Further, when $i$ is even (resp. odd), $(R_{v_{i+1}})^*_\bB$ acts on blocking classes of staircases with $b_\infty>1/3$ (resp. $b_\infty<1/3$).  For example,
$(R_{v_{2}})^*_\bB$ takes the blocking classes of $\Ss^L$ to those of $\Ss^E=S(\Ss^U))$.

The matrix

(R_{v_{i+1}})^*_\bB: = \bigl((S^{i})_\bB^*)\bigr)\circ (S^{i-1}R)_\bB^*\bigr)^{-1}

has eigenvectors $(s_i,r_i)^{\vee}, (3,1)^{\vee}$ with corresponding eigenvalues $1,-1$.  Thus it has order two, and hence also takes the degree components of the blocking classes in $(S^{i})^\sharp(\Ss^U)$ to those of $(S^{i-1}R)^\sharp(\Ss^U)$.

Proposition <ref> shows that
(S^{i-1}R)^*_\bB\bigl((1,1)^{\vee}\bigr) = (s_i,r_i) =   (S^{i})^*_\bB\bigl((1,1)^{\vee}\bigr).

and that $\det( (S^{i-1}R)^*) = -\det ((S^{i})^*)$.  The result now follows from Lemma <ref> (ii).

There turns out to be a second natural action of the reflections $R_{v_{j}}, j\ge 0,$ on blocking class degree corresponding to its action on the other component of $[0,1/3)\cup (1/3,\infty)$.
Indeed, as illustrated in Figure <ref> and Remark <ref> (i), this reflection acts in two ways on our staircases depending on whether one takes $b>1/3$ or $b<1/3$.  The action discussed above (with $j = i+2\ge 2$) interchanges  the steps of the two staircase families $(S^{i+1})^\sharp(\Ss^U)$ and
$(S^{i}R)^\sharp(\Ss^U) = (S^{i})^\sharp(\Ss^L)$ the centers of whose blocking classes converge to
$v_{i+2}$, while, as we will see,  the second action of $R_{v_{i+2}}$
fixes the center $v_{i+2}$ of the  blocking class
$(S^{i})^\sharp(\bB^U_0)$ and  takes the ascending blocking classes in $(S^{i})^\sharp(\Ss^U)$ with centers in $[v_{i+3}, w_{i+2}]$
to the descending blocking classes in $(S^{i+1}R)^\sharp(\Ss^U) = (S^{i+1})^\sharp(\Ss^L)$
with centers in $[w_{i+1}, v_{i+1}]$.  Note, when $i$ is odd (resp. even), the principal blocking class $(S^i)^\sharp(\bB^U_0)$ blocks a $b$-region with $b<1/3$ (resp. $b>1/3$).

We will denote the matrix that gives this second action on degree by $(R_{v_{i+1}})^*_\bP$.  It is again defined by its action on relevant blocking classes, but the family used is different  than before.

Notice that the matrix $(R_{v_{i+1}})^*_\bP$ obtained this way does not have order two, so that it matters that we define it via (<ref>) applied to its action on
the blocking classes in $(S^{i})^\sharp(\Ss^U)$.
Besides the eigenvector
$(3,1)^{\vee}$, its  second eigenvector (with eigenvalue $1$) must be given by the  degree components of the class
$(S^{i})^\sharp(\bB^U_0)$.  (We do need to check that all these conditions are compatible.)
Note also that  the resulting matrix $(R_{v_{i+2}})^*_\bP$ is not in general  integral.

Here we illustrate the different actions of the reflection $R_{v_3}$ that acts on the $z$ axis by fixing
fixing $v_3$.
The reflection $(R_{v_3})^\sharp$, represented by the light blue arrow, interchanges the upper and lower staircase families of $(S^2)^\sharp(\Ss^U)$ and $S^\sharp(\Ss^L)$, and the corresponding matrix $(R_{v_3})^*_{\bB}$ acts as a reflection on the degrees of the blocking classes associated to these staircases.
On the other hand, on the lower number line, we have the principal blocking class $S^\sharp(\bB^U_0)$ with center $v_3$ blocking the corresponding blue interval with $b<1/3$. On either side of this blocked region live the centers of the blocking classes of $(S^2)^\sharp(\Ss^L)$ and $S^\sharp(\Ss^U)$. The matrix $(R_{v_3})^*_{\bP}$ maps the degree coordinates of $S^\sharp(\Ss^U)$ to those of $(S^2)^\sharp(\Ss^L)$. This is represented by the green arrow. Note: this is not a reflection, so the arrow only goes in one direction.

We call the blocking classes $(S^{i})^\sharp(\bB^U_0), i\ge 0,$ principal blocking classes.
The next result spells out their main properties.

(i)  The principal  blocking class
$(S^{i})^\sharp(\bB^U_0)$ has  components

\bigl(\tfrac 38(y_{i+2}+y_{i+1}+ (-1)^{i})  , \tfrac 18(y_{i+2}+y_{i+1}+ 9(-1)^{i}), y_{i+2},y_{i+1},3, (-1)^i\bigr).

(ii)   The  transformation $(R_{v_{i+2}})_\bP^\sharp: (S^{i})^\sharp(\Ss^U) \to (S^{i+1})^\sharp(\Ss^L)$
acts on the degrees of the blocking classes  by the matrix
(R_{v_{i+2}})^*_\bP: = (S^{i+1}R)^*_\bB\ \bigl((S^{i})^*_\bB)\bigr)^{-1}

with determinant $\frac{y_{i+3}-y_{i+2}}{y_{i+2}-y_{i+1}}$.  Its eigenvalues are $\det\bigl((R_{v_{i+2}})^*_\bP\bigr)$ and $ 1$ with corresponding eigenvectors given by  $(3,1)^{\vee}$ and the degree components of $(S^{i})^\sharp(\bB^U_0)$.

The formula for the degree components of the principal blocking class $\bB: = (S^{i})^\sharp(\bB^U_0)$ follows from (<ref>).
The claim in   (ii)  follows immediately from Proposition <ref>.

The reflection $R_{v_2}$ that fixes $6$ has two extensions to an action on blocking classes:

\begin{align}
(R_{v_2})^*_\bB = \left(\begin{array}{cc} 4 & -15\\ 1& -4\end{array}\right),\qquad (R_{v_2})^*_\bP= \left(\begin{array}{cc} -59& 90\\-20& 31\end{array}\right).
\end{align}

Here $(R_{v_2})^*_\bB$, with determinant $-1$, interchanges the blocking classes of $\Ss^E = S(\Ss^U)$ with those of  $\Ss^L$. Both of these familes have $b_\infty<1/3$ and their accumulation points limit to $v_2$. On the other hand, $(R_{v_2})^*_\bP$, with determinant $-29$, fixes the degree components $(3,2)$ of $\bB^U_0$ and takes the blocking classes of $\Ss^U$ to those of $(S^2R)^\sharp (\Ss^U)$.  Both of these staircases have $b_\infty>1/3$ and lie on different sides of the interval blocked by the principal blocking class $B^U_0$, see Fig <ref>. Thus $(R_{v_2})^*_\bB= (S)^*_\bB\circ ((R)^*_\bB)^{-1}$, and one can check that $(R_{v_2})^*_\bP = (SR)^*_\bB$.

