fact that, if $t_{1}\in\neg t_{2}$, $t_{2}\in\neg t_{3}$, and $t_{4}\in\neg
t_{3}$, $t_{1}\bigcup t_{4}$ will be a partition of $D$, but will generally
not be an ip.

When t-algebras are not simple, they can get fairly opaque. For example, it’s
possible to have $t_{1}\in[t_{2}]$, and $t=t_{1}\bigcap t_{2}$, but
$t_{1}-t\notin[t_{2}-t]$, even though it’s clear that for any probability
function $P$, $P(t_{1}-t)=P(t_{2}-t)$. This could never happen for a simple
t-algebra. Fortunately, starting with any t-algebra, it’s always possible to
use $\neg$ and $[t]$ to build an equivalent simple t-algebra. This will be
done iteratively, the first iteration being the t-algebra $T_{1}$:

######  Definition 99.

For t-algebra $(X,T)$,

For $t,t^{\prime}\in T$, $t\perp t^{\prime}$ if for some $t_{1},t_{2}\in T$,
$t_{1}\in\neg[t_{2}]$, $t\subset t_{1}$ and $t^{\prime}\subset t_{2}$

$T_{1}\equiv\\{t\bigcup t^{\prime}\,:\,t\perp t^{\prime}\\}$

$X_{1}\equiv\\{(\cup t)\bigcup(\cup t^{\prime})\,:\,t\in
T\>and\>t^{\prime}\in\neg[t]\\}$

######  Theorem 100.

If $(X,T)$ is a t-algebra then $(X_{1},T_{1})$ is a t-algebra

###### Proof.

Axioms (1) and (2) hold by the definitions of $\perp$ and $X_{1}$.

For axiom (3), given any $t_{1},t_{2}\in T_{1}$, there exist
$t_{1}^{\prime},t_{1}^{\prime\prime},t_{2}^{\prime},t_{2}^{\prime\prime}\in T$
s.t. $t_{i}^{\prime}\perp t_{i}^{\prime\prime}$ and
$t_{i}=t_{i}^{\prime}\bigcup t_{i}^{\prime\prime}$. Define
$t_{3}\equiv(t_{1}^{\prime}-t_{2}^{\prime})-t_{2}^{\prime\prime}$ and
$t_{4}\equiv(t_{1}^{\prime\prime}-t_{2}^{\prime})-t_{2}^{\prime\prime}$;
$t_{3},t_{4}\in T$; since $t_{3}\subset t_{1}^{\prime}$ and $t_{4}\subset
t_{1}^{\prime\prime}$, $t_{3}\perp t_{4}$, so $t_{3}\bigcup t_{4}\in T_{1}$.
$t_{1}-t_{2}=t_{3}\bigcup t_{4}$ so $t_{1}-t_{2}\in T_{1}$. ∎

We now turn to constructing a probability function for $T_{1}$.

######  Theorem 101.

If $(X,T,P)$ is a gps, $t_{1}\perp t_{2}$, $t_{3}\perp t_{4}$, and
$t_{1}\bigcup t_{2}=t_{3}\bigcup t_{4}$, then
$P(t_{1})+P(t_{2})=P(t_{3})+P(t_{4})\in[0,1]$; further, if
$t_{1}\in\neg[t_{2}]$ then $P(t_{1})+P(t_{2})=1$

###### Proof.

Take $t_{1}^{\prime}=t_{1}\bigcap t_{3}$, $t_{2}^{\prime}=t_{1}\bigcap t_{4}$,
$t_{3}^{\prime}=t_{2}\bigcap t_{3}$, $t_{4}^{\prime}=t_{2}\bigcap t_{4}$. All
$t_{i}^{\prime}$ are disjoint, and each $t_{i}$ is a union of two of the
$t_{j}^{\prime}$, so
$P(t_{1})+P(t_{2})=P(t_{1}^{\prime})+P(t_{2}^{\prime})+P(t_{3}^{\prime})+P(t_{4}^{\prime})=P(t_{3})+P(t_{4})$.

If $t_{1}\in\neg[t_{2}]$ then $P(t_{1})+P(t_{2})=1$ by Thm 91.1.

More generally, if $t_{1}\perp t_{2}$, for some $t_{5},t_{6}\in T$,
$t_{6}\in\neg[t_{5}]$, $t_{1}\subset t_{5}$, and $t_{2}\subset t_{6}$.
$P(t_{5})+P(t_{6})=1$, so $P(t_{1})+P(t_{2})\in[0,1]$.∎

######  Definition 102.

If $(X,T,P)$ is a gps, $P_{1}:T_{1}\rightarrow[0,1]$ s.t. for $t_{1},t_{2}\in
T$, $t_{1}\perp t_{2}$, $P_{1}(t_{1}\bigcup t_{2})=P(t_{1})+P(t_{2})$.

######  Theorem 103.

If $(X,T,P)$ is a gps then $(X_{1},T_{1},P_{1})$ is a gps.

###### Proof.

Follows from Thm 100 and Thm 101. ∎

This process can now be repeated; starting with $(X_{1},T_{1},P_{1})$,
$(X_{2},T_{2},P_{2})$ can be constructed as
$((X_{1})_{1},(T_{1})_{1},(P_{1})_{1})$.

######  Definition 104.

If $(X,T,P)$ is a gps, $(X_{0},T_{0},P_{0})\equiv(X,T,P)$ and
$(X_{n+1},T_{n+1},P_{n+1})\equiv((X_{n})_{1},(T_{n})_{1},(P_{n})_{1})$.

######  Theorem 105.

If $(X,T,P)$ is a gps

1) $(X_{n},T_{n},P_{n})$ is a gps

2) $t\in T_{n}$ iff for some finite, pairwise disjoint $A\subset T$ with not
more than $2^{n}$ members s.t. $\bigcup A\in G_{Tn}$, some $B\subset A$,
$t=\bigcup B$

3) If $t\in T_{n}$ and $A\subset T$ is finite, pairwise disjoint, and
$t=\bigcup A$ then $P_{n}(t)=\sum_{x\in A}P(x)$

4) If $t\in T_{n}$ and $t\in T_{m}$ then $P_{n}(t)=P_{m}(t)$

###### Proof.

For 1-3, Straightforward induction. (4) follows from (2) & (3).∎

######  Definition 106.

If $(X,T)$ is a t-algebra

$X_{\mathbb{N}}\equiv\bigcup_{n\in\mathbb{N}^{+}}X_{n}$

$T_{\mathbb{N}}\equiv\bigcup_{n\in\mathbb{N}^{+}}T_{n}$

If $(X,T,P)$ is a dps, $P_{\mathbb{N}}:T_{\mathbb{N}}\rightarrow[0,1]$ s.t. if
$t\in T_{n}$ then $P_{\mathbb{N}}(t)=P_{n}(t)$

######  Theorem 107.

If $(X,T,P)$ is a gps then $(X_{\mathbb{N}},T_{\mathbb{N}},P_{\mathbb{N}})$ is
a simple gps

###### Proof.

That $(X_{\mathbb{N}},T_{\mathbb{N}})$ is a t-algebra follows from the fact
that if $t\in T_{\mathbb{N}}$ then for some $m\in\mathbb{N}^{+}$, all $n>m$,
$t$ is an element of $T_{n}$, and all $T_{n}$ are t-algebras. That
$(X_{\mathbb{N}},T_{\mathbb{N}},P_{\mathbb{N}})$ is a gps follows similarly.

It remains to show that $T_{\mathbb{N}}$ is simple. Take “$\neg_{n}$” to be
“$\neg$” defined on $T_{n}$ and “$\neg_{\mathbb{N}}$” to be “$\neg$” defined
on $T_{\mathbb{N}}$. For $t\in T_{\mathbb{N}}$, if
$t^{\prime}\in\neg_{\mathbb{N}}\neg_{\mathbb{N}}\neg_{\mathbb{N}}t$ then for
some $t_{1},t_{2}\in T_{\mathbb{N}}$, $t^{\prime}\in\neg_{\mathbb{N}}t_{2}$,
$t_{2}\in\neg_{\mathbb{N}}t_{1}$, and $t_{1}\in\neg_{\mathbb{N}}t$, so for
some $m,n,p\in\mathbb{N}$, $t^{\prime}\in\neg_{m}t_{2}$,
$t_{2}\in\neg_{n}t_{1}$, and $t_{1}\in\neg_{n}t$, in which case, with
$q=lub(\\{m,n,p\\})$, $t^{\prime}\in\neg_{q+1}t$, and so
$t^{\prime}\in\neg_{\mathbb{N}}t$ ∎

The following will be of occasional use.

######  Definition 108.

$[t]_{n}$ and $\neg_{n}t$ may be use to refer to $[t]$ and $\neg t$ on
$(X_{n},T_{n})$. $[t]_{\mathbb{N}}$ and $\neg_{\mathbb{N}}t$ may be use to
refer to $[t]$ and $\neg t$ on $(X_{\mathbb{N}},T_{\mathbb{N}})$.

######  Theorem 109.

If $(X,T,P)$ is a gps

1) $t\in T_{\mathbb{N}}$ iff for some finite, pairwise disjoint $A\subset T$
s.t. $\bigcup A\in G_{T\mathbb{N}}$, some $B\subset A$, $t=\bigcup B$

2) If $A\subset T$ is finite, pairwise disjoint, and $\bigcup A\in
T_{\mathbb{N}}$ then $P_{\mathbb{N}}(\bigcup A)=\Sigma_{t\in A}P(t)$

###### Proof.

Follows from Thm 105, ∎

####  IV.2.3 Convergence on a GPS

Earlier, the question of countable convergence was deferred. Some basic
concepts will now be presented.

######  Definition 110.

For t-algebra $(X,T)$, $\mathcal{A}$ is a _c-set_ if it is a countable
partition of some $y\in X$, and there exists a sequence on $\mathcal{A}$,
$(A_{n})_{n\in\mathbb{N}}$, $(\mathcal{A}=\\{A_{n}:n\in\mathbb{N}\\})$ s.t.
for all $n\in\mathbb{N}$, $\bigcup_{i\leq n}A_{i}\in T$.

######  Theorem 111.

If $(X,T)$ is a t-algebra, $\mathcal{A}$ is a c-set iff it is a countable
partition of some $y\in X$ and for any finite $B\subset\mathcal{A}$, $\bigcup
B\in T$

###### Proof.

$\Rightarrow$ There exists a sequence $(A_{n})_{n\in\mathbb{N}}$ s.t.
$\mathcal{A}=\\{A_{n}:n\in\mathbb{N}\\}$ and for all $n\in\mathbb{N}$,
$\bigcup_{i\leq n}A_{i}\in T$.

For all $A_{i}\in\mathcal{A}$, $A_{i}=\bigcup_{j\leq i}A_{j}-\bigcup_{k\leq
i-1}A_{k}$, so $A_{i}\in T$.

For each $b\in B$ there exists a unique $i\in\mathbb{N}$ s.t. $A_{i}=b$; take
$k$ to be the largest such number. $\bigcup B\subset\bigcup_{i\leq k}A_{i}\in
T$. Since $(X,T)$ is a t-algebra, $B$ is a finite subset of $T$, and $\bigcup
B$ is a subset of an element of $T$, $\bigcup B\in T$.

$\Leftarrow$ Immediate. ∎

If $(X,T)$ is a t-algebra and $(t_{n})_{n\in\mathbb{N}}$ is a sequence of
elements of $T$ s.t. for all $i\in\mathbb{N}$, $t_{i}\subset t_{i+1}$ and
$\bigcup_{i\in\mathbb{N}}t_{i}$ is the partition of some $S\in X$, then
$\mathcal{A}\equiv\\{t_{i+1}-t_{i}:i\in\mathbb{N}\\}$ is a c-set. Further, if
$\mathcal{A}$ is a c-set and $(A_{n})_{n\in\mathbb{N}}$ is any sequence of
it’s elements then $t_{i}=\bigcup_{j\leq i}A_{j}$ form such a sequence of
elements of $T$, so these two notions are equivalent.

######  Theorem 112.

If $(X,T,P)$ is a gps and $\mathcal{A}$ is a c-set then $\sum_{x\in A}P(x)\leq
1$.

###### Proof.

If $(A_{n})_{n\in\mathbb{N}}$ is any sequence on $\mathcal{A}$, for all
$N\in\mathbb{N}$, $\sum_{n\leq N}P(A_{n})=P(\bigcup_{n\leq N}A_{n})\leq 1$.∎

######  Definition 113.

A gps, $(X,T,P)$, is _convergent_ if for every c-set, $\mathcal{A}$,
$\sum_{x\in A}P(x)=1$.

######  Theorem 114.

If $(X,T,P)$ is a convergent dps, $\mathcal{A}$ and $\mathcal{B}$ are c-sets,
$Z\subset\mathcal{A}$, $Y\subset\mathcal{B}$, and $\bigcup Z=\bigcup Y$ then
$\sum_{x\in Z}P(x)=\sum_{x\in Y}P(x)$.

###### Proof.

Take $C\equiv\\{\alpha:for\>some\>a\in Y,\>b\in Z,\>\alpha=a\bigcap
b\>and\>\alpha\neq\emptyset\\}$ and $V\equiv(\mathcal{B}-Y)\bigcup C$. $V$ has
the following properties: every element of $V$ is an element of $T$; for every
finite $\alpha\subset V$ there exits a finite $b\subset B$ s.t.
$\bigcup\alpha\subset\bigcup b$, and so $\bigcup\alpha\in T$; and finally $V$
is pairwise disjoint & $\bigcup V=\bigcup B$. Therefore $V$ is a c-set.

$\sum_{x\in B-Y}P(x)+\sum_{x\in C}P(x)=\sum_{x\in V}P(x)=1=\sum_{x\in
B}P(x)=\sum_{x\in B-Y}P(x)+\sum_{x\in Y}P(x)$, so $\sum_{x\in
C}P(x)=\sum_{x\in Y}P(x)$. Proceeding similarly with $(A-Z)\bigcup C$ yields
$\sum_{x\in C}P(x)=\sum_{x\in Z}P(x)$. Therefore $\sum_{x\in Z}P(x)=\sum_{x\in
Y}P(x)$.∎

######  Definition 115.

If $(X,T)$ is a t-algebra,
$T_{c}\equiv\\{x:For\>some\>c\textrm{-}set,\>A,\>some\>Z\subset A,\>x=\bigcup
Z\\}$.

If $(X,T,P)$ is a convergent dps, $P_{c}:T_{c}\rightarrow[0,1]$ s.t. if $A$ is
a c-set and $B\subset A$ then $P_{c}(\bigcup B)=\sum_{x\in B}P(x)$.

######  Theorem 116.

If $(X,T,P)$ is convergent $A\subset T_{c}$ is countable, pairwise disjoint,
and $\bigcup A\in T_{c}$ then $\sum_{x\in A}P_{c}(x)=P_{c}(\bigcup A)$.

###### Proof.

For every $x\in A$ take $Y_{x}$ to a be a c-set s.t. for $B_{x}\subset Y_{x}$,
$\bigcup B_{x}=x$. Take $Z$ to be a c-set s.t. for $D\subset Z$, $\bigcup
D=\bigcup A$. Finally, for each $x\in A$, take
$C_{x}\equiv\\{\alpha:for\>some\>a\in B_{x},\>b\in D,\>\alpha=a\bigcap
b\>\&\>\alpha\neq\emptyset\\}$. Note that $\bigcup C_{x}=x$.

Each $(Y_{x}-B_{x})\bigcup C_{x}$ is a c-set, so $\sum_{t\in
C_{x}}P_{c}(t)=P_{c}(x)$. With $C\equiv\bigcup_{x\in A}C_{x}$, $(Z-D)\bigcup
C$ is also a c-set, so $P_{c}(\bigcup A)=\sum_{x\in A}\sum_{t\in
C_{x}}P_{c}(t)=\sum_{x\in A}P_{c}(x)$. ∎

If $(X,T)$ is a t-algebra, there’s no guarantee that $(X,T_{c})$ will be a
t-algebra; in particular, if $A$ and $B$ are c-sets, $(\bigcup
A)\bigcap(\bigcup B)$ might not be an element of $T_{c}$.555$(X,T_{c},P_{c})$
does, however, meet all the other requirements for being a gps; that is, if
for all pairs of c-sets, $A$ and $B$, $(\bigcup A)\bigcap(\bigcup B)\in T_{c}$
then $(X,T_{c},P_{c})$ is a gps Also, if $(X,T,P)$ is convergent,
$(X_{\mathbb{N}},T_{\mathbb{N}},P_{\mathbb{N}})$ may not be (though if
$(X_{\mathbb{N}},T_{\mathbb{N}},P_{\mathbb{N}})$ is convergent then $(X,T,P)$
is). Because convergence on $(X_{\mathbb{N}},T_{\mathbb{N}},P_{\mathbb{N}})$
is a useful quality for a gps to posses, the following simplified notation
will be used.

######  Definition 117.

If $(X,T)$ is a t-algebra, $T_{\Upsilon}\equiv(T_{\mathbb{N}})_{c}$. If
$(X,T,P)$ is a dps, it is $\Upsilon$-convergent if
$(X_{\mathbb{N}},T_{\mathbb{N}},P_{\mathbb{N}})$ is convergent. If $(X,T,P)$
is an $\Upsilon$-convergent dps then $P_{\Upsilon}\equiv(P_{\mathbb{N}})_{c}$.

The easiest way to remember this is that $\Upsilon$ has absolutely nothing to
do with these properties.

####  IV.2.4 $T_{S}$, $T_{S\mathbb{N}}$, and $T_{\mathbb{N}}^{S}$

######  Definition 118.

If $(X,T)$ is a t-algebra and $S\in X$ then

$T_{S}\equiv\\{t\in T:for\>some\>\gamma\in
G_{T},\>\bigcup\gamma=S\>and\>t\subset\gamma\\}$

$T_{S\mathbb{N}}\equiv(T_{S})_{\mathbb{N}}$ and
$T_{\mathbb{N}}^{S}\equiv(T_{\mathbb{N}})_{S}$.

It follows readily that $(\\{S\\},T_{S})$ is a t-algebra, both
$(\\{S\\},T_{S\mathbb{N}})$ and $(\\{S\\},T_{\mathbb{N}}^{S})$ are simple
t-algebras, and $T_{S\mathbb{N}}\subset T_{\mathbb{N}}^{S}$. One question that
arises is, under what conditions will $T_{S\mathbb{N}}=T_{\mathbb{N}}^{S}$;
that is, when does the rest of $T$ yield no further information about the
probabilities on $S$? There are a number of ways in which this can occur; the
most obvious is: for all $S^{\prime}\in X$, $S^{\prime}\neq S$, $S\bigcap
S^{\prime}=\emptyset$ (or more generally, for all $\gamma\in G_{TS}$,
$\gamma^{\prime}\in G_{TS^{\prime}}$, if $S\neq S^{\prime}$ then
$\gamma\bigcap\gamma^{\prime}=\emptyset$). In such cases, $T_{S}$ is
completely independent of the rest of $T$. Another way in which
$T_{S\mathbb{N}}=T_{\mathbb{N}}^{S}$ is if the structure of $T$ as a whole can
be mapped onto $T_{S}$; this possibility will now be sketched.

(In the following definition, $\mathbf{P}(S)$ is the power set of $S$)

######  Definition 119.

If $(X,T)$ and $(\\{S\\},T^{\prime})$ are t-algebras, $f:\bigcup
T\rightarrow\mathbf{P}(\bigcup T^{\prime})$ _reduces $T$ to $T^{\prime}$_ if

1) If $\gamma\in G_{T}$ then $\bigcup f[\gamma]\in G_{T^{\prime}}$

2) If $\gamma\in G_{T}$, $\alpha,\beta\in\gamma$, and $\alpha\neq\beta$ then
$f(\alpha)\bigcap f(\beta)=\emptyset$

3) If $\alpha\in\bigcup T$ and $\alpha\subset S$ then $f(\alpha)=\\{\alpha\\}$

######  Theorem 120.

If $f$ reduces $T$ to $T^{\prime}$ then it reduces $T_{1}$ to $T_{1}^{\prime}$

###### Proof.

For $t_{1},t_{2}\in T$, assume $t_{1}\in\neg t_{2}$; by (2), $f[t_{1}]\bigcap
f[t_{2}]=\emptyset$ and by (1) $\bigcup f[t_{1}\bigcup t_{2}]\in
G_{T^{\prime}}$, so $\bigcup f[t_{1}]\in\neg\bigcup f[t_{2}]$. It follows from
the definition of $\neg^{(n)}$ that if $t_{1}\in\neg^{(n)}t_{2}$ then $\bigcup
f[t_{1}]\in\neg^{(n)}\bigcup f[t_{2}]$. (1) & (2) in the definition of
reduction follow immediately from this. (3) continues to hold because $\bigcup
T_{1}=\bigcup T$.∎

######  Theorem 121.

If $f$ reduces $T$ to $T^{\prime}$ then it reduces $T_{\mathbb{N}}$ to
$T_{\mathbb{N}}^{\prime}$

###### Proof.

By Thm 120, $f$ reduces $T_{n}$ to $T_{n}^{\prime}$.

