\in [n+1]$ such that $z_i = 0$ and show that ${{\sf distance}}({\psi}^{-1}(z), {\psi}^{-1}(z+e_i)) \leq 5$. By the proof of Claim \[claim:f is a bijection\], if $i=n+1$ then ${\psi}^{-1}(z)$ and ${\psi}^{-1}(z+e_i)$ are consecutive vertices in some monotone symmetric chain, and thus ${{\sf distance}}({\psi}^{-1}(z), {\psi}^{-1}(z+e_i)) = 1$.

Therefore, we shall assume henceforth that $i \neq n+1$. Let $z = x \circ z_{n+1}$ and $z+e_i = (x+e_i) \circ z_{n+1}$ for some $x \in {\{0,1\}^{n}}$ and $z_{n+1} \in \{0,1\}$. Let $y,y' \in {\{0,1,{\sqcup}\}^{n}}$ be the encodings of the chains $C_y, C_{y'}$ that contain $x,x+e_i$ respectively. Similarly to the proof for ${{\sf maxStretch}}({\psi}) \le 4$, we perform the marking stage by first performing the marking stage on the prefix of $x$ of length $i-1$, then perform the marking stage on the suffix of $x$ of length $n-i$, and finally, perform the marking stage on the resulting, partially marked string. Denote by $s \in {\{0,1,\hat{0}, \hat{1}\}^{i-1}}$ and $t \in {\{0,1,\hat{0}, \hat{1}\}^{n-i}}$ the two partially marked strings such that the resulted strings after the second step on inputs $x$ and $x+e_i$ are $s \circ 0 \circ t$ and $s \circ 1 \circ t$ respectively. Suppose again for concreteness that the string $s$ contains $a$ unmarked zeros and 