§ THE PRE-STAIRCASES ARE LIVE

This section completes the proof of Theorem <ref>.  We first show that the pre-staircase families define in Section 3 are perfect pre-staircase families. Establishing that all staircase classes are perfect implies that
each such class $\bE_\ka = (d_\ka,m_\ka,p_\ka,q_\ka,t_\ka,\eps)$  is live at its center $p_\ka/q_\ka$ when
$b = m_\ka/d_\ka$.
Therefore, by (<ref>),  the capacity function $c_{H_{m_\ka/d_\ka}}$ takes the value $\frac {p_\ka d_\ka}{d^2_\ka - m^2_\ka}$  at the point $p_\ka/q_\ka$, and this implies by continuity[
This holds because for any quasi-perfect class $\bE = (d,m,p,q)$ we have $\mu_{\bE,b}(p/q) \le V_b(p/q)\bigl(\sqrt{1+\frac1{d^2-m^2}}\bigr)$ and here $d_\ka\to \infty$; see <cit.>.]
that  the limiting value $b_\infty$  is unobstructed, i.e.  $c_{H_{b_\infty}}(a_\infty) =  V_{b_\infty}(a_\infty)$, where $a_\infty=\lim(p_\ka/q_\ka)$.

What we have to show is that, at least for sufficiently large $\ka$, the class $\bE_\ka$ remains live at the limiting $b$ value $b_\infty: = \lim m_k/d_k$.    In [1] we established this  in two steps, first showing that if the ratios $m_\ka/d_\ka$     satisfy a bound such as that in (<ref>) below then, by the positivity of the intersections of exceptional classes,  the degree of any class $\bE$ such that
$\mu_{\bE, b_\infty}(p_\ka/q_\ka) \ge \mu_{\bE_\ka, b_\infty}(p_\ka/q_\ka)$ is bounded above by a constant that is independent of $\ka$.  This means that there can be only finitely many such classes.  In particular, there must be one class $\bE_{ov}$  that  dominates infinitely many of the steps. This is possible only if  the obstruction $\mu_{\bE_{ov}, b_\infty}(z)$
given by this class  goes through the accumulation point $(a_\infty, V_{b_\infty}(a_\infty))$ of the
Indeed, we know that the accumulation point is unobstructed so that $\bE_{ov}$ cannot be obstructive at this point, and if (for a descending staircase) the obstruction $\mu_{\bE_{ov}, b_\infty}$ crossed the volume curve to the right of the accumulation point then it could at best overshadow only a finite number of steps.]
We call such a class an overshadowing class because its obstruction $\mu_{\bE_{ov}, b_\infty}(z)$ overshadows
the staircase steps so that they cease to be visible at $b = b_\infty$.

In [1]  we were able to find rather good bounds for the degree of a potentially overshadowing class, and hence could show that they do not exist by a case by case analysis.
This method is not feasible here since the bounds on the degree of any overshadowing class of a staircase in the family $(S^iR^\de)^\sharp(\Ss^U)$ increase too rapidly with $i$.    However it turns out that we can exploit the fact that the obstruction $\mu_{\bE_{ov}, b_\infty}(z)$
goes through the accumulation point to obtain powerful arithmetic information about the degree components $d_{ov},m_{ov}$ of $\bE_{ov}$, which is enough to rule out the existence of such a class.
This argument hinges on the results of Lemma <ref>, namely, that the two functions

\begin{align}\label{eq:same}
b\mapsto  V_b(\acc(b)) \quad\mbox{ and} \qquad   b\mapsto \frac{1+\acc(b)}{3-b}
\end{align}

are the same.

§.§ The pre-staircase classes are perfect

We first prove that all the classes in the families $T^\sharp(\Ss^U)$ are perfect, that is, they are exceptional classes,
and then use this fact in Corollary <ref>
to gain information about the $z$-intervals that are blocked by the blocking classes.

We use the
following recognition principle, which is explained  for example in <cit.>.

An integral class $\bE: = dL-\sum_{i=1}^N n_i E_i$ in the $N$ fold blowup $\C P^2\# N \ov{\C P}^2$ represents an exceptional divisor if and only if it may be reduced to $E_1$ by repeated application of  Cremona transformations.

Here, a Cremona transformation is a composition of the transformation

\[c_{xyz}(dL-\sum_{i=1}n_iE_i)=(d+\delta_{xyz})L-\sum_{i\in \{x,y,z\}}(n_i+\delta_{x,y,z})E_i-\sum_{i \notin \{x,y,z\}} n_iE_i\]

and a reordering operation. Writing $\bE$ in coordinates $(d;n_1,\hdots,n_N),$ $c_{xyz}$ adds $\delta_{xyz}$ to the coordinates $d,n_x,n_y,n_z$ and the reordering can reorder any of the $n_i$.
Because Cremona moves are reversible, to verify $\bE_k$ is exceptional, we just need to show it reduces to some other $\bE_j$ that we know to be exceptional. We say $\bE_k$ and $\bE_j$ are Cremona equivalent if one can be reduced to the other.

Our staircase classes are quasi-perfect and hence have the form $(d,m, W(p/q))$ where $W(p/q) = (W_1,\dots,W_N)$ is the integral weight expansion of $p/q.$   We  denote such a  class by the tuple $(d,m,p,q)$, and consider the Cremona moves as acting on the corresponding sequence    $(d,m, W_1,W_2,W_3,\hdots)$.
Thus, for example,  we have

\begin{align*}& c_{012}((d,m,W_1,W_2,W_3,\hdots)) = \\
& \qquad\quad (2d-m-W_1
\end{align*}

Here is the key lemma.

Suppose $(d,m,p,q)$ satisfies  $3d=m+p+q$, and that $d,m$ are defined by (<ref>).
Then, $(d,m,p,q)$ is Cremona equivalent to $S^{\sharp}(d,m,p,q)$.

Let $S^\sharp(d,m,p,q)=(D,M,P,Q)$. By Lemma <ref>, when $p/q=[5+k,CF(x)]$ we have $S(p/q)=[5;1,4+k,CF(x)]$. Thus, the integral weight expansion of $S(p/q)$ will always have five more integers than the integral weight expansion of $p/q.$
As $S(p/q)=P/Q=(6p-q)/p,$ we have that

\begin{align*}
W(P/Q)&=W((6p-q)/p)=(p^{\times 5},p-q) \sqcup W((p-q)/q)\\
&=(Q^{\times 5},P-5Q) \sqcup W((p-q)/q)
\end{align*}

where $W((p-q)/q)$ by definition is $W(p/q)$ with the first entry $q$ removed.
Thus, to reduce the class $(D,M,P,Q)$ to the class $(d,m,p,q)$, it suffices to  show that we can reduce

\[(D,M,Q^{\times 5},P-5Q) \sqcup W((p-q)/q) \quad \text{to} \quad
(d,m,q) \sqcup W((p-q)/q).\]

Note that in this reduction we must get rid of five terms because as mentioned
$W(S(p/q))$ has five more terms than $W(p/q).$

Next, observe that

\begin{align*}& c_{256}c_{234}c_{016}c_{345}c_{012}(D;Q^{\times 5},M,P-5Q)=
\\ &\qquad\quad (8D-3(M+P);0^{\times 2},3D-M-2P+5Q,0^{\times 3},3D-2M-P).
\end{align*}

This can be seen by direct computation, where the zeros come from the linear Diophantine condition $3D-M-P-Q=0.$ The first three steps give the two zeros in positions 1 and 2, the fourth step results in the two zeros in positions 4 and 5, and the fifth step results in one zero in position 6.
Furthermore, since $P=6p-q$ and $Q=p$, it follows from (<ref>) that

D=\tfrac{1}{8}(3(7p-q)+\varepsilon t), \qquad M=\tfrac{1}{8}((7p-q)+3\varepsilon t).