If $\gamma\in G_{T\mathbb{N}}$ then for some $n$, $\gamma\in G_{Tn}$, and so
(1) and (2) hold. (3) hold because$\bigcup T_{\mathbb{N}}=\bigcup T$.∎

######  Theorem 122.

If $(X,T)$ is a t-algebra, $S\in X$, and there exists an $f$ that reduces $T$
to $T_{S}$ then $T_{\mathbb{N}}^{S}=T_{S\mathbb{N}}$.

###### Proof.

In all cases, $T_{S\mathbb{N}}\subset T_{\mathbb{N}}^{S}$

Take any $t\in T_{\mathbb{N}}^{S}$ and any $\gamma\in G_{T\mathbb{N}}$ s.t.
$t\subset\gamma$. By Thm 121 $f$ reduces $T_{\mathbb{N}}$ to
$T_{S\mathbb{N}}$, so by (1) and (3) in the definition of a reduction,
$t\subset\bigcup f[\gamma]\in G_{T_{S\mathbb{N}}}$; therefore $t\in
T_{S\mathbb{N}}$ and so $T_{\mathbb{N}}^{S}\subset T_{S\mathbb{N}}$. ∎

###  IV.3 Nearly Compatible Sets

Now to return to dps’s and, in particular, the
$(X_{\mathbb{N}},T_{\mathbb{N}},P_{\mathbb{N}})$ construction for dps’s.
Because the only additional requirement imposed on dps’s is that all
$\gamma\in G_{T}$ must be ip’s, and because this only places an indirect
restriction on the makeup of “$X$” in dps $(X,T,P)$, dps’s of the form
$(\\{S\\},T,P)$ will be of greatest interest. Naturally, any dps may be
decomposed into a set such dps’s, one for each $S\in X$, and the results of
this section may be applied to those component parts.

######  Definition 123.

A set of companionable sets, $A$, is _nearly compatible_ if it is pairwise
disjoint and for every $\alpha,\beta\in A$, every $(\lambda,p)\in Uni(\bigcup
A)$ either $\alpha_{\rightarrow(\lambda,p)}=\beta_{\rightarrow(\lambda,p)}$ or
$\alpha_{\rightarrow(\lambda,p)}\bigcap\beta_{\rightarrow(\lambda,p)}=\emptyset$.

All compatible sets are nearly compatible, but the converse does not hold. The
basic result of this section is that for a dps, $(\\{S\\},T,P)$, while
elements of $G_{T\mathbb{N}}$ need not be compatible, they must be nearly
compatible.

######  Definition 124.

$\mathcal{A}$ is a _nearly compatible collection_(_ncc_) if it is pairwise
disjoint and $\bigcup\mathcal{A}$ is nearly compatible. $\mathcal{A}$ is an
_$S$ -nccp_ if it is an ncc and $\bigcup\mathcal{A}$ is a partition of $S$.

######  Theorem 125.

If $S$ is a dynamic set and $\\{t_{1},t_{2}\\}$, $\\{t_{2},t_{3}\\}$, and
$\\{t_{3},t_{4}\\}$ are $S$-nccp’s, then so is $\\{t_{1},t_{4}\\}$.

###### Proof.

Take $\alpha\in t_{1}$, $\beta\in t_{4}$ and assume
$\alpha_{\rightarrow(\lambda,p)}\bigcap\beta_{\rightarrow(\lambda,p)}\neq\emptyset$.
With
$\bar{s}[-\infty,\lambda]\in\alpha_{\rightarrow(\lambda,p)}\bigcap\beta_{\rightarrow(\lambda,p)}$,
there must be a $\eta\in t_{2}$ and a $\nu\in t_{3}$ s.t.
$\bar{s}[-\infty,\lambda]\in\eta_{\rightarrow(\lambda,p)}$ and
$\bar{s}[-\infty,\lambda]\in\nu_{\rightarrow(\lambda,p)}$. Since $t_{1}\bigcup
t_{2}$, $t_{2}\bigcup t_{3}$, and $t_{3}\bigcup t_{4}$ are nearly compatible,
$\alpha_{\rightarrow(\lambda,p)}=\eta_{\rightarrow(\lambda,p)}=\nu_{\rightarrow(\lambda,p)}=\beta_{\rightarrow(\lambda,p)}$.
∎

Thm 125 is the key property of nearly compatible sets. Essentially, if $Y$ is
an attribute s.t. all ip’s are $Y$, and $Y$ is transitive in the sense of Thm
125, then all $\gamma\in G_{T\mathbb{N}}$ are $Y$. Near compatibility is a
close approximation of compatibility that possesses this property.

######  Theorem 126.

If $(\\{S\\},T,P)$ is a dps, $t_{1}\in T$, and $t_{2}\in\neg[t_{1}]$ then
$\\{t_{1},t_{2}\\}$ is an $S$-nccp.

###### Proof.

If $t_{1}\in\neg t_{2}$ then $t_{1}$ and $t_{2}$ are disjoint and
$t_{1}\bigcup t_{2}$ is a compatible (and so nearly compatible) partition, so
$\\{t_{1},t_{2}\\}$ is an $T$-nccp. The result is now immediate from Thm 125.
∎

Note that if $t_{1}\in\neg[t_{2}]$ there are no grounds to conclude that
$t_{1}\bigcup t_{2}$ is compatible, and in general it won’t be.

######  Theorem 127.

If $(\\{S\\},T,P)$ is a dps, $x_{1}\in T_{n}$, and
$x_{2}\in\neg_{n}[x_{1}]_{n}$ then $\\{x_{1},x_{2}\\}$ is a $S$-nccp.

###### Proof.

Holds for $n=0$ by Thm 126. Assume it holds for $n=m$. For $n=m+1$, if
$x_{2}\in\neg_{m+1}x_{1}$ then $x_{1}\bigcup x_{2}\in G_{T(m+1)}$; therefore
there must be a $y_{1},y_{2}\in T_{m}$ s.t. $y_{2}\in\neg_{m}[y_{1}]_{m}$ and
$y_{1}\bigcup y_{2}=x_{1}\bigcup x_{2}$. By assumption on $n=m$, $x_{1}\bigcup
x_{2}=y_{1}\bigcup y_{2}$ is nearly compatible, so $\\{x_{1},x_{2}\\}$ is a
$S$-nccp. From Thm 125 it then follows that if
$x_{2}\in\neg_{m+1}[x_{1}]_{m+1}$ then $\\{x_{1},x_{2}\\}$ is an $S$-nccp.∎

######  Theorem 128.

If $(\\{S\\},T,P)$ is a dps then

1) If $\gamma\in G_{T\mathbb{N}}$ then for some finite $S$-nccp $A\subset T$,
$\gamma=\bigcup A$

2) If $t\in T_{\mathbb{N}}$ then for some finite ncc, $A\subset T$, $t=\bigcup
A$

###### Proof.

Immediate from Thm 127. ∎

The richer a dps’s t-algebra is, the more information the dps carries. A
t-algebra contains minimal information if $G_{T\mathbb{N}}$ contains only the
original ip’s; in that case, no information can be derived from the dps as a
whole that isn’t known by considering the individual
$(\gamma,T_{\gamma},P_{\gamma})$ in isolation. As the t-algebra grows richer,
more information can be derived from it. Thm 128 indicates the outer limit on
the information that can be derived from a dps on a single dynamic set.

######  Definition 129.

If $(\\{S\\},T,P)$ is a dps, it is _maximal_ if $\gamma\in G_{T\mathbb{N}}$
iff there’s a finite $S$-nccp, $A\subset T$, s.t. $\bigcup A=\gamma$.

If $(\\{S\\},T,P)$ and $(\\{S\\},T^{\prime},P^{\prime})$ are dps’s, $T\subset
T^{\prime}$, and $P=P^{\prime}|_{T}$, then if $T$ is maximal,
$(\\{S\\},T^{\prime},P^{\prime})$ will give no further information about
$(\\{S\\},T,P)$. However, if $(\\{S\\},T,P)$ is not maximal, but
$(\\{S\\},T^{\prime},P^{\prime})$ is, this immediately yields a wealth of
information about $(\\{S\\},T,P)$, indeed more information than
$(\\{S\\},T_{\mathbb{N}},P_{\mathbb{N}})$ yields; it means that the following
construct forms a simple gps:

$G_{Te}\equiv\\{\gamma:for\>some\>finite\>S\textrm{-}nccp,\>A\subset
T,\>\gamma\equiv\bigcup A\\}$

$T_{e}\equiv\\{t:for\>some\>finite\>ncc,\>A\subset T,\>some\>\gamma\in
G_{Te},\>t=\bigcup A\subset\gamma\\}$

$P_{e}:T_{e}\rightarrow[0,1]$ s.t. if $A\subset T$ is a finite ncc and
$\bigcup A\equiv t\in T_{e}$ then
$P_{e}(t)=P_{\mathbb{N}}^{\prime}(t)=\sum_{t^{\prime}\in A}P(t^{\prime})$.

Note that $T_{e}=T_{\mathbb{N}}$ only if $(\\{S\\},T,P)$ is maximal. In all
other cases, $(\\{S\\},T_{e},P_{e})$ contains more information than
$(\\{S\\},T_{\mathbb{N}},P_{\mathbb{N}})$.

For $\Upsilon$-convergent dps’s, this can be pushed further.

######  Definition 130.

If $(\\{S\\},T,P)$ is a dps, it is _$\omega$ -maximal_ if for every countable
$S$-nccp, $A\subset T$, every finite $B\subset A$, $\bigcup B\in
T_{\mathbb{N}}$.

$(\\{S\\},T,P)$ is _$\Upsilon$ -maximal_ if it is $\omega$-maximal and
$\Upsilon$-convergent.

For an $\omega$-maximal dps, every countable $S$-nccp composed of elements of
$T$ is a c-set in $T_{\mathbb{N}}$. This means that for a $\Upsilon$-maximal
dps, $\gamma\in G{}_{T\Upsilon}$ iff there’s a countable $S$-nccp, $A\subset
T$, s.t. $\bigcup A=\gamma$. It was mentioned previously that, in general,
$(\\{S\\},T_{\Upsilon},P_{\Upsilon})$ is not a gps; however it can be seen
that for $\Upsilon$-maximal dps’s, $(\\{S\\},T_{\Upsilon},P_{\Upsilon})$ is
not only a gps, it’s a simple gps.

Returning to the case where $(\\{S\\},T,P)$ and
$(\\{S\\},T^{\prime},P^{\prime})$ are dps’s, $T\subset T^{\prime}$, and
$P=P^{\prime}|_{T}$, if $(\\{S\\},T^{\prime},P^{\prime})$ is
$\Upsilon$-maximal then the following construct forms a simple gps:

$G_{T\varepsilon}\equiv\\{\gamma:for\>some\>countable\>S\textrm{-}nccp,\>A\subset
T,\>\gamma\equiv\bigcup A\\}$

$T_{\varepsilon}\equiv\\{t:for\>some\>countable\>ncc,\>A\subset
T,\>some\>\gamma\in G_{Te},\>t=\bigcup A\subset\gamma\\}$

$P_{\varepsilon}:T_{\varepsilon}\rightarrow[0,1]$ s.t. if $A$ is a countable
ncc and $\bigcup A\equiv t\in T_{\varepsilon}$ then
$P_{\varepsilon}(t)=P_{\Upsilon}^{\prime}(t)=\sum_{t^{\prime}\in
A}P(t^{\prime})$.

When it’s applicable, this is, of course, a very useful construct.

###  IV.4 Deterministic & Herodotistic Spaces

A system is deterministic if a complete knowledge of the present yields a
complete knowledge of the future; it is herodotistic if complete knowledge of
the present yields a complete knowledge of the past. More formally:

######  Definition 131.

A dynamic set, $S$, is _deterministic_ if for every $(\lambda,p)\in Uni(S)$,
$S_{(\lambda,p)\rightarrow}$ is a singleton (that is,
$S_{(\lambda,p)\rightarrow}$ has only one element).

$S$ is _herodotistic_ if for every $(\lambda,p)\in Uni(S)$,
$S_{\rightarrow(\lambda,p)}$ is a singleton.

In this section deterministic and herodotistic dps’s will be investigated, as
will deterministic/herodotistic “universes”, where the e-automata as whole is
deterministic/herodotistic though the system being measured might not be. It
will follow from the various results that for the probabilities seen in
quantum physics to hold, neither the systems being investigated nor the
universe as a whole can be either deterministic or herodotistic.

No other material in this article will depend on the material in this section.

####  IV.4.1 DPS’s on Deterministic & Herodotistic Spaces

######  Theorem 132.

If a dynamic set, $S$, is deterministic or herodotistic then $S$ is a dynamic
space all its subsets are subspaces.

###### Proof.

For $A\subset S$, $\bar{p}_{1},\bar{p}_{2}\in A$,
$\bar{p}_{1}(\lambda)=\bar{p}_{2}(\lambda)$, if $S$ is herodotistic then
$\bar{p}_{1}[-\infty,\lambda]=\bar{p}_{2}[-\infty,\lambda]$ so
$\bar{p}_{1}[-\infty,\lambda]\circ\bar{p}_{2}[\lambda,\infty]=\bar{p}_{2}\in
A$. Similarly, if $S$ is deterministic then
$\bar{p}_{1}[-\infty,\lambda]\circ\bar{p}_{2}[\lambda,\infty]=\bar{p}_{1}\in
A$∎

######  Theorem 133.

1) If $D$ is deterministic then for any $X\subset D$, $\lambda\in\Lambda_{D}$,
$X=+X[-\infty,\lambda]$

2) If $D$ is herodotistic then for any $X\subset D$, $\lambda\in\Lambda_{D}$,
$X=+X[\lambda,\infty]$

###### Proof.

1) For any $\bar{p}[-\infty,\lambda]\in X[-\infty,\lambda]$ there exists only
one $\bar{p}^{\prime}\in D$ s.t.
$\bar{p}^{\prime}[-\infty,\lambda]=\bar{p}[-\infty,\lambda]$.

2) Similar∎

######  Theorem 134.

1) If $D$ is deterministic then $A,B\subset D$ are compatible iff they are
disjoint and bounded from below

2) If $D$ is herodotistic then $A,B\subset D$ are compatible iff they are
disjoint and bounded from above

###### Proof.

1) If $A$ and $B$ are disjoint then $A[-\infty,\lambda]\bigcap
B[-\infty,\lambda]=\emptyset$ for all $\lambda$.

2) If $p\in A(\lambda)\bigcap B(\lambda)$ then since $\rightarrow(\lambda,p)$
has only one element,
$A_{\rightarrow(\lambda,p)}=B_{\rightarrow(\lambda,p)}$.∎

######  Definition 135.

If $(\\{S\\},T,P)$ is a dps, it is _closed under combination_ if for all
pairwise disjoint $A\subset T$ s.t. $\bigcup A$ is an ip, $\bigcup A\in
G_{T}$.

######  Theorem 136.

If $(X,T,P)$ is a dps and for all $S\in X$, $S$ is either deterministic or
herodotistic and $(\\{S\\},T_{S},P_{s})$ is closed under combination, then for
all $t_{1},t_{2}\in T$ s.t $\bigcup t_{1}=\bigcup t_{2}$

1) If $\gamma\in G_{T}$ and $t_{1}\subset\gamma$ then $(\gamma-t_{1})\bigcup
t_{2}\in G_{T}$

2) $P(t_{1})=P(t_{2})$

###### Proof.

(1) follows from Thm 134 and (2) follows from (1). ∎

Note the importance of closure under combination. One can map any dps,
$(X,T,P)$, it onto a herodotistic dps as follows. For every $S\in X$,
$\bar{p}\in S$, define $\overline{p}_{h}$ by
$\overline{p}_{h}(\lambda)=(\bar{p}(\lambda),\bar{p}[0,\lambda])$. With
$S_{h}\equiv\\{\overline{p}_{h}:\bar{p}\in S\\}$, $S_{h}$ is herodotistic. One
can then readily construct $(X_{h},T_{h},P_{h})$, which will be a dps, but in
most cases will not be closed under combination; indeed, since $\bigcup
t_{h}=\bigcup t_{h}^{\prime}$ iff $\bigcup t=\bigcup t^{\prime}$ and
$P(t)=P_{h}(t_{h})$, the conclusion of above theorem will generally not apply
to $(X_{h},T_{h},P_{h})$.

####  IV.4.2 DPS’s in Deterministic Universes

######  Theorem 137.

If partition $\gamma$ is decided by an idea e-automata, $Z=(D,I,F)$ s.t. $D$
is deterministic, then for some $\lambda_{0}\in\Lambda$, all
$\alpha\in\gamma$, $\alpha$ is $sbb\lambda_{0}$ and $wba\lambda_{0}$.

###### Proof.

A: For every $(s,e),(s,e^{\prime})\in I$,
$\Sigma_{(s,e)\rightarrow}=\Sigma_{(s,e^{\prime})\rightarrow}$

\- With $\bar{s}[0,0]$ s.t. $\bar{s}[0,0](0)=s$,
$\bar{s}[0,0]\in\Sigma_{(s,e)}\bigcap\Sigma_{(s,e^{\prime})}$ (since
$(s,e),(s,e^{\prime})\in I$), so the result follows from Thm 39. -

B: Taking $S\equiv\bigcup\gamma$, for any $s\in S(0)$,
$(\mathcal{O}_{F})_{(0,s)\rightarrow}$ is either a singleton or empty

\- Follows from definition of “deterministic” and (A) -

C: For some $\lambda_{0}\in\Lambda$, all $\alpha\in\gamma$, there’s a
$\lambda_{\alpha}\in\Lambda$, s.t. $\alpha$ is $sbb\lambda_{0}$,
$wba\lambda_{\alpha}$, and for all $s\in\alpha(\lambda_{0})$,
$\alpha_{s\rightarrow}[\lambda_{0},\lambda_{\alpha}]$ is a singleton

\- Follows from (B) and the definition of being decided by an e-automata -

D: For all $\alpha,\beta\in\gamma$, if $\alpha\neq\beta$ then
$\alpha(\lambda_{0})\bigcap\beta(\lambda_{0})=\emptyset$

\- Follows from (B) and the definition of being decided by an e-automata -

The theorem is immediate from (C) & (D).∎

######  Definition 138.

If $Y$ is a collection subsets of dynamic set $S$ then $Y$ is _cross-section
(at $\lambda$)_ if it is pairwise disjoint and for all $y\in Y$, $y$ is a
subspace, $sbb\lambda$, and $wba\lambda$.

######  Theorem 139.

1) If $Y$ is a cross-section then it’s compatible.

2) If $Y$ is cross-section at $\lambda$ then $\\{\bigcup Y\\}$ is a cross-
section at $\lambda$

###### Proof.

Clear∎

######  Definition 140.

A t-algebra is _deterministically decidable_ if for all $t\in T$, $t$ is a
cross-section.

It is _deterministically normal_ if it is deterministically decidable and

1) If $t\in T$ then $\\{\bigcup t\\}\in T$

2) If $A\subset T$ is finite and $\bigcup A$ is a partition & a cross-section
then $\bigcup A\in G_{T}$.

$(X,T,P)$ is a _deterministically decidable dps_ if $(X,T)$ is
deterministically decidable & _deterministically normal_ if $(X,T)$ is
deterministically normal.

######  Theorem 141.

If $(X,T,P)$ is deterministically normal dps

1) For any $t,t^{\prime}\in T$, if $\\{\bigcup t,\bigcup t^{\prime}\\}$ is a
cross-section and a partition of some $S\in X$ then $t^{\prime}\in\neg[t]$.

2) For all $t_{1},t_{2}\in T$ s.t. $\bigcup t_{1}=\bigcup t_{2}$,
$P(t_{1})=P(t_{2})$

###### Proof.

1) For any $\gamma\in G_{T}$ s.t. $t\subset\gamma$,
$t\bigcup\\{\bigcup(\gamma-t)\\}$ is a cross-section. Since $\bigcup
t^{\prime}=\bigcup(\gamma-t)$, $t\bigcup\\{\cup t^{\prime}\\}$ is a cross-
section. Therefore $t\bigcup\\{\cup t^{\prime}\\}$, $\\{\cup t\\}\bigcup
t^{\prime}$, and $\\{\cup t\\}\bigcup\\{\cup t^{\prime}\\}$ are all elements
of $G_{T}$.

2) Take any $\gamma\in G_{T}$ s.t. $t_{1}\subset\gamma$. It follows from (1)
that $\gamma-t_{1}\in\neg[t_{2}]$, so $t_{1}\in[t_{2}]$ ∎

If $X=\\{S\\}$ then (1) becomes - For any $t,t^{\prime}\in T$,
$t^{\prime}\in\neg[t]$ iff $\\{\bigcup t,\bigcup t^{\prime}\\}$ is a cross-
section and a partition of $S$.

