow\mathbb{N}$ where
$f^{\prime}(x)=f(x)$ for all $x\in X$ and $g(e)\leq
f^{\prime}(e)\leq\prod_{j=1}^{n}f(e_{j})^{l}$ and
$(\mathfrak{p},I)\models(f^{\prime},\emptyset,\overline{a},\gamma)$. Since the
set of $f^{\prime}$ is finite and computable, we are now in the situation of
case $Y=\emptyset$.

Case $|Y|\geq 2$: We show how to reduce to the situation where $|Y|\leq 1$. By
4.3, we may assume that $g$ is a constant function. Take $e_{1},e_{2}\in Y$
non-equal. Let $\alpha,r,s\in R$ be such that $e_{1}\alpha=e_{2}r$ and
$e_{2}(\alpha-1)=e_{1}s$. Since for all prime ideals $\mathfrak{p}\lhd R$,
either $\alpha\notin\mathfrak{p}$ or $\alpha-1\notin\mathfrak{p}$, by 4.1,
$(f,g,\overline{a},\gamma)\in V$ if and only if

$\bigsqcup_{(f_{1},f_{2},g_{1},g_{2})\in\Omega_{f,g,2}}(f_{1},g_{1},\overline{a},\alpha\gamma)\sqcap(f_{2},g_{2},\overline{a},(\alpha-1)\gamma)\in\mathbb{V}.$

Note that if $g$ is constant then $g_{1}$ and $g_{2}$ are constant for all
$(f_{1},f_{2},g_{1},g_{2})\in\Omega_{f,g}$.

For each $(f_{1},f_{2},g_{1},g_{2})\in\Omega_{f,g}$, either $|Y_{1}|<|Y|$ or
$e_{1},e_{2}\in Y_{1}$. In the first case we are done. In the second,
$(f_{1},g_{1},\overline{a},\alpha\gamma)\in V$ if and only if
$(f_{1},g_{1}|_{Y_{1}\backslash\\{e_{1}\\}},\overline{a},\alpha\gamma)\in V$.
This is because, for all prime ideals $\mathfrak{p}$ with
$\alpha\notin\mathfrak{p}$,
$|R_{\mathfrak{p}}/e_{1}R_{\mathfrak{p}}|\geq|R_{\mathfrak{p}}/e_{2}R_{\mathfrak{p}}|$
and hence if $|R_{\mathfrak{p}}/e_{2}R_{\mathfrak{p}}|\geq g_{1}(e_{2})$ then
$|R_{\mathfrak{p}}/e_{1}R_{\mathfrak{p}}|\geq g_{1}(e_{2})=g_{1}(e_{1})$. So
we may replace $(f_{1},g_{1},\overline{a},\alpha\gamma)$ by
$(f_{1},g_{1}|_{Y_{1}\backslash\\{e_{1}\\}},\overline{a},\alpha\gamma)$. A
similar argument shows that either $|Y_{2}|<|Y|$ or we can replace
$(f_{2},g_{2},\overline{a},(\alpha-1)\gamma)$ by
$(f_{2},g_{2}|_{Y_{1}\backslash\\{e_{2}\\}},\overline{a},(\alpha-1)\gamma)$. ∎

######  Corollary 7.5.

Let $R$ be a recursive Prüfer domain with the radical relation and
$\mathrm{EPP}(R)$ recursive. There is an algorithm which given
$r,a,\gamma,\delta\in R$, $n,n^{\prime}\in\mathbb{N}$, $e_{j}\in R$ for $1\leq
j\leq n$, $e_{j}^{\prime}\in R$ for $1\leq j\leq n^{\prime}$,
$N\in\mathbb{N}$, $N_{j}\in\mathbb{N}$ for $1\leq j\leq n$ and
$N_{j}^{\prime}\in\mathbb{N}$ for $1\leq j\leq n^{\prime}$, answers whether
there exist $h\in\mathbb{N}_{0}$, prime ideals $\mathfrak{p}_{i}\lhd R$ and
ideals $I_{i}\lhd R_{\mathfrak{p}_{i}}$ for $1\leq i\leq h$ such that

  1. (1)

$(\mathfrak{p}_{i},I_{i})\models(r,a,\gamma,\delta)$ for $1\leq i\leq h$,

  2. (2)

$\left|\oplus_{i=1}^{h}R_{\mathfrak{p}_{i}}/I_{i}\right|=N$,

  3. (3)

$\left|\oplus_{i=1}^{h}R_{\mathfrak{p}_{i}}/e_{j}R_{\mathfrak{p}_{i}}\right|=N_{j}$
for $1\leq j\leq n$, and

  4. (4)

$\left|\oplus_{i=1}^{h}R_{\mathfrak{p}_{i}}/e_{j}R_{\mathfrak{p}_{i}}\right|\geq
N_{j}^{\prime}$ for $1\leq j\leq n^{\prime}$.

###### Proof.

The proof is as in 7.3 but we use 7.4 in place of 7.2. ∎

######  Theorem 7.6.

Let $R$ be a recursive Prüfer domain. The theory of $R$-modules of size $n$ is
decidable uniformly in $n$ if and only if $\mathrm{EPP}(R)$ is recursive.

###### Proof.

The forward direction is 3.19.

Standard arguments using the Baur-Monk theorem and the fact that $T_{R}$ is
recursively axiomatisable imply that the theory of $R$-modules of size $n$ is
decidable uniformly in $n$ if and only if there is an algorithm which, given
$N\in\mathbb{N}$, pairs of pp-$1$-formulae $\varphi_{i}/\psi_{i}$ and
$A_{i}\in\mathbb{N}$ for $1\leq i\leq m$, answers whether there exists
$M\in\textnormal{Mod}\text{-}R$ such that satisfying

$\left|\nicefrac{{x=x}}{{x=0}}\right|=N\wedge\bigwedge_{i=1}^{m}\left|\nicefrac{{\varphi_{i}}}{{\psi_{i}}}\right|=A_{i}.$

Unfortunately, we can’t directly apply the statement of [GLT19, 4.1] to reduce
to the case that the pp-pairs $\nicefrac{{\varphi_{i}}}{{\psi_{i}}}$ are of
the form $\nicefrac{{d|x}}{{x=0}}$ or $\nicefrac{{xb=0}}{{c|x}}$. However, the
proof of [GLT19, 4.1] can be easily modified to allow us to do this. Roughly
speaking, starting with a sentence $\chi$ of the form

$\bigwedge_{i=1}^{l}\left|\nicefrac{{\varphi_{i}}}{{\psi_{i}}}\right|=A_{i}\wedge\bigwedge^{m}_{i=l+1}\left|\nicefrac{{\varphi_{i}}}{{\psi_{i}}}\right|\geq
A_{i}$

where $A_{i}\in\mathbb{N}$ and $\nicefrac{{\varphi_{i}}}{{\psi_{i}}}$ is an
arbitrary pp-pair for $1\leq i\leq m$, each step of the proof of [GLT19, 4.1]
produces a finite set $S$ of finite tuples of sentences
$({\raisebox{1.5pt}{$\chi$}}_{1},\ldots,{\raisebox{1.5pt}{$\chi$}}_{k})$ as in
(7), but with the form of the pp-pairs involved progressively improved, such
that there is an $R$-module satisfying $\chi$ if and only there exist
$({\raisebox{1.5pt}{$\chi$}}_{1},\ldots,{\raisebox{1.5pt}{$\chi$}}_{k})\in S$
and $R$-modules $M_{i}\in\textnormal{Mod}\text{-}R$ with
$M_{i}\models{\raisebox{1.5pt}{$\chi$}}_{i}$ for $1\leq i\leq k$. In order to
adapt the proof to our situation, the reader just needs to note that, for each
of the steps of the proof of [GLT19, 4.1], if the initial sentence $\chi$ is
as in (7), then the sentences ${\raisebox{1.5pt}{$\chi$}}_{i}$ in the finite
tuples
$({\raisebox{1.5pt}{$\chi$}}_{1},\ldots,{\raisebox{1.5pt}{$\chi$}}_{k})\in S$
are of the same form as in (7).

Let $N\in\mathbb{N}$, $Z$ be a finite set of pp-pairs of the form
$\nicefrac{{d|x}}{{x=0}}$ and $\nicefrac{{xb=0}}{{c|x}}$, and
$f:Z\rightarrow\mathbb{N}$. Let $\chi$ be the sentence

$\left|\nicefrac{{x=x}}{{x=0}}\right|=N\wedge\bigwedge_{\nicefrac{{\varphi}}{{\psi}}\in
Z}\left|\nicefrac{{\varphi}}{{\psi}}\right|=f(\nicefrac{{\varphi}}{{\psi}}).$

Recall that every finite module over a Prüfer domain is isomorphic to a direct
sum of modules of the form $R_{\mathfrak{p}}/I$ for some prime ideal
$\mathfrak{p}\lhd R$ and ideal $I\lhd R_{\mathfrak{p}}$.

Let $S_{Z}$, $\rho_{Z}$ and $s_{Z}$ be as in 6.13 with $\lambda:=1$. Enumerate
$S_{Z}:=\\{q_{1},\ldots,q_{m}\\}$ and let
$q_{i}:=(r_{i},r_{i}a_{i},\gamma_{i},\delta_{i})$. By definition, for all
prime ideals $\mathfrak{p}\lhd R$ and ideals $I\lhd R_{\mathfrak{p}}$ there
exists $1\leq i\leq m$ such that $(\mathfrak{p},I)\models q_{i}$. Therefore,
there exists $M\in\textnormal{Mod}\text{-}R$ satisfying $\chi$ if and only if
there exist $N_{i}\in\mathbb{N}$ and $f_{i}:Z\rightarrow\mathbb{N}$ for $1\leq
i\leq m$ such that $N=\prod_{i=1}^{m}N_{i}$, for all
$\nicefrac{{\varphi}}{{\psi}}\in Z$,
$f(\nicefrac{{\varphi}}{{\psi}})=\prod_{i=1}^{m}f_{i}(\nicefrac{{\varphi}}{{\psi}})$
and there exist $h_{i}\in\mathbb{N}_{0}$, prime ideals $\mathfrak{p}_{ij}\lhd
R$ and ideals $I_{ij}\lhd R_{\mathfrak{p}_{ij}}$ for $1\leq j\leq h_{i}$ such
that

  * (a)i

$(\mathfrak{p}_{ij},I_{ij})\models q_{i}$ for $1\leq j\leq h_{i}$ and

  * (b)i

$\bigoplus_{j=1}^{h_{j}}R_{\mathfrak{p}_{ij}}/I_{ij}\models\left|\nicefrac{{x=x}}{{x=0}}\right|=N_{i}\wedge\bigwedge_{\nicefrac{{\varphi}}{{\psi}}\in
Z}\left|\nicefrac{{\varphi}}{{\psi}}\right|=f_{i}(\nicefrac{{\varphi}}{{\psi}}).$

Fix $1\leq i\leq m$, $N_{i}\in\mathbb{N}$ and $f_{i}$ as above. If $r_{i}=0$
and $(\mathfrak{p},I)\models(r_{i},r_{i}a_{i},\gamma_{i},\delta_{i})$ then
$R_{\mathfrak{p}}/I=R_{\mathfrak{p}}$, which is not finite. Hence, (a)i and
(b)i holds if and only if $h_{i}=0$, $N_{i}=1$ and
$f_{i}(\nicefrac{{\varphi}}{{\psi}})=1$ for all
$\nicefrac{{\varphi}}{{\psi}}\in Z$. So, we may assume that $r_{i}\neq 0$.
Note, by 2.14iv, if $R_{\mathfrak{p}}/I$ is finite then
$I^{\\#}=\mathfrak{p}$. So,
$(\mathfrak{p},I)\models(r_{i},r_{i}a_{i},\gamma_{i},\delta_{i})$ if and only
if $I=r_{i}J$ for some $J\lhd R_{\mathfrak{p}}$, $a\in J$,
$\gamma\notin\mathfrak{p}$ and $\delta\notin\mathfrak{p}$.

Therefore $h_{i}\in\mathbb{N}_{0}$, $\mathfrak{p}_{ij}\lhd R$ and $I_{ij}\lhd
R_{\mathfrak{p}_{ij}}$ for $1\leq j\leq h_{i}$ satisfy (a)i and (b)i if and
only if for each $1\leq j\leq h_{i}$, there exists $J_{ij}\lhd
R_{\mathfrak{p}_{ij}}$ such that $I_{ij}=r_{i}J_{ij}$, $a\in J_{ij}$,
$\gamma\delta\notin\mathfrak{p}$,
$\left|\bigoplus_{j=1}^{h_{i}}R_{\mathfrak{p}_{ij}}/I_{ij}\right|=N_{i}$, and,
for all $\nicefrac{{d|x}}{{x=0}}\in Z$,

  1. (1)

if $\rho_{Z}(q_{i})(\nicefrac{{d|x}}{{x=0}})=1$ then
$\left|\oplus_{j=1}^{h_{i}}R_{\mathfrak{p}_{ij}}/s_{Z}(q_{i},\nicefrac{{d|x}}{{x=0}})R_{\mathfrak{p}_{ij}}\right|=f_{i}(\nicefrac{{d|x}}{{x=0}})\cdot
N_{i}^{-1}$

  2. (2)

if $\rho_{Z}(q_{i})(\nicefrac{{d|x}}{{x=0}})=2$ then
$\left|\oplus_{j=1}^{h_{i}}R_{\mathfrak{p}_{ij}}/s_{Z}(q_{i},\nicefrac{{d|x}}{{x=0}})R_{\mathfrak{p}_{ij}}\right|=f_{i}(\nicefrac{{d|x}}{{x=0}})\cdot
N_{i}$

  3. (3)

if $\rho_{Z}(q_{i})(\nicefrac{{d|x}}{{x=0}})=3$ then
$f_{i}(\nicefrac{{d|x}}{{x=0}})=1$

and for all $\nicefrac{{xb=0}}{{c|x}}\in Z$,

  1. (1)

if $\rho_{Z}(q_{i})(\nicefrac{{xb=0}}{{c|x}})=1$ then
$f_{i}(\nicefrac{{xb=0}}{{c|x}})=N_{i}$

  2. (2)

if $\rho_{Z}(q_{i})(\nicefrac{{xb=0}}{{c|x}})=2$ then
$\left|\oplus_{j=1}^{h_{i}}R_{\mathfrak{p_{ij}}}/cR_{\mathfrak{p_{ij}}}\right|=f_{i}(\nicefrac{{xb=0}}{{c|x}})$

  3. (3)

if $\rho_{Z}(q_{i})(\nicefrac{{xb=0}}{{c|x}})=3$ then
$\left|\oplus_{j=1}^{h_{i}}I_{ij}/bI_{ij}\right|=f_{i}(\nicefrac{{xb=0}}{{c|x}})$

  4. (4)

if $\rho_{Z}(q_{i})(\nicefrac{{xb=0}}{{c|x}})=4$ then
$f_{i}(\nicefrac{{xb=0}}{{c|x}})=1$

  5. (5)

if $\rho_{Z}(q_{i})(\nicefrac{{xb=0}}{{c|x}})=5$ then
$\left|\oplus_{j=1}^{h_{i}}R_{\mathfrak{p}_{ij}}/s_{Z}(q_{i})(\nicefrac{{xb=0}}{{c|x}})R_{\mathfrak{p}_{ij}}\right|=f_{i}(\nicefrac{{xb=0}}{{c|x}})\cdot
N_{i}$.

Since $R_{\mathfrak{p}_{ij}}/I_{ij}$ is finite, by 2.16, if $b\neq 0$ then
$\left|I_{ij}/bI_{ij}\right|=\left|R_{\mathfrak{p}_{ij}}/bR_{\mathfrak{p}_{ij}}\right|$.
By definition, $\rho_{Z}(\nicefrac{{xb=0}}{{c|x}})=3$ implies $b\neq 0$. So
($3$) in the second list of conditions may be replaced by

  * ($3$’)

if $\rho_{Z}(q_{i})(\nicefrac{{xb=0}}{{c|x}})=3$ then
$\left|\oplus_{j=1}^{h_{i}}R_{\mathfrak{p}_{ij}}/bR_{\mathfrak{p}_{ij}}\right|=f_{i}(\nicefrac{{xb=0}}{{c|x}})$.