Performing these substitutions, we get

\begin{align*}
& 8D-3(M+P)\;=\;\tfrac{1}{8}(3(p+q)-\varepsilon t)=d\\
& 3D-M-2P+5Q\;=\;q \\
& 3D-2M-P\;=\; \tfrac{1}{8}(p+q-\varepsilon t)=m
\end{align*}

We conclude that

\[c_{367}c_{345}c_{127}c_{456}c_{123}(D;Q^{\times 5},M,P-5Q)=(d;0^{\times 2},q,0^{\times 3},m).\]

Thus these five Cremona moves and an appropriate reordering reduces

\[(D,M,Q^{\times 5},P-5Q) \sqcup W((p-q)/q) \quad \text{to} \quad
(d,m,q) \sqcup W((p-q)/q).\]

Hence, the class $(D,M,P,Q)$ is Cremona equivalent to  $(d,m,p,q)$, as claimed.

In [1], it was shown that both $\Ss^U$ and $\Ss^L$ are perfect, so it is enough to show that $S^\sharp$ preserves Cremona equivalence, but there is an equally nice argument that $R^\sharp$ preserves Cremona equivalence.

Suppose $(d,m,p,q)$ are such that $d,m$ are defined by (<ref>), and $p/q>7.$ Then, $(d,m,p,q)$ is Cremona equivalent to $R^\sharp(d,m,p,q).$

Let $R^\sharp(d,m,p,q)=(D,M,P,Q)$.
Assume $p/q=[6+k;CF(x)]$ for some $k \geq 1$ and $x \geq 1$. Then, we have

\begin{align*}
\end{align*}

By Lemma <ref> (ii), $R(p/q)=(6p-35q)/(p-6q)= [6,k,CF(x)],$ and thus

\begin{align*}
W(P/Q)&=(Q^{\times 6},P-6Q^{\times k},Q-k(P-6Q),\hdots)\\
&=(p-6q^{\times 6},q^{\times k},p-(6+k)q,\hdots)
\end{align*}

Only the first 6 terms of the weight expansions $W(P/Q)$ and $W(p/q)$ differ. Thus, to show that $(D,M,P,Q)$ is Cremona equivalent to $(d,m,p,q)$, we need to consider the degree coordinates and the first 6 terms of the weight sequence for each.

We use the notation $\mapsto_{ijk}$ to represent applying $c_{ijk}$ to the previous tuple.
Applying two Cremona moves to $(D,M,P,Q)$ gives:

\begin{align}\label{eq:RPQ} (D,M,Q^{\times 6}) &\mapsto_{123} (2D-3Q,M,D-2Q^{\times 3}, Q^{\times 3})\nonumber \\ &\mapsto_{456} (4D-9Q,M,D-2Q^{\times 3},2D-5Q^{\times 3}).
\end{align}

Applying three Cremona moves to $(d,m,p,q)$ gives:

\begin{align*}
(d,m,q^{\times 6}) & \mapsto_{012} (2d-m-2q,d-2q,d-m-q^{\times 2},q^{\times 4}) \\
&\mapsto_{034} (3d-2(m+2q),2d-m-4q,d-m-q^{\times 4},q^{\times 2})\\
&\mapsto_{156} (5d-3(m+3q),2d-m-4q,3d-2(m+3q),d-m-q^{\times 3},2d-m-4q^{\times 2}),
\end{align*}

which we can reorder to get

\begin{align} \label{eq:Rpq}
(5d-3(m+3q),3d-2(m+3q),2d-m-4q^{\times 3},d-m-q^{\times 3}).
\end{align}

We claim this reordered tuple is precisely (<ref>). To see this, we write each term in each of the tuples in terms of $p$ and $q$.
We are going to assume that for $(d,m,p,q)$, $\eps=1.$ By by (<ref>), Lemma <ref> (iii), and the definition of $R$, we can write $(d,m,p,q)$ and $(D,M,P,Q)$ completely in terms of $p$ and $q$ as follows:

\begin{align*}
(d,m,p,q)&=(\frac18(3(p+q)+t),\frac18(p+q+3t),p,q) \\
\end{align*}

Now, we expand all of the entries in (<ref>) and (<ref>) in terms of $p$ and $q$ to get the following equalities:

\begin{align*}
5d-3(m+3q)&=\frac12(3p-15q-t)=4D-9Q \\
2d-m-4q&=\frac18(5p-27q-t)=D-2Q \\
\end{align*}

This completes the proof.

For each $T \in \Gg$, the classes in the pre-staircase family $T^\sharp(\Ss^U)$ are perfect, that is, they are exceptional classes.

By <cit.>, this holds for $\Ss^U$ and $\Ss^L$.   Hence this is an immediate consequence of
Lemma <ref>.

Here is a typical corollary.  For simplicity we only consider the principal blocking classes mentioned in Lemma <ref>, but a similar argument applies to all blocking classes that have associated perfect staircases.

For each $i\ge 0$, the $z$-interval blocked by
the principal blocking class $(S^i)^\sharp(\bB^U_0)$ contains $[w_{i+2},w_{i+1}]$.

Because the pre-staircases $(S^{i+1}R)^\sharp(\Ss^U_0), (S^{i})^\sharp(\Ss^U_0), $
that are associated to $\bB$ consist of perfect classes, we know from <cit.> that their $z$-limit points
$\al_{\bB,\ell}, \al_{\bB,u}$ are unobstructed and that the interval $(\al_{\bB,\ell}, \al_{\bB,\ell})$ lying between them
is precisely the $z$-interval blocked by $\bB$.
Thus it suffices to show that

\al_{\bB,\ell}< w_{i+2} < w_{i+1} <  \al_{\bB,u}.

Since $S$ preserves order and takes $w_i$ to $w_{i+1}$ for all $i$,  we only need  check this for $i=0$, and this can be done either by direct evaluation or by comparing the continued fraction expansions of these quantities.

§.§ Recognizing staircases

We proved the following staircase recognition theorem in <cit.>.

Let  $\Ss = (\bE_\ka )$  be a perfect pre-staircase, let $\la$ be as in Lemma <ref>, and denote by
$D,M,P,Q$ the constants $X$ defined by (<ref>), where $x_\ka  = d_\ka , m_{\ka }, p_\ka , q_\ka $ respectively.
Suppose in addition
that at least one of the following conditions holds:

(i) There is $r/s>0$ such that $ M/D<r/s$,

\frac{m_\ka ^2-1}{d_\ka m_\ka } <  \frac MD<  \frac{s + m_\ka (rd_\ka -sm_\ka )}{r+d_\ka (rd_\ka -sm_\ka )},\qquad \forall \
\ka \ge \ka _0,

and there is no
overshadowing class
at $(z,b)=(P/Q,M/D)$ of degree $d'< s/(r-sb_\infty)$ and with $m'/d' >r/s$.

(ii)  There is $r/s>0$ such that $M/D >r/s$,

\frac{m_\ka (sm_\ka  - rd_\ka )-s}{d_\ka (sm_\ka -rd_\ka ) - r} <  \frac MD< \frac{m_\ka }{d_\ka } \qquad \forall \
\ka \ge \ka _0,

and there is no
overshadowing class
at $(z,b)=(P/Q,M/D)$ of degree $d'< s/(sb_\infty - r)$ and with $m'_\ka /d' <r/s$.