####  IV.4.3 DPS’s in Herodotistic Universes

######  Definition 142.

If $t$ is a compatible set, it is _herodotistically decidable_ if for some
$\lambda_{0}\in\Lambda$, all $\alpha\in t$, there’s a
$\lambda_{\alpha}\in\Lambda$ s.t.
$\alpha=+\alpha[\lambda_{0},\lambda_{\alpha}]$, and for all
$s\in\alpha(\lambda_{\alpha})$, $\alpha_{\rightarrow
s}[\lambda_{0},\lambda_{\alpha}]$ is a singleton.

Under these conditions, $t$ is said to be _born at $\lambda_{0}$_

A t-algebra $(X.T)$ is _herodotistically decidable_ if every $t\in T$ is
herodotistically decidable (which holds iff every $\gamma\in G_{T}$ is
herodotistically decidable).

######  Theorem 143.

If ip $\gamma$ is decided by an ideal e-automata, $(D,I,F)$, and $D$ is
herodotistic, then $\gamma$ is herodotistically decidable.

###### Proof.

If $D$ is herodotistic then for any $e\in F$, any
$(s,e^{\prime})\in\mathcal{\theta}_{e}(\Lambda(\mathcal{O}_{e}))$,
$\Sigma_{(s,e^{\prime})}$ is a singleton, and by Thm 55.3 for any
$(s,e^{\prime}),(s,e^{\prime\prime})\in\mathcal{\theta}_{e}(\Lambda(\mathcal{O}_{e}))$,
$\Sigma_{(s,e^{\prime})}=\Sigma_{(s,e^{\prime\prime})}$. The theorem then
follows immediately from the definition of being decided by an ideal
e-automata.∎

######  Theorem 144.

If $(\cup t_{1})\bigcap(\cup t_{2})=\emptyset$, both $t_{1}$ and $t_{2}$ are
herodotistically decidable, and there exists a $\lambda$ s.t. both $t_{1}$ and
$t_{2}$ are born at $\lambda$, then $t_{1}$ and $t_{2}$ are nearly compatible

###### Proof.

Immediate from the definitions of herodotistically decidable and nearly
compatible. ∎

Let’s say that for herodotistically decidable t-algebra, $(X,T)$,
$t_{1},t_{2}\in T$ are _sympathetic_ if $\bigcup t_{1}=\bigcup t_{2}$ and
there exists a $t^{\prime}\in\neg t_{1}$, $\lambda\in\Lambda$ s.t. both
$t_{2}$ and $t^{\prime}$are born at $\lambda$. Thm 144 means that for a
maximal dps (or a dps that can be embedded in a maximal dps), if $t_{1}$ and
$t_{2}$ are sympathetic then $P(t_{1})=P(t_{2})$. It will soon be seen that
this is incompatible with quantum probabilities.

Note that being herodotistic has a less dramatic effect on an ideal e-automata
than being deterministic. This is because, for ideal e-automata, the
environment is assumed to remember something about the past (due being all-
reet), but know little about the future (due to being unbiased).

##  V Application to Quantum Measurement

###  V.1 Preliminary Matters

In this part, results from the prior sections will be applied to quantum
physics as described by the Hilbert Space formalism. In order to do this, a
few preliminary matters need to be addressed.

####  V.1.1 A Note On Paths

Quantum physics is often viewed in terms of transitions between states, rather
than in terms of system paths; an obvious exception being the path integration
formalism. Before proceeding, it may be helpful to start by describing
conditions under which state transitions can be re-represented as paths.

Start by taking $(\lambda_{1},s_{1})\Rightarrow(\lambda_{2},s_{2})$ to mean
that state $s_{1}$ at time $\lambda_{1}$ can transition to state $s_{2}$ at
time $\lambda_{2}$. Paths may then be defined as any parametrized function,
$\bar{s}$, s.t. for all $\lambda_{1}<\lambda_{2}$,
$(\lambda_{1},\bar{s}(\lambda_{1}))\Rightarrow(\lambda_{2},\bar{s}(\lambda_{2}))$.
To show that this set of paths is equivalent to the transition relation, it
must be shown that if $(\lambda_{1},s_{1})\Rightarrow(\lambda_{2},s_{2})$ then
there exists a path, $\bar{s}$, s.t. $\bar{s}(\lambda_{1})=s_{1}$ and
$\bar{s}(\lambda_{2})=s_{2}$ (the converse clearly holds).

In quantum mechanics, the $\Rightarrow$ relation has the following properties:

1) If $(\lambda_{1},s_{1})\Rightarrow(\lambda_{2},s_{2})$ then
$\lambda_{1}<\lambda_{2}$

2) For all $(\lambda,s)$, all $\lambda^{\prime}<\lambda$, there’s a
$s^{\prime}$ s.t. $(\lambda^{\prime},s^{\prime})\Rightarrow(\lambda,s)$

3) For all $(\lambda,s)$, all $\lambda^{\prime}>\lambda$, there’s a
$s^{\prime}$ s.t. $(\lambda,s)\Rightarrow(\lambda^{\prime},s^{\prime})$

4) For all $(\lambda_{1},s_{1})$, $(\lambda_{2},s_{2})$ s.t.
$(\lambda_{1},s_{1})\Rightarrow(\lambda_{2},s_{2})$, all
$\lambda_{1}<\lambda^{\prime}<\lambda_{2}$, there’s a $s^{\prime}$ s.t.
$(\lambda_{1},s_{1})\Rightarrow(\lambda^{\prime},s^{\prime})$ and
$(\lambda^{\prime},s^{\prime})\Rightarrow(\lambda_{2},s_{2})$.

If the measurements are strongly unbiased, then the following holds:

5) If $(\lambda_{1},s_{1})\Rightarrow(\lambda_{2},s_{2})$, and
$(\lambda_{2},s_{2})\Rightarrow(\lambda_{3},s_{3})$, then
$(\lambda_{1},s_{1})\Rightarrow(\lambda_{3},s_{3})$.666If
$<\lambda_{1},s_{1}|\lambda_{2},s_{2}>\neq 0$, then it’s clear that
$(\lambda_{1},s_{1})\Rightarrow(\lambda_{2},s_{2})$. However, there is some
ambiguity as to whether $<\lambda_{1},s_{1}|\lambda_{2},s_{2}>=0$ necessarily
implies that the transition can not take place, or if it simply demands that
the transition occurs with probability $0$. This question grows acute if
there’s a $(\lambda,s)$ s.t. $\lambda_{1}<\lambda<\lambda_{2}$,
$<\lambda_{1},s_{1}|\lambda,s>\neq 0$, and $<\lambda,s|\lambda_{2},s_{2}>\neq
0$. In that case, in order for (5) to hold, we would have to allow the
transition $(\lambda_{1},s_{1})\Rightarrow(\lambda_{2},s_{2})$, but say that
it occurs with probability $0$. There’s no necessity to define the
$\Rightarrow$ relation in this manner, but doing so is in keeping both with
the path integral formalism and the orthodox interpretations of quantum
mechanics. There one would say that there exist possible paths from
$(\lambda_{1},s_{1})$ to $(\lambda_{2},s_{2})$, but they interfere with each
other in such a way as to keep the total amplitude of the transition $0$;
however, if a further measurement caused only a subset of these paths to be
taken, then the probability can become non-zero.

If measurements are unbiased, but not strongly unbiased, there still exists a
covering of $\Rightarrow$ s.t. within each element of the covering 1-5 hold.
To establish $(\lambda_{1},s_{1})\Rightarrow(\lambda_{2},s_{2})$ iff there
exists a path, $\bar{s}$, s.t. $\bar{s}(\lambda_{1})=s_{1}$ and
$\bar{s}(\lambda_{2})=s_{2}$, it is sufficient to establish it within each
element of the covering. As it turns out, statements 1-5 are sufficient for
accomplishing this.

The key step is to show that if
$(\lambda_{0},s_{0})\Rightarrow(\lambda_{1},s_{1})$ then there’s a partial
path, $\bar{s}[\lambda_{0},\lambda_{1}]$ s.t.
$\bar{s}[\lambda_{0},\lambda_{1}](\lambda_{0})=s_{0}$ and
$\bar{s}[\lambda_{0},\lambda_{1}](\lambda_{1})=s_{1}$. If $\Lambda$ is
discrete this can be readily shown using statement 5 and induction. The proof
when $\Lambda$ a continuum will now be briefly sketched. First, for each
$\lambda_{0}<\lambda<\lambda_{1}$, form the set of all $s$ s.t.
$(\lambda_{0},s_{0})\Rightarrow(\lambda,s)$ and
$(\lambda,s)\Rightarrow(\lambda_{1},s_{1})$. For
$\lambda_{1/2}\equiv\frac{1}{2}(\lambda_{0}+\lambda_{1})$ choose a $s_{1/2}$
from $\lambda_{1/2}$’s set. Now for each $\lambda_{0}<\lambda<\lambda_{1/2}$
form the set of all $s$ s.t. $(\lambda_{0},s_{0})\Rightarrow(\lambda,s)$ and
$(\lambda,s)\Rightarrow(\lambda_{1/2},s_{1/2})$, and for each
$\lambda_{1/2}<\lambda<\lambda_{1}$ form the set of all $s$ s.t.
$(\lambda_{1/2},s_{1/2})\Rightarrow(\lambda,s)$ and
$(\lambda,s)\Rightarrow(\lambda_{1},s_{1})$. For
$\lambda_{1/4}\equiv\frac{1}{2}(\lambda_{0}+\lambda_{1/2})$ and
$\lambda_{3/4}\equiv\frac{1}{2}(\lambda_{1/2}+\lambda_{1})$, choose a
$s_{1/4}$ and a $s_{3/4}$ from the newly formed sets and repeat the process.
When this has been done for all $\lambda=m/2^{n}$, at each of the other
$\lambda$’s take the intersection of all the formed sets, and select one
element (the intersection must be non-empty). The set of all the selected
$(\lambda_{m/2^{n}},s_{m/2^{n}})$ and all the $(\lambda,s)$’s chosen from the
intersections is the graph of a partial path running from
$(\lambda_{0},s_{0})$ to $(\lambda_{1},s_{1})$.

Just as as one can start with the transition relation and use it to define a
set of paths, one can also start with a set of paths, and from it derive the
transition relation: If $S$ is a dynamic set,
$(\lambda_{1},s_{1})\Rightarrow(\lambda_{2},s_{2})$ if
$S_{(\lambda_{1},s_{1})\rightarrow(\lambda_{2},s_{2})}\neq\emptyset$. With the
$\Rightarrow$ relation defined in this way, 1-4 above will always hold; if $S$
is a dynamic space, 5 will also hold. However, if the dynamic set is not a
dynamic space, the two representations may not be equivalent, paths can
contain more information. To see this, consider the case where
$(\lambda_{1},s_{1})\Rightarrow(\lambda_{2},s_{2})$,
$(\lambda_{2},s_{2})\Rightarrow(\lambda_{3},s_{3})$, and
$(\lambda_{1},s_{1})\Rightarrow(\lambda_{3},s_{3})$; this says that a path
runs from $(\lambda_{1},s_{1})$ to $(\lambda_{2},s_{2})$, a path runs from
$(\lambda_{2},s_{2})$ to $(\lambda_{3},s_{3})$, and a path run from
$(\lambda_{1},s_{1})$ to $(\lambda_{3},s_{3})$, but it doesn’t guarantee that
any individual path runs through all three points. However, when constructing
paths from the transition relation, it might be possible to construct such a
path. Therefore, starting with a dynamic set $S$, using $S$ to construct the
$\Rightarrow$ relation, then using $\Rightarrow$ to construct the set of
paths, $S^{\prime}$, you can have $S\varsubsetneq S^{\prime}$, the elements of
$S^{\prime}-S$ being non-existent paths that can’t be ruled out based on the
transition relation alone.

Form here on out we will revert to the path formalism.

####  V.1.2 Discretely Determined Partitions

The mathematical framework employed in quantum mechanics limits the types of
measurements that the theory can talk about. These limitations are of
essentially two types. First, because the probabilities are defined directly
on the outcomes, rather than on the sets of outcomes, partitions have to be at
most countable. Second, because measurements are described by projection
operators on the state space, experimental outcomes correspond to sequences of
measurements of $S(\lambda)$ at discrete values of $\Lambda_{S}$ ($S$ being
the system’s dynamic set). The following definitions will allow us to work
within these constraints.

######  Definition 145.

For dynamic set, $S$, and $L\subset\Lambda_{S}$:

$\alpha\subset S$ is _determined on $L$_ if $\alpha=\bigcap_{\lambda\in
L}S_{\rightarrow(\lambda,\alpha(\lambda))\rightarrow}=\\{\bar{p}\in
S:for\,all\>\lambda\in L,\>\bar{p}(\lambda)\in\alpha(\lambda)\\}$
(essentially,
$\alpha=S_{\rightarrow(\lambda_{1},\alpha(\lambda_{1}))\rightarrow...\rightarrow(\lambda_{i},\alpha(\lambda_{i}))\rightarrow_{...}}$)

If $\gamma$ is a partition of $S$, $\gamma$ is determined on $L$ if all
$\alpha\in\gamma$ are determined on $L$.

$\alpha$ is _discretely determined_ if it is determined on some discrete $L$;
similarly for partitions.

Note that if $\alpha$ is determined on $L$, and $L\subset L^{\prime}$, then
$\alpha$ is determined on $L^{\prime}$.

Quantum probabilities are calculated using equations of the form
$P(A_{1},A_{2},...|S)=<S,\lambda_{0}|\mathbb{P}(A_{1};\lambda_{1})\mathbb{P}(A_{2};\lambda_{2})...\mathbb{P}(A_{n};\lambda_{n})\mathbb{P}(A_{n};\lambda_{n})...\mathbb{P}(A_{2};\lambda_{2})\mathbb{P}(A_{1};\lambda_{1})|S,\lambda_{0}>$
where $S$ is the initial system state and $\mathbb{P}(A_{i};\lambda_{i})$ is
the projection operator onto state space region $A_{i}$ at $\lambda_{i}$. This
manner of calculation necessarily limits the formalism to partitions composed
of discretely determined measurements. Before moving on, let’s briefly take a
closer look at the nature of this discreteness.

To be able to calculate (or even represent)
$|\psi>\equiv...\mathbb{P}(A_{i};\lambda_{i})...\mathbb{P}(A_{2};\lambda_{2})\mathbb{P}(A_{1};\lambda_{1})|S,\lambda_{0}>$,
$L\equiv\\{\lambda_{1},\lambda_{2},...\lambda_{i},...\\}$ must have a least
element (which should not be less than $\lambda_{0}$), and for every
$\lambda\in L$, the set of elements of $L$ that are greater than $\lambda$
must have a least element; in other words, $L$ must be well ordered. Under
these conditions, we have
$|\psi_{1}>\equiv\mathbb{P}(A_{1};\lambda_{1})|S,\lambda_{0}>$,
$|\psi_{2}>\equiv\mathbb{P}(A_{2};\lambda_{2})|\psi_{1}>$, etc. If the
sequence is finite, terminating at $n$, then $|\psi>=|\psi_{n}>$. Otherwise,
$|\psi>$ is the limit of the sequence. The required probability is then
$<\psi|\psi>$.

Similarly, to be able to calculate
$O\equiv\mathbb{P}(A_{1};\lambda_{1})\mathbb{P}(A_{2};\lambda_{2})...\mathbb{P}(A_{i};\lambda_{i})...\mathbb{P}(A_{i};\lambda_{i})...\mathbb{P}(A_{2};\lambda_{2})\mathbb{P}(A_{1};\lambda_{1})$,
$L\equiv\\{\lambda_{1},\lambda_{2},...\lambda_{i},...\\}$ must have a greatest
element, and for every $\lambda\in L$, the set of elements of $L$ that are
less than $\lambda$ must have a greatest element; in other words, $L$ must be
“upwardly well ordered”. Under these conditions, the probability is
$<S,\lambda_{0}|O|S,\lambda_{0}>$.

Because we are interested in partitions with total probability that’s
guaranteed to be $1$ based only on the structure of the measurements, and
independent of the details of the inner-products, we are limited to partitions
that are determined on parameter sets of this second type. It should be noted,
however, that this further qualification is of little consequence, because all
finitely determined partitions are allowed, and countable cases can be taken
as the limit of a sequence of finite cases.

######  Definition 146.

If $L$ is the subset of a parameter, it is _upwardly well-ordered_ if every
non-empty subset of $L$ has a greatest element.

If $\gamma$ is a partition of $S$, $\gamma\in\daleth_{S}$ if $\gamma$ is
countable and for some upwardly well-ordered, bounded from below
$L\subset\Lambda_{S}$, $\gamma$ is determined on $L$.

If $\gamma\in\daleth_{S}$, $L(\gamma)$ is the set of upwardly well-ordered,
bounded from below sets, $l\subset\Lambda$, s.t.$\gamma$ is determined on $l$.

From here on out, “discretely determined” will refer to being determined on an
$L$ that’s upwardly well-ordered and bounded from below.

Regardless of how “discrete” is interpreted, being limited to discretely
determined measurements is a surprisingly strong constraint. To see this,
consider the following simple type of measurement:

######  Definition 147.

If $S$ is a dynamic space and $\alpha\subset S$, $\alpha$ is a _moment_ if for
all $\lambda\in\Lambda_{S}$, $\alpha$ is either $bb\lambda$ or $ba\lambda$.

Assume that $\alpha$ is $bb\lambda$ and for all $\lambda^{\prime}>\lambda$,
$\alpha$ is $ba\lambda^{\prime}$. This makes $\alpha$ a moment. If $\alpha$ is
$ba\lambda$ then $\alpha$ is a measurement of the system state at $\lambda$.
Otherwise, if $\alpha$ is not $ba\lambda$, $\alpha$ may be thought of as a
measurement of the system’s state and its rate of change. Colloquially, it may
be thought of as a measurement of position and velocity777The
$\sim|\alpha|_{\lambda}^{+}$ encountered in Section III.5 are examples of
measurements of rate of change; the $\sim|\alpha|_{\lambda}^{+}$ contain paths
that are in some compatible set, $t$, as of $\lambda$, but whose “velocities”
ensure that they will exit $t$ immediately after $\lambda$. One limitation of
the Hilbert Space formulation of quantum physics is that it can only be used
to describe measurements of system state alone.

######  Theorem 148.

If $X$ is a pairwise-disjoint collection of moments of a dynamic space, then
$X$ is compatible.

###### Proof.

Moments of a dynamic space are clearly companionable. Since $X$ is pairwise-
disjoint, it only remains to show that for all $\alpha\in X$, if
$\beta\in(\alpha)_{\lambda}^{X}$ and
$p\in\alpha(\lambda)\bigcap\beta(\lambda)$ then
$\alpha_{\rightarrow(\lambda,p)}=\beta_{\rightarrow(\lambda,p)}$.

A: If $(\alpha)_{\lambda}^{X}\neq\\{\alpha\\}$ then $\alpha$ is $bb\lambda$.

\- Assume $\alpha[-\infty,\lambda]\bigcap\beta[-\infty,\lambda]\neq\emptyset$
and $\alpha\neq\beta$. If $\alpha$ is $ba\lambda$ then
$\alpha\bigcap\beta\neq\emptyset$ which contradicts $X$ being pairwise-
disjoint. Since $\alpha$ is a moment, and is not $ba\lambda$, it is
$bb\lambda$. -

Assume $\beta\in(\alpha)_{\lambda}^{X}$ and $\beta\neq\alpha$, then by (A),
$\alpha$ is $bb\lambda$. Similarly, because $\alpha\in(\beta)_{\lambda}^{X}$,
$\beta$ is $bb\lambda$. Therefore, for any
$p\in\alpha(\lambda)\bigcap\beta(\lambda)$,
$\alpha_{\rightarrow(\lambda,p)}=\beta_{\rightarrow(\lambda,p)}$. ∎

It follows that any partition composed of moments is an ip. We can reasonably
assume that all moment measurements can be performed; indeed, they appear to
be quite useful. However, they can not all be represented using the quantum
formalism. This represents a rather severe limitation in the mathematical
language of quantum physics.

In spite of such limitations, there is one way in which $\daleth_{S}$ may be
considered overly inclusive. Dynamic sets are a very broad concept, which can
make them unwieldy to use. To reign in their unruliness, we generally assume
that they can be decomposed into ip’s, and that these ip’s are related to the
possible measurement on the set. The subset of $\daleth_{S}$ containing
elements that support to such decompositions will prove useful.

######  Definition 149.

If $S$ is a dynamic set, $\Gamma_{S}$ is the set of $\gamma\in\daleth_{S}$
s.t. for every $\alpha\in\gamma$ there’s an ip of $S$, $\varsigma$, s.t. for
some $\nu\in\varsigma$, $\alpha\subset\nu$.

