Pullback and pushout pasting, with cancellation of the square adjacent to the outer edge

Suppose first that both small squares are pullbacks. Given $r:W\to D$ and $s:W\to C$ with $edr=zs$, the right pullback gives a unique $t:W\to B$ with $yt=dr$ and $bt=s$. The left pullback then gives a unique $u:W\to A$ with $xu=r$ and $au=t$.

For the cancellation law, assume the right square and outer rectangle are pullbacks. A compatible pair $r:W\to D$, $t:W\to B$ for the left square gives $bt:W\to C$ and $edr=zbt$. The outer property yields a unique $u:W\to A$ with $xu=r$ and $bau=bt$. Since $yau=dxu=dr=yt$, right-pullback uniqueness gives $au=t$. Outer uniqueness gives uniqueness of $u$.

This $u$ satisfies $bau=s$. If $u^{\mathrm{\prime }}$ has $x{u}^{\mathrm{\prime }}=r$ and $ba{u}^{\mathrm{\prime }}=s$, then the right pullback gives $a{u}^{\mathrm{\prime }}=t$, and the left pullback gives $u^{\mathrm{\prime }}=u$. Hence the outer rectangle is a pullback.

Reversing steps 1.1, 1.2, and 2.1 by [L1] proves the corresponding composition and cancellation laws for pushouts.