Then  $\Ss$ is  live, and it is a staircase for $H_{M/D}$ that accumulates at $P/Q$.

The next result states the estimates that we must establish in order to apply the above theorem.

Consider a pre-staircase with classes $\bigl(d_\ka ,m_\ka ;q_\ka \bw(p_\ka /q_\ka )\bigr)$, where the ratios $b_\ka : =m_\ka /d_\ka $
have  limit $b_\infty$, and let the constants $D,D',D'',M,M',M'', \si$ be as in (<ref>), with $x_\ka  = d_\ka , m_\ka $ respectively.

(i)  Suppose that $M\ov D - \ov M D\ne 0$.  Then the $b_\ka $ are strictly increasing iff

M\ov D - \ov M D = 2\sqrt{\si}(M''D'-M'D'') = \frac{m_1d_0 -m_0 d_1}{\sqrt{\si}}> 0,

and otherwise they are strictly decreasing.

(ii)  if $m_1d_0 - d_1 m_0 > 0$, $b_\infty<r/s \le 1$,  and

\begin{align}\label{eq:DMineq}
|m_1d_0 - d_1 m_0| \le \sqrt {\si} \frac{sD-rM}{|rD-sM|},
\end{align}

there is $\ka _0$ such that

\frac{m_\ka }{d_\ka } \le b_\infty = \frac MD  \le
\frac{s+ m_\ka (rd_\ka -sm_\ka )}{r+d_{\ka }(rd_\ka -sm_\ka )},\quad \mbox { for } \;\; \ka \ge \ka _0.

if $M''D'-M'D'' < 0$, $b_\infty>r/s > 0$, and   (<ref>) holds,
then  there is $\ka _0$ such that

%b_\ka : =
\frac{m_\ka (sm_\ka -rd_\ka )-s}{d_\ka (sm_\ka -rd_\ka ) -r}\le  b_\infty = \frac MD  \le \frac{m_\ka }{d_\ka }, \quad  \mbox { for } \;\; \ka \ge \ka _0,

This is a reformulation of <cit.> in which (i) incorporates the calculation

M''D'-M'D'' = \frac{m_1d_0 - d_1 m_0}{2\si}

that follows from (<ref>).

In <cit.>, an $r/s$ was carefully chosen in order to reduce the number of potential overshadowing classes that needed to be ruled out. For our purposes, we simply need to know an $r/s$ exists since we use an arithmetic argument to rule out overshadowing classes. The following corollary shows that some $r/s$ does indeed exist.

There exists an $r/s$ such that either the condition (ii) or (iii) in Lemma <ref> is satisfied.

For condition (ii), assume $m_1d_0-d_1m_0$ is positive. As $r/s$ approaches $b_\infty$ from the right $\frac{sD-rM}{|rD-sM|}=\frac{1-\tfrac{r}{s}b_\infty}{r/s-b_{\infty}}$ approaches infinity. Hence, there always exists some $b_\infty<r/s$ such that condition (ii) is satisfied. A similar argument applies for condition (iii).

The statement in Lemma <ref> is not identical to that in  <cit.>
because we have changed notation, now indexing by $\ka$ rather than by $k$.  We now explain
the relation between the two statements.
In <cit.>, we calculated that

M''D'-M'D'' = \frac 1{2(2n+3)\sqrt {\si}}(m_1d_0 - m_0 d_1).

there the staircases were divided into two parts according to the different endings, and (as in Lemma <ref>) the recursion parameter was $\si + 2$ where $\si = (2n+1)(2n+5) = \nu^2-4$.
This reformulation should not change the limiting values $M,D$ and hence $M',M'', D', D''$.  However  in our current notation $m_{k=1},d_{k=1}$ are denoted $m_{\ka=2}, d_{\ka=2}$, since the staircases in [1] are indexed by $k$ while in the current paper we combine the two strands and  index via $\ka$, which is (approximately) $2k$.  Thus if we index by $\ka$ and take
$2n+3 = \nu$ we have

\begin{align*}
& m_2d_0 - m_0 d_2 = (\nu m_1 - m_0)d_0 - m_0(\nu d_1 - d_0) = \nu(m_1d_0 - m_0 d_1),\\
& \frac 1{2(2n+3)\sqrt {\si}}(m_2d_0 - m_0 d_2) = \frac 1{2\sqrt {\si}} (m_1d_0 - m_0 d_1).
\end{align*}

which is consistent with the identities in Lemma <ref>. $\er$

Corollary <ref> establishes that we do not need to compute $m_1d_0-d_1m_0$ for estimating $r/s$. By Lemma <ref> (i), the sign of this quantity determines whether the terms $b_\ka=m_\ka/d_\ka$ strictly increase or decrease for a pre-staircase, which is important for the overshadowing argument.

While the quantities $m_\kappa,d_\kappa$ do depend on both $i$ and $n$, we suppress those indices for simplicity.

(i) For $S^i(\Ss^U_{\ell,n})$, $m_1d_0-m_0d_1= -\eps (2ny_i+y_{i+1})$, where $\eps = (-1)^i$.

(ii) For $S^i(\Ss^U_{u,n})$, $m_1d_0-m_0d_1=  -\eps(2ny_{i+1}+y_{i+2})$, where $\eps = (-1)^i$.

(iii) For $S^i(\Ss^L_{\ell,n})$, $m_1d_0-m_0d_1= -\eps(2ny_{i+1}+y_i)$, where $\eps = (-1)^{i+1}$.

(iv)  For $S^i(\Ss^L_{u,n})$, $m_1d_0-m_0d_1= -\eps(2ny_{i+2}+y_{i+1})$, where $\eps = (-1)^{i+1}$.

Let  $\bullet$ be one of  $(U,\ell),(U,u),(L,\ell),(L,u)$.
For two seeds $(p_{0},q_{0},t_{0},\eps)$ and $(p_{1},q_{1},t_{1},\eps)$
of $S^i(\Ss_{\bullet,n})$, the degree formulas from  (<ref>) imply

\begin{align} \label{eq:det1}
\end{align}

Denote the seeds of  the initial staircase
in  $\Ss^U$ or $\Ss^L$ by $(p_0^\bullet,q_0^\bullet,t_0^\bullet)$ and $(p_1^\bullet,q_1^\bullet,t_1^\bullet)$, and
note that $S^i=\begin{pmatrix} y_{i+1} & -y_{i} \\ y_i & -y_{i-1} \end{pmatrix}$, where $y_{i-1} =6y_i-y_{i+1}$.   Using  the invariance of $t_0^\bullet,t_1^\bullet$ under $S$ we obtain that

\begin{align*} m_1d_0-m_0d_1&=\frac{\eps}{8}\Bigl(\bigl(t_1^\bullet(p_0^\bullet+q_0^\bullet)-t_0^\bullet(p_1^\bullet+q_1^\bullet)\bigr)y_{i+1}\\
&\qquad +\bigl(t_1^\bullet(p_0^\bullet-q_0^\bullet)-t_0^\bullet(p_1^\bullet-q_1^\bullet)+6(t_0^\bullet q_1^\bullet-t_1^\bullet q_0^\bullet)\bigr)y_i\Bigr).
\end{align*}

Then, by substituting the relevant formulas for the seeds from Lemma <ref> and Lemma  <ref>, we get the desired results.