If $D$ is a dynamic space then $\Gamma_{S}=\daleth_{S}$. In the quantum
formalism, when $\Gamma_{S}\subsetneq\daleth_{S}$, it is $\Gamma_{S}$ that is
of interest, for if $\alpha$ is an outcome of a quantum measurement, then it
must be an element of an ip. To see why, start with the a time ordered product
of projection operators corresponding to the measurement,
$...\mathbb{P}(A_{i};\lambda_{i})...\mathbb{P}(A_{2};\lambda_{2})\mathbb{P}(A_{1};\lambda_{1})$.
Define $\mathbb{P}(\neg
A_{j};\lambda_{j})\equiv\mathbb{I}-\mathbb{P}(A_{j};\lambda_{j})$; with
$X_{j}$ equal to either $A_{j}$ or $\neg A_{j}$, form the set of all time
ordered products of the form
$...\mathbb{P}(X_{i};\lambda_{i})...\mathbb{P}(X_{2};\lambda_{2})\mathbb{P}(X_{1};\lambda_{1})$.
This set corresponds to an ip.

######  Theorem 150.

If $S$ is a dynamic set and $\alpha\in\gamma\in\Gamma_{S}$ then $\alpha$ is
companionable

###### Proof.

Take $\varsigma$ to be an ip and $\alpha\subset\nu\in\varsigma$. If
$\bar{p}_{1},\bar{p}_{2}\in\alpha$ and
$\bar{p}_{1}(\lambda)=\bar{p}_{2}(\lambda)$ then
$\bar{p}=\bar{p}_{1}[-\infty,\lambda]\circ\bar{p}_{2}[\lambda,\infty]\in\nu$.
Take any $L\in L(\alpha)$. For all $\lambda\in L$,
$\bar{p}(\lambda)\in\alpha(\lambda)$, so $\bar{p}\in\alpha$, and so $\alpha$
is a subspace. With $\lambda_{0}=glb(L)$, there must be a
$\lambda\leq\lambda_{0}$ s.t. $\nu$ is $sbb\lambda$. Since $\alpha$ is
determined on $L$, $\alpha$ is then also $sbb\lambda$. Taking $\lambda_{1}$ to
be $L$’s greatest element, $\alpha$ is $wba\lambda_{1}$.∎

######  Theorem 151.

If $S$ is a dynamic set then given any $\alpha,\beta\in\bigcup\Gamma_{S}$ s.t.
$\alpha_{\rightarrow(\lambda,p)}\bigcap\beta_{\rightarrow(\lambda,p)}\neq\emptyset$,
if $\bar{p}_{1}[-\infty,\lambda]\in\alpha_{\rightarrow(\lambda,p)}$ and
$\bar{p}_{2}[\lambda,\infty]\in\beta_{(\lambda,p)\rightarrow}$ then
$\bar{p}_{1}[-\infty,\lambda]\circ\bar{p}_{2}[\lambda,\infty]\in S$.

###### Proof.

Take
$\bar{p}[-\infty,\lambda]\in\alpha_{\rightarrow(\lambda,p)}\bigcap\beta_{\rightarrow(\lambda,p)}$,
$\bar{p}\equiv\bar{p}[-\infty,\lambda]\circ\bar{p}_{2}[\lambda,\infty]$. Since
$\beta$ is a subspace, $\bar{p}\in S$. Take $\varsigma$ to be an ip and
$\alpha\subset\nu\in\varsigma$. For some $\eta\in\gamma$, $\bar{p}\in\eta$.
$\bar{p}[-\infty,\lambda]\in\nu_{\rightarrow(\lambda,p)}\bigcap\eta_{\rightarrow(\lambda,p)}$,
so $\eta_{\rightarrow(\lambda,p)}=\nu_{\rightarrow(\lambda,p)}$, and so
$\bar{p}_{1}[-\infty,\lambda]\in\eta_{\rightarrow(\lambda,p)}$. Since $\eta$
is a subspace and
$\bar{p}[\lambda,\infty]=\bar{p}_{2}[\lambda,\infty]\in\eta_{(\lambda,p)\rightarrow}$,
$\bar{p}_{1}[-\infty,\lambda]\circ\bar{p}_{2}[\lambda,\infty]\in\eta\subset
S$. ∎

####  V.1.3 Interconnected Dynamic Sets

An all-reet e-automata can not forget anything its ever known about the
system. Under the right conditions, however, an e-automata can discover
something about the system’s past that is not implied by the current state.
This is most readily seen if the system parameter is discrete.

For an e-automata with a discrete parameter, assume that at $\lambda$ the
e-automata is in state $(s_{\lambda},e_{\lambda})$ and at $\lambda+1$ it’s in
state $(s_{\lambda+1},e_{\lambda+1})$. It would be reasonable (though not
necessary) to assume that the system state of $s_{\lambda}$ is not reflected
in $e_{\lambda}$, and doesn’t get reflected in the environment until
$e_{\lambda+1}$. In this case, it takes one step in time for the environment
to learn about the system. More generally, of course, the set of allowed
environmental states at $\lambda+1$ will be a function of both $s_{\lambda}$
and $s_{\lambda+1}$, as well as $e_{\lambda}$, but this still asserts that it
is possible to learn something about the system’s past that’s not reflected in
the current state of the system.

If the parameter is continuous and the measurements are at discrete times we
wouldn’t expect this to happen, though it can; the environment may gain
information at $\lambda_{2}$ about the state of the system at $\lambda_{1}$
that isn’t implied by the state of the system at $\lambda_{2}$. In order for
this to occur the system itself would have to remember something about its
state at $\lambda_{1}$ until $\lambda_{2}$, at which point the knowledge is
simultaneously passed to the environment and forgotten by the system. This is
clearly an edge case, but a pernicious one, as it allows for anomalous ip’s
with discretely determined measurements; in order to regularize this set of
ip’s, this edge case needs to be eliminated. In order to be eliminated, the
system dynamics need to be interconnected, a property that will defined first
for dynamic spaces, and then for dynamic sets.

######  Definition 152.

A dynamic space, $D$, is _interconnected_ if

for all $\lambda_{1}<\lambda_{2}$, every $p_{1},p_{1}^{\prime}\in
D(\lambda_{1})$, $p_{2},p_{2}^{\prime}\in D(\lambda_{2})$ s.t.
$(\lambda_{1},p_{1})\rightarrow(\lambda_{2},p_{2})\neq\emptyset$,
$(\lambda_{1},p_{1})\rightarrow(\lambda_{2},p_{2}^{\prime})\neq\emptyset$,
$(\lambda_{1},p_{1}^{\prime})\rightarrow(\lambda_{2},p_{2})\neq\emptyset$, and
$(\lambda_{1},p_{1}^{\prime})\rightarrow(\lambda_{2},p_{2}^{\prime})\neq\emptyset$,

there exists a $\lambda\in[\lambda_{1},\lambda_{2}]$ and a $p\in D(\lambda)$
s.t. $(\lambda_{1},p_{1})\rightarrow(\lambda,p)\neq\emptyset$,
$(\lambda_{1},p_{1}^{\prime})\rightarrow(\lambda,p)\neq\emptyset$,
$(\lambda,p)\rightarrow(\lambda_{2},p_{2})\neq\emptyset$, and
$(\lambda,p)\rightarrow(\lambda_{2},p_{2}^{\prime})\neq\emptyset$.

Being interconnected is equivalent to saying that if $p_{1}\neq
p_{1}^{\prime}$ then
$\rightarrow\\{(\lambda_{1},p_{1}),(\lambda_{1},p_{1}^{\prime})\\}\rightarrow(\lambda_{2},p_{2})\rightarrow$,
$\rightarrow(\lambda_{1},p_{1})\rightarrow(\lambda_{2},p_{2}^{\prime})\rightarrow$,
and
$\rightarrow(\lambda_{1},p_{1}^{\prime})\rightarrow(\lambda_{2},p_{2}^{\prime})\rightarrow$
can not be mutually compatible. Essentially, for any
$\bar{p}[-\infty,\lambda_{1}]\in\,\rightarrow(\lambda_{1},p_{1})$,
$\bar{p}^{\prime}[-\infty,\lambda_{1}]\in\,\rightarrow(\lambda_{1},p_{1}^{\prime})$,
if $\bar{p}[-\infty,\lambda_{1}]$ and $\bar{p}^{\prime}[-\infty,\lambda_{1}]$
have not been distinguished by $\lambda_{1}$, and there is no measurement
between $\lambda_{1}$ and $\lambda_{2}$, then they can not be distinguished at
$\lambda_{2}$ because the point $(\lambda,p)$ destroys the ability to
distinguish them. Nearly all actively studied dynamic systems are
interconnected, and quantum systems always are.

Now to generalize interconnectedness for dynamic sets.

######  Definition 153.

If $S$ is a dynamic set and
$\bar{p}_{11},\bar{p}_{21},\bar{p}_{12},\bar{p}_{22}\in S$,
$(\bar{p}_{11},\bar{p}_{21},\bar{p}_{12},\bar{p}_{22})$ is an _interconnect on
$[\lambda_{1},\lambda_{2}]$_ if

$\bar{p}_{11}[-\infty,\lambda_{1}]=\bar{p}_{12}[-\infty,\lambda_{1}]$

$\bar{p}_{21}[-\infty,\lambda_{1}]=\bar{p}_{22}[-\infty,\lambda_{1}]$

$\bar{p}_{11}[\lambda_{2},\infty]=\bar{p}_{21}[\lambda_{2},\infty]$

$\bar{p}_{12}[\lambda_{2},\infty]=\bar{p}_{22}[\lambda_{2},\infty]$.

The paths $\bar{p}_{11},\bar{p}_{21},\bar{p}_{12},\bar{p}_{22}$ can be thought
of as sample paths from the sets
$\rightarrow(\lambda_{1},p_{1})\rightarrow(\lambda_{2},p_{2})\rightarrow$,
etc., that were used in the dynamic space definition of interconnectedness.
Because e-automata are unbiased, if there are no measurements on these paths
between $\lambda_{1}$ and $\lambda_{2}$ then
$+\\{\bar{p}_{11}[-\infty,\lambda_{1}],\bar{p}_{21}[-\infty,\lambda_{1}]\\}$
should behave like a dynamic space in $[\lambda_{1},\lambda_{2}]$.

######  Definition 154.

If $A$ is a dynamic set, $[\lambda_{1},\lambda_{2}]$ is a _space-segment of
$A$_ if for all $\bar{p}\in A$,
$[\lambda,\lambda^{\prime}]\subset[\lambda_{1},\lambda_{2}]$,
$\bar{p}^{\prime}[\lambda,\lambda^{\prime}]\in
A_{(\lambda,\bar{p}(\lambda))\rightarrow(\lambda^{\prime},\bar{p}(\lambda^{\prime}))}$,
$\bar{p}[-\infty,\lambda]\circ\bar{p}^{\prime}[\lambda,\lambda^{\prime}]\circ\bar{p}[\lambda^{\prime},\infty]=A$.

If an experiment has not distinguished between the paths of an interconnect on
_$[\lambda_{1},\lambda_{2}]$_ by $\lambda_{1}$, and there is no measurement
between $\lambda_{1}$ and $\lambda_{2}$, then interconnectedness will insure
that they can not be distinguished at $\lambda_{2}$:

######  Definition 155.

A dynamic set, $S$, is _interconnected_ if for every
$\lambda_{1}<\lambda_{2}$, every interconnect on _$[\lambda_{1},\lambda_{2}]$
, _$(\bar{p}_{11},\bar{p}_{21},\bar{p}_{12},\bar{p}_{22})$, s.t.
$[\lambda_{1},\lambda_{2}]$ is a space-segment of
$+\\{\bar{p}_{11}[-\infty,\lambda_{1}],\bar{p}_{21}[-\infty,\lambda_{1}]\\}$

there exists a $\lambda\in[\lambda_{1},\lambda_{2}]$ and an interconnect on
$[\lambda,\lambda_{2}]$,
$(\bar{p}_{11}^{\prime},\bar{p}_{21}^{\prime},\bar{p}_{12}^{\prime},\bar{p}_{22}^{\prime})$,
s.t.
$\bar{p}_{ij}^{\prime}[-\infty,\lambda_{1}]=\bar{p}_{ij}[-\infty,\lambda_{1}]$,
$\bar{p}_{ij}^{\prime}[\lambda_{2},\infty]=\bar{p}_{ij}[\lambda_{2},\infty]$,
and all $\bar{p}_{ij}^{\prime}(\lambda)=\bar{p}_{mn}^{\prime}(\lambda)$.

$[\lambda_{1},\lambda_{2}]$ being a space-segment of
$+\\{\bar{p}_{11}[-\infty,\lambda_{1}],\bar{p}_{21}[-\infty,\lambda_{1}]\\}$
helps to insure that there exists an interconnect,
$(\bar{p}_{11}^{\prime},\bar{p}_{21}^{\prime},\bar{p}_{12}^{\prime},\bar{p}_{22}^{\prime})$,
rather than just four paths that share the same point. If $S$ is a dynamic
space, this definition is equivalent to the prior one.

Finally, to make interconnectedness more readily applicable to $\Gamma_{S}$
for the case when the system is not a dynamic space.

######  Definition 156.

If $L\subset\Lambda$ is upwardly well-ordered, and $\lambda$ is not a lower-
bound of $L$, $pred_{L}(\lambda)\equiv lub(L\bigcap[-\infty,\lambda))$.

$pred_{L}(\lambda)$ is short for “predecessor of $\lambda$ on $L$”.

######  Definition 157.

For $\gamma\in\Gamma_{S}$, $\mathcal{L}(\gamma)\subset L(\gamma)$ s.t. if
$L\in\mathcal{L}(\gamma)$ then for all $\alpha\in\gamma$, $\lambda\in
L-\\{glb(L)\\}$, $[pred_{L}(\lambda),\lambda]$ is a space-segment of
$+\alpha[-\infty,pred_{L}(\lambda)]$.

$\gamma\in\Gamma_{S}^{s}$ if $\gamma\in\Gamma_{S}$ and
$\mathcal{L}(\gamma)\neq\emptyset$.

For quantum systems, the dynamics between successive measurements is always a
space-segment, and so when discussing measurements only the elements of
$\Gamma_{S}^{s}$ are of interest. For dynamic spaces,
$\Gamma_{S}^{s}=\Gamma_{S}=\daleth_{S}$.

###  V.2 Partitions of Unity for Quantum Systems

A partition with total probability of $1$ will be given the fancy title of
being a _partition of unity_. In this section, we consider the quantum
partitions of unity.

When discussing quantum probabilities there are two cases to be considered:
the conditional case, where an initial state is known, and the non-conditional
case, where no initial state is known & everything about the system is
discovered via the experiment. We will start by considering the conditional
case.

####  V.2.1 The Conditional Case

######  Definition 158.

If $\gamma$ is a partition of dynamic set $S$ and $\bar{p}_{1},\bar{p}_{2}\in
S$, $\bar{p}_{1}$ and $\bar{p}_{2}$ are _co-located on $\gamma$_ if there’s a
$\alpha\in\gamma$ s.t. $\bar{p}_{1},\bar{p}_{2}\in\alpha$; $\bar{p}_{1}$ and
$\bar{p}_{2}$ _converge at $\lambda$_ if
$\bar{p}_{1}[\lambda,\infty]=\bar{p}_{2}[\lambda,\infty]$

The central object of study in this section will be the set of partitions
$q_{S}$:

######  Definition 159.

For dynamic set $S$, $\gamma\in q_{S}$ if $\gamma\in\Gamma_{S}^{s}$ and for
some $L\in\mathcal{L}(\gamma)$, all $\lambda_{1},\lambda_{2}\in L$ s.t.
$\lambda_{1}\equiv pred_{L}(\lambda_{2})$, all
$\bar{p}_{1}[-\infty,\lambda_{1}],\bar{p}_{2}[-\infty,\lambda_{1}]\in
S[-\infty,\lambda_{1}]$, either all
$\bar{p}\in+\bar{p}_{1}[-\infty,\lambda_{1}]$,
$\bar{p}^{\prime}\in+\bar{p}_{2}[-\infty,\lambda_{1}]$ that converge at
$\lambda_{2}$ are co-located on $\gamma$, or none are.

In the following claim, total probabilities of a set of outcomes are
“guaranteed to to equal $1$” if it’s known to be $1$ based only on the
structure of the outcomes, and independent of the details of the transition
probabilities, including the choice of initial state.

######  Claim 160.

For quantum probabilities in the conditional case, the sum over probabilities
of a set of outcomes is guaranteed to equal $1$ iff the the set of outcomes is
an element of $q_{S}$.

######  Remark.

It has already been argued that quantum outcomes are elements of
$\bigcup\Gamma_{S}^{s}$. The rest of the claim may be justified as follows.

Quantum outcomes are represented by time ordered products of projection
operators. Let’s take $L$ to be the set of times at which a projection
operator is applied, and define $\lambda_{1}$ to be the largest element of
$L$, $\lambda_{2}$ to be the largest element of $L-\\{\lambda_{0}\\}$, etc.
(Note that as the subscript increases, the time decreases.) An outcome is then
represented by
$\Pi_{i}=...\mathbb{P}(A_{ij};\lambda_{j})...\mathbb{P}(A_{i1};\lambda_{1})$,
where $A_{ij}\subset S(\lambda_{j})$, the “$i$” subscript identifies the
outcome, and the “$j$” subscript identifies the time. If the set of outcomes
form a partition then $\sum_{i}\Pi_{i}=\mathbb{I}$ ($\mathbb{I}$ being the
identity operator). This yields a total probability of $1$ irregardless of the
choice of the initial state iff
$\sum_{i}<\psi,\lambda_{0}|\Pi_{i}\Pi_{i}^{\dagger}|\psi,\lambda_{0}>=1$ for
all initial states $|\psi,\lambda_{0}>$, which holds iff
$\sum_{i}\Pi_{i}\Pi_{i}^{\dagger}=\mathbb{I}$.

For any $\lambda_{j}\in L$, define $\chi_{j}$ by
$(...,s_{k},...,s_{j})\in\chi_{j}$ if for all
$\lambda_{q}<\lambda_{r}\leq\lambda_{j}$ ($\lambda_{q},\lambda_{r}\in L$) the
transition $(s_{q},\lambda_{q})\Rightarrow(s_{r},\lambda_{r})$ is allowed;

Further, define $\Xi_{j}$ by
$(...,s_{k},s_{k}^{\prime},...,s_{j+1},s_{j+1}^{\prime},s_{j})\in\Xi_{j}$ if

1) $(...,s_{k},...,s_{j+1},s_{j})\in\chi_{j}$ and
$(...,s_{k}^{\prime},...,s_{j+1}^{\prime},s_{j})\in\chi_{j}$

2) For some outcome, $\Pi_{i}$, $s_{j}\in A_{ij}$ and for all $k>j$ (s.t.
$\lambda_{k},\in L$) $s_{k},s_{k}^{\prime}\in A_{ik}$.

$\sum_{i}\Pi_{i}\Pi_{i}^{\dagger}=\mathbb{I}$ can then be expanded to

Condition 1:

$\displaystyle\int_{\Xi_{1}}ds_{1}ds_{2}ds_{2}^{\prime}$
$\displaystyle...|s_{2},\lambda_{2}><s_{2},\lambda_{2}|s_{1},\lambda_{1}>$
$\displaystyle<s_{1},\lambda_{1}|s_{1}^{\prime},\lambda_{1}><s_{2}^{\prime},\lambda_{2}|...=\mathbb{I}$

Start by integrating over $s_{1}$; the above identity can not hold unless for
every $(...,s_{2},s_{2}^{\prime})=(\tilde{s},\tilde{s}^{\prime})\in
Dom(\Xi_{1})$ either
$<s_{2},\lambda_{2}|(\int_{(\tilde{s},\tilde{s}^{\prime},s_{n})\in\Xi_{1}}ds_{1}|s_{1},\lambda_{1}><s_{1},\lambda_{1}|)|s_{2}^{\prime},\lambda_{2}>=1$
or
$<s_{2},\lambda_{2}|(\int_{(\tilde{s},\tilde{s}^{\prime},s_{n})\in\Xi_{1}}ds_{1}|s_{1},\lambda_{1}><s_{1},\lambda_{1}|)|s_{1}^{\prime},\lambda_{2}>=0$.
When $\tilde{s}=\tilde{s}^{\prime}$ the first equality clearly holds. For
$\tilde{s}\neq\tilde{s}^{\prime}$, the only way that we could have an $s_{1}$
s.t. $(\tilde{s},s_{1})\in\chi_{1}$, and
$(\tilde{s}^{\prime},s_{1})\in\chi_{1}$, but
$(\tilde{s},\tilde{s}^{\prime},s_{1})\notin\Xi_{1}$, and still have one of
these equalities hold is if either $<s_{2},\lambda_{2}|s_{1},\lambda_{1}>=0$
or $<s_{2}^{\prime},\lambda_{2}|s_{1},\lambda_{1}>=0$ (this makes the claim of
$(\tilde{s},s_{1})\in\chi_{1}$, and $(\tilde{s}^{\prime},s_{1})\in\chi_{1}$
uncomfortable, but the situation can not be ruled out). Since we are
interested in the conditions under which the probabilities are guaranteed to
sum to $1$ regardless of the details of inner-products, we eliminate this last
case and are lead to:

Condition 2: If $(\tilde{s},\tilde{s}^{\prime})\in Dom(\Xi_{1})$ then for all
$s_{1}$ s.t. $(\tilde{s},s_{1})\in\chi_{1}$ and
$(\tilde{s}^{\prime},s_{1})\in\chi_{1}$,
$(\tilde{s},\tilde{s}^{\prime},s_{1})\in\Xi_{1}$.