Finally, the condition that
$\left|\bigoplus_{j=1}^{h_{i}}R_{\mathfrak{p}_{ij}}/I_{ij}\right|=N_{i}$ can
be replaced by the condition that there exist
$N_{i}^{\prime},N_{i}^{\prime\prime}\in\mathbb{N}$ such that
$N_{i}=N_{i}^{\prime}N_{i}^{\prime\prime}$,
$\left|\bigoplus_{j=1}^{h_{i}}R_{\mathfrak{p}_{ij}}/r_{i}R_{\mathfrak{p}_{ij}}\right|=N_{i}^{\prime}$
and
$\left|\bigoplus_{j=1}^{h_{i}}R_{\mathfrak{p}_{ij}}/J_{ij}\right|=N_{i}^{\prime\prime}$.
The proof can now be finished using 7.3. ∎

##  8\. Removing $|\nicefrac{{d|x}}{{x=0}}|=D$ and
$|\nicefrac{{xb=0}}{{c|x}}|=G$.

This section uses results from sections 6 and 7. For $d\in
R\backslash\\{0\\}$, respectively $b,c\in R\backslash\\{0\\}$, we show that
there is an algorithm answering whether there exists a sum of modules
$\oplus_{i=1}^{h}R_{\mathfrak{p}_{i}}/dI_{i}$, respectively
$\oplus_{i=1}^{h}I_{i}/bcR_{\mathfrak{p}_{i}}$, satisfying a sentence as in
5.1 under the assumption that one of the conjuncts is
$\left|d|x/x=0\right|=D$, respectively $\left|xb=0/c|x\right|=G$. These
results will be used in section 10 to eliminate expressions of the form
$|\nicefrac{{d|x}}{{x=0}}|=D$ and $|\nicefrac{{xb=0}}{{c|x}}|=G$, where
$D,G\geq 2$.

######  Proposition 8.1.

Let $R$ be a recursive Prüfer domain such that $\mathrm{EPP}(R)$ and the
radical relation are recursive. There exists an algorithm which, given a
sentence $\chi$ of the form

$\left|\nicefrac{{d|x}}{{x=0}}\right|=\
D\wedge\bigwedge_{\nicefrac{{\varphi}}{{\psi}}\in
X}\left|\nicefrac{{\varphi}}{{\psi}}\right|=f(\nicefrac{{\varphi}}{{\psi}})\wedge\bigwedge_{\nicefrac{{\varphi}}{{\psi}}\in
Y}\left|\nicefrac{{\varphi}}{{\psi}}\right|\geq
g(\nicefrac{{\varphi}}{{\psi}}),$

where $d\in R\backslash\\{0\\}$, $D\in\mathbb{N}$, $f:X\rightarrow\mathbb{N}$,
$g:Y\rightarrow\mathbb{N}$ and $X,Y$ are disjoint finite sets of pp-pairs of
the form $\nicefrac{{xb=0}}{{c|x}}$ and $\nicefrac{{a|x}}{{x=0}}$, answers
whether there exist $h\in\mathbb{N}_{0}$, prime ideals $\mathfrak{p}_{i}\lhd
R$ and ideals $I_{i}\lhd R_{\mathfrak{p}_{i}}$ for $1\leq i\leq h$ such that
$\oplus_{i=1}^{h}R_{\mathfrak{p}_{i}}/dI_{i}\models{\raisebox{1.5pt}{$\chi$}}$.

###### Proof.

Let $Z:=X\cup Y$ and let $S_{Z}$, $\rho_{Z}$ and $s_{Z}$ be as in 6.13 with
$\lambda:=d$. Enumerate $S_{Z}:=\\{q_{1},\ldots,q_{m}\\}$ and let
$q_{i}:=(r_{i},r_{i}a_{i},\gamma_{i},\delta_{i})$. By definition, for all
prime ideals $\mathfrak{p}\lhd R$ and ideals $I\lhd R_{\mathfrak{p}}$, there
exists $1\leq i\leq m$ such that $(\mathfrak{p},I)\models q_{i}$.

Therefore, there exist $h\in\mathbb{N}_{0}$, prime ideals
$\mathfrak{p}_{i}\lhd R$ and ideals $I_{i}\lhd R_{\mathfrak{p}_{i}}$ for
$1\leq i\leq h$ such that
$\oplus_{i=1}^{h}R_{\mathfrak{p}_{i}}/dI_{i}\models{\raisebox{1.5pt}{$\chi$}}$
if and only if there exist $D_{i}\in\mathbb{N}$ for $1\leq i\leq m$ such that
$\prod_{i=1}^{m}D_{i}=D$, $(\overline{f},\overline{g})\in\Omega_{f,g,m}$ and,
for $1\leq i\leq m$, $h_{i}\in\mathbb{N}_{0}$, prime ideals
$\mathfrak{p}_{ij}\lhd R$ and ideals $I_{ij}\lhd R_{\mathfrak{p}_{ij}}$ such
that, for $1\leq i\leq m$,

  * (a)i

$(\mathfrak{p}_{ij},I_{ij})\models q_{i}$ for $1\leq j\leq h_{i}$

  * (b)i

$\oplus_{j=1}^{h_{j}}R_{\mathfrak{p}_{ij}}/dI_{ij}$ satisfies

$\left|\nicefrac{{d|x}}{{x=0}}\right|=\
D_{i}\wedge\bigwedge_{\nicefrac{{\varphi}}{{\psi}}\in
X_{i}}\left|\nicefrac{{\varphi}}{{\psi}}\right|=f_{i}(\nicefrac{{\varphi}}{{\psi}})\wedge\bigwedge_{\nicefrac{{\varphi}}{{\psi}}\in
Y_{i}}\left|\nicefrac{{\varphi}}{{\psi}}\right|\geq
g_{i}(\nicefrac{{\varphi}}{{\psi}}).$

Fix $1\leq i\leq m$, $D_{i}\in\mathbb{N}$ and
$(\overline{f},\overline{g})\in\Omega_{f,g,m}$ as above. Note that

$\left|\nicefrac{{d|x}}{{x=0}}(\oplus_{j=1}^{h_{i}}R_{\mathfrak{p}_{ij}}/dI_{ij})\right|=D_{i}\
\text{ if and only if }\
\left|\oplus_{j=1}^{h_{i}}R_{\mathfrak{p}_{ij}}/I_{ij}\right|=D_{i}.$

So, exactly as in 7.6, we may assume $r_{i}\neq 0$ for $1\leq i\leq m$.
Moreover, by 2.14iv, $I_{ij}^{\\#}=\mathfrak{p}_{ij}$.

Now $h_{i}\in\mathbb{N}_{0}$, $\mathfrak{p}_{ij}\lhd R$ and $I_{ij}\lhd
R_{\mathfrak{p}_{ij}}$ for $1\leq j\leq h_{i}$ satisfy (a)i if and only if for
$1\leq j\leq h_{i}$, there exists $J_{ij}\lhd R_{\mathfrak{p}_{i}}$ such that
$I_{ij}=r_{i}J_{ij}$, $a_{i}\in J_{ij}$, $\gamma_{i}\notin\mathfrak{p}_{ij}$
and $\delta_{i}\notin J_{ij}^{\\#}=I_{ij}^{\\#}=\mathfrak{p}_{ij}$.

Now $h_{i}\in\mathbb{N}_{0}$, $\mathfrak{p}_{ij}\lhd R$ and $J_{ij}\lhd
R_{\mathfrak{p}_{ij}}$ with $I_{ij}:=r_{i}J_{ij}$ for $1\leq j\leq h_{i}$
satisfy (b)i if and only if there exists
$D^{\prime}_{i},D^{\prime\prime}_{i}\in\mathbb{N}$ with
$D^{\prime}_{i}D^{\prime\prime}_{i}=D_{i}$ such that

$\left|\oplus_{j=1}^{h_{i}}R_{\mathfrak{p}_{ij}}/J_{ij}\right|=D^{\prime}_{i}\
\text{ and }\
\left|\oplus_{j=1}^{h_{i}}R_{\mathfrak{p}_{ij}}/r_{i}R_{\mathfrak{p}_{ij}}\right|=D^{\prime\prime}_{i},$

for all $\nicefrac{{xb=0}}{{c|x}}\in X_{i}$ (respectively
$\nicefrac{{xb=0}}{{c|x}}\in Y_{i}$),

  1. (1)

if $\rho_{Z}(q_{i})(\nicefrac{{xb=0}}{{c|x}})=1$ then
$f_{i}(\nicefrac{{xb=0}}{{c|x}})=D_{i}$ (respectively $D_{i}\geq
g_{i}(\nicefrac{{xb=0}}{{c|x}})$)

  2. (2)

if $\rho_{Z}(q_{i})(\nicefrac{{xb=0}}{{c|x}})=2$ then
$\left|\oplus_{j=1}^{h_{i}}R_{\mathfrak{p}_{ij}}/cR_{\mathfrak{p}_{ij}}\right|=f_{i}(\nicefrac{{xb=0}}{{c|x}})$
(respectively
$\left|\oplus_{j=1}^{h_{i}}R_{\mathfrak{p}_{ij}}/cR_{\mathfrak{p}_{ij}}\right|\geq
g_{i}(\nicefrac{{xb=0}}{{c|x}})$ )

  3. (3)

if $\rho_{Z}(q_{i})(\nicefrac{{xb=0}}{{c|x}})=3$ then
$\left|\oplus_{j=1}^{h_{i}}R_{\mathfrak{p}_{ij}}/bR_{\mathfrak{p}_{ij}}\right|=f_{i}(\nicefrac{{xb=0}}{{c|x}})$
(respectively
$\left|\oplus_{j=1}^{h_{i}}R_{\mathfrak{p}_{ij}}/bR_{\mathfrak{p}_{ij}}\right|\geq
g_{i}(\nicefrac{{xb=0}}{{c|x}})$)

  4. (4)

if $\rho_{Z}(q_{i})(\nicefrac{{xb=0}}{{c|x}})=4$ then
$f_{i}(\nicefrac{{xb=0}}{{c|x}})=1$ (respectively
$g_{i}(\nicefrac{{xb=0}}{{c|x}})=1$)

  5. (5)

if $\rho_{Z}(q_{i})(\nicefrac{{xb=0}}{{c|x}})=5$ then
$\left|\oplus_{j=1}^{h_{i}}R_{\mathfrak{p}_{ij}}/s_{Z}(q_{i},\nicefrac{{xb=0}}{{c|x}})R_{\mathfrak{p}_{ij}}\right|=f_{i}(\nicefrac{{xb=0}}{{c|x}})\cdot
D_{i}$ (respectively
$\left|\oplus_{j=1}^{h_{i}}R_{\mathfrak{p}_{ij}}/s_{Z}(q_{i},\nicefrac{{xb=0}}{{c|x}})R_{\mathfrak{p}_{ij}}\right|\geq
g_{i}(\nicefrac{{xb=0}}{{c|x}})\cdot D_{i}$).

and for all $\nicefrac{{a|x}}{{x=0}}\in X_{i}$ (respectively
$\nicefrac{{a|x}}{{x=0}}\in Y_{i}$),

  1. (1)

if $\rho_{Z}(q_{i})(\nicefrac{{a|x}}{{x=0}})=1$ then
$\left|\oplus_{j=1}^{h_{i}}R_{\mathfrak{p}_{ij}}/s_{Z}(q_{i},\nicefrac{{a|x}}{{x=0}})R_{\mathfrak{p}_{ij}}\right|=f_{i}(\nicefrac{{a|x}}{{x=0}})\cdot
D_{i}^{-1}$ (respectively
$\left|\oplus_{j=1}^{h_{i}}R_{\mathfrak{p}_{ij}}/s_{Z}(q_{i},\nicefrac{{a|x}}{{x=0}})R_{\mathfrak{p}_{ij}}\right|\geq
g_{i}(\nicefrac{{a|x}}{{x=0}})\cdot D_{i}^{-1}$),

  2. (2)

if $\rho_{Z}(q_{i})(\nicefrac{{a|x}}{{x=0}})=2$ then
$\left|\oplus_{j=1}^{h_{i}}R_{\mathfrak{p}_{ij}}/s_{Z}(q_{i},\nicefrac{{a|x}}{{x=0}})R_{\mathfrak{p}_{ij}}\right|=f_{i}(\nicefrac{{a|x}}{{x=0}})\cdot
D_{i}$ (respectively
$\left|\oplus_{j=1}^{h_{i}}R_{\mathfrak{p}_{ij}}/s_{Z}(q_{i},\nicefrac{{a|x}}{{x=0}})R_{\mathfrak{p}_{ij}}\right|\geq
g_{i}(\nicefrac{{a|x}}{{x=0}})\cdot D_{i}$), and,

  3. (3)

if $\rho_{Z}(q_{i})(\nicefrac{{d|x}}{{x=0}})=3$ then
$f_{i}(\nicefrac{{a|x}}{{x=0}})=1$ (respectively
$g_{i}(\nicefrac{{a|x}}{{x=0}})=1$).

So we are now done by 7.5. ∎

######  Proposition 8.2.

Let $R$ be a recursive Prüfer domain such that $\mathrm{EPP}(R)$ and the
radical relation are recursive. There exists an algorithm which, given a
sentence $\chi$ of the form

$\left|\nicefrac{{xb=0}}{{c|x}}\right|=\
G\wedge\bigwedge_{\nicefrac{{\varphi}}{{\psi}}\in
X}\left|\nicefrac{{\varphi}}{{\psi}}\right|=f(\nicefrac{{\varphi}}{{\psi}})\wedge\bigwedge_{\nicefrac{{\varphi}}{{\psi}}\in
Y}\left|\nicefrac{{\varphi}}{{\psi}}\right|\geq
g(\nicefrac{{\varphi}}{{\psi}}),$

where $b,c\in R\backslash\\{0\\}$, $G\in\mathbb{N}$,
$f:X\rightarrow\mathbb{N}$, $g:Y\rightarrow\mathbb{N}$ and $X,Y$ are disjoint
finite sets of pp-pairs of the form
$\nicefrac{{xb^{\prime}=0}}{{c^{\prime}|x}}$ and $\nicefrac{{d|x}}{{x=0}}$,
answers whether there exist $h\in\mathbb{N}_{0}$, prime ideals
$\mathfrak{p}_{i}\lhd R$ and ideals $I\lhd R_{\mathfrak{p}_{i}}$ with $b,c\in
I_{i}$ for $1\leq i\leq h$ such that
$\oplus_{i=1}^{h}I_{i}/bcR_{\mathfrak{p}_{i}}\models{\raisebox{1.5pt}{$\chi$}}$.

###### Proof.

The proof, which we leave to the reader, is very similar to 8.1, except we use
6.15 in place of 6.13 and the fact that
$\left|\nicefrac{{xb=0}}{{c|x}}(I/bcR_{\mathfrak{p}})\right|=\left|R_{\mathfrak{p}}/I\right|$
in place of
$\left|\nicefrac{{d|x}}{{x=0}}(R_{\mathfrak{p}}/dI)\right|=\left|R_{\mathfrak{p}}/I\right|$.
∎

##  9\. Further syntactic reductions

In this section we continue work of section 5 to improve the form of the
conjunction in 5.1. Some of these reductions use results in sections 7 and 8,
which only apply to Prüfer domains (i.e. they do not apply to arbitrary
arithmetical rings).

Let $W$ be the set of $\mathcal{L}_{R}$-sentences of the form

$\left|\nicefrac{{d|x}}{{x=0}}\right|\square_{1}D\wedge\left|\nicefrac{{xb=0}}{{c|x}}\right|\square_{2}E\wedge\bigwedge_{\nicefrac{{\varphi}}{{\psi}}\in
X}\left|\nicefrac{{\varphi}}{{\psi}}\right|=f(\nicefrac{{\varphi}}{{\psi}})\wedge\bigwedge_{\nicefrac{{\varphi}}{{\psi}}\in
Y}\left|\nicefrac{{\varphi}}{{\psi}}\right|\geq
g(\nicefrac{{\varphi}}{{\psi}})\wedge\Xi$

where $\square_{1},\square_{2}\in\\{\geq,=,\emptyset\\}$, $d,c,b\in
R\backslash\\{0\\}$, $D,E\in\mathbb{N}$, $f:X\rightarrow\mathbb{N}$,
$g:Y\rightarrow\mathbb{N}$, $X,Y$ are finite subsets of pp-pairs of the form
$\nicefrac{{xb^{\prime}=0}}{{x=0}}$ and $\nicefrac{{x=x}}{{c^{\prime}|x}}$,
and $\Xi$ an auxiliary sentence.

Let $V$ be the set of $w\in W$ such that there exists an $R$-module which
satisfies $w$.

######  Definition 9.1.

Let $w\in W$ be as in (9). Define

$\displaystyle z_{1}$
$\displaystyle:=\left|\\{\nicefrac{{xb^{\prime}=0}}{{x=0}}\in X\ |\
f(\nicefrac{{xb^{\prime}=0}}{{x=0}})>1\\}\right|,$ $\displaystyle z_{2}$
$\displaystyle:=\left|\\{\nicefrac{{xb^{\prime}=0}}{{x=0}}\in Y\ |\
g(\nicefrac{{xb^{\prime}=0}}{{x=0}})>1\\}\right|,$ $\displaystyle z_{3}$
$\displaystyle:=\left|\\{\nicefrac{{x=x}}{{c^{\prime}|x}}\in X\ |\
f(\nicefrac{{x=x}}{{c^{\prime}|x}})>1\\}\right|,\text{ and }$ $\displaystyle
z_{4}$ $\displaystyle:=\left|\\{\nicefrac{{x=x}}{{c^{\prime}|x}}\in Y\ |\
g(\nicefrac{{x=x}}{{c^{\prime}|x}})>1\\}\right|.$

The short signature is defined to be the tuple $(\square_{1},\square_{2})$ and
the extended signature, $\mathtt{exsig}\,w$, is defined to be the tuple
$((\square_{1},\square_{2}),(z_{1},z_{2}),(z_{3},z_{4}))$.

