simple permutations are contained in more complicated geometric grid classes.

Example \#2: Avoiding $4312$ and $3142$ {#sec-4312-3142}
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We begin our next example with a characterisation of its simple permutations.

\[prop-4312-3142-simples\] The simple permutations of ${\operatorname{Av}}(4312,3142)$ and ${\operatorname{Geom}}{\mbox{\begin{footnotesize}$\left(\begin{array}{rrr}0&1&1\\1&0&-1\end{array}\right)$\end{footnotesize}}}$ coincide.

First observe that $${\operatorname{Grid}}{\mbox{\begin{footnotesize}$\left(\begin{array}{rrr}0&1&1\\1&0&-1\end{array}\right)$\end{footnotesize}}}
=
{\operatorname{Geom}}{\mbox{\begin{footnotesize}$\left(\begin{array}{rrr}0&1&1\\1&0&-1\end{array}\right)$\end{footnotesize}}}
\subseteq
{\operatorname{Av}}(4312, 3142),$$ so it suffices to prove that the simple permutations in ${\operatorname{Av}}(4312, 3142)$ are contained this grid class. Specifically, we will show that in any simple permutation of ${\operatorname{Av}}(4312, 3142)$ the entries that follow the maximum make up a “wedge permutation” oriented as $<$ (which is equivalent to avoiding both 132 and 312), and those preceding the maximum form an increasing sequence.

So, let a simple permutation $\pi \in {\operatorname{Av}}(4312, 3142)$ be given. Because $\pi$ avoids 4312 there can be no 312 pattern after its maximum so, for the sake of contradiction, assume that there is a $132$ pattern. Specifically, choose such a pattern $acb$ where $a$ is as low as possible, and $c$ is as high as possible (for the chosen $a$). This yields the situation depicted on the left in Figure \[fig-ex2-simple-no-132-after-max\]. Now, in order that the cell bounded by $\{b,c\}$ not form an interval, there must be some entry $d$ 