Which is equivalent to applying the $q_{S}$ condition at $\lambda_{1}$.

Apply Condition 2 to Condition 1, Condition 1 is reduced to:

Condition 3:

$\displaystyle\int_{\Xi_{1}}ds_{2}ds_{3}ds_{3}^{\prime}$
$\displaystyle...|s_{3},\lambda_{3}><s_{3},\lambda_{3}|s_{2},\lambda_{2}>$
$\displaystyle<s_{2},\lambda_{2}|s_{3}^{\prime},\lambda_{3}><s_{3}^{\prime},\lambda_{3}|...=\mathbb{I}$

Which is identical in form to the Condition 1. Therefore, to satisfy Condition
$1$, it is sufficient to satisfy Condition 1, for all elements of $L$ to
satisfy Condition 2 (with the “$1$” subscript replaced with “$i$”). Note that
Condition 2 was imposed as a necessary condition for Condition 1 to hold; we
now see that it’s a necessary and sufficient condition. Finally, note that
requiring Condition 2 on all $\lambda_{i}$ is equivalent to requiring that the
set of outcomes is an element of $q_{S}$.

######  Theorem 161.

For dynamic set $S$

1) If $\gamma\in q_{S}$ then $\gamma$ is nearly compatible

2) If $S$ is interconnected, $\gamma\in\Gamma_{S}^{s}$, and $\gamma$ is nearly
compatible then $\gamma\in q_{S}$

###### Proof.

By Thm 150, for all $\gamma\in\Gamma_{S}^{s}$, all $\alpha\in\gamma$ are
companionable

1) For $\alpha,\beta\in\gamma$ assume
$\alpha_{\rightarrow(\lambda,p)}\bigcap\beta_{\rightarrow(\lambda,p)}\neq\emptyset$,
and $\alpha\neq\beta$. $\lambda$ can not be an upper-bound on any $L(\gamma)$.
Take any
$\bar{p}[-\infty,\lambda]\in\alpha_{\rightarrow(\lambda,p)}\bigcap\beta_{\rightarrow(\lambda,p)}$,
$\bar{p}^{\prime}[-\infty,\lambda]\in\alpha_{\rightarrow(\lambda,p)}$,
$\bar{p}_{1}[\lambda,\infty]\in\alpha_{(\lambda,p)\rightarrow}$,
$\bar{p}_{2}[\lambda,\infty]\in\beta_{(\lambda,p)\rightarrow}$.

$\bar{p}_{1}\equiv\bar{p}[-\infty,\lambda]\circ\bar{p}_{1}[\lambda,\infty]\in\alpha$

$\bar{p}_{1}^{\prime}\equiv\bar{p}^{\prime}[-\infty,\lambda]\circ\bar{p}_{1}[\lambda,\infty]\in\alpha$

$\bar{p}_{2}\equiv\bar{p}[-\infty,\lambda]\circ\bar{p}_{2}[\lambda,\infty]\in\beta$

$\bar{p}_{2}^{\prime}\equiv\bar{p}^{\prime}[-\infty,\lambda]\circ\bar{p}_{2}[\lambda,\infty]$
(Exists by Thm 151)

Applying the definition of $q_{S}$ to any $\lambda^{\prime}\in L$ s.t.
$\lambda^{\prime}>\lambda$, it follows that $\bar{p}_{2}^{\prime}\in\beta$.
Therefore $\alpha_{\rightarrow(\lambda,p)}=\beta_{\rightarrow(\lambda,p)}$.

2) Assume $\gamma\in\Gamma_{S}^{s}$ is nearly compatible. Take any
$L\in\mathcal{L}(\gamma)$, any $\lambda_{1}\in L$ s.t. $\lambda_{1}\neq
glb(L)$, and define $\lambda_{0}\equiv pred_{L}(\lambda)$. Take any
$\alpha\in\gamma$ and any $\bar{p}_{1},\bar{p}_{2}\in\alpha$ that converge at
$\lambda_{1}$ (if they exist). Take $\bar{p}_{3},\bar{p}_{4}\in S$ such that
$\bar{p}_{3}[-\infty,\lambda_{0}]=\bar{p}_{1}[-\infty,\lambda_{0}]$,
$\bar{p}_{4}[-\infty,\lambda_{0}]=\bar{p}_{2}[-\infty,\lambda_{0}]$, and
$\bar{p}_{3}$ and $\bar{p}_{4}$ converge at $\lambda_{1}$. Further take
$\bar{p}_{3}\in\beta\in\gamma$. The theorem is proved if
$\bar{p}_{4}\in\beta$.

Because $(\bar{p}_{1},\bar{p}_{2},\bar{p}_{3},\bar{p}_{4})$ is an interconnect
on $[\lambda_{0},\lambda_{1}]$, and $S$ is interconnected, there’s a
$\lambda^{\prime}\in[\lambda_{0},\lambda_{1}]$ and
$\bar{p}_{1}^{\prime},\bar{p}_{2}^{\prime},\bar{p}_{3}^{\prime},\bar{p}_{4}^{\prime}\in
S$ s.t.
$\bar{p}_{i}^{\prime}[-\infty,\lambda_{0}]=\bar{p}_{i}[-\infty,\lambda_{0}]$,
$\bar{p}_{i}^{\prime}[\lambda_{1},\infty]=\bar{p}_{i}[\lambda_{1},\infty]$,
$\bar{p}_{1}^{\prime}[-\infty,\lambda^{\prime}]=\bar{p}_{3}^{\prime}[-\infty,\lambda^{\prime}]$,
$\bar{p}_{2}^{\prime}[-\infty,\lambda^{\prime}]=\bar{p}_{4}^{\prime}[-\infty,\lambda^{\prime}]$,
and all
$\bar{p}_{i}^{\prime}(\lambda^{\prime})=\bar{p}_{j}^{\prime}(\lambda^{\prime})=p$.

Since $\bar{p}_{i}^{\prime}$ and $\bar{p}_{i}$ are equal on $L$, and all
elements of $\gamma$ are determined on $L$, they are co-located on $\gamma$.
Therefore $\bar{p}_{1}^{\prime},\bar{p}_{2}^{\prime}\in\alpha$ and
$\bar{p}_{3}^{\prime}\in\beta$. Since
$\bar{p}_{1}^{\prime}[-\infty,\lambda^{\prime}]=\bar{p}_{3}^{\prime}[-\infty,\lambda^{\prime}]\in\alpha_{\rightarrow(\lambda^{\prime},p)}\bigcap\beta_{\rightarrow(\lambda^{\prime},p)}$
and $\gamma$ is nearly compatible,
$\alpha_{\rightarrow(\lambda^{\prime},p)}=\beta_{\rightarrow(\lambda^{\prime},p)}$.
Since
$\bar{p}_{4}^{\prime}[-\infty,\lambda^{\prime}]=\bar{p}_{2}^{\prime}[-\infty,\lambda^{\prime}]\in\alpha_{\rightarrow(\lambda^{\prime},p)}$,
$\bar{p}_{4}^{\prime}[-\infty,\lambda^{\prime}]\in\beta_{\rightarrow(\lambda^{\prime},p)}$.
Since
$\bar{p}_{4}^{\prime}[\lambda_{1},\infty]=\bar{p}_{4}[\lambda_{1},\infty]=\bar{p}_{3}[\lambda_{1},\infty]$,
for all $\lambda\in L$, $\bar{p}_{4}^{\prime}(\lambda)\in\beta(\lambda)$, so
$\bar{p}_{4}^{\prime}\in\beta$. Since $\bar{p}_{4}$ and $\bar{p}_{4}^{\prime}$
are equal on all $\lambda\in L$, $\bar{p}_{4}\in\beta$. ∎

####  V.2.2 The Non-Conditional Case

For the non-conditional case, no initial state is assumed. Once again
representing each outcome, $\Pi_{i}$, as a product of projection operators,
$\Pi_{i}=...\mathbb{P}(A_{ij};\lambda_{j})...\mathbb{P}(A_{i1};\lambda_{1})$,
the probability of a given outcome is
$P_{i}=\frac{1}{Tr(\mathbb{I})}Tr(\Pi_{i}\Pi_{i}^{\dagger})$ (ignoring
complications that arise if $Tr(\mathbb{I})$ is infinite). This can be arrived
at by assuming that, if the initial state is unknown, then the probability is
the average for all possible initial states: Given any initial state,
$(s,\lambda_{0})$, the probability for obtaining outcome $i$ is
$P_{i|(s,\lambda_{0})}=<s,\lambda_{0}|\Pi_{i}\Pi_{i}^{\dagger}|s,\lambda_{0}>$.
With $V$ the volume of $S(\lambda_{0})$, the average probability is then
$\frac{1}{V}\Sigma_{s\in
S(\lambda_{0})}<s,\lambda_{0}|\Pi_{i}\Pi_{i}^{\dagger}|s,\lambda_{0}>$. Since
$V=Tr(\mathbb{I})$ and $\Sigma_{s\in
S(\lambda_{0})}<s,\lambda_{0}|\Pi_{i}\Pi_{i}^{\dagger}|s,\lambda_{0}>=Tr(\Pi_{i}\Pi_{i}^{\dagger})$,
this average probability is equal to $P_{i}$ given above.

In the non-conditional case, the total probability is therefore guaranteed to
be $1$ iff $\sum_{i}Tr(\Pi_{i}\Pi_{i}^{\dagger})=Tr(\mathbb{I})$.

Note that the assumption that the non-conditional probability is equal to the
average conditional probabilities need not hold for a dps; there is nothing in
the nature of the dps that demands it. The assumption is equivalent to a
certain kind of additivity: Take $\alpha$ to be an outcome that’s $bb\lambda$,
define $x\equiv\alpha(\lambda)$, and select $x_{1},x_{2}$ s.t. $x_{1}\bigcup
x_{2}=x$ and $x_{1}\bigcap x_{2}=\emptyset$; define
$\alpha_{1}\equiv\alpha\bigcap S_{\rightarrow(\lambda,x_{1})}$ and
$\alpha_{2}\equiv\alpha\bigcap S_{\rightarrow(\lambda,x_{2})}$; the assumption
being made is that under such circumstances,
$P(\alpha)=P(\alpha_{1})+P(\alpha_{2})$.

We would expect this to hold if $\alpha$ is bounded from above by $\lambda$,
but not if it’s bounded from below. To see that it will hold when bounded from
above, assume $\alpha$ is $ba\lambda$ (so the measurement of
$\\{x_{1},x_{2}\\}$ would be the final measurement, rather than the initial
one), take $\gamma$ to be any ip s.t. $\alpha\in\gamma$, and define
$A\equiv\gamma-\\{\alpha\\}.$ $A\bigcup\\{\alpha_{1},\alpha_{2}\\}$ is an ip.
If $P(A)$ is unchanged by whether $A$ is paired with $\alpha$ or
$\\{\alpha_{1},\alpha_{2}\\}$ then $P(\alpha)=P(\alpha_{1})+P(\alpha_{2})$.
Call this property “additivity of final state”; call the bounded from below
case “additivity of initial state”.

Achieving additivity of initial state is not quite as straightforward as
additivity final of state. One obvious way to do so is to assume additivity of
initial state and reversibility. There are, however, other ways for the
behavior to be realized. (See Appendix C for the definition of reversibility
on a DPS.)

###  V.3 Maximal Quantum Systems

The foundations of quantum physics may be thought of as being composed of
three interconnected parts: measurement theory, probability theory, and
probability dynamics. This article has largely been concerned with creating a
mathematical language sufficient for utilization in the measurement theory &
that portion of the probability theory implied by the measurement theory. One
does not expect a scientific theory to follow immediately from the
mathematical language that it uses, unless the theory is fairly trivial. For
this reason, Thm 161 is quite evocative.

Thm 161 says an interconnected dynamic system will posses the basic character
of a quantum system if the finite, discretely-determined portion of its dps
can be embedded in a $\Upsilon$-maximal dps. The question now is, under what
conditions is such an embedding possible?

All countable, discretely-determined ip’s will have consistent probabilities
if the system is countably additive in its final state. As mentioned earlier,
it may be assumed that a dynamic system has this property. To see why it is
sufficient, start by choosing any $\lambda$, and any countable partition of
$S(\lambda)$. By assumption, the total probability of the associated outcomes
will be $1$. For any of these outcome, optionally choose any
$\lambda^{\prime}>\lambda,$ and any countable partition of
$S(\lambda^{\prime})$; by assumption, the total probability of these new
outcomes will equal the probability of the original outcome. Continuing in
this way, one can create all countable, discretely-determined ip’s of the
dynamic space.

To understand when such a dps could be maximal, consider a simple case of a
set of outcomes that are nearly compatible, but not compatible. Assume there
are $p_{1,0},p_{1,1,}p_{1,2}\in S(\lambda_{1})$ and
$p_{2,0},p_{2,1,}p_{2,2}\in S(\lambda_{2})$ ($\lambda_{1}<\lambda_{2}$) s.t.
each of the $p_{2,j}$’s can be reached from $(\lambda_{1},p_{1,j})$ and
$(\lambda_{1},p{}_{1,(j+1)\,mod\,3})$, but not
$(\lambda_{1},p{}_{1,(j+2)\,mod\,3})$. For $p_{2,0}$ and $p_{2,1}$ form
outcomes consisting of both of the $p_{1,i}$ that can reach them:
$O_{0}=(\lambda_{1},\\{p{}_{1,0},p{}_{1,1}\\})\rightarrow(\lambda_{2},p_{2,0})$
and
$O_{1}=(\lambda_{1},\\{p{}_{1,1},p{}_{1,2}\\})\rightarrow(\lambda_{2},p_{2,1})$.
For $p_{2,2}$ form outcomes from $p_{1,2}$ and $p_{1,0}$ individually:
$O_{21}=(\lambda_{1},p{}_{1,2})\rightarrow(\lambda_{2},p_{2,2})$ and
$O_{22}=(\lambda_{1},p{}_{1,0})\rightarrow(\lambda_{2},p_{2,2})$. The
collection of these $4$ outcomes is nearly compatible, but not compatible.
That they are nearly compatible can be seen from the fact that they satisfy
the $q_{S}$ condition; that they are not compatible can be seen from the fact
that while the combination of the first two outcomes imply that only
$\\{p_{1,0},p_{1,1,}p_{1,2}\\}$ was measured at $\lambda_{1}$, the second two
imply that $p_{1,2}$ was distinguished from $p_{1,0}$ at $\lambda_{1}$.

The inclusion of such sets of outcomes into a t-algebra would make
probabilities more highly additive; in particular, it would weaken the non-
additivity of double-slit experiments. To see this, take the simplest case of
$S(\lambda_{i})=\\{p_{i,0},p_{i,1,}p_{i,2}\\}$; because
$\\{O_{0},O_{1},O_{21}\bigcup O_{22}\\}$ is compatible, and would certainly be
on the t-algebra, we’d then have $P(O_{21}\bigcup
O_{22})=P(O_{21})+P(O_{22})$, rendering the probabilities of this particular
double-slit experiment entirely additive. However, if there exists a $p\in
S(\lambda_{2})$ that can be reached by $p_{1,0}$, $p_{1,1}$, and $p_{1,2}$,
then the $q_{S}$ condition entails that there can be no element of $q_{S}$
that contains $O_{0}$, $O_{1}$, $O_{21}$, and $O_{22}$; in this case a maximal
t-algebra can not include $\\{O_{0},O_{1},O_{21},O_{22}\\}$, and so will not
demand $P(O_{21}\bigcup O_{22})=P(O_{21})+P(O_{22})$.

This illustrates how the finite & finitely determined portion of a dps can be
maximal even if its probabilities are highly non-additive. If the paths of a
dynamic system form a richly interlocking network, then the elements of
$q_{S}$ that are not ip’s will be limited, making it more likely that the
system will be maximal. A rich network of paths will not limit the kinds of
countable, discretely determined experiments that can be performed, it simply
limits the amount of probabilistic information that can be extracted from
them. A similar effect was seen previously when considering
interconnectedness.

For non-deterministic systems, it is difficult to distinguish sets of paths
that can not occur from sets of paths that occur with probability $0$. If the
approach to system dynamics is to include paths unless they can be ruled out
in principal and let the probability function handle the rest, then the
network of interlocking paths will be enriched. This will generally cause the
discretely determined portion of such systems to be maximal. Let’s see this in
detail.

######  Definition 162.

If $\gamma_{1}$ and $\gamma_{2}$ are partitions of some set, $S$,
$\gamma_{1}\prec\gamma_{2}$ if for all $\alpha\in\gamma_{1}$ there’s a
$\beta\in\gamma_{2}$ s.t. $\alpha\subset\beta$

As mentioned above, partitions with total probability of $1$ can be
iteratively created by taking any $\gamma\in\Gamma_{S}^{s}$ with probability
known to be $1$, and for each $\alpha\in\gamma$, select some $\lambda$ s.t.
$\alpha$ is $ba\lambda$, and slice up $\alpha$ by partitioning
$\alpha(\lambda)$. However, not all partitions can be formed in this manner.
If the partition is not compatible, and so can not be decided by an ideal
e-automata, it is allowed to forget at $\lambda$ some of what happened prior
to $\lambda$. In such cases, to form the new partition, you wouldn’t simply
take the elements of $\gamma$ and append measurements at $\lambda$, a further
step would also be possible: for each measurement outcome at $\lambda$,
multiple elements of $\gamma$ may be combined into a single outcome. More
precisely, select some countable set of $\gamma_{i}\succ\gamma$ such that each
$\gamma_{i}\in\Gamma_{S}^{s}$. For every $p\in S(\lambda_{n})$, select a
$\gamma_{i}$. For every $\gamma_{i}$, every $\alpha\in\gamma_{i}$, create a
countable partition of the set of $p\in S(\lambda_{n+1})$ that “selected”
$\gamma_{i}$, $X_{i,\alpha}$, and for every $A\in X_{i,\alpha}$ form the new
outcome $\alpha_{\rightarrow(\lambda_{n},A)\rightarrow}$. The set of all such
outcomes forms a new element of $\Gamma_{S}^{s}$. Under these general
conditions, we can not comfortably assume that our new partition has a total
probability of $1$.

The $q_{S}$ condition may be seen as a constraint on this formation process;
whenever a $\gamma_{i}\succ\gamma$ is selected for one $p\in S(\lambda)$, the
$q_{S}$ condition constrains what may happen at the other elements of
$S(\lambda)$. If this constraint implies that for all $\gamma\in q_{S}$, all
$\lambda$ s.t. $\gamma$ is $ba\lambda$, all $p\in S(\lambda)$ must select the
same $\gamma_{i}\succ\gamma$, and the selected $\gamma_{i}$ must itself be an
element of $q_{S}$, then we can expect all elements of $q_{S}$ to have a total
probability of $1$.

To understand the effect that combining outcomes at one $p\in S(\lambda)$ has
at the other points in $S(\lambda)$, it is sufficient to consider the
combination of two outcomes. Since the two outcomes, $\alpha,\beta\in\gamma$,
must form a discretely-determined set when combined into $\alpha\bigcup\beta$,
they must be chosen so that for all but one element of $L$,
$\alpha=\alpha_{\rightarrow(\lambda,\alpha(\lambda)\bigcup\beta(\lambda))\rightarrow}$
and
$\beta=\beta_{\rightarrow(\lambda,\alpha(\lambda)\bigcup\beta(\lambda))\rightarrow}$;
if $\alpha\neq\beta$, then $\alpha(\lambda)$ and $\beta(\lambda)$ must be
disjoint at the one remaining $\lambda\in L$.

Now to see the effect of the $q_{S}$ condition. Let’s start by considering a
pair of outcomes determined at a single $\lambda$,
$S_{\rightarrow(\lambda_{1},A)\rightarrow}$ and
$S_{\rightarrow(\lambda_{1},B)\rightarrow}$, and assume that at $p\in
S(\lambda_{2})$ these two are combined to form
$S_{\rightarrow(\lambda_{1},A\bigcup
B)\rightarrow(\lambda_{2},p)\rightarrow}$. The $q_{S}$ condition demands that
$A$ and $B$ must then also be combined for the outcomes at other elements of
$S(\lambda)$. To see which ones, some further definitions will be helpful.
(Note that $S_{(\lambda_{1},A)\rightarrow(\lambda_{2},p)}(\lambda_{1})$ is the
set of elements of $A$ that can reach $p$.)