We equip the set $\\{\geq,=,\emptyset\\}$ with a total order $\succ$ by
putting $\geq\ \succ\ =\ \succ\ \emptyset$. We partially order the set of
short signatures $\\{\geq,=,\emptyset\\}^{2}$ by setting
$(\square_{1},\square_{2})\geq(\square^{\prime}_{1},\square^{\prime}_{2})$
whenever $\square_{1}\succeq\square^{\prime}_{1}$ and
$\square_{2}\succeq\square^{\prime}_{2}$. Finally, we partially order on the
set of extended signatures by setting

$((\square_{1},\square_{2}),(z_{1},z_{2}),(z_{3},z_{4}))\geq((\square^{\prime}_{1},\square^{\prime}_{2}),(z^{\prime}_{1},z^{\prime}_{2}),(z^{\prime}_{3},z^{\prime}_{4}))$

if and only if
$(\square_{1},\square_{2})>(\square^{\prime}_{1},\square^{\prime}_{2})$ or
$(\square_{1},\square_{2})=(\square^{\prime}_{1},\square^{\prime}_{2})$ and

  * •

$z_{1}+z_{2}>z^{\prime}_{1}+z^{\prime}_{2}$ or
$z_{1}+z_{2}=z^{\prime}_{1}+z^{\prime}_{2}$ and $z_{2}\geq z^{\prime}_{2}$,
and

  * •

$z_{3}+z_{4}>z^{\prime}_{3}+z^{\prime}_{4}$ or
$z_{3}+z_{4}=z^{\prime}_{3}+z^{\prime}_{4}$ and $z_{4}\geq z^{\prime}_{4}$.

We now present various algorithms which, given $w\in W$ of a particular form,
output $\underline{w}\in\mathbb{W}$ such that $w\in V$ if and only if
$\underline{w}\in\mathbb{V}$, and,
$\mathtt{exsig}\,\underline{w}<\mathtt{exsig}\,w$. By convention, we give both
$\top\in\mathbb{W}$ and $\bot\in\mathbb{W}$ extended signature
$((\emptyset,\emptyset),(0,0),(0,0))$.

######  Remark 9.2.

The order on the set of extended signatures is artinian.

As in §5, given $w\in W$, we may always assume that $w$ is of the form

${\raisebox{1.5pt}{$\chi$}}_{f,g}\wedge\Xi$

where $X,Y$ are disjoint finite sets of pp-pairs of the form
$\nicefrac{{d|x}}{{x=0}}$ or $\nicefrac{{xb=0}}{{c|x}}$,
$f:X\rightarrow\mathbb{N}_{2}$, $g:Y\rightarrow\mathbb{N}_{2}$ and $\Xi$ is an
auxiliary sentence.

######  Remark 9.3.

If $w\in W$ has extended signature
$((\square_{1},\square_{2}),(z_{1},z_{2}),(z_{3},z_{4}))$ then $Dw$ has
extended signature $((\square_{1},\square_{2}),(z_{3},z_{4}),(z_{1},z_{2}))$
where $D$ denotes the duality defined in 2.3.

######  Remark 9.4.

Let $X,Y$ be disjoint finite subsets of pp-pairs,
$f:X\rightarrow\mathbb{N}_{2}$, $g:Y\rightarrow\mathbb{N}_{2}$ and $\Xi$ an
auxiliary sentence be such that

${\raisebox{1.5pt}{$\chi$}}_{f,g}\wedge\Xi=\bigwedge_{\nicefrac{{\varphi}}{{\psi}}\in
X}\left|\nicefrac{{\varphi}}{{\psi}}\right|=f(\nicefrac{{\varphi}}{{\psi}})\wedge\bigwedge_{\nicefrac{{\varphi}}{{\psi}}\in
Y}\left|\nicefrac{{\varphi}}{{\psi}}\right|\geq
g(\nicefrac{{\varphi}}{{\psi}})\wedge\Xi\in W.$

Then, for each $(f_{1},\ldots,f_{n},g_{1},\ldots,g_{n})\in\Omega_{f,g,n}$ and
$1\leq i\leq n$,

${\raisebox{1.5pt}{$\chi$}}_{f_{i},g_{i}}\wedge\Xi=\bigwedge_{\nicefrac{{\varphi}}{{\psi}}\in
X_{i}}\left|\nicefrac{{\varphi}}{{\psi}}\right|=f_{i}(\nicefrac{{\varphi}}{{\psi}})\wedge\bigwedge_{\nicefrac{{\varphi}}{{\psi}}\in
Y_{i}}\left|\nicefrac{{\varphi}}{{\psi}}\right|\geq
g_{i}(\nicefrac{{\varphi}}{{\psi}})\wedge\Xi\in W,$

and,
$\mathtt{exsig}\,{\raisebox{1.5pt}{$\chi$}}_{f_{i},g_{i}}\wedge\Xi\leq\mathtt{exsig}\,{\raisebox{1.5pt}{$\chi$}}_{f,g}\wedge\Xi$.
Moreover, for each $(\overline{f},\overline{g})\in\Omega_{f,g,n}$ and $1\leq
i\leq n$, either $X=X_{i}$ and $Y=Y_{i}$, or,
$\mathtt{exsig}\,{\raisebox{1.5pt}{$\chi$}}_{f_{i},g_{i}}\wedge\Xi<{\raisebox{1.5pt}{$\chi$}}_{f,g}\wedge\Xi$.

###### Proof.

Fix $(\overline{f},\overline{g})\in\Omega_{f,g,n}$ and $1\leq i\leq n$. Let
$((\square_{1},\square_{2}),(z_{1},z_{2}),(z_{3},z_{4}))$ be the extended
signature of ${\raisebox{1.5pt}{$\chi$}}_{f,g}\wedge\Xi$ and
$((\square^{\prime}_{1},\square^{\prime}_{2}),(z^{\prime}_{1},z^{\prime}_{2}),(z^{\prime}_{3},z^{\prime}_{4}))$
the extended signature of ${\raisebox{1.5pt}{$\chi$}}_{f_{i},g_{i}}\wedge\Xi$.
Note that since $X$ and $Y$ are disjoint, so are $X_{i}=X\cup(Y\backslash
Y_{i})$ and $Y_{i}$.

That the short signature of ${\raisebox{1.5pt}{$\chi$}}_{f_{i},g_{i}}$ is less
than the short signature of ${\raisebox{1.5pt}{$\chi$}}_{f,g}$ follows from
the fact that $Y_{i}\subseteq Y$ and $X_{i}=X\cup(Y\backslash Y_{i})$.

We show that $z_{1}+z_{2}\geq z_{1}^{\prime}+z_{2}^{\prime}$ and $z_{2}\geq
z_{2}^{\prime}$. By definition, $z_{1}+z_{2}$ is the number of
$\nicefrac{{x=x}}{{c|x}}\in X\cup Y$ and $z_{2}$ is the number of
$\nicefrac{{x=x}}{{c|x}}\in Y$. Since
$g_{i}(\nicefrac{{x=x}}{{c|x}})=g(\nicefrac{{x=x}}{{c|x}})>1$ for all
$\nicefrac{{x=x}}{{c|x}}\in Y_{i}$, we see that $z_{2}^{\prime}$ is the number
of $\nicefrac{{x=x}}{{c|x}}\in Y_{i}$. So $z_{2}^{\prime}\leq z_{2}$ because
$Y_{i}\subseteq Y$. Since $X$ and $Y$ are disjoint, so are
$X_{i}=X\cup(Y\backslash Y_{i})$ and $Y_{i}$. Therefore
$z_{1}^{\prime}+z_{2}^{\prime}$ is equal to the number of
$\nicefrac{{x=x}}{{c|x}}\in X_{i}\cup Y_{i}=X\cup Y$ with
$\nicefrac{{x=x}}{{c|x}}\in Y_{i}$, or, $\nicefrac{{x=x}}{{c|x}}\in X_{i}$ and
$f_{i}(\nicefrac{{x=x}}{{c|x}})>1$. Since $X\cup Y=X_{i}\cup Y_{i}$,
$z_{1}+z_{2}\geq z_{1}^{\prime}+z_{2}^{\prime}$. A similar argument shows that
$z_{3}+z_{4}\geq z_{3}^{\prime}+z_{4}^{\prime}$ and $z_{4}\geq
z_{4}^{\prime}$. So the extended signature of
${\raisebox{1.5pt}{$\chi$}}_{f,g}$ is less than the extended signature of
${\raisebox{1.5pt}{$\chi$}}_{f_{i},g_{i}}$.

We now prove the moreover. Since $X,Y$ are disjoint, $X\neq X_{i}$ if and only
if $Y\neq Y_{i}$. Suppose $X\neq X_{i}$. Then there exists $\varphi/\psi\in
Y\backslash Y_{i}$. By assumption, $g(\varphi/\psi)>1$.

If $\varphi/\psi$ is $d|x/x=0$ then the short signature of
${\raisebox{1.5pt}{$\chi$}}_{f,g}$ is $(\geq,\square)$ and the short signature
of ${\raisebox{1.5pt}{$\chi$}}_{f_{i},g_{i}}$ is $(=,\square^{\prime})$ or
$(\emptyset,\square^{\prime})$, and, by what we have already proved,
$\square^{\prime}\leq\square$. So the short signature of
${\raisebox{1.5pt}{$\chi$}}_{f_{i},g_{i}}$ is strictly less than the short
signature of ${\raisebox{1.5pt}{$\chi$}}_{f,g}$. The case of $xb=0/c|x$ for
$b,c\neq 0$ is similar.

If $\varphi/\psi$ is $x=x/c|x$ then

$\left|\\{x=x/c^{\prime}|x\in Y_{i}\ |\
g_{i}(x=x/c^{\prime}|x)>1\\}\right|<\left|\\{x=x/c^{\prime}|x\in Y\\}\right|.$

So
$\mathtt{exsig}\,{\raisebox{1.5pt}{$\chi$}}_{f_{i},g_{i}}<\mathtt{exsig}\,{\raisebox{1.5pt}{$\chi$}}_{f,g}$.
The case of $\varphi/\psi$ equal to $xb=0/x=0$ is similar. ∎

###  9.1. Reducing the short signature

######  Proposition 9.5.

Let $R$ be a recursive Prüfer domain with $\mathrm{EPP}(R)$ recursive. There
is an algorithm which given $w\in W$ with short signature $(=,\square)$ or
$(\square,=)$, for some $\square\in\\{\emptyset,=,\geq\\}$, outputs
$\underline{w}\in\mathbb{W}$ such that $w\in V$ if and only if
$\underline{w}\in\mathbb{V}$, and, $\underline{w}$ is a lattice combination of
elements $w^{\prime}\in W$ such that the short signature of $w^{\prime}$ is
strictly less than the short signature of $w$.

###### Proof.

Suppose $w$ has short signature $(=,\square)$. Then $w$ is of the form

$\left|d|x/x=0\right|=D\wedge{\raisebox{1.5pt}{$\chi$}}_{f,g}\wedge\Xi$

where $d\in R\backslash\\{0\\}$, $D\in\mathbb{N}_{2}$,
$f:X\rightarrow\mathbb{N}_{2}$, $g:Y\rightarrow\mathbb{N}_{2}$, $X$ and $Y$
are disjoint finite subsets of $\\{x=x/c^{\prime}|x,xb^{\prime}=0/x=0\ |\
c^{\prime},b^{\prime}\in R\\}\cup\\{xb=0/c|x\\}$ for some $b,c\in
R\backslash\\{0\\}$, and $\Xi$ is an auxiliary sentence.

If $M\models w$ then, by 2.11 and 6.1, there exist $h\in\mathbb{N}$, prime
ideals $\mathfrak{p}_{i}\lhd R$ and ideals $I_{i}\lhd R_{\mathfrak{p}_{i}}$
for $1\leq i\leq h$ and $M^{\prime}\in\textnormal{Mod}\text{-}R$ such that
$M\equiv M^{\prime}\oplus\oplus_{i=1}^{h}R_{\mathfrak{p}_{i}}/dI_{i}$ and
$\left|d|x/x=0(M^{\prime})\right|=1$.

Thus there exists $M\in\textnormal{Mod}\text{-}R$ such that $M\models w$ if
and only if there exist $(f_{1},f_{2},g_{1},g_{2})\in\Omega_{f,g,2}$,
$h\in\mathbb{N}$, prime ideals $\mathfrak{p}_{i}\lhd R$ and $I_{i}\lhd
R_{\mathfrak{p}_{i}}$ for $1\leq i\leq h$, such that
$\oplus_{i=1}^{h}R_{\mathfrak{p}_{i}}/dI_{i}$ satisfies

$w_{(f_{1},f_{2},g_{1},g_{2})}:=\left|d|x/x=0\right|=D\wedge{\raisebox{1.5pt}{$\chi$}}_{f_{1},g_{1}}\wedge\Xi$

and $M^{\prime}$ which satisfies

$w^{\prime}_{(f_{1},f_{2},g_{1},g_{2})}:=\left|d|x/x=0\right|=1\wedge{\raisebox{1.5pt}{$\chi$}}_{f_{2},g_{2}}\wedge\Xi.$

Let $\Omega\subseteq\Omega_{f,g,2}$ be the set of
$(f_{1},f_{2},g_{1},g_{2})\in\Omega_{f,g,2}$, such that there exist
$h\in\mathbb{N}_{0}$, prime ideals $\mathfrak{p}_{i}\lhd R$ and $I_{i}\lhd
R_{\mathfrak{p}_{i}}$ for $1\leq i\leq h$, such that
$\oplus_{i=1}^{h}R_{\mathfrak{p}_{i}}/dI_{i}$ satisfies
$w_{(f_{1},f_{2},g_{1},g_{2})}$. So $w\in V$ if and only if

$\underline{w}:=\bigsqcup_{(f_{1},f_{2},g_{1},g_{2})\in\Omega}w^{\prime}_{(f_{1},f_{2},g_{1},g_{2})}\in\mathbb{V}.$

By 8.1, given $w$, we can compute $\Omega$, and, so, we can compute
$\underline{w}$. The short signature of each
$w^{\prime}_{(f_{1},f_{2},g_{1},g_{2})}$ is $(\emptyset,\square^{\prime})$ for
some $\square^{\prime}\in\\{\emptyset,=,\leq\\}$ and, by 9.4,
$\square\succeq\square^{\prime}$. So
$(=,\square)>(\emptyset,\square^{\prime})$ as required.

Suppose $w$ has short signature $(\square,=)$. Then $w$ is

$\left|xb=0/c|x\right|=E\wedge{\raisebox{1.5pt}{$\chi$}}_{f,g}\wedge\Xi$

where $b,c\in R\backslash\\{0\\}$, $E\in\mathbb{N}_{2}$,
$f:X\rightarrow\mathbb{N}_{2}$, $g:Y\rightarrow\mathbb{N}_{2}$, $X$ and $Y$
are disjoint finite subsets of $\\{x=x/c^{\prime}|x,xb^{\prime}=0/x=0\ |\
c^{\prime},b^{\prime}\in R\\}\cup\\{d|x/x=0\\}$ for some $d\in
R\backslash\\{0\\}$, and $\Xi$ is an auxiliary sentence.

If $M\models w$ then, by 2.11 and 6.2, there exist
$M_{1},M_{2},M^{\prime}\in\textnormal{Mod}\text{-}R$ such that
$b\in\textnormal{ann}_{R}M_{1}$, $c\in\textnormal{ann}_{R}M_{2}$ and
$\left|xb=0/c|x(M^{\prime})\right|=1$ and there exist $h\in\mathbb{N}_{0}$,
prime ideals $\mathfrak{p}_{i}$ and ideals $I_{i}\lhd R_{\mathfrak{p}_{i}}$
for $1\leq i\leq h$ such that $b,c\in I_{i}$ and

$\oplus_{i=1}^{h}I_{i}/bcR_{\mathfrak{p}_{i}}\oplus M_{1}\oplus M_{2}\oplus
M^{\prime}\models w.$

Therefore, $w\in V$ if and only if there exist
$E_{1},E_{2},E_{4}\in\mathbb{N}$ with $E_{1}E_{2}E_{4}=E$ and
$(\overline{f},\overline{g})\in\Omega_{f,g,4}$ such that

$w_{1,E_{1},f_{1},g_{1}}:=\left|x=x/c|x\right|=E_{1}\wedge{\raisebox{1.5pt}{$\chi$}}_{f_{1},g_{1}}\wedge\Xi\wedge\left|b|x/x=0\right|=1\in
V,$
$w_{2,E_{2},f_{2},g_{2}}:=\left|xb=0/x=0\right|=E_{2}\wedge{\raisebox{1.5pt}{$\chi$}}_{f_{2},g_{2}}\wedge\Xi\wedge\left|c|x/x=0\right|=1\in
V,$
$w_{3,f_{3},g_{3}}:={\raisebox{1.5pt}{$\chi$}}_{f_{3},g_{3}}\wedge\Xi\wedge\left|xb=0/c|x\right|=1\in
V$

and there exist $h\in\mathbb{N}_{0}$, prime ideals $\mathfrak{p}_{i}$ and
ideals $I_{i}\lhd R_{\mathfrak{p}_{i}}$ for $1\leq i\leq h$ such that $b,c\in
I_{i}$ and

$\oplus_{i=1}^{h}I_{i}/bcR_{\mathfrak{p}_{i}}\models\left|xb=0/c|x\right|=E_{4}\wedge{\raisebox{1.5pt}{$\chi$}}_{f_{4},g_{4}}\wedge\Xi.$

Let $\mathcal{H}$ be the set of
$(E_{1},E_{2},(\overline{f},\overline{g}))\in\mathbb{N}^{2}\times\Omega_{f,g,4}$
such that there exists $E_{4}$ with $E_{1}E_{2}E_{4}=E$ and there exist
$h\in\mathbb{N}_{0}$, prime ideals $\mathfrak{p}_{i}$ and ideals $I_{i}\lhd
R_{\mathfrak{p}_{i}}$ for $1\leq i\leq h$ such that $b,c\in I_{i}$ and

$\oplus_{i=1}^{h}I_{i}/bcR_{\mathfrak{p}_{i}}\models\left|xb=0/c|x\right|=E_{4}\wedge{\raisebox{1.5pt}{$\chi$}}_{f_{4},g_{4}}\wedge\Xi.$

Then $w\in V$ if and only if

$\underline{w}:=\bigsqcup_{(E_{1},E_{2},(\overline{f},\overline{g}))\in\mathcal{H}}w_{1,E_{1},f_{1},g_{1}}\sqcap
w_{2,E_{2},f_{3},g_{3}}\sqcap w_{3,f_{3},g_{3}}\in V.$

By 8.2, given $w$, we can compute $\mathcal{H}$ and so we can compute
$\underline{w}$.