For example, for (ii), we have

\begin{align}\label{eq:seedU0}
(p^U_{0,u},q^U_{0,u},t^U_{0,u})=(-5,-1,2), \qquad (p^U_{1,u},q^U_{1,u},
t^U_{1,u}) =(2n+8,1,2n+5)
\end{align}

\begin{align*}
& =-\eps\bigl((2n+6)y_{i+1} - y_{i}\bigr)\ =\ 2n y_{i+1} + y_{i+2}
\end{align*}
as claimed.

For a pre-staircase $T^\sharp(\Ss^U)$, if $b_\infty>1/3$, then $m_\ka/d_\ka$ strictly decrease. If $b_\infty<1/3$, then $m_\ka/d_\ka$ strictly increase.

Lemma <ref> implies that  $m_1d_0-m_0d_1$ always has the opposite sign to $\eps$.
As $\eps$ is positive if and only if $b_\infty>1/3$, it follows that  $m_1d_0-m_0d_1<0$ if and only if $b_\infty>1/3$. Now use Lemma <ref> (i).

Finally, we prove an estimate that will be useful below.

Let $\Ss$ be one of the descending stairs $S^i(\Ss^U_{u,n})$
for all  $i,n\ge 0$ except $(i,n) = (0,0)$ or one of
the stairs $S^i(\Ss^L_{u,n})$ for all  $i\ge 0, n\ge 1$.  Denote the steps of $\Ss$ by $(d_\ka ,m_\ka ,p_\ka ,q_\ka ,t_\ka ,\eps), \ka \ge 0,$ and
let $b_\infty = \lim m_\ka /d_\ka $.

Then there is $\ka _0$ such that

\begin{align}\label{eq:MDTineq}
\frac{p_\ka  q_\ka }{p_\ka  + q_\ka } > \frac{d_\ka  - m_\ka b_\infty}{3-b_\infty},\quad \ka \ge \ka _0.
\end{align}

$(3d_\ka -m_\ka )p_\ka q_\ka  = (p_\ka +q_\ka )(d^2_\ka  - m^2_\ka  + 1)$, we have

\begin{align*}
\frac{p_\ka +q_\ka }{d_\ka } & = \Bigl(3- \frac{m_\ka }{d_\ka }\Bigr) p_\ka q_\ka  - (p_\ka +q_\ka )\bigl(d_\ka  - \frac{m^2_\ka }{d_\ka }\bigr)\\
& = (3-b_\infty)p_\ka q_\ka  - (p_\ka  + q_\ka )(d- m_\ka  b_\infty) - \bigl( \frac{m_\ka }{d_\ka } - b_\infty\bigr)\bigl(p_\ka q_\ka -m_\ka  (p_\ka +q_\ka )\bigr).
\end{align*}

Therefore (<ref>) will hold provided that

\begin{align*}
\frac{p_\ka +q_\ka }{d_\ka }> \bigl( \frac{m_\ka }{d_\ka } - b_\infty\bigr)\bigl(m_\ka  (p_\ka +q_\ka )-p_\ka q_\ka \bigr), \quad \ka \ge \ka _0.
\end{align*}

m_\ka  (p_\ka +q_\ka )-p_\ka q_\ka  = m_\ka  (3d_\ka -m_\ka )- (d^2_\ka  - m^2_\ka  + 1)  < m_\ka  (3d_\ka -m_\ka )- (d^2_\ka  - m^2_\ka )

it suffices to show that

\begin{align*}
\frac{3d_\ka -m_\ka }{d_\ka }> \bigl( \frac{m_\ka }{d_\ka } - b_\infty\bigr)\bigl(m_\ka  (3d_\ka -m_\ka )- (d^2_\ka  - m^2_\ka )\bigr), \quad \ka \ge \ka _0.
\end{align*}

Since by Lemma <ref> we have $m_\ka  = M\la^\ka  + \ov M \la^{-\ka }, \ d_\ka  = D\la^\ka  + \ov D \la^{-\ka }$ for some $\la>1$, one easily checks that
this will hold if

\begin{align}\label{eq:MDT}
\frac{3D-M}{3M-D} >  |\ov M D - \ov D M| = \frac{|m_1d_0 - d_1 m_0|}{\sqrt \si}, \quad \si = \nu^2-4,
\end{align}

where $\nu$ is the recursion parameter of $\Ss$.

Claim 1: (<ref>) holds when $\Ss  = (S^i)^{\sharp}\bigl( \Ss^U_{u,n}\bigr)$ for all $ i, n \ge 0$ except $(i,n) = (0,0)$.

First consider the case $\Ss = \Ss^U_{u,n}$.   When $X = P, Q ,T$, Lemma <ref> implies that  $X = \frac 1{2\sqrt \si}\bigl( 2x_1 - x_0(\nu-\sqrt\si)\bigr)$,
where $\nu = 2n+3$ is the recursion variable, and $\si = \nu^2 - 4 = (2n+1)(2n+5)$.
The seeds for this staircase are given in (<ref>).
Since  $\nu-\sqrt\si > 0$  and the seed class has $p_0+q_0 < 0$ while $t_0 > 0$, we have

\begin{align*}
\frac{3D-M}{3M-D} & = \frac{P+Q}{T} = \frac{ 2(p_1+q_1) - (p_0+q_0)(\nu-\sqrt\si)}{ 2t_1 - t_0(\nu-\sqrt\si)} \\
& > \frac{ (p_1+q_1)}{ t_1}  = \frac{2n+9}{2n+5} \\
& > \frac {m_0d_1 - m_1d_0}{\sqrt \si} = \frac{2n + 6}{\sqrt{(2n+1)(2n+5)}}
\end{align*}

provided that $(2n+9)^2(2n+1) > (2n+6)^2(2n+5)$.  But this  holds unless $n = 0,1,2,3$.  If
$n=1,2,3$ then
one can check by direct calculation that (<ref>) holds.  Moreover, although it does not hold when $n=0$ one can check that in this case we have

\begin{align}\label{eq:case0}
\frac{3D-M}{3M-D} >  \frac{34}{35}   \ \frac {m_0d_1 - m_1d_0}{\sqrt \si} =  \frac{y_4}{y_2 y_3}\
\frac {6}{\sqrt \si}.
\end{align}

Now suppose that $\Ss  = (S^i)^{\sharp}\bigl( \Ss^U_{u,n}\bigr)$.
Notice that $\si$ and $\nu$ are invariant under the shift.  The quantity  $3M-D$
is also invariant under the shift since it is the limit of
$(3m_\ka  - d_\ka) \la^{-\ka}  = \eps t_\ka \la^{-\ka} $.  To consider how much the right hand side of (<ref>) increases under iterations of the shift, we again use Lemma <ref>.  Because
$y_{i+2} < 6y_{i+1}$ for all $i > 0$

we can estimate
\frac{ 2ny_{i+1} + y_{i+2}}{2ny_{1} + y_{2}} =    \frac{ 2ny_{i+1} + y_{i+2}}{2n + 6}  < y_{i+1}, \quad \forall \  i>0, \ n \ge 0

Therefore the result will hold in the case $n>0$ if we show that when we apply $S^i$ the quantity $3D-M$ increases by a factor of at least $y_{i+1}$.
But by Lemma <ref>, $3D-M = P+Q$ is taken by $S^i$ to $(y_{i+1} + y_i)P- (y_i+ y_{i-1}) Q$.
Therefore we need $(y_{i+1} + y_i)P- (y_i+ y_{i-1}) Q > y_{i+1}(P+Q)$, or equivalently

y_iP  > (y_{i+1} + y_i+ y_{i-1})Q = 7 y_iQ.