######  Definition 163.

If $S$ is a dynamic set, $\lambda_{1}\leq\lambda_{2}$, $A\subset
S(\lambda_{1})$, $p\in S(\lambda_{2})$ then

$(A,\lambda_{1})\downarrow(p,\lambda_{2})\equiv
S_{(\lambda_{1},A)\rightarrow(\lambda_{2},p)}(\lambda_{1})$

For $X\subset S(\lambda_{2})$ ,
$(A,\lambda_{1})\downarrow(X,\lambda_{2})\equiv
S_{(\lambda_{1},A)\rightarrow(\lambda_{2},X)}(\lambda_{1})$.

$(A,\lambda_{1})\downarrow(p,\lambda_{2})$ may be thought of as $p$’s
footprint in $A$.

######  Definition 164.

$p^{\prime}\in[p,\lambda_{2};A,\lambda_{1}]$ if
$(A,\lambda_{1})\downarrow(p^{\prime},\lambda_{2})\bigcap(A,\lambda_{1})\downarrow(p,\lambda_{2})\neq\emptyset$.

$p^{\prime}\in[p,\lambda_{2};A,\lambda_{1}]$ if $p$ and $p^{\prime}$’s
footprints overlap, so $p^{\prime}\in[p,\lambda_{2};A,\lambda_{1}]$ if there
exists an element of $A$ that can reach both $p$ and $p^{\prime}$. According
to the $q_{S}$ condition, if outcomes $A$ and $B$ combine at $p$, and
$p^{\prime}\in[p,\lambda_{2};A,\lambda_{1}]\bigcap[p,\lambda_{2};B,\lambda_{1}]$,
then they must combine at $p^{\prime}$. If
$p^{\prime\prime}\in[p^{\prime},\lambda_{2};A,\lambda_{1}]\bigcap[p^{\prime},\lambda_{2};B,\lambda_{1}]$,
they then must also combine at $p^{\prime\prime}$. This leads to:

######  Definition 165.

If $S$ is a dynamic set, $\lambda_{1}\leq\lambda_{2}$, $p\in S(\lambda_{2})$,
and $A$ and $B$ are disjoint subsets of $S(\lambda_{1})$:

$\left\|p,\lambda_{2};A,B,\lambda_{1}\right\|_{0}\equiv[\lambda_{2},p;\lambda_{1},A]\bigcap[\lambda_{2},p;\lambda_{1},B]$

$\left\|p,\lambda_{2};A,B,\lambda_{1}\right\|_{n+1}\equiv\bigcup_{p^{\prime}\in\left\|p,\lambda_{2};A,B,\lambda_{1}\right\|_{n}}\left\|p^{\prime},\lambda_{2};A,B,\lambda_{1}\right\|_{0}$

$\left\|p,\lambda_{2};A,B,\lambda_{1}\right\|\equiv\bigcup_{n\in\mathbb{N}}\left\|p,\lambda_{2};A,B,\lambda_{1}\right\|_{n}$

If $A$ and $B$ combine at $p$, then the $q_{S}$ condition demands that they
must combine at all elements of
$\left\|p,\lambda_{2};A,B,\lambda_{1}\right\|$.

There may be cases of $p^{\prime}\in S(\lambda_{2})$ s.t. for some
$p_{1}\in\left\|p,\lambda_{2};A,B,\lambda_{1}\right\|$,
$(A,\lambda_{1})\downarrow(p^{\prime},\lambda_{2})\bigcap(A,\lambda_{1})\downarrow(p_{1},\lambda_{2})\neq\emptyset$,
and for some $p_{2}\in\left\|p,\lambda_{2};A,B,\lambda_{1}\right\|$,
$(B,\lambda_{1})\downarrow(p^{\prime},\lambda_{2})\bigcap(B,\lambda_{1})\downarrow(p_{2},\lambda_{2})\neq\emptyset$,
but there are no elements of $\left\|p,\lambda_{2};A,B,\lambda_{1}\right\|$
for which both hold, and so $p^{\prime}$ is not an element of
$\left\|p,\lambda_{2};A,B,\lambda_{1}\right\|$. This is a generalization of
the example seen earlier. If such cases occur, the $q_{S}$ condition will be
insufficient; they are eliminated if paths form a densely interlocking
network. More precisely:

1) Form the footprints of $\left\|p,\lambda_{2};A,B,\lambda_{1}\right\|$ in
$A$ and $B:$
$x\equiv(A,\lambda_{1})\downarrow(\left\|p,\lambda_{2};A,B,\lambda_{1}\right\|,\lambda_{2})$
and
$y\equiv(B,\lambda_{1})\downarrow(\left\|p,\lambda_{2};A,B,\lambda_{1}\right\|,\lambda_{2})$

2) Close the footprints with all intersecting footprints. That is, define:

$X_{0}\equiv x$

$X_{n+1}\equiv\bigcup\\{(A,\lambda_{1})\downarrow(p^{\prime},\lambda_{2}):(A,\lambda_{1})\downarrow(p^{\prime},\lambda_{2})\bigcap
X_{n}\neq\emptyset\\}$

$X\equiv\bigcup_{n\in\mathbb{N}}X_{n}$

Similarly, starting with $y$, construct $Y$.

3) For the $q_{S}$ condition to be sufficient: For any
$p^{\prime}\notin\left\|p,\lambda_{2};A,B,\lambda_{1}\right\|$, if
$(A,\lambda_{1})\downarrow(p^{\prime},\lambda_{2})$ is in $X$ then
$(B,\lambda_{1})\downarrow(p^{\prime},\lambda_{2})$ must be disjoint from $Y$,
and if $(B,\lambda_{1})\downarrow(p^{\prime},\lambda_{2})$ is in $Y$ then
$(A,\lambda_{1})\downarrow(p^{\prime},\lambda_{2})$ must be disjoint from $X$.

With $a=A-X$ and $b=B-Y$, combining
$S_{\rightarrow(\lambda_{1},A)\rightarrow(\lambda_{2},p)\rightarrow}$ and
$S_{\rightarrow(\lambda_{1},B)\rightarrow(\lambda_{2},p)\rightarrow}$ under
the $q_{S}$ condition then has the effect of replacing $\\{A,B\\}$ with
$\\{a,b,X\bigcup Y\\}$; the resulting partition will therefore have a total
probability of $1$. It is interesting to note that statement (3) will be
satisfied if the paths are either quite dense or quite sparse; only the
intermediate case may cause difficulty.

Combining pairs of outcomes that are determined at multiple $\lambda$ lead to
similar conclusions. Once again, in order for the combination to form a
discretely-determined set, the two outcomes can only disagree at a single
$\lambda$. If the two outcomes disagree on the last measurement then the
analysis is little changed from above, except that we need only consider the
subset of points in $A$ & $B$ that can be reached from the prior measurements.
The trickier case is when a sequence of further measurements come after the
two being combined; for example, the case where
$S_{\rightarrow(\lambda_{1},A)\rightarrow(\lambda_{2},C)\rightarrow(\lambda_{3},p)\rightarrow}$
and
$S_{\rightarrow(\lambda_{1},B)\rightarrow(\lambda_{2},C)\rightarrow(\lambda_{3},p)\rightarrow}$
are combined. To see what the $q_{S}$ condition demands of the other
$p^{\prime}\in S(\lambda_{3})$ in such cases, it will be helpful to expand
some the above definitions:

######  Definition 166.

If $S$ is a dynamic set, $\lambda_{1}\leq\lambda_{2}$, $A\subset
S(\lambda_{1})$, $\mathcal{Z}$ is a set of subsets of $S(\lambda_{2})$, and
$Z\in\mathcal{Z}$:

$Z^{\prime}\in[Z,\mathcal{Z},\lambda_{2};A,\lambda_{1}]$ if
$Z^{\prime}\in\mathcal{Z}$ and
$(A,\lambda_{1})\downarrow(Z^{\prime},\lambda_{2})\bigcap(A,\lambda_{1})\downarrow(Z,\lambda_{2})\neq\emptyset$

If, further, $A$ and $B$ are disjoint subsets of $S(\lambda_{0})$:

$\left\|Z,\mathcal{Z},\lambda_{2};A,B,\lambda_{1}\right\|_{0}\equiv[\lambda_{2},\mathcal{Z},Z;\lambda_{1},A]\bigcap[\lambda_{2},\mathcal{Z},Z;\lambda_{1},A]$

$\left\|Z,\mathcal{Z},\lambda_{2};A,B,\lambda_{1}\right\|_{n+1}\equiv\bigcup_{Z^{\prime}\in\left\|Z,\mathcal{Z},\lambda_{2};A,B,\lambda_{1}\right\|_{n}}\left\|Z^{\prime},\mathcal{Z},\lambda_{2};A,B,\lambda_{1}\right\|_{0}$

$\left\|Z,\mathcal{Z},\lambda_{2};A,B,\lambda_{1}\right\|\equiv\bigcup_{n\in\mathbb{N}^{+}}\left\|Z,\mathcal{Z},\lambda_{2};A,B,\lambda_{1}\right\|_{n}$

If for each $p\in S(\lambda_{3})$, $Z_{p}\equiv
S_{(\lambda_{1},A)\rightarrow(\lambda_{3},p)}(\lambda_{2})$, and
$\mathcal{Z}\equiv\\{Z_{p}:p\in S(\lambda_{3})\\}$ then
$Z_{p^{\prime}}\in[Z_{p},\mathcal{Z},\lambda_{2};A,\lambda_{1}]$ iff
$p^{\prime}\in[p,\lambda_{3};A,\lambda_{1}]$; from this it follows that
$p^{\prime}\in\left\|p,\lambda_{3};A,B,\lambda_{1}\right\|$ iff
$Z_{p^{\prime}}\in\left\|Z_{p},\mathcal{Z},\lambda_{2};A,B,\lambda_{1}\right\|$.
This allows us to take our earlier analysis on combining
$S_{\rightarrow(\lambda_{1},A)\rightarrow(\lambda_{3},p)\rightarrow}$ and
$S_{\rightarrow(\lambda_{1},B)\rightarrow(\lambda_{3},p)\rightarrow}$, and
project it onto any $\lambda_{2}\in(\lambda_{1},\lambda_{3})$: if
$S_{\rightarrow(\lambda_{1},A)\rightarrow(\lambda_{3},p)\rightarrow}$ and
$S_{\rightarrow(\lambda_{10},B)\rightarrow(\lambda_{3},p)\rightarrow}$ are
combined, then
$S_{\rightarrow(\lambda_{1},A)\rightarrow(\lambda_{3},p^{\prime})\rightarrow}$
and
$S_{\rightarrow(\lambda_{1},B)\rightarrow(\lambda_{3},p^{\prime})\rightarrow}$
must be combined if
$Z_{p^{\prime}}\in\left\|Z_{p},\mathcal{Z},\lambda_{2};A,B,\lambda_{1}\right\|$.
To extend this to combining
$S_{\rightarrow(\lambda_{1},A)\rightarrow(\lambda_{2},C)\rightarrow(\lambda_{3},p)\rightarrow}$
and
$S_{\rightarrow(\lambda_{1},B)\rightarrow(\lambda_{2},C)\rightarrow(\lambda_{3},p)\rightarrow}$,
simply replace $Z_{p}$ with $Z_{p,C}\equiv Z_{p}\bigcap C$ and $\mathcal{Z}$
with $\mathcal{Z}_{C}\equiv\\{Z_{p,C}:p\in S(\lambda_{3})\\}$. If
$S_{\rightarrow(\lambda_{1},A)\rightarrow(\lambda_{2},C)\rightarrow(\lambda_{3},p)\rightarrow}$
and
$S_{\rightarrow(\lambda_{1},B)\rightarrow(\lambda_{2},C)\rightarrow(\lambda_{3},p)\rightarrow}$
are combined then
$S_{\rightarrow(\lambda_{1},A)\rightarrow(\lambda_{2},C)\rightarrow(\lambda_{3},p^{\prime})\rightarrow}$
and
$S_{\rightarrow(\lambda_{1},B)\rightarrow(\lambda_{2},C)\rightarrow(\lambda_{3},p^{\prime})\rightarrow}$
must be combined if
$Z_{p^{\prime},C}\in\left\|Z_{p,C},\mathcal{Z}_{C},\lambda_{2};A,B,\lambda_{1}\right\|$.
This leads by the same reasoning to the same conclusion: if the network of
paths is sufficiently dense (or sparse), the finite and finitely determined
partitions that satisfy the $q_{S}$ condition can be expected to have a total
probability of $1$.

It follows that if the dynamic set in a dps follows the rule that paths are
included unless they can be excluded in principle, then we can reasonably
expect the discretely-determined portion of the dps to be maximal.

###  V.4 Terminus & Exordium

A word or two is in order about dps’s, $(X,T,P)$, for which $X$ is not a
singleton. In such cases $T_{\mathbb{N}}^{S}$ may be larger than
$T_{S\mathbb{N}}$, and so admit partitions that are not nearly-compatible (and
therefore not elements of $q_{S}$). If the various $(\\{S\\},T_{S})$ are
copies of one another, in the sense of the reductions introduced in Sec.
IV.2.4, then $T_{\mathbb{N}}^{S}=T_{S\mathbb{N}}$. For quantum systems, if we
consider the various Hilbert space bases as generating the various elements of
$X$, we can partially assert that this is the case. From these bases, each
$S(\lambda)$ is equipped with a $\Sigma$-algebra, and outcomes are of the form
$S_{\rightarrow(\lambda_{1},\sigma_{1})\rightarrow...\rightarrow(\lambda_{n},\sigma_{n})\rightarrow}$
where the $\sigma$’s are elements of their respective $\Sigma$-algebra’s. For
any $S,S^{\prime}\in X$, $\lambda\in\Lambda$, the $\Sigma$-algebra’s on
$S(\lambda)$ and $S^{\prime}(\lambda)$ are isomorphic. This gets us part of
the way there. One further piece is missing: while the outcomes are
isomorphic, the dynamics may not be. In particular, it’s possible to have a
case where two points, $(\lambda,p_{1})$ and $(\lambda,p_{2})$, can transition
to $(\lambda^{\prime},p^{\prime})$, but under isomorphisms from
$(S(\lambda),\Sigma)$ to $(S^{\prime}(\lambda),\Sigma^{\prime})$ only one of
them can. It would then be possible to have $\gamma\in G_{TS^{\prime}}$, but
its isomorphic image not be an element of $G_{TS}$; $T_{\mathbb{N}}^{S}$ may
then be larger than $T_{S\mathbb{N}}$.888Then again it may not. For example,
if the transitions are different because the states are deterministic, or
conserved, in $S_{1}$ but not in $S_{2}$, this will not effect the existence
of a reduction. That is because, when states are deterministic, the order in
which measurements occur is irrelevant, so many different measurement
sequences will result in the same ip.

However, in quantum mechanics we would not expect any such differing dynamics
to effect the existence of a reduction between bases; for a given element of
$q_{S}$ we would expect to be able to select sequences of projection operators
s.t., under a given basis transformation, the transformed representation will
also be an element of $q_{S}$. More generally, the discretely determined
probabilities ought to be sufficient to determine all inner-products in a
given bases, which then entail the discretely determined probabilities in all
other bases. Dps’s in the various bases are therefore, in a very real sense,
copies of each other.

Moreover, it’s not clear that all bases are experimentally relevant;
naturally, if a basis is not experimentally relevant, then it can’t contribute
any information to $T_{\mathbb{N}}^{S}$. Unsurprisingly, given the subject
matter of physics, when a basis is given an experimental interpretation, it’s
generally in terms of paths though space. For example, when the conjugate-
coordinate basis is interpreted as momentum, it is necessarily being given a
spacial path interpretation; an interpretation predicated on the conjugate-
coordinate’s mathematical relation to aspects of spacial path probabilities
such as the probability flux. This connection between the conjugate-coordinate
basis and the particle’s spacial path is put on display when the value of the
conjugate-coordinate is determined by tracking particle paths in a detector.

However, because quantum systems satisfy all the dps axioms, even if
$T_{\mathbb{N}}^{S}$ is larger than $T_{S\mathbb{N}}$, this fact will have to
be reflected in the quantum probabilities. We don’t need results like Thm 161
to conclude that quantum systems are dps’s, we know that without them. What
such results do tell us is that, at least with regard to their measurement
theory and a central core of their probability theory, quantum systems appear
to be nothing more than fairly generic dps’s.

Indeed, one may start to suspect that quantum mechanics, as it currently
stands, is a phenomenological theory. It certainly is not holy writ, handed
down from on high, which we can’t possibly hope to truly understand, but which
we must none the less accept in all detail. Rather, it evolved over time by
trial and error as a means of calculating results to match newly discovered
experimental phenomenon. It has been quite successful in achieving that goal;
however it has also historically yielded little understanding of why its
calculational methods work. This lack of understanding has often led to claims
that we don’t understand these matters because we can’t understand them; they
lie outside the sphere of human comprehension. While we can not state with
certainty that such bold claims of necessary ignorance are false, we can say
that they are scientifically unfounded, and so ought to be approached with
skepticism. With all humility, it is hoped that this article has lent strength
to such skepticism, and shed light on some matters that have heretofore been
allowed to lie in darkness.

##  Appendix A $\Delta$-Additivity

Quantum probabilities posses an interesting algebraic property that is not a
direct consequence of their t-algebra.

For any $t\in T$ s.t. $\\{\bigcup t\\}$ is also an element of $T$, define
$\Delta(t)\equiv P(t)-P(\\{\bigcup t\\})$. The $\Delta$ function measures the
additivity of a gps; if a gps has additive probabilities, then $\Delta(t)=0$
for all $t$. Quantum probabilities are not additive, but their $\Delta$
function is. For any $t\in T$ s.t. $\\{\bigcup t\\}\in T$ and for all pairs
$\alpha,\beta\in t$, $\\{\alpha,\beta\\},\\{\alpha\bigcup\beta\\}\in T$,

$\Delta(t)=\sum_{pairs\,of\,\alpha,\beta\in t}\Delta(\\{\alpha,\beta\\})$

This is an immediate consequence of quantum probabilities being calculated
from expressions of the form
$<\psi,\lambda_{0}|\Pi\Pi^{\dagger}|\psi,\lambda_{0}>$, where $\Pi$ is a
product of projection operators.

Any gps that obeys this rule is _$\Delta$ -additive_.

It follows from the definition of $\Delta$ that for disjoint $t_{1}$ and
$t_{2}$, $\Delta(t_{1}\cup t_{2})=\Delta(t_{1})+\Delta(\\{\bigcup t_{1}\\}\cup
t_{2})$. Applying this to the above formula, we get
$\Delta(\\{\alpha_{1}\cup\alpha_{2},\alpha_{3}\\})=\Delta(\\{\alpha_{1},\alpha_{3}\\})+\Delta(\\{\alpha_{2},\alpha_{3}\\})$,
which is directly analogous to the additivity seen in classic probabilities.
If $t$ has more than $3$ elements then
$\Delta(\\{\alpha_{1}\cup\alpha_{2},\alpha_{3},...\\})=\Delta(\\{\alpha_{1},\alpha_{3},...\\})+\Delta(\\{\alpha_{2},\alpha_{3},...\\})-\Delta(\\{\alpha_{3},...\\})$.

In the orthodox interpretation, $\Delta$-additivity is generally viewed as
being due to path interference possessing wave-like properties . An
interesting question is, under what conditions will a system obeying the
intuitive interpretation display $\Delta$-additivity?

Start by defining a _moment partition_ to be any partition, $\gamma$, s.t. for
any $\lambda$, either all $\alpha\in\gamma$ are $bb\lambda$, or all
$\alpha\in\gamma$ are $ba\lambda$. (See Defn 147 for the definition of
“moment”, and Thm 148 for proof that a partition composed of moments is an
ip). Moment partitions have some very useful properties. First, for any
$\alpha,\beta\in\gamma$,
$(\gamma-\\{\alpha,\beta\\})\bigcup(\alpha\bigcup\beta)$ is also a moment
partition. Second, if $\gamma$ is a moment partition, and $\nu\in\gamma$, then
$\gamma_{\nu}\equiv\\{\nu,S-\nu\\}$ is also a moment partition. Indeed if
$\gamma$ is an ip, and for all $\nu\in\gamma$, $\\{\nu,S-\nu\\}$ is also an
ip, then $\gamma$ is a moment partition.

Now take any $\alpha\in t\in T$, any moment partition, $\gamma$, and define
$\alpha/\gamma\equiv\\{\alpha\bigcap\beta:\beta\in\gamma\>and\>\alpha\bigcap\beta\neq\emptyset\\}$.
$\Delta(\alpha/\gamma)$ is the effect the measurement $\gamma$ has on the
probability that $\alpha$ occurs.