Now if $w$ has short signature $(\square,=)$ then
$w_{1,E_{1},f_{1},g_{1}},w_{2,E_{2},f_{3},g_{3}}$ and $w_{3,f_{3},g_{3}}$ have
short signature $(\square^{\prime},\emptyset)$ and by 9.4,
$\square^{\prime}\preceq\square$. So
$(\square^{\prime},\emptyset)<(\square,=)$, as required. ∎

######  Lemma 9.6.

Let $R$ be an arithmetical ring. Let $c,d\in R$ and $D\in\mathbb{N}$. For all
$C\in\mathbb{N}_{2}$,

$T_{R}\models\left|\nicefrac{{xc^{D-1}d=0}}{{c|x}}\right|=1\wedge\left|\nicefrac{{d|x}}{{x=0}}\right|\geq
D\wedge\left|\nicefrac{{x=x}}{{c|x}}\right|=C\\\
\leftrightarrow\left|\nicefrac{{xc^{D-1}d=0}}{{c|x}}\right|=1\wedge\left|\nicefrac{{x=x}}{{c|x}}\right|=C$

and

$T_{R}\models\left|\nicefrac{{c^{D}d|x}}{{x=0}}\right|=1\wedge\left|\nicefrac{{d|x}}{{x=0}}\right|\geq
D\wedge\left|\nicefrac{{x=x}}{{c|x}}\right|=C\\\ \leftrightarrow\bigvee_{D\leq
E\leq
C^{D}}\left|\nicefrac{{d|x}}{{x=0}}\right|=E\wedge\left|\nicefrac{{x=x}}{{c|x}}\right|=C.$

###### Proof.

Suppose $M\models\left|\nicefrac{{xc^{D-1}d=0}}{{c|x}}\right|=1$. Then
$xc^{i}d=0\leq_{M}c|x$ for $0\leq i\leq D-1$. So, by 5.7, for $0\leq i\leq
D-1$,

$\left|\nicefrac{{c^{i}d|x}}{{c^{i+1}d|x}}(M)\right|=\left|\nicefrac{{x=x}}{{c|x}}(M)\right|.$

Hence

$\left|\nicefrac{{d|x}}{{x=0}}(M)\right|=\left|\nicefrac{{x=x}}{{c|x}}(M)\right|^{D}\left|\nicefrac{{c^{D}d|x}}{{x=0}}(M)\right|.$

Therefore, if $C\geq 2$ then

$T_{R}\models\left|\nicefrac{{xc^{D-1}d=0}}{{c|x}}\right|=1\wedge\left|\nicefrac{{d|x}}{{x=0}}\right|\geq
D\wedge\left|\nicefrac{{x=x}}{{c|x}}\right|=C\leftrightarrow\\\
\left|\nicefrac{{xc^{D-1}d=0}}{{c|x}}\right|=1\wedge\left|\nicefrac{{x=x}}{{c|x}}\right|=C.$

Now suppose that $c^{D}d\in\textnormal{ann}_{R}M$. Then $c^{D}d|x$ is
equivalent to $x=0$ in $M$. Therefore, by 5.7,

$\left|\nicefrac{{d|x}}{{c^{D}d|x}}(M)\right|=\left|\nicefrac{{x=x}}{{c^{D}|x+xd=0}}(M)\right|\leq\left|\nicefrac{{x=x}}{{c^{D}|x}}(M)\right|\leq\left|\nicefrac{{x=x}}{{c|x}}\right|^{D}.$

Hence

$T_{R}\models\left|c^{D}d|x/x=0\right|=1\wedge\left|d|x/x=0\right|\geq
D\wedge\left|x=x/c|x\right|=C\leftrightarrow\\\ \bigvee_{D\leq E\leq
C^{D}}\left|d|x/x=0\right|=E\wedge\left|x=x/c|x\right|=C.\qed$

######  Proposition 9.7.

Let $R$ be a recursive arithmetical ring. There is an algorithm, which given
$w\in W$ with extended signature
$((\geq,\square),(z_{1},z_{2}),(z_{3},z_{4}))$ with $z_{1}\geq 1$ or
$z_{3}\geq 1$, returns $\underline{w}\in\mathbb{W}$ such that $w\in V$ if and
only if $\underline{w}\in\mathbb{V}$, and,
$\mathtt{exsig}\,\underline{w}<\mathtt{exsig}\,w$.

###### Proof.

Suppose $w:={\raisebox{1.5pt}{$\chi$}}_{f,g}\wedge\Xi\in W$, where
$f:X\rightarrow\mathbb{N}_{2}$, $g:Y\rightarrow\mathbb{N}_{2}$, $X,Y$ are
disjoint finite sets of appropriate pp-pairs and $\Xi$ is an auxiliary
sentence, has extended signature
$((\geq,\square),(z_{1},z_{2}),(z_{3},z_{4}))$ with $z_{1}\geq 1$. Then
$\nicefrac{{d|x}}{{x=0}}\in Y$ for some $d\in R\backslash\\{0\\}$ and
$\nicefrac{{x=x}}{{c|x}}\in X$ for some $c\in R$. Let
$D=g(\nicefrac{{x=x}}{{c|x}})$.

Then, by 2.9, $w\in V$ if and only if

$\bigsqcup_{(\overline{f},\overline{g})\in\Omega_{f,g,2}}\left({\raisebox{1.5pt}{$\chi$}}_{f_{1},g_{1}}\wedge\left|\nicefrac{{xc^{D-1}d=0}}{{c|x}}\right|=1\wedge\Xi\right)\sqcap\left({\raisebox{1.5pt}{$\chi$}}_{f_{2},g_{2}}\wedge\left|\nicefrac{{c^{D}d|x}}{{x=0}}\right|=1\wedge\Xi\right)\in\mathbb{V}.$

For each $(\overline{f},\overline{g})\in\Omega_{f,g,2}$, we define
$\underline{w}_{\overline{f},\overline{g}},\underline{w}^{\prime}_{\overline{f},\overline{g}}\in\mathbb{W}$
such that

$\mathtt{exsig}\,\underline{w}_{\overline{f},\overline{g}},\mathtt{exsig}\,\underline{w}^{\prime}_{\overline{f},\overline{g}}<\mathtt{exsig}\,w,$
$\underline{w}_{\overline{f},\overline{g}}\in\mathbb{V}\ \text{ if and only if
}\
{\raisebox{1.5pt}{$\chi$}}_{f_{1},g_{1}}\wedge\left|\nicefrac{{xc^{D-1}d=0}}{{c|x}}\right|=1\wedge\Xi\in
V,\text{ and,}$
$\underline{w}^{\prime}_{\overline{f},\overline{g}}\in\mathbb{V}\ \text{ if
and only if }\
{\raisebox{1.5pt}{$\chi$}}_{f_{2},g_{2}}\wedge\left|\nicefrac{{c^{D}d|x}}{{x=0}}\right|=1\wedge\Xi\in
V.$

Once we have done this, the proof is complete since then

$\underline{w}:=\bigsqcup_{(\overline{f},\overline{g})\in\Omega_{f,g,2}}\underline{w}_{\overline{f},\overline{g}}\sqcap\underline{w}^{\prime}_{\overline{f},\overline{g}}$

has the properties required by the statement.

If
$\mathtt{exsig}\,{\raisebox{1.5pt}{$\chi$}}_{f_{1},g_{1}}<\mathtt{exsig}\,{\raisebox{1.5pt}{$\chi$}}_{f,g}$
then set
$\underline{w}_{\overline{f},\overline{g}}:={\raisebox{1.5pt}{$\chi$}}_{f_{1},g_{1}}\wedge\left|\nicefrac{{xc^{D-1}d=0}}{{c|x}}\right|=1\wedge\Xi$.
Otherwise, by 9.4, $X=X_{1}$ and $Y=Y_{1}$. Further,
$f_{1}(\nicefrac{{x=x}}{{c|x}})>1$ and, by definition of $\Omega_{f,g,2}$,
$g_{1}(\nicefrac{{d|x}}{{x=0}})=g(\nicefrac{{d|x}}{{x=0}})=D$. Let
$Y_{1}^{\prime}:=Y_{1}\backslash\\{\nicefrac{{d|x}}{{x=0}}\\}$ and
$g_{1}^{\prime}:=g_{1}|_{Y_{1}^{\prime}}$. Then, by 9.6,

${\raisebox{1.5pt}{$\chi$}}_{f_{1},g_{1}}\wedge\left|\nicefrac{{xc^{D-1}d=0}}{{c|x}}\right|=1\wedge\Xi\in
V$

if and only if

$\underline{w}_{\overline{f},\overline{g}}:={\raisebox{1.5pt}{$\chi$}}_{f_{1},g_{1}^{\prime}}\wedge\left|\nicefrac{{xc^{D-1}d=0}}{{c|x}}\right|=1\wedge\Xi\in\mathbb{V}.$

Moreover, $\underline{w}_{\overline{f},\overline{g}}$ has short signature
$(\emptyset,\square^{\prime})$, where, by 9.4,
$\square^{\prime}\preceq\square$. Therefore
$\mathtt{exsig}\,\underline{w}_{\overline{f},\overline{g}}<\mathtt{exsig}\,w$
because $(\emptyset,\square^{\prime})<(\emptyset,\square)$.

If
$\mathtt{exsig}\,{\raisebox{1.5pt}{$\chi$}}_{f_{2},g_{2}}<\mathtt{exsig}\,{\raisebox{1.5pt}{$\chi$}}_{f,g}$
then let
$\underline{w}^{\prime}_{\overline{f},\overline{g}}:={\raisebox{1.5pt}{$\chi$}}_{f_{2},g_{2}}\wedge\left|\nicefrac{{c^{D}d|x}}{{x=0}}\right|=1\wedge\Xi$.
Otherwise, by 9.4, $X=X_{1}$ and $Y=Y_{1}$. Further,
$f_{2}(\nicefrac{{x=x}}{{c|x}})>1$ and, by definition of $\Omega_{f,g,2}$,
$g_{2}(\nicefrac{{d|x}}{{x=0}})=g(\nicefrac{{d|x}}{{x=0}})=D$.

Let $X_{2}^{\prime}:=X_{2}\backslash\\{\nicefrac{{x=x}}{{c|x}}\\}$ and
$Y_{2}^{\prime}:=Y_{2}\backslash\\{\nicefrac{{d|x}}{{x=0}}\\}$. Let
$f_{2}^{\prime}:=f_{2}|_{X_{2}^{\prime}}$,
$g_{2}^{\prime}:=g_{2}|_{Y_{2}^{\prime}}$ and
$C:=f_{2}(\nicefrac{{x=x}}{{c|x}})$. Then
${\raisebox{1.5pt}{$\chi$}}_{f_{2},g_{2}}\wedge\left|\nicefrac{{c^{D}d|x}}{{x=0}}\right|=1\wedge\Xi$
is

$\left|\nicefrac{{d|x}}{{x=0}}\right|\geq
D\wedge\left|\nicefrac{{x=x}}{{c|x}}\right|=C\wedge{\raisebox{1.5pt}{$\chi$}}_{f_{2}^{\prime},g_{2}^{\prime}}\wedge\Xi\wedge\left|\nicefrac{{c^{D}d|x}}{{x=0}}\right|=1.$

Let

$\underline{w}^{\prime}_{\overline{f},\overline{g}}:=\bigsqcup_{D\leq E\leq
C^{D}}\left|\nicefrac{{d|x}}{{x=0}}\right|=E\wedge\left|\nicefrac{{x=x}}{{c|x}}\right|=C\wedge{\raisebox{1.5pt}{$\chi$}}_{f_{2}^{\prime},g_{2}^{\prime}}\wedge\left|\nicefrac{{c^{D}d|x}}{{x=0}}\right|=1\wedge\Xi.$

By 9.6, $\underline{w}^{\prime}_{\overline{f},\overline{g}}\in\mathbb{V}$ if
and only if
${\raisebox{1.5pt}{$\chi$}}_{f_{2},g_{2}}\wedge\left|\nicefrac{{c^{D}d|x}}{{x=0}}\right|=1\wedge\Xi\in
V$. The short signature of each component of the join defining
$\underline{w}^{\prime}_{\overline{f},\overline{g}}$ is $(=,\square^{\prime})$
where, by 9.4, $\square^{\prime}\preceq\square$. Therefore
$\mathtt{exsig}\,\underline{w}^{\prime}_{\overline{f},\overline{g}}<\mathtt{exsig}\,w$
because $(=,\square^{\prime})<(\geq,\square)$.

Suppose $w\in W$ has extended signature
$((\geq,\square),(z_{1},z_{2}),(z_{3},z_{4}))$ with $z_{3}\geq 1$. Then $Dw$
has extended signature $((\geq,\square),(z_{3},z_{4}),(z_{1},z_{2}))$. By the
previous case, we can compute $\underline{w}$, a lattice combination of
elements of $W$ with extended signatures strictly less than $Dw$ such that
$Dw\in V$ if and only if $\underline{w}\in\mathbb{V}$. Now $w\in V$ if and
only if $D\underline{w}\in\mathbb{V}$ and
$\mathtt{exsig}\,D\underline{w}<\mathtt{exsig}\,w$. ∎

######  Lemma 9.8.

Let $R$ be a Prüfer domain. Let $a,b,c\in R$ and $E,C\in\mathbb{N}_{2}$.
Suppose that $r,s,\alpha\in R$ are such that $c\alpha=ar$ and
$a(\alpha-1)=cs$. Define

  1. (1)

$\Sigma_{1}$ to be the formula $\left|\nicefrac{{x=x}}{{\alpha-1|x}}\right|=1$

  2. (2)

$\Sigma_{2}$ to be the formula
$\left|\nicefrac{{x=x}}{{\alpha|x}}\right|=1\wedge\left|\nicefrac{{ar|x}}{{x=0}}\right|=1$

  3. (3)

$\Sigma_{3}$ to be the formula
$\left|\nicefrac{{x=x}}{{\alpha|x}}\right|=1\wedge\left|\nicefrac{{xr=0}}{{a|x}}\right|=1\wedge\left|\nicefrac{{rb|x}}{{x=0}}\right|=1$

  4. (4)

$\Sigma_{4}$ to be the formula
$\left|\nicefrac{{x=x}}{{\alpha|x}}\right|=1\wedge\left|\nicefrac{{xr=0}}{{a|x}}\right|=1\wedge\left|\nicefrac{{xb=0}}{{r|x}}\right|=1$.

Then for all $M\in\textnormal{Mod}\text{-}R$ there exist
$M_{1},\ldots,M_{4}\in\textnormal{Mod}\text{-}R$ such that
$M_{i}\models\Sigma_{i}$ for $1\leq i\leq 4$ and $M\equiv
M_{1}\oplus\ldots\oplus M_{4}$.

  1. (i)

If $i\in\\{1,4\\}$ and $C<E$ then

$T_{R}\models\lnot(\Sigma_{i}\wedge\left|\nicefrac{{x=x}}{{a|x}}\right|=C\wedge\left|\nicefrac{{xb=0}}{{c|x}}\right|\geq
E),$

and if $i\in\\{1,4\\}$ and $E\leq C$ then

$T_{R}\models\Sigma_{i}\wedge\left|\nicefrac{{x=x}}{{a|x}}\right|=C\wedge\left|\nicefrac{{xb=0}}{{c|x}}\right|\geq
E\leftrightarrow\\\ \bigvee_{E\leq E^{\prime}\leq
C}\Sigma_{i}\wedge\left|\nicefrac{{x=x}}{{a|x}}\right|=C\wedge\left|\nicefrac{{xb=0}}{{c|x}}\right|=E^{\prime}.$

  2. (ii)

In $T_{R}$ the following equivalence holds.