But this holds because $P/Q$ is the accumulation point of a staircase in $\Ss^U$ and so satisfies  $P/Q> 7$.  Indeed $w_1 = 7$ is blocked by $\bB^U_0$ by Corollary <ref>.

In the case $n = 0$, the the right hand side of (<ref>) increases by the factor $y_{i+2}/6$.
it suffices to show that when we apply $S^i$ the quantity $3D-M$ increases by a factor of at least $\frac{35}{34}\ \frac{y_{i+2}}{6} $.  Since $\frac{35}{34}\ \frac{y_{i+2}}{6} \le y_{i+1}$ when $i\ge 2$, the previous argument applies to show that this holds.   In the case $i = 1$, we must check that

7P - Q>  \frac{35}{34}\ \frac{35}{6} (P+Q),

which holds because
$P/Q = [7;\{5,1\}^\infty]  > [7;6] = 43/6.$
This completes the proof.
Claim 2: (<ref>) holds when $\Ss  = (S^i)^{\sharp}\bigl( \Ss^L_{u,n}\bigr)$, $n\ge 1$ and $i\ge 0$
Let $\Ss =  \Ss^L_{u,n}$.
The initial seeds are now
$(p^L_{0,u},q^L_{0,u},t^L_{0,u})=(-29,-5,2 )$ and $(p^L_{1,u},q^L_{1,u},
t^L_{1,u}) = (12n-1, 2n-2, 2n+1)$.  Further $ \nu =  2n+3$ so that $\si = \sqrt{(2n+1)(2n+5)}$ as before.
If we simplify the inequality as before and use the result in Lemma <ref> (iv), it follows that it suffices to check that

\frac{ (p_1+q_1)}{ t_1}  > \frac {m_1d_0 - m_0d_1}{\sqrt \si} =\frac{ 2n y_2 + y_1}{\sqrt \si} =
\frac{ 12n + 1}{\sqrt \si}.

But this holds for all $n\ge1$.

Next consider the case $i>0$. We argue as before, noting that for each $n$ the accumulation point  $P/Q$ is larger than the center of the blocking class $\bB^L_n$, which is $\frac {12n+1}{2n}$ by Theorem <ref>.
By Lemma <ref> (iv), when we apply $S^i$  the right hand side grows by the factor
\frac{2n y_{i+2} + y_{i+1}}{2n y_2 + y_1} = \frac{ 2ny_{i+2} + y_{i+1}}{12n+1} , \qquad n\ge 1.
%= \frac{ 2y_{i+2} + y_{i+1}}{13}.

Therefore it suffices to check that

\begin{align*}
(y_{i+1} + y_i)P- (y_i+ y_{i-1}) Q& > \frac{2n y_{i+2} + y_{i+1}}{12n+1}(P+Q)\\&
= \Bigl(y_{i+1} - \frac{2n}{12n+1}y_i \Bigr)(P+Q).
\end{align*}

Because $y_{i+1} + y_i + y_{i-1} = 7 y_i$, this  simplifies to

\frac{14n+1}{12n+1}y_i P > y_i\Bigl(7-\frac {2n}{12n+1}\Bigr)Q.

Thus we need $P/Q> \frac{82n+7}{14n+1}$.  But this holds for all $n\ge 1$ since $P/Q> \frac {12n+1}{2n}$.

The proof of the lemma is now complete.

§.§ Overshadowing classes

To utilize the Staircase Recognition Theorem <ref>, it remains to show there are no overshadowing classes.
Here, we not only show that there are no overshadowing classes for $S^iR^\delta(\Ss^U)$, but prove a more general result about overshadowing classes. Namely, given a general perfect pre-staircase family with recursion parameter $\ge 3$, if $b_\infty<1/3$ and $m_\ka/d_\ka$ strictly increase (resp. $b_\infty>1/3$ and $m_\ka/d_\ka$ strictly decrease), then the staircase family is live provided only that the
staircase is not overshadowed by the obstruction from the exceptional class $\bE_1:=3L-E_0-2E_1-E_{2\ldots 6}$.   Recall from
Remark <ref> (ii) and Lemma <ref> that when  $b \in (1/5,5/11)$ this obstruction is given by the formula $z\mapsto  \frac{1+z}{3-b}$, and hence always passes through the accumulation point
$\bigl(\acc(b),V_b(\acc(b))\bigr)$. Hence, it could overshadow the descending staircases for these $b$.
Since the $\Ss^U$ staircases have $b> 5/11$ while the $\Ss^L$ staircases have $b< 1/5$,
we only need be concerned about  their images under a shift $S^i$.
The next lemma shows that these staircases are not overshadowed in this way,
the slope of any overshadowing class must be greater than the slope of the obstruction  $\mu_{\bE_1,b}$.

Let $\Ss$ be any descending pre-staircase in one of the families $(S^iR^\de)^{\sharp} (\Ss^U),$ where $i>0, \de\in \{0,1\}$.
Then the slope of an overshadowing class
must be larger than $\frac{1}{3-b_\infty}$.

The slope of an overshadowing class must be larger than the slope of the line segments from the accumulation point $(z_\infty,b_\infty)$ to the outer corners $\bigl(\frac{p_\ka}{q_\ka}, \frac{p_{\ka}}{d_{\ka}-m_{\ka}b_{\infty}}\bigr)$ of the staircase.
Therefore we must check that

\[
\frac{\frac{p_{\ka}}{d_{\ka}-m_{\ka}b_{\infty}}-\frac{1+z_{\infty}}{3-b_{\infty}}}{\frac{p_{\ka}}{q_{\ka}}-z_{\infty}} > \frac{1}{3-b_\infty}.
\]

When we simplify $z_\infty$ cancels from the inequality, and we get

\[
(3-b_\infty)p_\ka q_\ka>(p_{\ka}+q_{\ka})(d_\ka-m_\ka b_\infty),
\]

which was proved in Lemma <ref>.

We now use an arithmetic argument to rule out the existence of any other overshadowing classes. The following lemma establishes properties about the common divisors of $(d,m,p,q)$ needed for the argument.

Given a quasi-perfect class $(d,m,p,q)$,
assume there is an integer $k$ such that $\frac{km}p$ and $\frac{kd}p$ (resp. $\frac{km}q$ and  $\frac{kd}q$) are both integers.  Then $p | k$ (resp. $q|k$).

Assume $\frac{km}p$ and  $\frac{kd}p$ are both integers. Let $c_m=\gcd(p,m)$ and $c_{d}=\gcd(p,d)$. As $q=3d-p-m$ and $\gcd(p,q)=1$, we must have $\gcd(p,d,m)=1$. Thus, $\gcd(c_m,d)=\gcd(c_d,m) = \gcd(c_m,c_d)= 1$.

Let $c_m'$ (resp. $c_d'$) be the largest product of primes occurring in $c_m$ (resp. $c_d$) that divides $p$. Then, by the above, we can write $p=c_m'c_d'p'$ where $\gcd(c'_m,c'_d)=1$ and $p'$ involves primes not dividing either $c_m$ or $c_d$, so that $\gcd(p',m)=gcd(p',d)=1$.

Then, by unique factorization $p | k$ if and only if $c_m', c_d',$ and $p'$ all divide $k$. But
because $\frac{km}p \in \Z$, both $c_d'$ and $p'$ divide $k$.  Similarly, $c_m'|k$.
Hence  $c_m', c_d',$ and $p'$ all divide $k$ as required.