For $\beta\in\alpha/\gamma$ take $P_{o}(\beta)$ to be the probability that
$\beta$ occurs if all interactions required to measure $\gamma$ are
vanishingly small (while all the interactions for measuring $\alpha$ are
unchanged). $P_{o}(\beta)$ may be thought of as the omniscient probability: we
simply know which element of $\gamma$ occurs without having to perform a
measurement. (In the intuitive interpretation, some $\beta$ occurs even if the
measurement does not take place.) It follows that
$P(\alpha)=\Sigma_{\beta\in\alpha/\gamma}P_{o}(\beta)$. $P_{o}$ is, of course,
simply a conceptual construct; it can only be experimentally determined if
$P_{o}=P$.

Now define $\delta(\beta)\equiv P(\beta)-P_{o}(\beta)$; $\delta(\beta)$
represents the amount of deflection into/out of $\beta$ due to the measurement
of $\gamma$. Since $P(\alpha)=\Sigma_{\beta\in\alpha/\gamma}P_{o}(\beta)$,
$\Delta(\alpha/\gamma)=\sum_{\beta\in\alpha/\gamma}\delta(\beta)$. Finally,
for $\beta\in\alpha/\gamma$, define $\neg\beta\equiv\alpha-\beta$.

######  Theorem 167.

1) For any outcome, $\alpha$, any countable moment partition, $\gamma$,
$\alpha/\gamma$ is $\Delta$-additive iff
$\sum_{\beta\in\alpha/\gamma}\delta(\beta)=\sum_{\beta\in\alpha/\gamma}\delta(\neg\beta)$.

2) A discretely determined dps is $\Delta$-additive iff for every outcome,
$\alpha$, every moment partition, $\gamma$, s.t. $\alpha/\gamma$$\in T$,
$\alpha/\gamma$ is $\Delta$-additive.

###### Proof.

1) Assume $\alpha/\gamma$ has $N\geq 3$ elements ($N=2$ is trivial). Note that
since probabilities $P_{o}$ are fully additive,
$(N-1)\sum_{\beta\in\alpha/\gamma}P_{o}(\beta)=$$\sum_{\beta\in\alpha/\gamma}\sum_{\eta\in\alpha/\gamma-\\{\beta\\}}P_{o}(\eta)=$$\sum_{\beta\in\alpha/\gamma}P_{o}(\neg\beta)$.
Adding $(N-1)\sum_{\beta\in\alpha/\gamma}P_{o}(\beta)$ to the left side of
$\sum_{\beta\in\alpha/\gamma}\delta(\beta)=\sum_{\beta\in\alpha/\gamma}\delta(\neg\beta)$,
and $\sum_{\beta\in\alpha/\gamma}P_{o}(\neg\beta)$ to the right, yields
$(N-2)P(\alpha)+\sum_{\beta\in\alpha/\gamma}P(\beta)=\sum_{\beta\in\alpha/\gamma}P(\neg\beta)$.
Now note that
$(N-1)\sum_{\beta\in\alpha/\gamma}P(\beta)=$$\sum_{\beta\in\alpha/\gamma}P(\alpha/\gamma-\\{\beta\\})$.
Subtracting $(N-1)\sum_{\beta\in\alpha/\gamma}P(\beta)$ from the left side,
and $\sum_{\beta\in\alpha/\gamma}P(\alpha/\gamma-\\{\beta\\})$ from the left,
yields
$(N-2)\Delta(\alpha/\gamma)=\sum_{\beta\in\alpha/\gamma}\Delta(\alpha/\gamma-\\{\beta\\})$.
It remains to show that this is equivalent to
$\Delta(\alpha/\gamma)=\sum_{pairs\,of\,\beta,\eta\in\alpha/\gamma}\Delta(\\{\beta,\eta\\})$.

They are clearly equivalent when $N=3$. Assume they are equivalent for $N=M$,
for $N=M+1$

$\Delta(\alpha/\gamma)=\frac{1}{N-2}\sum_{\beta\in\alpha/t}\Delta(\alpha/\gamma-\\{\beta\\})$

$=\frac{1}{N-2}\sum_{\beta\in\alpha/t}\Delta((\alpha-\beta)/\gamma)$

$=\frac{1}{N-2}\sum_{\beta\in\alpha/\gamma}\sum_{pairs\,of\,\kappa,\eta\in\alpha/\gamma-\\{\beta\\}}\Delta(\\{\kappa,\eta\\})$

$=\sum_{pairs\,of\,\beta,\eta\in\alpha/\gamma}\Delta(\\{\beta,\eta\\})$

2) Let’s say that $T$ is discretely determined, and for $t\in T$, all
$\alpha,\beta\in t$, $\\{\alpha\bigcup\beta\\}\in T$. This means that for some
$L\subset\Lambda$ s.t $t$ is determined on $L$, at all elements of $L$ except
one, all $\alpha\in T$ correspond to the same outcome.

Take $\lambda$ to be the element of $L$ at which the various $\alpha\in t$
correspond to different outcomes. Taking $\nu=\bigcup t$, there exists a
moment partition at $\lambda$, $\gamma$, s.t. $t=\nu/\gamma$. ∎

So the probabilities of an intuitive system will have the algebraic properties
of quantum theory if for all outcomes, $\alpha$, all countable moment
partitions, $\gamma$,
$\sum_{\beta\in\alpha/\gamma}\delta(\beta)=\sum_{\beta\in\alpha/\gamma}\delta(\neg\beta)$.
Let’s now see how this requirement may hold.

First note that, since $\delta$ measures the effect of the environment on the
system, experimental methods should be chosen so as to minimize $\delta$. One
way to do this is to perform “passive measurements”, a type of measurement
that corresponds to how double-slit experiments are generally pictured. In a
passive measurement, to determine the probabilities for elements of
$\alpha/\gamma$, start with some $\gamma_{0}\in G_{T}$ such that
$\alpha\in\gamma_{0}$. Perform the measurement of $\gamma_{0}$ as before, but
for $\eta\in\gamma$ block $\neg\eta$ from occurring; if $\neg\eta$ would have
occurred, you get a null result. Now if $\alpha$ occurs it means that
$\alpha\bigcap\eta$ occurred. The proportion of $\alpha$’s to all trials run
(including null results) is then $P(\alpha\bigcap\eta)$. Doing this in turn
for each $\eta\in\gamma$ yields the probabilities for $\alpha/\gamma$.

Imagine that position measurements on particles are performed in this manner.
In the intuitive interpretation, if $\delta(\eta)\neq 0$ it’s because the
blocking off of $\neg\eta$ creates some minimum but non-vanishing ambient
field in $\eta$, which deflects the particles. The probabilities on
$\gamma_{0}/\gamma$ will sum to $1$ if these ambient fields cause the
particle’s path to be deflected among the outcomes of $\gamma_{0}$, but not
among outcomes of $\gamma$. Because $\gamma$ is a moment partition, this means
that paths are not deflected prior to the measurement taking place; this is
also a sufficient condition for additivity of final state to hold.

For these passive measurements,
$\sum_{\beta\in\alpha/\gamma}\delta(\beta)=\sum_{\beta\in\alpha/\gamma}\delta(\neg\beta)$
if the deflections in $\eta\in\gamma$ due to blocking off $\neg\eta$ are equal
to the sum of the deflections in $\eta$ caused by blocking off each element of
$\gamma-\\{\eta\\}$ individually. So the dps describing particle position will
be $\Delta$-additive if the ambient fields in $\eta$ caused by blocking
$\neg\eta$ is equal to the superposition of the ambient fields in $\eta$
caused by blocking the individual $\nu\in\gamma-\\{\eta\\}$. There are other
ways for $\Delta$-additivity to hold, but this one is conceptually simple, as
well as plausible.

##  Appendix B Conditional Probabilities & Probability Dynamics

Conditional probabilities are a central concept of probability theory. They
are particularly interesting for dps’s because probability dynamics are
expressed in terms of conditional probabilities.

There are two ways in which a dps’s probabilities may be considered to be
dynamic. First, the probabilities of what will happen changes as our knowledge
of what has happened continues to unfold. Second, given that we measured the
system to be in state $s$ at time $\lambda_{0}$, we may be interested in the
probability of measuring the system in state as being in state $x$ at time
$\lambda$ as $\lambda$ varies. The Schrodinger equation deals with dynamics of
this second sort. Conditional probabilities are required for exploring both
kinds of dynamics.

The notion of conditional probability is the same in a gps as it is in a
classic probability space; essentially, if $P(B)\neq 0$, $P(A|B)=P(A\bigcap
B)/P(B)$. For a dps, the conditional probability of particular interest is the
probability that $t$ occurs given that, as of $\lambda$, the measurement is
consistent with $t$. In order to delineate this, a few preliminary definitions
will prove helpful.

######  Definition 168.

If $(X,T,P)$ is a dps, $t\in T$, and $t\subset\gamma\in G_{T}$,

$|t[-\infty,\lambda]|_{\gamma}\equiv\\{|\alpha[-\infty,\lambda]|_{\gamma}:\alpha\in
t\\}$

$(t)_{\lambda}^{\gamma}\equiv\bigcup_{\alpha\in t}(\alpha)_{\lambda}^{\gamma}$

$(X,T,P)$ is _$\Lambda$ -complete_ if for all $t\in T$, $\gamma\in G_{T}$ s.t.
$t\subset\gamma$, $\lambda\in\Lambda$, $(t)_{\lambda}^{\gamma}\in T$

Because $(t)_{\lambda}^{\gamma}\subset\gamma$, $\Lambda$-completeness places
no restrictions on the make-up of $G_{T}$, it only places a restriction on the
$\Sigma$-algebras of the constituent ip probability spaces. It is therefore a
fairly weak condition.

$P(t|(t)_{\lambda}^{\gamma})$ is the probability that $t$ occurs given that,
as of $\lambda$, the measurement is consistent with $t$. Because probabilities
of this type are of interest, it’s useful to extend the definition of
consistent probabilities (Defn 84) to insure that the they are independent of
$\gamma$.

######  Definition 169.

A dps, $(X,T,P)$, is _conditionally consistent_ if it is $\Lambda$-complete
and for all $t,t^{\prime}\in T$, all $\lambda\in\Lambda$, all
$t\subset\gamma\in G_{T}$, $t^{\prime}\subset\gamma^{\prime}\in G_{T}$ s.t.
$|t[-\infty,\lambda]|_{\gamma}=|t^{\prime}[-\infty,\lambda]|_{\gamma^{\prime}}$,
$P((t)_{\lambda}^{\gamma})=P((t^{\prime})_{\lambda}^{\gamma^{\prime}})$.

Conceptually, this demand is met if probabilities are consistent at all times
as the experiments unfold. Note that since
$|t[-\infty,\lambda]|_{\gamma}=|t^{\prime}[-\infty,\lambda]|_{\gamma^{\prime}}$,
the “all-reet nots” (introduced in Sec. III.5) of $(t)_{\lambda}^{\gamma}$ and
$(t^{\prime})_{\lambda}^{\gamma^{\prime}}$ are the same; as a result, if a
t-algebra is sufficiently rich, it ought to be conditionally consistent.

######  Definition 170.

If $(X,T,P)$ is a conditionally consistent dps, $t,t^{\prime}\in T$, and
$t^{\prime}\subset\gamma\in G_{T}$, then with
$y\equiv|t^{\prime}[-\infty,\lambda]|_{\gamma}$, $P(t||y)\equiv
P(t|(t^{\prime})_{\lambda}^{\gamma})$.

This is somewhat more intuitive notation for conditional probabilities. The
$P(t||y)$ allow us to see how dps probabilities unfold with time. Note that
$y$ does not have to be an element of $T$ in order for $P(t||y)$ to be
defined.

A common assumption with regard to conditional probabilities on stochastic
processes is that they satisfy the Markov property. The equivalent property
for dps is:

######  Definition 171.

A dps, $(X,T,P)$, is _point-Markovian_ if for all $t\in T$, $(\lambda,p)\in
Uni(\bigcup t)$, if $t=t_{\rightarrow(\lambda,p)}\circ
t_{(\lambda,p)\rightarrow}$ then
$+t_{\rightarrow(\lambda,p)},+t_{(\lambda,p)\rightarrow}\in T$ and
$P(t)=P(+t_{\rightarrow(\lambda,p)})\cdot P(+t_{(\lambda,p)\rightarrow})$.

(In the above definition, notation that has been used on sets of dynamic paths
have been applied to collections of sets. For example,
$t_{\rightarrow(\lambda,p)}$, which is understood to mean
$\\{\alpha_{\rightarrow(\lambda,p)}:\alpha\in t\\}$, and
$t_{\rightarrow(\lambda,p)}\circ t_{(\lambda,p)\rightarrow}$, which is
understood to mean
$\\{\alpha_{\rightarrow(\lambda,p)}\circ\beta_{(\lambda,p)\rightarrow}:\alpha,\beta\in
t,\>\alpha_{\rightarrow(\lambda,p)}\circ\beta_{(\lambda,p)\rightarrow}\neq\emptyset\\}$.
It is hoped that this notation has not caused confusion.)

When the probability function is not additive, this property looses much of
its power. None-the-less, the property can generally be assumed to hold, and
does have some interesting consequences.

One interesting property of point-Markovian dps’s with discrete parameters is
that, if the t-algebra is sufficiently rich, then their probabilities tend to
be additive. This can be seen for the case of $t\in T$ s.t. for some
$\lambda_{0}<\lambda$, $t$ is $sbb\lambda_{0}$, $wba\lambda$, and $(\bigcup
t)[\lambda_{0},\lambda]$ is finite.999Finite spin systems are point-Markovian,
and all their $t\in T$ satisfy these restrictions; however their parameters
are generally assumed to be non-discrete, which does allow them to have non-
additive probabilities. Start with any such $t\in T$ and define
$t_{1}\equiv\\{\alpha_{\rightarrow(\lambda-1,q)\rightarrow(\lambda,p)\rightarrow}:\alpha\in
t,\>q\in\alpha(\lambda-1),\>and\>p\in\alpha(\lambda)\\}$; since $t$ is
$wba\lambda$, for any $\gamma\in G_{T}$ s.t. $t\subset\gamma$,
$(\gamma-t)\bigcup t_{1}$ is an ip, so if it is an element of $G_{T}$ then
$P(t)=P(t_{1})$. If, further, the dps is point-Markovian, then
$P(\\{\alpha_{\rightarrow(\lambda-1,q)\rightarrow(\lambda,p)\rightarrow}\\})=P(\\{+\alpha_{\rightarrow(\lambda-1,q)}\\})\cdot
P(\\{+\alpha_{(\lambda-1,q)\rightarrow(\lambda,p)\rightarrow}\\})$. The same
manipulations can now be performed on the
$+\alpha_{\rightarrow(\lambda-1,q)}$’s. Defining $X_{\bigcup
t}\equiv\\{+\\{\bar{s}[\lambda_{0},\lambda]\\}:\bar{s}\in\bigcup t\\}$, these
iterations eventually yield $P(t)=P(X_{\bigcup t})$. Therefore, for any
$t^{\prime}$ s.t. $t^{\prime}$ is bounded by $\lambda_{0}$ & $\lambda$, and
$\bigcup t^{\prime}=\bigcup t$, $P(t^{\prime})=P(X_{\bigcup t})=P(t)$.

##  Appendix C Invariance On Dynamic Sets and DPS’s

###  C.1 Invariance On Dynamic Sets

Invariance is among the most fundamental concepts in the study of dynamic
systems. We start by establishing the concept for dynamic sets. (In what
follows, if $f$ & $g$ are functions, “$f\cdot g$” is the function s.t. $f\cdot
g(x)=f(g(x))$)

######  Definition 172.

A _global invariant_ , $I$, on a dynamic set, $S$, is a pair of functions,
$I_{\Lambda}:\Lambda_{S}\rightarrow\Lambda_{S}$ and
$I_{P}:\mathcal{P}_{S}\rightarrow\mathcal{P}_{S}$ s.t.
$\\{I_{\mathcal{P}}\cdot\bar{p}\cdot I_{\Lambda}:\bar{p}\in S\\}=S$.

For any $\bar{p}\in S$, $I(\bar{p})\equiv I_{\mathcal{P}}\cdot\bar{p}\cdot
I_{\Lambda}$; for $A\subset S$, $I[A]=\\{I(\bar{p})\,:\,\bar{p}\in A\\}$ (so
$I$ is an invariant iff $I[S]=S$).

Global invariants are overly restrictive when $\Lambda$ is bounded by $0$. To
handle that case equitably, the following definition of invariance will be
employed.

######  Definition 173.

If $S$ is a dynamic set and $\Lambda_{S}$ is unbounded from below, then $I$ is
an _invariant_ on $S$ if it is a global invariant on $S$.

If $\Lambda_{S}$ is bounded from below then $I$ is an _invariant_ on $S$ if
there exists a dynamic set, $S^{*}$ s.t. $\Lambda_{S^{*}}$ is unbounded from
below, $S=S^{*}[0,\infty]$, and $I$ is an invariant on $S^{*}$.

Unless stated otherwise, for the remainder of this section dynamic sets will
be assumed to be unbounded from below.

######  Theorem 174.

If $I$ is an invariant on $S$

1) $I_{\mathcal{P}}$ is a surjection

2) $I_{\Lambda}$ and $I_{\mathcal{P}}$ are invertible iff $I_{\Lambda}$ is an
injection and $I_{\mathcal{P}}$ is a bijection

3) $I_{\Lambda}$ and $I_{\mathcal{P}}$ are invertible and
$I^{-1}\equiv(I_{\Lambda}^{-1},I_{P}^{-1})$ is an invariant on $S$ iff
$I_{\Lambda}$ and $I_{\mathcal{P}}$ are bijections

###### Proof.

1) If it is not then $\mathcal{P}_{I[S]}\subset
Ran(I_{\mathcal{P}})\subsetneq\mathcal{P}_{S}$ and so $I[S]\neq S$.

2) Follows from (1) and that functions are invertible iff they are injections.

3) $\Rightarrow$ Follows from (2) and that for $I^{-1}$ to be an invariant,
the domain of $I_{\Lambda}^{-1}$ needs to be $\Lambda$.

$\Leftarrow$ Because $I_{\Lambda}$ and $I_{\mathcal{P}}$ are bijections, for
all $\bar{p}\in S$ $I^{-1}(I(\bar{p}))=I_{P}^{-1}\cdot
I_{\mathcal{P}}\cdot\bar{p}\cdot I_{\Lambda}\cdot I_{\Lambda}^{-1}=\bar{p}$.
Therefore $I^{-1}[I[S]]=S$. Since $I[S]=S$, $I^{-1}[S]=S$.∎

######  Definition 175.

Dynamic set $S$ is _weakly $\Lambda$-invariant_ if for all $\lambda\in\Lambda$
there’s an invariant, $L^{\lambda}$, s.t. for all
$\lambda^{\prime}\in\Lambda_{S}$,
$L_{\Lambda}^{\lambda}(\lambda^{\prime})=\lambda^{\prime}+\lambda$.

$S$ is _strongly $\Lambda$-invariant_ if it is weakly $\Lambda$-invariant and
for all $\lambda\in\Lambda$, $L_{\mathcal{P}}^{\lambda}$ is the identity on
$\mathcal{P}_{S}$.

######  Theorem 176.

A dynamic space $D$ is strongly $\Lambda$-invariant iff it homogeneous and all
of $\mathcal{P}_{S}$ is homogeneously realized

###### Proof.

When $\Lambda_{D}$ is unbounded from below, this is fairly clear.

If $\Lambda_{D}$ is bounded from below:

$\Rightarrow$ follows immediately from the unbounded case.

For $\Leftarrow$, it’s necessary to construct a homogeneous, homogeneously
realized, unbounded from below $D^{*}$ s.t. $D^{*}[0,\lambda]=D$. This is
relatively easy. Note that if $D$ is homogeneous & homogeneously realized then
for any $\triangle\lambda>0$, any $\lambda,\lambda^{\prime}\in\Lambda_{D}$,
$D[\lambda,\lambda+\triangle\lambda]$ and
$D[\lambda^{\prime},\lambda^{\prime}+\triangle\lambda]$ are copies of each
other, in that if you move
$D[\lambda^{\prime},\lambda^{\prime}+\triangle\lambda]$ to $\lambda$, the two
are equal. So to construct $D^{*}$, take any $D[0,\triangle\lambda]$, append
it to the beginning of $D$, then append it to the beginning of the resulting
set, and so forth. Only one set, $D^{*}$, will equal this construction for all
intervals, $[-n\triangle\lambda,\infty]$; $D^{*}$ is a homogeneous,
homogeneously dynamic space, and $\Lambda_{D^{*}}$ is unbounded from below, so
$D^{*}$ is strongly $\Lambda$-invariant.∎

######  Definition 177.

$S$ is _reversible_ if for all $\lambda\in\Lambda$ there’s and invariant,
$R^{\lambda}$, s.t. for all $\lambda^{\prime}\in\Lambda_{S}$,
$R_{\Lambda}^{\lambda}(\lambda^{\prime})=\lambda-\lambda^{\prime}$, and
$R_{\mathcal{P}}^{\lambda}\cdot R_{\mathcal{P}}^{\lambda}$ is the identity on
$\mathcal{P}_{S}$.