$\Sigma_{2}\wedge\left|\nicefrac{{x=x}}{{a|x}}\right|=C\wedge\left|\nicefrac{{xb=0}}{{c|x}}\right|\geq
E\leftrightarrow\Sigma_{2}\wedge\left|\nicefrac{{x=x}}{{a|x}}\right|=C\wedge\left|\nicefrac{{xb=0}}{{x=0}}\right|\geq
E$

  3. (iii)

In $T_{R}$ the following equivalence holds.

$\Sigma_{3}\wedge\left|\nicefrac{{x=x}}{{a|x}}\right|=C\wedge\left|\nicefrac{{xb=0}}{{c|x}}\right|\geq
E\leftrightarrow\Sigma_{3}\wedge\left|\nicefrac{{x=x}}{{a|x}}\right|=C\wedge\left|\nicefrac{{xb=0}}{{r|x}}\right|\geq\lceil
E/C\rceil$

###### Proof.

The first claim follows from 2.9.

(i) Suppose $M\models\Sigma_{1}$. Then

$\left|x=x/a|x(M)\right|=\left|x=x/cs|x(M)\right|\geq\left|xb=0/c|x(M)\right|.$

Suppose $M\models\Sigma_{4}$. Then

$\left|r|x/ar|x(M)\right|=\left|x=x/xr=0+a|x(M)\right|=\left|x=x/a|x(M)\right|$

because $\left|xr=0/a|x(M)\right|=1$ and hence $xr=0\leq_{M}a|x$. So

$\left|xb=0/c|x(M)\right|=\left|xb=0/ar|x(M)\right|\leq\left|r|x/ar|x(M)\right|=\left|x=x/a|x(M)\right|.$

The first equality holds because $\left|x=x/\alpha|x(M)\right|=1$ and the
inequality holds because $\left|xb=0/r|x\right|=1$ and hence
$xb=0\leq_{M}r|x$.

Therefore, if $M\models\Sigma_{i}$ for $i\in\\{1,4\\}$ then

$\left|xb=0/c|x(M)\right|\leq\left|x=x/a|x(M)\right|.$

The first claim follows from this.

(ii) Suppose $M\models\Sigma_{2}$. Then $c|x$ is equivalent to $ar|x$ in $M$
and $ar|x$ is equivalent to $x=0$ in $M$. So
$\left|xb=0/c|x(M)\right|=\left|xb=0/x=0(M)\right|$.

(iii) Suppose $M\models\Sigma_{3}$. Then

$\left|xb=0/c|x(M)\right|=\left|xb=0/ar|x(M)\right|=\left|xb=0/r|x(M)\right|\cdot\left|r|x/ar|x(M)\right|.$

Since $\left|xr=0/a|x(M)\right|=1$, by 5.7,
$\left|r|x/ar|x(M)\right|=\left|x=x/a|x(M)\right|$. The claim now follows. ∎

######  Proposition 9.9.

Let $R$ be a recursive Prüfer domain. There is algorithm which given $w\in W$
with extended signature $((\square,\geq),(z_{1},z_{2}),(z_{3},z_{4}))$ with
$z_{1}\geq 1$ or $z_{3}\geq 1$ outputs $\underline{w}\in\mathbb{W}$ such that
$w\in V$ if and only if $\underline{w}\in\mathbb{V}$ and
$\mathtt{exsig}\,\underline{w}<\mathtt{exsig}\,w$.

###### Proof.

For $w\in W$ as in (9), define $\deg w=E$. Given $w\in W$ with extended
signature $((\square,\geq),(z_{1},z_{2}),(z_{3},z_{4}))$ with $z_{1}\geq 1$,
we will show how to compute $\underline{w}\in\mathbb{W}$, a lattice
combination of $w^{\prime}\in W$, such that for each $w^{\prime}$, either
$\mathtt{exsig}\,w^{\prime}<\mathtt{exsig}\,w$ or
$\mathtt{exsig}\,w^{\prime}=\mathtt{exsig}\,w$ and $\deg w^{\prime}<\deg w$,
and, such that $w\in V$ if and only if $\underline{w}\in\mathbb{V}$. Since
$\deg$ takes values in $\mathbb{N}$, by iterating this process we will
eventually compute $\underline{w}\in\mathbb{W}$ such that
$\mathtt{exsig}\,\underline{w}<\mathtt{exsig}\,w$.

Suppose $w:={\raisebox{1.5pt}{$\chi$}}_{f,g}\wedge\Xi\in W$, where
$f:X\rightarrow\mathbb{N}_{2}$, $g:Y\rightarrow\mathbb{N}_{2}$, $X,Y$ are
disjoint finite sets of appropriate pp-pairs and $\Xi$ is an auxiliary
sentence, has extended signature
$((\square,\geq),(z_{1},z_{2}),(z_{3},z_{4}))$ with $z_{1}\geq 1$. Since $w$
has extended signature $((\square,\geq),(z_{1},z_{2}),(z_{3},z_{4}))$ with
$z_{1}\geq 1$, there exist $\nicefrac{{x=x}}{{a|x}}\in X$ and
$\nicefrac{{xb=0}}{{c|x}}\in Y$.

Let $\Sigma_{1},\Sigma_{2},\Sigma_{3},\Sigma_{4}$ be as in 9.8. Then $w\in V$
if and only if

$\bigsqcup_{(\overline{f},\overline{g})\in\Omega_{f,g,4}}\bigsqcap_{i=1}^{4}{\raisebox{1.5pt}{$\chi$}}_{f_{i},g_{i}}\wedge\Xi\wedge\Sigma_{i}\in\mathbb{V}.$

We show that for each $(\overline{f},\overline{g})\in\Omega_{f,g,4}$, either
the extended signature of ${\raisebox{1.5pt}{$\chi$}}_{f_{i},g_{i}}\wedge\Xi$
is strictly less than the extended signature of
${\raisebox{1.5pt}{$\chi$}}_{f,g}\wedge\Xi$ or we will show that we can
compute $\underline{w}_{i,(\overline{f},\overline{g})}\in\mathbb{W}$ such that
$\underline{w}_{i,(\overline{f},\overline{g})}$ is a lattice combination of
$w^{\prime}\in W$ with either $\mathtt{exsig}\,w^{\prime}<\mathtt{exsig}\,w$,
or, $\mathtt{exsig}\,w^{\prime}\leq\mathtt{exsig}\,w$ and $\deg w^{\prime}<w$.

Fix $(\overline{f},\overline{g})\in\Omega_{f,g,4}$ and $1\leq i\leq 4$. By
9.4,
$\mathtt{exsig}\,{\raisebox{1.5pt}{$\chi$}}_{f_{i},g_{i}}\wedge\Xi<\mathtt{exsig}\,{\raisebox{1.5pt}{$\chi$}}_{f,g}\wedge\Xi$
unless $X_{i}=X$ and $Y_{i}=Y$. Moreover, if
$f_{i}(\nicefrac{{x=x}}{{a|x}})=1$ then
$\mathtt{exsig}\,{\raisebox{1.5pt}{$\chi$}}_{f_{i},g_{i}}\wedge\Xi<\mathtt{exsig}\,{\raisebox{1.5pt}{$\chi$}}_{f,g}\wedge\Xi$.
So we may assume $X_{i}=X$, $Y_{i}=Y$ and $f_{i}(\nicefrac{{x=x}}{{a|x}})>1$.

Let $X^{\prime}:=X\backslash\\{\nicefrac{{x=x}}{{a|x}}\\}$,
$Y^{\prime}:=Y\backslash\\{\nicefrac{{xb=0}}{{c|X}}\\}$,
$f_{i}^{\prime}:=f_{i}|_{X^{\prime}}$, $g_{i}^{\prime}:=g_{i}|_{Y^{\prime}}$,
$C:=f_{i}(\nicefrac{{x=x}}{{a|x}})$ and
$E:=g_{i}(\nicefrac{{xb=0}}{{c|x}})=g(\nicefrac{{xb=0}}{{c|x}})$. Then
${\raisebox{1.5pt}{$\chi$}}_{f_{i},g_{i}}\wedge\Xi$ is

$\left|\nicefrac{{x=x}}{{a|x}}\right|=C\wedge\left|\nicefrac{{xb=0}}{{c|x}}\right|\geq
E\wedge{\raisebox{1.5pt}{$\chi$}}_{f_{i}^{\prime},g_{i}^{\prime}}\wedge\Xi.$

Case $i=1$ or $i=4$: If
$f_{1}(\nicefrac{{x=x}}{{a|x}})<g_{1}(\nicefrac{{xb=0}}{{c|x}})$ then, by 9.8,

$T_{R}\models\lnot({\raisebox{1.5pt}{$\chi$}}_{f_{i},g_{i}}\wedge\Xi\wedge\Sigma_{i}).$

So set $\underline{w}_{i,(\overline{f},\overline{g})}:=\bot$.

Suppose $f_{i}(\nicefrac{{x=x}}{{a|x}})\geq g_{i}(\nicefrac{{xb=0}}{{c|x}})$.
Then, by 9.8,

$\left|\nicefrac{{x=x}}{{a|x}}\right|=C\wedge\left|\nicefrac{{xb=0}}{{c|x}}\right|\geq
E\wedge{\raisebox{1.5pt}{$\chi$}}_{f_{i}^{\prime},g_{i}^{\prime}}\wedge\Xi\in
V$

if and only

$\underline{w}_{i,(\overline{f},\overline{g})}:=\bigsqcup_{E\leq
E^{\prime}\leq
C}\left|\nicefrac{{x=x}}{{a|x}}\right|=C\wedge\left|\nicefrac{{xb=0}}{{c|x}}\right|=E^{\prime}\wedge{\raisebox{1.5pt}{$\chi$}}_{f_{i}^{\prime},g_{i}^{\prime}}\wedge\Xi\wedge\Sigma_{i}\in\mathbb{V}.$

So we are done since the short signatures of the components of the join
defining $\underline{w}_{i,(\overline{f},\overline{g})}$ are
$(\square^{\prime},=)$ where, by 9.4, $\square^{\prime}\preceq\square$.

Case $i=2$: By 9.8,

$\left|\nicefrac{{x=x}}{{a|x}}\right|=C\wedge\left|\nicefrac{{xb=0}}{{c|x}}\right|\geq
E\wedge{\raisebox{1.5pt}{$\chi$}}_{f_{2}^{\prime},g_{2}^{\prime}}\wedge\Xi\wedge\Sigma_{2}\in
V$

if and only

$\underline{w}_{i,(\overline{f},\overline{g})}:=\left|\nicefrac{{x=x}}{{a|x}}\right|=C\wedge\left|\nicefrac{{xb=0}}{{x=0}}\right|\geq
E\wedge{\raisebox{1.5pt}{$\chi$}}_{f_{2}^{\prime},g_{2}^{\prime}}\wedge\Xi\wedge\Sigma_{2}\in
V.$

So we are done since the short signature of
$\underline{w}_{i,(\overline{f},\overline{g})}$ is
$(\square^{\prime},\emptyset)$ where, by 9.4,
$\square^{\prime}\preceq\square$.

Case $i=3$: By 9.8,

$\left|\nicefrac{{x=x}}{{a|x}}\right|=C\wedge\left|\nicefrac{{xb=0}}{{c|x}}\right|\geq
E\wedge{\raisebox{1.5pt}{$\chi$}}_{f_{2}^{\prime},g_{2}^{\prime}}\wedge\Xi\wedge\Sigma_{2}\in
V$

if and only

$\underline{w}_{i,(\overline{f},\overline{g})}:=\left|\nicefrac{{x=x}}{{a|x}}\right|=C\wedge\left|\nicefrac{{xb=0}}{{r|x}}\right|\geq\lceil
E/C\rceil\wedge{\raisebox{1.5pt}{$\chi$}}_{f_{2}^{\prime},g_{2}^{\prime}}\wedge\Xi\wedge\Sigma_{3}\in
V.$

Since $\lceil E/C\rceil<E$, we have
$\deg\underline{w}_{i,(\overline{f},\overline{g})}<E=\deg{\raisebox{1.5pt}{$\chi$}}_{f,g}\wedge\Xi$.

So we have proved the lemma for the case $z_{1}\geq 1$. Suppose $w\in W$ has
extended signature $((\square,\geq),(z_{1},z_{2}),(z_{3},z_{4}))$ with
$z_{3}\geq 1$. Then $Dw$ has extended signature
$((\square,\geq),(z_{3},z_{4}),(z_{1},z_{2}))$. So by the version of the lemma
just proved, we can compute $\underline{w}\in W$ with
$\mathtt{exsig}\,\underline{w}<\mathtt{exsig}\,Dw$ such that $Dw\in V$ if and
only if $\underline{w}\in\mathbb{V}$. Now, $w\in V$ if and only if
$D\underline{w}\in\mathbb{V}$ and
$\mathtt{exsig}\,D\underline{w}<\mathtt{exsig}\,w$, as required. ∎

###  9.2. Reducing the extended signature

######  Lemma 9.10.

Let $a,b\in R$ and $A,B\in\mathbb{N}_{2}$. Suppose that $r,s,\alpha\in R$ are
such that $a\alpha=br$ and $b(\alpha-1)=as$. Define

  1. (1)

$\Sigma_{1}$ to be the sentence
$|\nicefrac{{x=x}}{{\alpha|x}}|=1\wedge|\nicefrac{{rb|x}}{{x=0}}|=1\wedge|\nicefrac{{b^{A}|x}}{{x=0}}|=1$,

  2. (2)

$\Sigma_{2}$ to be the sentence
$|\nicefrac{{x=x}}{{\alpha|x}}|=1\wedge|\nicefrac{{rb|x}}{{x=0}}|=1\wedge|\nicefrac{{xb^{A-1}=0}}{{b|x}}|=1$,

  3. (3)

$\Sigma_{3}$ to be the sentence
$|\nicefrac{{x=x}}{{\alpha|x}}|=1\wedge|\nicefrac{{xb=0}}{{r|x}}|=1$,

  4. (4)

$\Sigma_{4}$ to be the sentence
$|\nicefrac{{x=x}}{{(\alpha-1)|x}}|=1\wedge|\nicefrac{{sa|x}}{{x=0}}|=1\wedge|\nicefrac{{a^{B}|x}}{{x=0}}|=1$,

  5. (5)

$\Sigma_{5}$ to be the sentence
$|\nicefrac{{x=x}}{{(\alpha-1)|x}}|=1\wedge|\nicefrac{{sa|x}}{{x=0}}|=1\wedge|\nicefrac{{xa^{B-1}=0}}{{a|x}}|=1$,
and

  6. (6)

$\Sigma_{6}$ to be the sentence
$|\nicefrac{{x=x}}{{(\alpha-1)|x}}|=1\wedge|\nicefrac{{xa=0}}{{s|x}}|=1$.

Then, for all $M\in\textnormal{Mod}\text{-}R$, there exist
$M_{i}\in\textnormal{Mod}\text{-}R$ for $1\leq i\leq 6$ such that
$M\equiv\oplus_{i=1}^{6}M_{i}$ and for $1\leq i\leq 6$,
$M_{i}\models\Sigma_{i}$.

Moreover, there is an algorithm which, given $a,b,r,s,\alpha,A,B$ as above,
$1\leq i\leq 6$ and $\square,\square^{\prime}\in\\{=,\geq\\}$, either returns
$\bot$, in which case,

$T_{R}\models\lnot(\left|\nicefrac{{x=x}}{{a|x}}\right|\square
A\wedge\left|\nicefrac{{x=x}}{{b|x}}\right|\square^{\prime}B\wedge\Sigma_{i})$

or returns $n\in\mathbb{N}$ and $\sigma_{1},\ldots,\sigma_{n}$ such that

$T_{R}\models\left|\nicefrac{{x=x}}{{a|x}}\right|\square
A\wedge\left|\nicefrac{{x=x}}{{b|x}}\right|\square^{\prime}B\wedge\Sigma_{i}\leftrightarrow\bigvee_{j=1}^{n}(\sigma_{j}\wedge\Sigma_{i})$

and each $\sigma_{j}$ is either of the form

$\left|\nicefrac{{x=x}}{{a^{\prime}|x}}\right|\square_{j}A^{\prime}\wedge\left|\nicefrac{{x=x}}{{b^{\prime}|x}}\right|\square^{\prime}_{j}B^{\prime},$

where $\square_{j},\square^{\prime}_{j}\in\\{\emptyset,=,\geq\\}$,
$\square_{j}\preceq\square$, $\square^{\prime}_{j}\preceq\square^{\prime}$,
$a^{\prime},b^{\prime}\in R$, $A^{\prime},B^{\prime}\in\mathbb{N}$ and
$A^{\prime}B^{\prime}<AB$ or of the form

$\left|\nicefrac{{x=x}}{{a^{\prime}|x}}\right|\square_{j}A^{\prime}\wedge\left|\nicefrac{{x=x}}{{b^{\prime}|x}}\right|\square^{\prime}_{j}B^{\prime}\wedge\left|\nicefrac{{x=x}}{{x=0}}\right|=N,$

where $\square_{j},\square^{\prime}_{j}\in\\{=,\geq\\}$,
$a^{\prime},b^{\prime}\in R$ and
$A^{\prime\prime},B^{\prime\prime},N\in\mathbb{N}.$

###### Proof.

Note that

$T_{R}\vdash|\nicefrac{{x=x}}{{\alpha|x}}|=1\wedge|\nicefrac{{rb|x}}{{x=0}}|=1\rightarrow(a|x\leftrightarrow
x=0).$

Case 1: $\Sigma\in\\{\Sigma_{1},\Sigma_{2}\\}$ and $\square$ is $=$.