The case for $q$ follows similarly.

Suppose that at $(z_\infty,b_\infty)$ there is an overshadowing class $\bE_{ov}= (d_{ov}, m_{ov}, \bbm)$ for some pre-staircase.  Recall from Corollary <ref> that  the limit points $z_\infty, b_\infty$ are irrational.
By <cit.>, there are integers $A\ge 0, C>0$ such that

\mu_{\bE_{ov}, b} (z) = \frac{A +Cz}{d_{ov} - m_{ov}b}, \qquad z\approx z_\infty.

Further, if $\bE_{ov}$ has break point $a_{ov} = p_{ov}/q_{ov}$ then[
In fact, the argument given below does NOT use the estimates for $A,C$, but rather arithmetic bounds obtained
because the obstruction $\mu_{\bE_{ov}, b}$   goes through the accumulation point.]

\begin{align}\label{eq:obstr}
A\ge 0,\;\; C\ge q_{ov} &\mbox{ when } z_\infty < a_{ov},\\ \notag
A\ge p_{ov},\;\; C\ge 0 &\mbox{ when } z_\infty > a_{ov}.
\end{align}

We denote by $\ell_b$ the line $z\mapsto \ell_b(z) = (1+z)/(3-b)$. By (<ref>), this has the property
that $\ell_b(\acc(b)) = V_b(\acc(b))$, i.e. for each $b$  its graph crosses the volume curve at the accumulation point.

Let $\Ss$ be a descending pre-staircase with irrational accumulation point $z_\infty$ that is associated to
a perfect blocking class
$\bB = (d_{\bB},m_{\bB},p_{\bB},q_{\bB},t_{\bB})$.  Suppose that
$\bE_{ov}= (d_{ov}, m_{ov}, \bbm)$ is an overshadowing class,
and denote by

\mu_{\bE_{ov}, b} (z) = \frac{A +Cz}{d_{ov} - m_{ov}b}.

the corresponding obstruction.
Assume the slope $C/(d_{ov} - m_{ov}b_\infty)$ of $\mu_{\bE_{ov}, b_\infty} (z) $ is $> 1/(3-b_\infty)$.
Then there is a positive integer $k$ such that

(i)  If $b_\infty>1/3$ and $m_{\bB}/d_{\bB}>1/3$, then $m_{ov}/d_{ov}<1/3$ and $d_{ov}-3m_{ov}=kt_{\bB}$

(ii) If $b_\infty<1/3$ and $m_{\bB}/d_{\bB}<1/3$, $m_{ov}/d_{ov}>1/3$ and $3m_{ov}-d_{ov}=kt_{\bB}$.

Note first that the function $\mu_{\bE_{ov},b_\infty}$ must obstruct an interval $(z_\infty, z_\infty + \eps)$  to the right of the limit point and hence have break point $a_{ov} > z_\infty$.  Hence (<ref>) implies that $A\ge 0$ and $C\ge q_{ov}$.
Further, the condition on the slope implies that  $C>A$.

Next note that

\frac {1+z_\infty}{3-b_\infty} = \frac {p_{\bB}}{d_{\bB}-m_{\bB}b_\infty} =  \frac{A +Cz_\infty}{d_{ov} - m_{ov}b_\infty}.

This implies

\begin{align*}
&(1+z_\infty)(d_{\bB}-m_{\bB}b_\infty) =  p_{\bB}(3-b_\infty), \\
&\qquad \Longrightarrow z_\infty =  \frac{p_{\bB}(3-b_\infty) - (d_{\bB}-m_{\bB}b_\infty)}{d_{\bB}-m_{\bB}b_\infty} = \frac{b_\infty(m_{\bB}-p_{\bB}) + 3p_{\bB}-d_{\bB}}{d_{\bB}-m_{\bB}b_\infty}\\
&\mbox{ and }\\
&(A+Cz_\infty)(d_{\bB}-m_{\bB}b_\infty) =  p_{\bB}(d_{ov} -  m_{ov} b_\infty )\\
&\qquad \Longrightarrow z_\infty =  \frac{p_{\bB}(d_{ov} - b_\infty m_{ov}) - A(d_{\bB}-m_{\bB}b_\infty)}{C(d_{\bB}-m_{\bB}b_\infty)} =
\frac{b_\infty(Am_{\bB}-p m_{ov}) + p_{\bB} d_{ov}- Ad_{\bB}}{C(d_{\bB}-m_{\bB}b_\infty)}
\end{align*}

b_\infty(Am_{\bB}-p_{\bB} m_{ov}) + p_{\bB} d_{ov}- Ad_{\bB} =C\bigl(b_\infty(m_{\bB}-p_{\bB}) + 3p_{\bB}-d_{\bB}\bigr).

All quantities here are integers except for $b_\infty$ which is irrational because $z_\infty$ is.  Therefore the coefficients of $b_\infty$ must be equal.  Thus we have

Am_{\bB}-p_{\bB} m_{ov} = C(m_{\bB}-p_{\bB}), \qquad p_{\bB} d_{ov}- Ad_{\bB} = C(3p_{\bB}-d_{\bB}).

We can solve these equations for $m_{ov}, d_{ov}$ to get

\begin{align}\label{eq:mdC}
&m_{ov} = \frac{(A-C)m_{\bB} + Cp_{\bB}}{p_{\bB}} = C + (A - C)m_{\bB}/p_{\bB},\\ \notag
& d_{ov} = \frac{3p_{\bB}C + (A-C)d_{\bB}}{p_{\bB}} = 3C +  (A - C)d_{\bB}/p_{\bB}.
\end{align}

As $m_{ov},d_{ov}$ are integers, $ (A - C)m_{\bB}/p_{\bB}$ and $(A - C)d_{\bB}/p_{\bB}$ are both integers.
Then, by Lemma <ref>, $p_{\bB}|(A-C)$.  This proves $k=(C-A)/p_{\bB}$ is an integer.

To see $k > 0$, we have an equation of the form $(A+Cz)/\la = (1 + z)/\la'$ and assume that $C/\la > 1/\la'$, i.e. $C\la'>\la$.
\la + \la z = A\la' + Cz\la' > A\la' + \la z,

so that $A \la'< \la < C\la'$.  In other words, $A<C$.

The formulas in (<ref>) then imply that if $b_\infty>1/3$, so that we also have $m_{\bB}/d_{\bB}>1/3$  by Corollary <ref>, then $3m_{ov}<  d_{ov}$.
We can compute

\[d_{ov}-3m_{ov}=3C-kd_{\bB}-3C+3km_{\bB}=k(3m_{\bB}-d_{\bB})=kt_{\bB}.\]
This proves (i).
The proof for (ii) follows similarly.

We prove a similar result for ascending staircases.

Let $\Ss$ be an ascending pre-staircase with irrational accumularion point $z_\infty$  that is associated to
a perfect blocking class
$\bB = (d_{\bB},m_{\bB},p_{\bB},q_{\bB},t_{\bB})$. Suppose that
$\bE_{ov}= (d_{ov}, m_{ov}, \bbm)$ is an overshadowing class,
and denote by

\mu_{\bE_{ov}, b} (z) = \frac{A +Cz}{d_{ov} - m_{ov}b}.

the corresponding obstruction. Then there is a positive integer $k$ such that

(i)  If $b_\infty>1/3$ and $m_{\bB}/d_{\bB}>1/3$, then $m_{ov}/d_{ov}<1/3$ and $d_{ov}-3m_{ov}=kt_{\bB}$

(ii) If $b_\infty<1/3$ and $m_{\bB}/d_{\bB}>1/3$, then $m_{ov}/d_{ov}>1/3$ and $3m_{ov}-d_{ov}=kt_{\bB}$.