######  Theorem 178.

1) If $S$ is reversible then it is weakly $\Lambda$-invariant.

2) If $S$ is reversible and for all $\lambda_{1},\lambda_{2}\in\Lambda$,
$R_{\mathcal{P}}^{\lambda_{1}}=R_{\mathcal{P}}^{\lambda_{2}}$ then it is
strongly $\Lambda$-invariant.

###### Proof.

1) For every $\lambda\in\Lambda_{S}$, $L^{\lambda}=R^{\lambda}\cdot R^{0}$ is
an invariant (because the composite of any two invariants is an invariant;
$R^{0}$ being $R^{\lambda}$ with $\lambda=0$), and for all
$\lambda^{\prime}\in\Lambda$,
$L_{\Lambda}^{\lambda}(\lambda^{\prime})=R_{\Lambda}^{\lambda}(R_{\Lambda}^{0}(\lambda^{\prime}))=\lambda^{\prime}+\lambda$.

2) Again with $L^{\lambda}=R^{\lambda}\cdot R^{0}$, if for all
$\lambda_{1},\lambda_{2}\in\Lambda$,
$R_{\mathcal{P}}^{\lambda_{1}}=R_{\mathcal{P}}^{\lambda_{2}}$ then
$R_{\mathcal{P}}^{\lambda}\cdot R_{\mathcal{P}}^{0}=R_{\mathcal{P}}^{0}\cdot
R_{\mathcal{P}}^{0}$, which is the identity on $\mathcal{P}_{S}$. ∎

###  C.2 Invariance on DPS’s

To apply the notion of invariance to dps’s, we’ll start by expanding the
definition to cover invariance on a collection of dynamic sets. (In the
interest of concision, previous considerations for the case where $\Lambda$ is
bounded by $0$ will not be explicitly mentioned, but they should be assumed to
continue to apply.)

######  Definition 179.

If $X$ is a collection of dynamic sets, $I$ is an _invariant_ on $X$ if it is
a pair of functions $I_{\Lambda}:\bigcup_{S\in
X}\Lambda_{S}\rightarrow\bigcup_{S\in X}\Lambda_{S}$ and
$I_{\mathcal{P}}:\bigcup_{S\in X}\mathcal{P}_{S}\rightarrow\bigcup_{S\in
X}\mathcal{P}_{S}$ s.t. for some bijection, $B_{I}:X\rightarrow X$, all $S\in
X$, $I[S]\equiv\\{\bar{p}\,:\,for\>some\>\bar{p}^{\prime}\in
S,\>\bar{p}=I_{\mathcal{P}}\cdot\bar{p}^{\prime}\cdot
I_{\Lambda}\\}=B_{I}(S)$.

In the case where $X=\\{S\\}$ this reduces to the prior definition of
invariance. When $B_{I}$ is simply the identity on $X$, $I$ represents a
collection of dynamic set invariants, one for each element of $X$. In the more
general case, where the dynamic sets in the collection are allowed to map onto
each other under the transformation, the collection can be invariant under the
transformation even when none of the individual dynamic sets are.

Generalizing types of invariance is straightforward. For example:

######  Definition 180.

$X$ is _reversible_ if for all $S,S^{\prime}\in X$,
$\Lambda_{S}=\Lambda_{S^{\prime}}=\Lambda$, and for all $\lambda\in\Lambda$
there’s and invariant on $X$, $R^{\lambda}$, s.t. for each $S\in X$,
$R_{\mathcal{P}}^{\lambda}\cdot R_{\mathcal{P}}^{\lambda}$ is the identity on
$\mathcal{P}_{S}$, and for all $\lambda^{\prime}\in\Lambda$,
$R_{\Lambda}^{\lambda}(\lambda^{\prime})=\lambda-\lambda^{\prime}$.

Invariance on a dps is now:

######  Definition 181.

If $(X,T,P)$ is a dps, and $I$ is an invariant on $X$, $I$ is an _invariant_
on $(X,T,P)$ if

1) If $\gamma\in G_{T}$ and $I[\gamma]$ is an ip on some $S\in X$ then
$I[\gamma]\in G_{T}$

2) If $t\in T$ and for some $\gamma\in G_{T}$, $I[t]\subset\gamma$ then
$I[t]\in T$

3) If $t\in T$ and $I[t]\in T$ then $P(t)=P(I[t])$

The dps versions of invariants such as reversibility follow immediately.

##  Appendix D Parameter Theory

######  Definition 182.

A set, $\Lambda$, together with a binary relation on $\Lambda$, $<$, a binary
function on $\Lambda$, $+$, and a constant $0\in\Lambda$, is an _open
parameter_ if the structure, $(\Lambda,<,+,0)$, satisfies the following:

Total Ordering:

1) For all $\lambda\in\Lambda$, $\lambda\nless\lambda$

2) For all $\lambda_{1},\lambda_{2},\lambda_{3}\in\Lambda$, if
$\lambda_{1}<\lambda_{2}$ and $\lambda_{2}<\lambda_{3}$ then
$\lambda_{1}<\lambda_{3}$

3) For all $\lambda_{1},\lambda_{2}\in\Lambda$, either
$\lambda_{1}=\lambda_{2}$ or $\lambda_{1}<\lambda_{2}$ or
$\lambda_{2}<\lambda_{1}$

Addition:

4) For all $\lambda\in\Lambda$, $\lambda+0=\lambda$

5) For all $\lambda_{1},\lambda_{2}\in\Lambda$,
$\lambda_{1}+\lambda_{2}=\lambda_{2}+\lambda_{1}$

6) For all $\lambda_{1},\lambda_{2},\lambda_{3}\in\Lambda$,
$(\lambda_{1}+\lambda_{2})+\lambda_{3}=\lambda_{1}+(\lambda_{2}+\lambda_{3})$

The standard interrelationship between ordering and addition:

7) For all $\lambda_{1},\lambda_{2},\lambda_{3}\in\Lambda$,
$\lambda_{1}<\lambda_{2}$ iff
$\lambda_{1}+\lambda_{3}<\lambda_{2}+\lambda_{3}$

8) For all $\lambda_{1},\lambda_{2}\in\Lambda$, if $\lambda_{1}<\lambda_{2}$
then there’s a $\lambda_{3}\in\Lambda$ s.t.
$\lambda_{1}+\lambda_{3}=\lambda_{2}$

Possesses a positive element:

9) There exists a $\lambda\in\Lambda$ s.t $\lambda>0$

The enumerated statements in this definition will be referred to as the “open
parameter axioms”, and the individual statements will be referred to by
number: opa 1 referring to “For all $\lambda\in\Lambda$,
$\lambda\nless\lambda$”, etc.

Because in most cases $<$, $+$, and $0$ will be immediately apparent given the
set $\Lambda$, parameters will often be referred to by simply referring to
$\Lambda$.

######  Definition 183.

For $\lambda\in\Lambda$, $\lambda^{\prime}\in\Lambda$ is an _immediate
successor_ to $\lambda$ if $\lambda^{\prime}>\lambda$ and there does not exist
a $\lambda^{\prime\prime}\in\Lambda$ s.t.
$\lambda^{\prime}>\lambda^{\prime\prime}>\lambda$.

A parameter is _discrete_ if every $\lambda\in\Lambda$ has an immediate
successor.

A parameter is _dense_ if no $\lambda\in\Lambda$ has an immediate successor.

######  Theorem 184.

An open parameter is either discrete or dense

###### Proof.

Follows from opa 8 and opa 7.∎

######  Definition 185.

For a discrete open parameter, the immediate successor to $0$ is
$\boldsymbol{1}$.

######  Theorem 186.

If $\Lambda$ is discrete, then for every $\lambda\in\Lambda$, the immediate
successor to $\lambda$ is $\lambda+\boldsymbol{1}$.

###### Proof.

This too follows from opa 8 and opa 7.∎

######  Definition 187.

If $\chi\subset\Lambda$, $\lambda\in\Lambda$ is an _upper-bound_ of $\chi$ if
for every $\lambda^{\prime}\in\chi$, $\lambda\geq\lambda^{\prime}$; $\lambda$
is the _least upper-bound_ if it is an upper-bound, and given any other upper-
bound, $\lambda_{+}$, $\lambda\leq\lambda_{+}$. If $\chi$ has no upper-bound,
then $\chi$ is _unbounded from above_.

Similarly, $\lambda\in\Lambda$ is an _lower-bound_ of $\chi$ if for every
$\lambda^{\prime}\in\chi$, $\lambda\leq\lambda^{\prime}$; $\lambda$ is the
_greatest lower-bound_ if it is a lower-bound, and given any other lower-
bound, $\lambda_{-}$, $\lambda_{-}\leq\lambda$. If $\chi$ has no lower-bound,
then $\chi$ is _unbounded from below_.

$\lambda^{\prime}$ is the _additive inverse_ of $\lambda$ if
$\lambda+\lambda^{\prime}=0$

######  Theorem 188.

If $\Lambda$ is an open parameter

1) $\Lambda$ is unbounded from above.

2) If $\Lambda$ is bounded from below, it’s greatest lower bound is $0$.

3) If $\Lambda$ does not have a least element, every $\lambda\in\Lambda$ has
an additive inverse; if it does have a least element, only $0$ has an additive
inverse

###### Proof.

1) Follows from opa 9 and opa 7 (with help from opa 4)

2) Also follows from opa 7 with help from opa 4

3) First take the case where $\Lambda$ does not have a least element. If
$\lambda<0$ then by opa 8 there’s a $\lambda^{\prime}$ s.t.
$\lambda+\lambda^{\prime}=0$. If $\lambda>0$ take any $\lambda^{\prime}$ s.t
$\lambda+\lambda^{\prime}<0$ (such a $\lambda^{\prime}$ must exist because
$\Lambda$ is unbounded from below). As just established, there must be a
$\Delta\lambda$ s.t. $\lambda+\lambda^{\prime}+\Delta\lambda=0$, so
$\lambda^{\prime}+\Delta\lambda$ is the additive inverse of $\lambda$.

Now assume $\Lambda$ does have a least element. By opa 7 and pt. 2 of this
theorem, for any $\lambda,\lambda^{\prime}\in\Lambda$ s.t. $\lambda\neq 0$,
$\lambda+\lambda^{\prime}\geq\lambda>0$. ∎

Open parameters admit models which, under most interpretations of “parameter”,
would not be considered admissible. For example, infinite ordinals (with the
expected interpretations of $<$, $+$, and $0$) are open parameters, as are the
extended reals. Rational numbers are also open parameters, as are numbers
which, in decimal notation, have only a finite number of non-zero digits.

To eliminate these less-than-standard models, parameters will be defined as
open parameters that are finite (sometimes called Archimedean), and either
discrete or continuous (that is, parameters have the crucial property that all
limits which tend toward fixed, finite values exist). This will be
accomplished through the well known method of adding a “completeness axiom”.

######  Definition 189.

An open parameter, $\Lambda$, is a _parameter_ if every non-empty subset of
$\Lambda$ that is bounded from above has a least upper bound.

From here on out, it will be assumed that “$\Lambda$” refers to a parameter.

######  Definition 190.

$\lambda$ added to itself $n$ times will be denoted $n\lambda$ (for example
$3\lambda\equiv\lambda+\lambda+\lambda$). $0\lambda\equiv 0$.

$\lambda-\lambda^{\prime}\equiv\begin{cases}0&if\,\Lambda
is\,bounded\,by\,0\,and\,\lambda<\lambda^{\prime}\\\
\Delta\lambda&where\,\Delta\lambda+\lambda=\lambda^{\prime},\>otherwise\end{cases}$

######  Theorem 191.

If $\Lambda$ is a parameter then for every
$\lambda_{1},\lambda_{2}\in\Lambda$, $0<\lambda_{1}<\lambda_{2}$, there exists
an $n\in\mathbb{N}$ s.t. $n\lambda_{1}>\lambda_{2}$.

###### Proof.

Assume that for all $n\in\mathbb{N}$, $n\lambda_{1}<\lambda_{2}$. Then the set
$\chi=\\{x\in\Lambda\,:\,x=n\lambda_{1}\\}$, is bounded from above. Therefore
it has a least upper bound, $\lambda_{M}$. Since $\lambda_{1}>0$,
$\lambda_{M}-\lambda_{1}<\lambda_{M}$, and so $\lambda_{M}-\lambda_{1}$ can’t
an upper bound of $\chi$. Therefore for some $i\in\mathbb{N}$,
$i\lambda_{1}>\lambda_{M}-\lambda_{1}$. But then
$(i+2)\lambda_{1}>\lambda_{M}$, so $\lambda_{M}$ can not be an upper bound.
Therefore $\chi$ is unbounded, and so for some $n\in\mathbb{N}$,
$n\lambda_{1}>\lambda_{2}$. ∎

Thm 191 is equivalent to saying that for all $\lambda\in\Lambda$, $\lambda$ is
finite.

######  Theorem 192.

For any $n\in\mathbb{N}^{+}$

1) $\lambda_{1}>\lambda_{2}$ iff $n\lambda_{1}>n\lambda_{2}$

2) $\lambda_{1}=\lambda_{2}$ iff $n\lambda_{1}=n\lambda_{2}$

###### Proof.

1) Follows from opa 7, opa 2

2) Follows from (1) and opa 3.∎

######  Theorem 193.

If $\Lambda$ is a discrete parameter then given any $\lambda\in\Lambda$,
$\lambda>0$ there’s an $n\in\mathbb{N}$ s.t. $\lambda=n\boldsymbol{1}$.

###### Proof.

Follows from Thms 186 and 191∎

######  Theorem 194.

For any $n,m\in\mathbb{N}$

1) $n\boldsymbol{1}+m\boldsymbol{1}=(n+m)\boldsymbol{1}$

2) If $n>m$ then $n\boldsymbol{1}>m\boldsymbol{1}$

###### Proof.

1) Follows from opa 6

2) Take $k=n-m$; By (1) and opa 7, (2) holds iff $k\boldsymbol{1}>0$, which
follows from Thm 192.1. ∎

Thms 193 and 194, together with Thm 188.3, create a complete characterization
of discrete parameters. A similar characterization for dense parameters will
now be sketched.

######  Definition 195.

For $\lambda,\lambda^{\prime}\in\Lambda$, $m\in\mathbb{N}$,
$n\in\mathbb{N}^{+}$, $\lambda^{\prime}=\frac{m}{n}\lambda$ if
$n\lambda^{\prime}=m\lambda$.

######  Theorem 196.

If $\Lambda$ is dense then for all $\lambda\in\Lambda$, $m\in\mathbb{N}$,
$n\in\mathbb{N}^{+}$

1) $\frac{m}{n}\lambda\in\Lambda$

2) If $\lambda^{\prime}=\frac{m}{n}\lambda$ and
$\lambda^{\prime\prime}=\frac{m}{n}\lambda$ then
$\lambda^{\prime}=\lambda^{\prime\prime}$

###### Proof.

1) Take $\chi\equiv\\{\lambda^{\prime}\in\Lambda\,:\,n\lambda^{\prime}\leq
m\lambda\\}$. $\chi$ is bounded from above, so take $\lambda_{\chi}$ to be the
least upper bound. If $n\lambda_{\chi}$ is either greater or less than
$m\lambda$ then density provides an example that contradicts $\lambda_{\chi}$
being the least upper bound, so by opa 3 $n\lambda_{\chi}=m\lambda$.

2) $n\lambda^{\prime}=m\lambda$ and $n\lambda^{\prime\prime}=m\lambda$, so
$\lambda=\lambda^{\prime}$ by Thm 192.2.∎

######  Theorem 197.

If $\Lambda$ is dense then for all $\lambda\in\Lambda$

1) If $m_{1},m_{2}\in\mathbb{N}$, $n_{1},n_{2}\in\mathbb{N}^{+}$ and
$\frac{m_{1}}{n_{1}}=\frac{m_{2}}{n_{2}}$ then
$\frac{m_{1}}{n_{1}}\lambda=\frac{m_{2}}{n_{2}}\lambda$

2) For $q_{1},q_{2}\in\mathbb{Q}$,
$q_{1}\lambda+q_{2}\lambda=(q_{1}+q_{2})\lambda$

3) For $q_{1},q_{2}\in\mathbb{Q}$, $\lambda>0$, if $q_{1}>q_{2}$ then
$q_{1}\lambda>q_{2}\lambda$

###### Proof.

1) If $\frac{m_{1}}{n_{1}}=\frac{m_{2}}{n_{2}}$ then either there exist a
$m,n,k_{1},k_{2}$ s.t. $m_{1}=k_{1}m$, $n_{1}=k_{1}n$, $m_{2}=k_{2}m$, and
$n_{2}=k_{2}n$. It is sufficient to show that
$\frac{m_{1}}{n_{1}}\lambda=\frac{m}{n}\lambda$. Taking
$\lambda^{\prime}\equiv\frac{m_{1}}{n_{1}}\lambda$,
$(k_{1}n)\lambda^{\prime}=(k_{1}m)\lambda$. By Thm 194.1,
$k_{1}(n\lambda^{\prime})=k_{1}(m\lambda)$. By Thm 192.2
$n\lambda^{\prime}=m\lambda$.

2) For some $m_{1},m_{2}\in\mathbb{N}$, $n\in\mathbb{N}^{+}$,
$q_{1}=\frac{m_{1}}{n}$ and $q_{2}=\frac{m_{2}}{n}$. With
$\lambda_{1}\equiv\frac{m_{1}}{n}\lambda$ and
$\lambda_{2}\equiv\frac{m_{2}}{n}\lambda$,
$m_{1}\lambda+m_{2}\lambda=n\lambda_{1}+n\lambda_{2}$ and so by opa 6,
$n(\lambda_{1}+\lambda_{2})=(m_{1}+m_{2})\lambda$, which mean
$\lambda_{1}+\lambda_{2}=\frac{(m_{1}+m_{2})}{n}\lambda$.

3) Follows from (2) and opa 7, and the fact that $q_{3}\equiv q_{2}-q_{1}$ is
a positive rational number (note that if $\lambda>0$ and $q>0$ then
$q\lambda>0$).∎

######  Definition 198.

A parameter sequence, $(\lambda_{n})_{n\in\mathbb{N}}$, is _convergent_ if
there exists a $\lambda^{\prime}\in\Lambda$ s.t. for any $\Delta\lambda>0$
there’s an $n\in\mathbb{N}$ s.t. for all $i>n$,
$\lambda_{i}\in(\lambda^{\prime}-\Delta\lambda,\lambda^{\prime}+\Delta\lambda)$.
In this case we say $\lambda=\lim\lambda_{n}$.

$(\lambda_{n})_{n\in\mathbb{N}}$, is _Cauchy-convergent_ if for any
$\Delta\lambda>0$ there’s an $n\in\mathbb{N}$ s.t. for all $i,j>n$,
$\lambda_{j}\in(\lambda_{i}-\Delta\lambda,\lambda_{i}+\Delta\lambda)$.

$(\lambda_{n})_{n\in\mathbb{N}}$ is _monotonic_ if either for all an
$i\in\mathbb{N}$, $\lambda_{i+1}\geq\lambda_{i+1}$, or for all an
$i\in\mathbb{N}$, $\lambda_{i+1}\leq\lambda_{i+1}$.

######  Theorem 199.

If $\Lambda$ is a dense parameter and $(q_{n})_{n\in\mathbb{N}}$ is a
monotonic, Cauchy-convergent sequence of rational numbers, then
$(q_{n}\lambda)_{n\in\mathbb{N}}$ is a convergent parameter sequence (with the
understanding that if $\Lambda$ is bounded from below then all $q_{i}$ are
non-negative).

###### Proof.

A: If $(q_{n})_{n\in\mathbb{N}}$ is a Cauchy-convergent sequence of rational
numbers, then $(q_{n}\lambda)_{n\in\mathbb{N}}$ is a Cauchy-convergent.

\- Follows from Thm 197 and the fact that the set of rational numbers is
dense. -

B: If $\Lambda$ is a parameter, and $A\subset\Lambda$ is bounded from below,
then $A$ must have a greatest lower bond

\- Take $B$ to be the set of lower bounds of $A$. $B$ is bounded from above by
every element of $A$, so take $b$ to be the least upper bound of $B$. $b$ must
be a lower bound of $A$ because if for any $a\in A$, $a<b$ then $b$ can not be
the least upper bound of $B$. It also must be the greatest lower bound,
because if any $x>b$ is a lower bound of $A$, then $x\in B$, in which case $b$