In this case, $\left|\nicefrac{{x=x}}{{a|x}}\right|\square
A\wedge\left|\nicefrac{{x=x}}{{b|x}}\right|\square^{\prime}B\wedge\Sigma$ is
equivalent to
$\left|\nicefrac{{x=x}}{{x=0}}\right|=A\wedge\left|\nicefrac{{x=x}}{{b|x}}\right|\square^{\prime}B\wedge\Sigma$
as required.

Case 2: $\Sigma\in\\{\Sigma_{1},\Sigma_{2}\\}$ and $\square$ is $\geq$ and
$B\geq A$.

In this case $\left|\nicefrac{{x=x}}{{a|x}}\right|\square
A\wedge\left|\nicefrac{{x=x}}{{b|x}}\right|\square^{\prime}B\wedge\Sigma$ is
equivalent to
$\left|\nicefrac{{x=x}}{{b|x}}\right|\square^{\prime}B\wedge\Sigma$.

Case 3: $\Sigma=\Sigma_{1}$, $\square$ is $\geq$ and $\square^{\prime}$ is
$=$.

If $b^{A}\in\textnormal{ann}_{R}M$ and
$\left|\nicefrac{{x=x}}{{b|x}}(M)\right|=B$ then $B\leq\left|M\right|\leq
B^{A}$. Thus $\left|\nicefrac{{x=x}}{{a|x}}\right|\geq
A\wedge\left|\nicefrac{{x=x}}{{b|x}}\right|=B\wedge\Sigma_{1}$ is equivalent
to

$\bigvee_{B^{A}\geq A^{\prime}\geq
A}\left|\nicefrac{{x=x}}{{x=0}}\right|=A^{\prime}\wedge\left|\nicefrac{{x=x}}{{b|x}}\right|\geq
B\wedge\Sigma_{1}$

Case 4: $\Sigma=\Sigma_{2}$, $\square$ is $\geq$ and $\square^{\prime}$ is
$=$.

Suppose $M$ satisfies $\left|\nicefrac{{xb^{A-1}=0}}{{b|x}}\right|=1$. Then
$\left|\nicefrac{{b^{i}|x}}{{b^{i+1}|x}}(M)\right|=\left|\nicefrac{{x=x}}{{b|x}}(M)\right|$
for $i\geq A-1$. So
$\left|\nicefrac{{x=x}}{{b^{A}|x}}(M)\right|=\left|\nicefrac{{x=x}}{{b|x}}(M)\right|^{A}$.
Thus if $\left|\nicefrac{{x=x}}{{b|x}}(M)\right|\geq 2$ then
$\left|M\right|\geq A$. Therefore $\left|\nicefrac{{x=x}}{{a|x}}\right|\geq
A\wedge\left|\nicefrac{{x=x}}{{b|x}}\right|=B\wedge\Sigma_{2}$ is equivalent
to $\left|\nicefrac{{x=x}}{{b|x}}\right|=B\wedge\Sigma_{2}$.

Case 5: $\Sigma\in\\{\Sigma_{1},\Sigma_{2}\\}$ and $\square$ is $\geq$ and
$B<A$.

If $\square^{\prime}$ is $\geq$ then

$\left|\nicefrac{{x=x}}{{a|x}}\right|\geq
A\wedge\left|\nicefrac{{x=x}}{{b|x}}\right|\geq B\wedge\Sigma$

is equivalent to

$(\left|\nicefrac{{x=x}}{{b|x}}\right|\geq A\vee\bigvee_{B\leq
B^{\prime}<A}\left|\nicefrac{{x=x}}{{x=0}}\right|\geq
A\wedge\left|\nicefrac{{x=x}}{{b|x}}\right|=B^{\prime})\wedge\Sigma$

So we may reduce to the case where $\square^{\prime}$ is $=$ and hence to
either case $3$ or $4$ at the expense of replacing $B$ by $B^{\prime}$ with
$B\leq B^{\prime}<A$. This is not a problem since in case $3$ we show that
$\sigma_{j}$ has the form
$\left|\nicefrac{{x=x}}{{a^{\prime}|x}}\right|\square_{j}A^{\prime}\wedge\left|\nicefrac{{x=x}}{{b^{\prime}|x}}\right|\square^{\prime}_{j}B^{\prime}\wedge\left|\nicefrac{{x=x}}{{x=0}}\right|=N$
i.e. there is no restriction on $A^{\prime}$ or $B^{\prime}$ and in case $4$
we show that $\left|\nicefrac{{x=x}}{{a|x}}\right|\geq
A\wedge\left|\nicefrac{{x=x}}{{b|x}}\right|=B^{\prime}\wedge\Sigma_{2}$ is
equivalent to
$\left|\nicefrac{{x=x}}{{b|x}}\right|=B^{\prime}\wedge\Sigma_{2}$ and
$B^{\prime}\leq A$.

Case 6: $\Sigma=\Sigma_{3}$.

Suppose that $M$ satisfies $\Sigma_{3}$. Then $a|x$ is equivalent to $br|x$ in
$M$ and

$\left|\nicefrac{{x=x}}{{br|x}}(M)\right|=\left|\nicefrac{{x=x}}{{b|x}}(M)\right|\cdot\left|\nicefrac{{b|x}}{{br|x}}(M)\right|=\left|\nicefrac{{x=x}}{{b|x}}(M)\right|\cdot\left|\nicefrac{{x=x}}{{r|x}}(M)\right|.$

So, in this case, the result now follows from 5.6.

The remaining cases follow from the cases we have already covered by
exchanging the roles of $\alpha$ and $\alpha-1$, $a$ and $b$, $r$ and $s$, and
$A$ and $B$. ∎

######  Proposition 9.11.

Let $R$ be a recursive Prüfer domain. There is an algorithm which, given $w\in
W$ with extended signature
$((\square_{1},\square_{2}),(z_{1},z_{2}),(z_{3},z_{4}))$ with $z_{1}+z_{2}>1$
or $z_{3}+z_{4}>1$, outputs $\underline{w}\in\mathbb{W}$ such that $w\in V$ if
and only if $\underline{w}\in\mathbb{V}$ and
$\mathtt{exsig}\,\underline{w}<\mathtt{exsig}\,w$.

###### Proof.

For $w\in W$, as in (9), define

$\deg_{1}w:=\prod_{\nicefrac{{x=x}}{{c|x}}\in
X}f(\nicefrac{{x=x}}{{c|x}})\cdot\prod_{\nicefrac{{x=x}}{{c|x}}\in
Y}g(\nicefrac{{x=x}}{{c|x}}).$

Given $w\in W$ with extended signature
$((\square_{1},\square_{2}),(z_{1},z_{2}),(z_{3},z_{4}))$ with
$z_{1}+z_{2}>1$, we will show how to compute $\underline{w}\in\mathbb{W}$, a
lattice combination of $w^{\prime}\in W$, such that $w\in V$ if and only if
$\underline{w}\in\mathbb{V}$ and such that for each $w^{\prime}$, either
$\mathtt{exsig}\,w^{\prime}<\mathtt{exsig}\,w$, or,
$\mathtt{exsig}\,w^{\prime}\leq\mathtt{exsig}\,w$ and
$\deg_{1}w^{\prime}<\deg_{1}w$. Since $\deg_{1}$ takes values in $\mathbb{N}$,
by iterating this process, we will eventually compute
$\underline{w}\in\mathbb{W}$ which is a lattice combination of $w^{\prime}\in
W$ such that $\mathtt{exsig}\,w^{\prime}<\mathtt{exsig}\,w$.

We start with a special case. Let $a,b,\alpha,r,s\in R$ be such that
$a\alpha=br$ and $b(\alpha-1)=as$. Let $\Sigma_{i}$, for $1\leq i\leq 6$, be
as in 9.10. Suppose that, for some $1\leq i\leq 6$, $w$ is

$\left|\nicefrac{{x=x}}{{a|x}}\right|\square
A\wedge\left|\nicefrac{{x=x}}{{b|x}}\right|\square^{\prime}B\wedge{\raisebox{1.5pt}{$\chi$}}_{f,g}\wedge\Sigma_{i}\wedge\Xi\in
W$

where $A,B\in\mathbb{N}_{2}$, $\square,\square^{\prime}\in\\{=,\geq\\}$, $X$
and $Y$ are finite sets of appropriate pp-pairs, $f:X\rightarrow\mathbb{N}$,
$g:Y\rightarrow\mathbb{N}$ and $\Xi$ is an auxiliary sentence.

We will compute $u=\bigsqcup_{j=1}^{n}u_{j}$ such that for each $1\leq j\leq
n$, either $\mathtt{exsig}\,u_{j}<\mathtt{exsig}\,w$, or,
$\mathtt{exsig}\,u_{j}\leq\mathtt{exsig}\,w$ and $\deg_{1}u_{j}<\deg_{1}w$.

If the algorithm from 9.10 returns $\bot$, then

$T_{R}\models\lnot(\left|\nicefrac{{x=x}}{{a|x}}\right|\square
A\wedge\left|\nicefrac{{x=x}}{{b|x}}\right|\square^{\prime}B\wedge{\raisebox{1.5pt}{$\chi$}}_{f,g}\wedge\Sigma_{i}\wedge\Xi).$

In this case, set $u:=\bot$. Otherwise, let $\sigma_{1},\ldots\sigma_{n}$ be,
as in 9.10, such that

$T_{R}\models\left|\nicefrac{{x=x}}{{a|x}}\right|\square
A\wedge\left|\nicefrac{{x=x}}{{b|x}}\right|\square^{\prime}B\wedge\Sigma_{i}\leftrightarrow\bigvee_{j=1}^{n}(\sigma_{j}\wedge\Sigma_{i}).$

Therefore

$T_{R}\models\left|\nicefrac{{x=x}}{{a|x}}\right|\square
A\wedge\left|\nicefrac{{x=x}}{{b|x}}\right|\square^{\prime}B\wedge{\raisebox{1.5pt}{$\chi$}}_{f,g}\wedge\Sigma_{i}\wedge\Xi\leftrightarrow\bigvee_{j=1}^{n}(\sigma_{j}\wedge{\raisebox{1.5pt}{$\chi$}}_{f,g}\wedge\Sigma_{i}\wedge\Xi)$

So $w\in V$ if and only if

$\bigsqcup_{j=1}^{n}(\sigma_{j}\wedge{\raisebox{1.5pt}{$\chi$}}_{f,g}\wedge\Sigma_{i}\wedge\Xi)\in\mathbb{V}.$

If $\sigma_{j}$ is of the form

$|x=x/a^{\prime}|x|\square_{j}A^{\prime}\wedge|x=x/b^{\prime}|x|\square^{\prime}_{j}B^{\prime}\wedge|x=x/x=0|=N,$

where $\square_{j},\square^{\prime}_{j}\in\\{=,\geq\\}$,
$a^{\prime},b^{\prime}\in R$ and
$A^{\prime\prime},B^{\prime\prime},N\in\mathbb{N}$ then
$\sigma_{j}\wedge{\raisebox{1.5pt}{$\chi$}}_{f,g}\wedge\Sigma_{i}\wedge\Xi$ is
a sentence about an $R$-module of fixed finite size. So, since
$\mathrm{EPP}(R)$ is recursive, by 7.6, we can effectively decide whether
$\sigma_{j}\wedge{\raisebox{1.5pt}{$\chi$}}_{f,g}\wedge\Sigma_{i}\wedge\Xi$
holds in some $R$-module. Set $u_{j}:=\top$ if
$\sigma_{j}\wedge{\raisebox{1.5pt}{$\chi$}}_{f,g}\wedge\Sigma_{i}\wedge\Xi$ is
true in some $R$-module and $u_{j}:=\bot$ otherwise. Finally, if $\sigma_{j}$
is of the form

$|x=x/a^{\prime}|x|\square_{j}A^{\prime}\wedge|x=x/b^{\prime}|x|\square^{\prime}_{j}B^{\prime},$

where $\square_{j},\square^{\prime}_{j}\in\\{\emptyset,=,\geq\\}$,
$a^{\prime},b^{\prime}\in R$ and $A^{\prime},B^{\prime}\in\mathbb{N}$ with
$\square_{j}\preceq\square$, $\square^{\prime}_{j}\preceq\square^{\prime}$ and
$A^{\prime}B^{\prime}<AB$ then set
$u_{j}:=\sigma_{j}\wedge{\raisebox{1.5pt}{$\chi$}}_{f,g}\wedge\Sigma_{i}\wedge\Xi$.
The condition that $\square_{j}\preceq\square$ and
$\square^{\prime}_{j}\preceq\square^{\prime}$ ensures that
$\mathtt{exsig}\,u_{j}\leq\mathtt{exsig}\,w$. The condition that
$A^{\prime}B^{\prime}<AB$ implies that $\deg_{1}u_{j}<\deg_{1}w$. So $w\in V$
if and only if $u:=\bigsqcup_{j=1}^{n}u_{j}\in\mathbb{V}$ and
$\deg_{1}u_{j}<\deg_{1}w$ for $1\leq j\leq n$ as required.

We now consider the general case. Suppose

$w:={\raisebox{1.5pt}{$\chi$}}_{f,g}\wedge\Xi,$

where $X$ and $Y$ are disjoint finite sets of appropriate pp-pairs,
$f:X\rightarrow\mathbb{N}_{2}$, $g:Y\rightarrow\mathbb{N}_{2}$ and $\Xi$ is an
auxiliary sentence, has extended signature
$((\square_{1},\square_{2}),(z_{1},z_{2}),(z_{3},z_{4}))$ with
$z_{1}+z_{2}>1$. There exist $a,b\in R$ with $a\neq b$ such that
$\nicefrac{{x=x}}{{a|x}},\nicefrac{{x=x}}{{b|x}}\in X\cup Y$. Then $w\in V$ if
and only if

$\bigsqcup_{(\overline{f},\overline{g})\in\Omega_{f,g,6}}\bigsqcap_{i=1}^{6}\Sigma_{i}\wedge{\raisebox{1.5pt}{$\chi$}}_{f_{i},g_{i}}\wedge\Xi\in\mathbb{V}.$

Claim: For all $(\overline{f},\overline{g})\in\Omega_{f,g,6}$ and $1\leq i\leq
6$, either

$\mathtt{exsig}\,\Sigma_{i}\wedge{\raisebox{1.5pt}{$\chi$}}_{f_{i},g_{i}}\wedge\Xi<\mathtt{exsig}\,{\raisebox{1.5pt}{$\chi$}}_{f,g}\wedge\Xi,\text{
or,
}\deg_{1}\Sigma_{i}\wedge{\raisebox{1.5pt}{$\chi$}}_{f_{i},g_{i}}\wedge\Xi<\deg_{1}{\raisebox{1.5pt}{$\chi$}}_{f,g},$

or, $X_{i}=X$, $Y_{i}=Y$,
$f_{i}(\nicefrac{{x=x}}{{c|x}})=f(\nicefrac{{x=x}}{{c|x}})$ for all
$\nicefrac{{x=x}}{{c|x}}\in X$ and
$g_{i}(\nicefrac{{x=x}}{{c|x}})=g(\nicefrac{{x=x}}{{c|x}})$ for all
$\nicefrac{{x=x}}{{c|x}}\in Y$.

By 9.4, if $X_{i}\neq X$ or $Y_{i}\neq Y$ then the extended signature of
${\raisebox{1.5pt}{$\chi$}}_{f_{i},g_{i}}$ is strictly less that the extended
signature of ${\raisebox{1.5pt}{$\chi$}}_{f,g}$. So suppose that $X=X_{i}$,
$Y=Y_{i}$. By definition,
$g(\nicefrac{{x=x}}{{c|x}})=g_{i}(\nicefrac{{x=x}}{{c|x}})$ for all
$\nicefrac{{x=x}}{{c|x}}\in Y=Y_{i}$ and $f_{i}(\nicefrac{{x=x}}{{c|x}})\leq
f(\nicefrac{{x=x}}{{c|x}})$ for all $\nicefrac{{x=x}}{{c|x}}\in X=X_{i}$. So,
if $f_{i}(\nicefrac{{x=x}}{{c|x}})<f(\nicefrac{{x=x}}{{c|x}})$, for some
$\nicefrac{{x=x}}{{c|x}}\in X$, then
$\deg_{1}\Sigma_{i}\wedge{\raisebox{1.5pt}{$\chi$}}_{f_{i},g_{i}}\wedge\Xi<\deg_{1}{\raisebox{1.5pt}{$\chi$}}_{f,g}$.
So we have proved the claim.