Notice first that we must have $A-C>0$ since if $A\le C$ the slope of the obstruction $\mu_{E_{ov},b_\infty}$ is at most that of the line $z\mapsto \frac{1+z}{3-b_\infty}$ which is $< V_{b_\infty}(z)$ for $z< z_\infty$, and so is not obstructive for $z< z_\infty$,  while the overshadowing class is obstructive.

Next, as
in Lemma <ref> we consider the equalities

\frac {1+z_\infty}{3-b_\infty} = \frac {q_{\bB}z_\infty}{d_{\bB}-m_{\bB}b_\infty} =  \frac{A +Cz_\infty}{d_{ov} - m_{ov}b_\infty}.

This implies

\begin{align*}
&(1+z_\infty)(d_{\bB}-m_{\bB}b_\infty) =  q_{\bB} z_\infty(3-b_\infty), \\
&\qquad \Longrightarrow {z_\infty}  =  \frac{d_{\bB}-m_{\bB}b_\infty} {q_{\bB} (3-b_\infty) - (d_{\bB}-m_{\bB}b_\infty)}
&\mbox{ and }\\
&(A+Cz_\infty)(d_{\bB}-m_{\bB}b_\infty) =  q_{\bB} z_\infty(d_{ov} -  m_{ov} b_\infty )\\
&\qquad \Longrightarrow {z_\infty}  = \frac {A(d_{\bB}-m_{\bB}b_\infty)}{ q_{\bB}(d_{ov} -  m_{ov} b_\infty )- C(d_{\bB}-m_{\bB}b_\infty)}
\end{align*}

Hence we must have

A (q_{\bB} (3-b_\infty) - (d_{\bB}-m_{\bB}b_\infty)) = q_{\bB}(d_{ov} -  m_{ov} b_\infty )- C(d_{\bB}-m_{\bB}b_\infty).

As before, the coefficients of $b_\infty$  on both sides must agree, which gives

\begin{align}\label{eq:dov}
d_{ov} = 3A - (A-C)d_{\bB}/q_{\bB}, \qquad m_{ov} = A - (A-C)m_{\bB}/q_{\bB}.
\end{align}

Since $A-C>0$ and $d_{ov},m_{ov}$ are integers, Lemma <ref>  imply $k=(A-C)/q_{\bB}$ is a positive integer.

Furthermore, these formulas prove that if $b_\infty>1/3$ and hence
$m_{\bB}/d_{\bB}> 1/3$ then $d_{ov} > 3 m_{ov}$. We can again compute

\[ d_{ov}-3m_{ov}=k\,t_{\bB}.\]

This proves (i). The proof for (ii) follows similarly.

We use the results of Lemma <ref> and Lemma <ref> to rule out overshadowing classes for a large set of pre-staircases, which implies that none of the pre-staircases considered in this paper have overshadowing classes.

Let $\Ss$ be a pre-staircase with irrational accumulation point $z_\infty$ that is associated to a perfect blocking class $\bB=(d_{\bB},m_{\bB},p_{\bB},q_{\bB},t_{\bB})$ such that $t_{\bB} \ge 3$. Suppose that either $b_\infty>1/3$ and the $m_\ka/d_\ka$ strictly decrease or $b_\infty<1/3$ and the $m_{\ka}/d_{\ka}$ strictly increase. Assume further that if $\Ss$ is descending any overshadowing class must have slope $> 1/(3-b_\infty)$.
Then, the pre-staircase $\Ss$ has no overshadowing classes at all.

These conditions and Lemma <ref> ensure that the conditions in either Lemma <ref> or Lemma <ref> hold for $\Ss$. For a class $\bE_{ov}=(d_{ov},m_{ov},\bbm_{ov})$ to be overshadowing, it must be obstructive for $b= b_\infty$ at its break point $a_{ov}$.
By <cit.>
this is possible only if $| d_{ov}b_\infty -m_{ov}|<1$.
This estimate and the results in Lemma <ref> and Lemma <ref> imply that for each staircase there is a positive integer $k$ such that

\[ 3>3|b_\infty d_{ov}-m_{ov}|>|d_{ov}-3m_{ov}|=t_{\bB}k\geq 3k.\]

As $k$ must be positive, no such $d_{ov}$ and $m_{ov}$ can exist.

For each $T \in \Gg$, the pre-staircase family $T^\sharp(\Ss^U)$ has no overshadowing classes.

Because $t_{\bB}=2n+3$ in all cases, the results of Corollary <ref> and Lemma <ref> show that all the staircases in $T^\sharp(\Ss^U)$ except for $\Ss^U_{u,0}$ satisfy the conditions in Lemma <ref> and hence have no overshadowing classes.    The proof for  $\Ss^U_{u,0}$ is given in <cit.>.

Now, we establish a straightforward  way to check if a perfect pre-staircase is live based on the above overshadowing arguments.

Let $\Ss$ be a perfect pre-staircase with irrational accumulation point $z_\infty$ associated to a blocking class $\bB$ with  recursion parameter $t_\bB\ge 3$. Suppose that either $b_\infty>1/3$ and the $m_\ka/d_\ka$ strictly decrease or $b_\infty<1/3$ and the $m_{\ka}/d_{\ka}$ strictly increase.  Assume further that if $\Ss$ is descending the slope of any overshadowing class must be $> 1/(3-b_\infty)$.   Then, $\Ss$ is a staircase, namely it is perfect and live.

By Corollary <ref> and Lemma <ref>, the perfect pre-staircase

satisfies the conditions of Theorem <ref> implying that $\Ss$ is live.

A consequence of this proposition is Theorem <ref>.

For each $T \in \Gg$, $T^\sharp(\Ss^U)$ is live.

For all staircases except $\Ss^U_{u,0}$ (which was shown to be live in [1]), the
conditions of Proposition <ref> are satisfied since $t_{\bB}=2n+3$, and Corollary <ref> and Lemma <ref> hold.

[1] M. Bertozzi, T. Holm, E. Maw, D. McDuff, G. Mwakyoma, A. R. Pires, and M. Weiler:  Infinite Staircases for Hirzebruch surfaces, arXiv:2010.08567

[2]  R. Casals and R. Vianna, Sharp ellipsoid embeddings and toric mutations, arXiv:2004.13232

[3]
D. Cristofaro-Gardiner, T. Holm, A. Mandini, and A. R. Pires,  On infinite staircases in toric symplectic four-manifolds, arXiv: 2004.07829

[4] D. Frenkel, D. Müller, Symplectic embeddings of $4$-dimensional ellipsoids into cubes, J. Symp. Geom. 13(4):765-847, 2015.

[5] N. Magill, D. McDuff, and M. Weiler, Staircase patterns  in Hirzebruch surfaces, in preparation.

[6]
D. McDuff and F. Schlenk, The embedding capacity of 4-dimensional symplectic ellipsoids, Ann. Math  (2) 175 (2012), no. 3, 1191–1282.

[7]
M. Usher, Infinite staircases in the symplectic embedding problem for four-dimensional ellipsoids into polydisks, Algebr. Geom. Topol. 19(4):1935-2022, 2019.