Therefore for each $(\overline{f},\overline{g})\in\Omega_{f,g,6}$ and $1\leq
i\leq 6$, either
$\mathtt{exsig}\,\Sigma_{i}\wedge{\raisebox{1.5pt}{$\chi$}}_{f_{i},g_{i}}\wedge\Xi<\mathtt{exsig}\,{\raisebox{1.5pt}{$\chi$}}_{f,g}\wedge\Xi$,
$\deg_{1}{\raisebox{1.5pt}{$\chi$}}_{f_{i},g_{i}}\wedge\Sigma_{i}\wedge\Xi<\deg_{1}{\raisebox{1.5pt}{$\chi$}}_{f,g}\wedge\Xi$
or ${\raisebox{1.5pt}{$\chi$}}_{f_{i},g_{i}}\wedge\Sigma_{i}\wedge\Xi$ is of
the form of the special case. Since, by 9.4, we always have
$\mathtt{exsig}\,\Sigma_{i}\wedge{\raisebox{1.5pt}{$\chi$}}_{f_{i},g_{i}}\wedge\Xi\leq\mathtt{exsig}\,{\raisebox{1.5pt}{$\chi$}}_{f,g}\wedge\Xi$,
we are done.

The version of the lemma with $z_{3}+z_{4}>1$ follows from the one we have
just proved by applying duality as in 9.9. ∎

Say $w\in W$ is reducible if $w$ has

  * •

short signature $(=,\square)$ or $(\square,=)$, or

  * •

extended signature $((\geq,\square),(z_{1},z_{2}),(z_{3},z_{4}))$ or
$((\square,\geq),(z_{1},z_{2}),(z_{3},z_{4}))$ with $z_{1}\geq 1$ or
$z_{3}\geq 1$, or

  * •

extended signature $((\square,\square^{\prime}),(z_{1},z_{2}),(z_{3},z_{4}))$
with $z_{1}+z_{2}>1$ or $z_{3}+z_{4}>1$.

######  Remark 9.12.

If $w\in W$ is reducible then $w$ is of the form required by one of the lemmas
9.5, 9.7, 9.9 or 9.11.

Thus $w\in W$ is not reducible if and only if $w$ has extended signature

  * •

$(\emptyset,\emptyset,(z_{1},z_{2}),(z_{3},z_{4}))$ with
$(z_{1},z_{2}),(z_{3},z_{4})\in\\{(1,0),(0,1),(0,0)\\}$ or

  * •

$((\square,\square^{\prime}),(z_{1},z_{2}),(z_{3},z_{4}))$ with
$(\square,\square^{\prime})\in\\{(\geq,\emptyset),(\emptyset,\geq),(\geq,\geq),(\emptyset,\emptyset)\\}$
and

$(z_{1},z_{2}),(z_{3},z_{4})\in\\{(0,1),(0,0)\\}$.

The next remark follows directly from 3.9.

######  Remark 9.13.

Let $R$ be a Prüfer domain. There is an algorithm which, given $w\in W$ with
extended signature $((\square_{1},\square_{2}),(z_{1},z_{2}),(z_{3},z_{4}))$
with
$(\square_{1},\square_{2})\in\\{(\geq,\emptyset),(\emptyset,\geq),(\geq,\geq),(\emptyset,\emptyset),\\}$
and $(z_{1},z_{2}),(z_{3},z_{4})\in\\{(0,1),(0,0)\\}$, answers whether $w\in
V$ or not.

##  10\. Algorithms for sentences which are not reducible

######  Lemma 10.1.

Let $R$ be a recursive Prüfer domain with $X(R)$ recursive. There is an
algorithm which, given $\lambda\in R$, $C\in\mathbb{N}$ and
$(r,ra,\gamma,\delta)\in R^{4}$, answers whether there exist
$h\in\mathbb{N}_{0}$, prime ideals $\mathfrak{p}_{i}\lhd R$ and ideals
$I_{i}\lhd R_{\mathfrak{p}_{i}}$ for $1\leq i\leq h$ such that
$(\mathfrak{p}_{i},I_{i})\models(r,ra,\gamma,\delta)$ and
$\left|\oplus_{i=1}^{h}R_{\mathfrak{p}_{i}}/\lambda
R_{\mathfrak{p}_{i}}\right|=C$.

###### Proof.

If $C=1$ then there is nothing to prove, so suppose that $C\neq 1$. Let
$p_{1},\ldots,p_{l}\in\mathbb{P}$ be distinct primes and
$n_{1},\ldots,n_{l}\in\mathbb{N}$ be such that
$C=\prod_{j=1}^{l}p_{j}^{n_{j}}$. For each prime ideal $\mathfrak{p}\lhd R$,
if $\left|R_{\mathfrak{p}}/\lambda R_{\mathfrak{p}}\right|$ is finite then it
is a prime power. Therefore, there exist prime ideals $\mathfrak{p}_{i}\lhd R$
and ideals $I_{i}\lhd R_{\mathfrak{p}_{i}}$ for $1\leq i\leq h$ such that
$(\mathfrak{p}_{i},I_{i})\models(r,ra,\gamma,\delta)$ and
$\left|\oplus_{i}R_{\mathfrak{p}_{i}}/\lambda R_{\mathfrak{p}_{i}}\right|=C$
if and only if for each $1\leq j\leq l$, there exist $h_{j}\in\mathbb{N}_{0}$,
prime ideals $\mathfrak{p}_{ij}\lhd R$ and ideals $I_{ij}\lhd
R_{\mathfrak{p}_{ij}}$ for $1\leq i\leq h_{j}$ such that
$(\mathfrak{p}_{ij},I_{ij})\models(r,ra,\gamma,\delta)$ and
$\left|\oplus_{i=1}^{h_{j}}R_{\mathfrak{p}_{ij}}/\lambda
R_{\mathfrak{p}_{ij}}\right|=p^{n_{j}}$. Thus we may reduce to the case that
$C=p^{n}$ for some $p\in\mathbb{P}$ and $n\in\mathbb{N}$.

We consider the cases $r=0$ and $r\neq 0$ separately.

Case $r=0$: For $\mathfrak{p}\lhd R$ prime and $I\lhd R_{\mathfrak{p}}$,
$(\mathfrak{p},I)\models(0,0,\gamma,\delta)$ if and only if $I=0$,
$\gamma\notin\mathfrak{p}$ and $\delta\neq 0$. So, there exist
$h\in\mathbb{N}_{0}$, prime ideals $\mathfrak{p}_{i}\lhd R$ and ideals
$I_{i}\lhd R_{\mathfrak{p}_{i}}$ for $1\leq i\leq h$ such that
$(\mathfrak{p}_{i},I_{i})\models(0,0,\gamma,\delta)$ and
$\left|\oplus_{i=1}^{h}R_{\mathfrak{p}_{i}}/\lambda
R_{\mathfrak{p}_{i}}\right|=p^{n}$ if and only if $(p,n;\lambda,\gamma,0,1)\in
X(R)$ and $\delta\neq 0$.

Case $r\neq 0$: If $r\neq 0$ then
$(\mathfrak{p},I)\models(r,ra,\gamma,\delta)$ if and only if $I=rJ$ for some
$J\lhd R_{\mathfrak{p}}$ and $(\mathfrak{p},J)\models(1,a,\gamma,\delta)$. So,
there exist $h\in\mathbb{N}_{0}$, prime ideals $\mathfrak{p}_{i}\lhd R$ and
ideals $I_{i}\lhd R_{\mathfrak{p}_{i}}$ for $1\leq i\leq h$ such that
$(\mathfrak{p}_{i},I_{i})\models(r,ra,\gamma,\delta)$ and
$\left|\oplus_{i=1}^{h}R_{\mathfrak{p}_{i}}/\lambda
R_{\mathfrak{p}_{i}}\right|=p^{n}$ if and only if there exist
$h\in\mathbb{N}_{0}$, prime ideals $\mathfrak{p}_{i}\lhd R$ and ideals
$J_{i}\lhd R_{\mathfrak{p}_{i}}$ for $1\leq i\leq h$ such that
$(\mathfrak{p}_{i},J_{i})\models(1,a,\gamma,\delta)$ and
$\left|\oplus_{i=1}^{h}R_{\mathfrak{p}_{i}}/\lambda
R_{\mathfrak{p}_{i}}\right|=p^{n}$. The last statement is equivalent to
$(p,n;\lambda,\gamma,a,\delta)\in X(R)$. ∎

######  Proposition 10.2.

Let $R$ be a recursive Prüfer domain with $\mathrm{EPP}(R)$ and $X(R)$
recursive. There is an algorithm which, given $c\in R$, $C\in\mathbb{N}$ and
$\Xi$ an auxiliary sentence, answers whether there exists
$M\in\textnormal{Mod}\text{-}R$ such that

$M\models\left|\nicefrac{{x=x}}{{c|x}}\right|=C\wedge\Xi.$

###### Proof.

Let $\chi$ be the sentence $\left|\nicefrac{{x=x}}{{c|x}}\right|=C\wedge\Xi$.
By 2.11, there exists $M\models{\raisebox{1.5pt}{$\chi$}}$ if and only if
there exist prime ideals $\mathfrak{p}_{i}\lhd R$ and uniserial
$R_{\mathfrak{p}_{i}}$-modules $U_{i}$ for $1\leq i\leq l$ such that
$\oplus_{i=1}^{l}U_{i}\models{\raisebox{1.5pt}{$\chi$}}$. Moreover, we may
assume that $U_{i}/U_{i}c\neq 0$ for all $1\leq i\leq l$. By 6.3, for each
$1\leq i\leq l$, either $c\in\textnormal{ann}_{R}U_{i}$ or, for some ideal
$I_{i}\lhd R_{\mathfrak{p}_{i}}$, $U_{i}\cong R_{\mathfrak{p}_{i}}/cI_{i}$.
Thus, there exists $M\models{\raisebox{1.5pt}{$\chi$}}$ if and only if there
exist $A,B\in\mathbb{N}$ with $AB=C$, $F\in\textnormal{Mod}\text{-}R$ such
that

$F\models\left|\nicefrac{{x=x}}{{x=0}}\right|=A\wedge\left|\nicefrac{{c|x}}{{x=0}}\right|=1\wedge\Xi$

and $h\in\mathbb{N}_{0}$, prime ideals $\mathfrak{p}_{i}\lhd R$ and ideals
$I_{i}\lhd R_{\mathfrak{p}_{i}}$ for $1\leq i\leq h$ such that
$\left|\oplus_{i=1}^{h}R_{\mathfrak{p}_{i}}/cR_{\mathfrak{p}_{i}}\right|=B$
and $R_{\mathfrak{p}_{i}}/cI_{i}\models\Xi$ for $1\leq i\leq h$. So, since
$\mathrm{EPP}(R)$ is recursive, by 7.6, it is enough to show that there is an
algorithm which answers whether there exist $h\in\mathbb{N}_{0}$, prime ideals
$\mathfrak{p}_{i}\lhd R$ and ideals $I_{i}\lhd R_{\mathfrak{p}_{i}}$ for
$1\leq i\leq h$ such that
$\left|\oplus_{i=1}^{h}R_{\mathfrak{p}_{i}}/cR_{\mathfrak{p}_{i}}\right|=B$
and $R_{\mathfrak{p}_{i}}/cI_{i}\models\Xi$ for $1\leq i\leq h$.

By 6.10, we can compute $(r_{j},r_{j}a_{j},\gamma_{j},\delta_{j})$ for $1\leq
j\leq n$ such that
$(\mathfrak{p},I)\models(r_{j},r_{j}a_{j},\gamma_{j},\delta_{j})$ for some
$1\leq j\leq n$ if and only if $R_{\mathfrak{p}}/cI\models\Xi$. Thus there
exist $h\in\mathbb{N}_{0}$, prime ideals $\mathfrak{p}_{i}\lhd R$ and ideals
$I_{i}\lhd R_{\mathfrak{p}_{i}}$ for $1\leq i\leq h$ such that
$\left|\oplus_{i=1}^{h}R_{\mathfrak{p}_{i}}/cR_{\mathfrak{p}_{i}}\right|=B$
and $R_{\mathfrak{p}_{i}}/cI_{i}\models\Xi$ for $1\leq i\leq h$ if and only if
there exist $B_{j}\in\mathbb{N}$ for $1\leq j\leq n$ such that
$B=\prod_{j=1}^{n}B_{j}$ and for $1\leq j\leq n$, there exist
$h_{j}\in\mathbb{N}_{0}$, prime ideals $\mathfrak{p}_{ij}\lhd R$ and ideals
$I_{ij}\lhd R_{\mathfrak{p}_{ij}}$ for $1\leq i\leq h_{j}$ such that
$\left|\oplus_{i=1}^{h_{j}}R_{\mathfrak{p}_{ij}}/cR_{\mathfrak{p}_{ij}}\right|=B_{j}$
and
$(\mathfrak{p}_{ij},I_{ij})\models(r_{j},r_{j}a_{j},\gamma_{j},\delta_{j})$
for $1\leq i\leq h_{j}$. The result now follows from 10.1. ∎

######  Corollary 10.3.

Let $R$ be a recursive Prüfer domain with $\mathrm{EPP}(R)$ and $X(R)$
recursive. There is an algorithm which, given $b\in R$, $B\in\mathbb{N}$ and
$\Xi$ an auxiliary sentence, answers whether there exists
$M\in\textnormal{Mod}\text{-}R$ such that

$M\models\left|\nicefrac{{xb=0}}{{x=0}}\right|=B\wedge\Xi.$

###### Proof.

Apply 2.3 to 10.2. ∎

######  Proposition 10.4.

Let $R$ be a recursive Prüfer domain with $\mathrm{EPP}(R)$ and $X(R)$
recursive. There is an algorithm which, given $c,b\in R$, $C,B\in\mathbb{N}$
and $\Xi$ an auxiliary sentence, answers whether there exists
$M\in\textnormal{Mod}\text{-}R$ such that

$M\models\left|\nicefrac{{x=x}}{{c|x}}\right|=C\wedge\left|\nicefrac{{xb=0}}{{x=0}}\right|=B\wedge\Xi.$

###### Proof.

For $\alpha\in R$, we write $\alpha\notin\textnormal{Att}$ for the sentence
$\left|\nicefrac{{x\alpha=0}}{{x=x}}\right|=1\wedge\left|\nicefrac{{x=x}}{{\alpha|x}}\right|=1$.
Recall, 2.9, that for all $M\in\textnormal{Mod}\text{-}R$, there are
$M_{1},M_{2}\in\textnormal{Mod}\text{-}R$ such that $M\equiv M_{1}\oplus
M_{2}$, $M_{1}$ satisfies $\alpha\notin\textnormal{Att}$ and $M_{2}$ satisfies
$\alpha-1\notin\textnormal{Att}$.

Let $\alpha,u,v\in R$ be such that $c\alpha=bu$ and $b(\alpha-1)=cv$. There
exists an $R$-module which satisfies
$\left|\nicefrac{{x=x}}{{c|x}}\right|=C\wedge\left|\nicefrac{{xb=0}}{{x=0}}\right|=B\wedge\Xi$
if and only if there exist $C_{1},C_{2},B_{1},B_{2}\in\mathbb{N}$ with
$C=C_{1}C_{2}$ and $B=B_{1}B_{2}$ and there exists an $R$-module satisfying

$\left|\nicefrac{{x=x}}{{bu|x}}\right|=C_{1}\wedge\left|\nicefrac{{xb=0}}{{x=0}}\right|=B_{1}\wedge\alpha\notin\textrm{Att}\wedge\Xi$

and an $R$-module satisfying

$\left|\nicefrac{{x=x}}{{c|x}}\right|=C_{2}\wedge\left|\nicefrac{{xcv=0}}{{x=0}}\right|=B_{2}\wedge(\alpha-1)\notin\textrm{Att}\wedge\Xi.$

By 2.3, there exists an $R$-module satisfying

$\left|\nicefrac{{x=x}}{{c|x}}\right|=C_{2}\wedge\left|\nicefrac{{xcv=0}}{{x=0}}\right|=B_{2}\wedge\alpha\notin\textrm{Att}\wedge\Xi$

if and only if there exists an $R$-module satisfying

$\left|\nicefrac{{xc=0}}{{x=0}}\right|=C_{2}\wedge\left|\nicefrac{{x=x}}{{cv|x}}\right|=B_{2}\wedge\alpha\notin\textrm{Att}\wedge
D\Xi.$

Thus, in order to prove the proposition, it is enough to show that there is an
algorithm which, given $b,u\in R$, $C,B\in\mathbb{N}$ and $\Xi$ an auxiliary
sentence, answers whether there exists an $R$-module satisfying the sentence
$\chi$ defined as

$\left|\nicefrac{{x=x}}{{bu|x}}\right|=C\wedge\left|\nicefrac{{xb=0}}{{x=0}}\right|=B\wedge\Xi.$

We may assume that $bu\neq 0$, for otherwise $\chi$ is a sentence about an
$R$-module of fixed finite size and since $\mathrm{EPP}(R)$ is recursive, we
can decide whether there exist $R$-modules satisfying such sentences.

By 2.11 and 6.3, there exists an $R$-module satisfying $\chi$ if and only if
there exists $F\in\textnormal{Mod}\text{-}R$ with
$bu\in\textnormal{ann}_{R}F$, $h\in\mathbb{N}_{0}$, prime ideals
$\mathfrak{p}_{i}\lhd R$ and ideals $I_{i}\lhd R_{\mathfrak{p}_{i}}$ for
$1\leq i\leq h$ and $M\in\textnormal{Mod}\text{-}R$ with $Mbu=M$ such that
$F\oplus\bigoplus_{i=1}^{h}R_{\mathfrak{p}_{i}}/buI_{i}\oplus M$ satisfies
$\chi$. Now, this happens if and only if there exist
$C_{1},C_{2}\in\mathbb{N}$ and $B_{1},B_{2},B_{3}\in\mathbb{N}$ with
$C=C_{1}C_{2}$ and $B=B_{1}B_{2}B_{3}$ such that

$F\models\left|\nicefrac{{x=x}}{{x=0}}\right|=C_{1}\wedge\left|\nicefrac{{xb=0}}{{x=0}}\right|=B_{1}\wedge\left|\nicefrac{{bu|x}}{{x=0}}\right|=1\wedge\Xi,$
$\oplus_{i=1}^{h}R_{\mathfrak{p}_{i}}/buI_{i}\models\left|\nicefrac{{x=x}}{{bu|x}}\right|=C_{2}\wedge\left|\nicefrac{{xb=0}}{{x=0}}\right|=B_{2}\wedge\Xi\text{,
and}$
$M\models\left|\nicefrac{{xb=0}}{{x=0}}\right|=B_{3}\wedge\left|\nicefrac{{x=x}}{{bu|x}}\right|=1\wedge\Xi.$

In view of 7.6 and 10.3, it is therefore enough to show that there is an
algorithm which answers whether there exists $h\in\mathbb{N}_{0}$, prime
ideals $\mathfrak{p}_{i}\lhd R$ and ideals $I_{i}\lhd R_{\mathfrak{p}_{i}}$
for $1\leq i\leq h$ such that

$\oplus_{i=1}^{h}R_{\mathfrak{p}_{i}}/buI_{i}\models\left|\nicefrac{{x=x}}{{bu|x}}\right|=C_{2}\wedge\left|\nicefrac{{xb=0}}{{x=0}}\right|=B_{2}\wedge\Xi.$

By 6.10, we can compute $n\in\mathbb{N}$ and
$(r_{j},r_{j}a_{j},\gamma_{j},\delta_{j})$ for $1\leq j\leq n$ such that
$R_{\mathfrak{p}}/buI\models\Xi$ if and only if
$(\mathfrak{p},I)\models(r_{j},r_{j}a_{j},\gamma_{j},\delta_{j})$ for some
$1\leq j\leq n$. It is therefore enough to show that there is an algorithm
which, given $(r,ra,\gamma,\delta)$, $b,u\in R$ and $C,B\in\mathbb{N}$,
answers whether there exist $h\in\mathbb{N}_{0}$, prime ideals
$\mathfrak{p}_{i}$ for $1\leq i\leq h$ and ideals $I_{i}\lhd
R_{\mathfrak{p}_{i}}$ for $1\leq i\leq h$ such that
$(\mathfrak{p}_{i},I_{i})\models(r,ra,\gamma,\delta)$ and

$\oplus_{i=1}^{h}R_{\mathfrak{p}_{i}}/buI_{i}\models\left|\nicefrac{{x=x}}{{bu|x}}\right|=C\wedge\left|\nicefrac{{xb=0}}{{x=0}}\right|=B.$

Case $r=0$: In this case $(\mathfrak{p},I)\models(r,ra,\gamma,\delta)$ implies
$I=0$. Moreover $(\mathfrak{p},0)\models(r,ra,\gamma,\delta)$ if and only if
$\gamma\notin\mathfrak{p}$ and $\delta\neq 0$. Thus, there exist prime ideals
$\mathfrak{p}_{i}$ for $1\leq i\leq h$ such that
$(\mathfrak{p}_{i},0)\models(r,ra,\gamma,\delta)$ and
$\oplus_{i=1}^{h}R_{\mathfrak{p}_{i}}\models\left|\nicefrac{{x=x}}{{bu|x}}\right|=C\wedge\left|\nicefrac{{xb=0}}{{x=0}}\right|=B$
if and only if $\delta\neq 0$, $B=1$ and there exist prime ideals
$\mathfrak{p}_{i}$ for $1\leq i\leq h$ such that
$\gamma\notin\mathfrak{p}_{i}$ and
$|\oplus_{i=1}^{h}R_{\mathfrak{p}_{i}}/buR_{\mathfrak{p}_{i}}|=C$. Such an
algorithm exists since $\mathrm{EPP}(R)$ is recursive.

Case $r\neq 0$: For all prime ideals $\mathfrak{p}$ and ideals $I\lhd
R_{\mathfrak{p}}$, since $bu\neq 0$,

$\left|xb=0/x=0(R_{\mathfrak{p}}/buI)\right|=\left|(buI:b)/buI\right|=\left|I/bI\right|.$

Now, if $\left|I/bI\right|$ is finite but not equal to $1$ then $I=\lambda
R_{\mathfrak{p}}$ for some $\lambda\neq 0$ and
$\left|I/bI\right|=\left|R_{\mathfrak{p}}/bR_{\mathfrak{p}}\right|$. Since
$b\neq 0$, $\left|I/bI\right|=1$ if and only if $b\notin I^{\\#}$.

Therefore, there exist prime ideals $\mathfrak{p}_{i}$ and ideals $I_{i}\lhd
R_{\mathfrak{p}_{i}}$ for $1\leq i\leq h$ such that
$\oplus_{i=1}^{h}R_{\mathfrak{p}_{i}}/buI_{i}\models\left|\nicefrac{{x=x}}{{bu|x}}\right|=C\wedge\left|\nicefrac{{xb=0}}{{x=0}}\right|=B\wedge\Xi$
if and only if there exist $C^{\prime},C^{\prime\prime}\in\mathbb{N}$ with
$C^{\prime}C^{\prime\prime}=C$ such that the following conditions hold.

  * (i)

There exist prime ideals $\mathfrak{p}_{i}\lhd R$ and $\lambda_{i}\in
R\backslash\\{0\\}$ for $1\leq i\leq h$ such that
$(\mathfrak{p}_{i},\lambda_{i}R_{\mathfrak{p}_{i}})\models(r,ra,\gamma,\delta)$,
$\left|\oplus_{i=1}^{h}R_{\mathfrak{p}_{i}}/buR_{\mathfrak{p}_{i}}\right|=C^{\prime}$
and
$\left|\oplus_{i=1}^{h}R_{\mathfrak{p}_{i}}/bR_{\mathfrak{p}_{i}}\right|=B$.

  * (ii)

There exist prime ideals $\mathfrak{p}_{i}\lhd R$ and ideals $I_{i}\lhd
R_{\mathfrak{p}_{i}}$ for $1\leq i\leq h$ such that
$(\mathfrak{p}_{i},I_{i})\models(r,ra,\gamma,\delta b)$ and
$\left|\oplus_{i=1}^{h}R_{\mathfrak{p}_{i}}/buR_{\mathfrak{p}_{i}}\right|=C^{\prime\prime}$.

Note that, if $(\mathfrak{p},\lambda
R_{\mathfrak{p}})\models(r,ra,\gamma,\delta)$ then
$(\mathfrak{p},rR_{\mathfrak{p}})\models(r,ra,\gamma,\delta)$. Since
$(rR_{\mathfrak{p}})^{\\#}=\mathfrak{p}R_{\mathfrak{p}}$,
$(\mathfrak{p},rR_{\mathfrak{p}})\models(r,ra,\gamma,\delta)$ if and only if
$\gamma\delta\notin\mathfrak{p}$. So $(i)$ holds if and only if there exist
prime ideals $\mathfrak{p}_{i}$ for $1\leq i\leq h^{\prime}$ such that
$\gamma\delta\notin\mathfrak{p}_{i}$ and
$\left|\oplus_{i=1}^{h^{\prime}}R_{\mathfrak{p}_{i}}/buR_{\mathfrak{p}_{i}}\right|=C^{\prime}$
and
$\left|\oplus_{i=1}^{h^{\prime}}R_{\mathfrak{p}_{i}}/bR_{\mathfrak{p}_{i}}\right|=B$.
So, since $\mathrm{EPP}(R)$ is recursive, by 7.2, there is an algorithm which
answers whether (i) holds or not.

By 10.1, since $X(R)$ is recursive, there is an algorithm which answers
whether (ii) holds or not. ∎

The rest of this section is spent proving the following proposition.

######  Proposition 10.5.

Let $R$ be a recursive Prüfer domain with $\mathrm{EPP}(R)$ and
$\mathrm{DPR}(R)$ recursive. There is an algorithm which, given $c,b\in R$,
$C,B\in\mathbb{N}$ and $\Xi$ an auxiliary sentence, answers whether there
exists $M\in\textnormal{Mod}\text{-}R$ such that

$M\models\left|\nicefrac{{x=x}}{{c|x}}\right|=C\wedge\left|\nicefrac{{xb=0}}{{x=0}}\right|\geq
B\wedge\Xi.$

We could choose the module in the following definition uniquely. For instance,
one can show that when $\mathfrak{q}\supsetneq\mathfrak{p}$ the uniserial
module
$[\mathfrak{q}R_{\mathfrak{q}}:\lambda\mathfrak{p}R_{\mathfrak{q}}]/\mathfrak{q}R_{\mathfrak{q}}$
has the required theory where
$[\mathfrak{q}R_{\mathfrak{q}}:\lambda\mathfrak{p}R_{\mathfrak{q}}]$ is the
set of elements $a\in Q$, the fraction field of $R$, such that
$a\lambda\mathfrak{q}R_{\mathfrak{q}}\subseteq\mathfrak{q}R_{\mathfrak{q}}$.
However, we are only ever interested in modules up to elementary equivalence.

######  Definition 10.6.

For $\lambda\in R\backslash\\{0\\}$ and prime ideals
$\mathfrak{p},\mathfrak{q}\lhd R$, let $M(\mathfrak{p},\mathfrak{q},\lambda)$
be $R_{\mathfrak{p}}/\lambda\mathfrak{q}R_{\mathfrak{p}}$ if
$\mathfrak{p}\supseteq\mathfrak{q}$ and a module with theory dual, in the
sense of [Her93, 6.6], to the theory of
$R_{\mathfrak{q}}/\lambda\mathfrak{p}R_{\mathfrak{q}}$ if
$\mathfrak{q}\supsetneq\mathfrak{p}$.

######  Lemma 10.7.

Suppose that $\lambda,a,\gamma,\delta\in R$ with $\lambda\neq 0$ are such that
if $(\mathfrak{p},I)\models(\lambda,\lambda a,\gamma,\delta)$ then
$R_{\mathfrak{p}}/I\models\Xi$ and if $(\mathfrak{p},I)\models(\lambda,\lambda
a,\delta,\gamma)$ then $R_{\mathfrak{p}}/I\models D\Xi$. Then
$\gamma\notin\mathfrak{p}$, $\delta\notin\mathfrak{q}$ and
$a\in\mathfrak{p}\cap\mathfrak{q}$ implies
$M(\mathfrak{p},\mathfrak{q},\lambda)\models\Xi$.

###### Proof.

Let $(\lambda,\lambda a,\gamma,\delta)$ be such that if
$(\mathfrak{p},I)\models(\lambda,\lambda a,\gamma,\delta)$ then
$R_{\mathfrak{p}}/I\models\Xi$ and if $(\mathfrak{p},I)\models(\lambda,\lambda
a,\delta,\gamma)$ then $R_{\mathfrak{p}}/I\models D\Xi$.

Suppose that $\gamma\notin\mathfrak{p}$, $\delta\notin\mathfrak{q}$ and
$a\in\mathfrak{p}\cap\mathfrak{q}$. If $\mathfrak{p}\supseteq\mathfrak{q}$
then
$(\mathfrak{p},\lambda\mathfrak{q}R_{\mathfrak{p}})\models(\lambda,\lambda
a,\gamma,\delta)$. So $M(\mathfrak{p},\mathfrak{q},\lambda)\models\Xi$. Now
suppose that $\mathfrak{q}\supsetneq\mathfrak{p}$. Then
$(\mathfrak{q},\lambda\mathfrak{p}R_{\mathfrak{q}})\models(\lambda,\lambda
a,\delta,\gamma)$. So
$R_{\mathfrak{q}}/\lambda\mathfrak{p}R_{\mathfrak{q}}\models D\Xi$. Hence, by
definition, $M(\mathfrak{p},\mathfrak{q},\lambda)\models\Xi$. ∎

######  Lemma 10.8.

Let $R$ be a recursive Prüfer domain with $\mathrm{EPP}(R)$ recursive. There
is an algorithm which, given $r,c\in R\backslash\\{0\\}$,
$a,b,\gamma,\delta\in R$ and $A,B,C\in\mathbb{N}$, answers whether there exist
$h\in\mathbb{N}_{0}$, prime ideals $\mathfrak{p}_{i}\lhd R$ and ideals
$I_{i}\lhd R_{\mathfrak{p}_{i}}$ for $1\leq i\leq h$ such that
$(\mathfrak{p}_{i},I_{i})\models(1,a,\gamma,\delta)$,
$\left|\oplus_{i=1}^{h}R_{\mathfrak{p}_{i}}/I_{i}\right|=A$ and

$\oplus_{i=1}^{h}R_{\mathfrak{p}_{i}}/cI_{i}\models\left|\nicefrac{{x=x}}{{c|x}}\right|=C\wedge\left|\nicefrac{{xb=0}}{{x=0}}\right|\geq
B.$

###### Proof.

Let $\mathfrak{p}\lhd R$ be a prime ideal and $I\lhd R_{\mathfrak{p}}$ be an
ideal. Then $a\in I_{i}$ if and only if $rca\in rcI_{i}$. So $a\in I_{i}$ if
and only if $\left|\nicefrac{{rca|x}}{{x=0}}(R_{\mathfrak{p}}/rcI)\right|=1$.
If $R_{\mathfrak{p}}/rcI\neq 0$ then $\gamma\notin\mathfrak{p}$ if and only if
$\left|\nicefrac{{x=x}}{{\gamma|x}}\right|=1$, and, $\delta\notin I^{\\#}$ if
and only if $\left|\nicefrac{{x\delta=0}}{{x=0}}\right|=1$. Note that

$\left|\nicefrac{{rc|x}}{{x=0}}(\oplus_{i=1}^{h}R_{\mathfrak{p}_{i}}/rcI_{i})\right|=\left|\oplus_{i=1}^{h}R_{\mathfrak{p}_{i}}/I_{i}\right|.$

Therefore, there exist $h\in\mathbb{N}_{0}$, prime ideals
$\mathfrak{p}_{i}\lhd R$ and ideals $I_{i}\lhd R_{\mathfrak{p}_{i}}$ for
$1\leq i\leq h$ as in the statement if and only if there exist
$h\in\mathbb{N}_{0}$, prime ideals $\mathfrak{p}_{i}\lhd R$ and ideals
$I_{i}\lhd R_{\mathfrak{p}_{i}}$ for $1\leq i\leq h$ such that
$\oplus_{i=1}^{h}R_{\mathfrak{p}_{i}}/rcI_{i}$ satisfies

${\raisebox{1.5pt}{$\chi$}}:=\left|\nicefrac{{rc|x}}{{x=0}}\right|=A\wedge\left|\nicefrac{{x=x}}{{c|x}}\right|=C\wedge\left|\nicefrac{{xb=0}}{{x=0}}\right|\geq
B\\\
\wedge\left|\nicefrac{{rca|x}}{{x=0}}\right|=1\wedge\left|\nicefrac{{x=x}}{{\gamma|x}}\right|=1\wedge\left|\nicefrac{{x\delta=0}}{{x=0}}\right|=1.$

By 8.1, there is an algorithm answering whether there exists
$\oplus_{i=1}^{h}R_{\mathfrak{p}_{i}}/rcI_{i}$ satisfying $\chi$. ∎

######  Proposition 10.9.

Let $R$ be a recursive Prüfer domain with $\mathrm{EPP}(R)$, $\mathrm{DPR}(R)$
and $X(R)$ recursive. There is an algorithm which, given $C,B\in\mathbb{N}$,
$c,b\in R$ with $c\neq 0$ and $r,a,\gamma,\delta\in R$ with $r\neq 0$, answers
whether there exists $M\in\textnormal{Mod}\text{-}R$ satisfying

$\left|\nicefrac{{x=x}}{{c|x}}\right|=C\wedge\left|\nicefrac{{xb=0}}{{x=0}}\right|\geq
B,$

such that $M$ is a direct sum of

  * •

modules of the form $R_{\mathfrak{p}}/rcI$ where $\mathfrak{p}$ is prime,
$I\lhd R_{\mathfrak{p}}$ is an ideal and
$(\mathfrak{p},I)\models(1,a,\gamma,\delta)$, and,

  * •

modules of the form $M(\mathfrak{p},\mathfrak{q},r)$ where
$\mathfrak{p},\mathfrak{q}\lhd R$ are prime ideals such that