Minimum-Weight Recovery Failure Yields a Half-Error Crossing Path

For a $Z$-error chain $E$, its commutation sign with the $X$ check at nonrough vertex $u$ is $(-1{)}^{({\mathrm{\partial }}_{1}E{)}_u}$: the exponent is the parity of incident error edges by [F1], and every overlapping $Z$ and $X$ factor anticommutes by [F3]. Thus the X-check component of the syndrome is ${\mathrm{\partial }}_{1}E$ by [F2]; every Z-check component is zero because Z-type Paulis commute. Equality of the perfect syndromes of $E$ and $R$ therefore gives ${\mathrm{\partial }}_1(E\mathrm{△}R)=0$.

Put $C=E\mathrm{△}R$. Logical failure means that $C$ is a nontrivial $Z$-logical class. By [F4], it contains a rough-to-rough path of length at least $d$. If that path repeats a vertex, delete the closed subpath between two occurrences; finitely many such deletions produce a self-avoiding path $P$ whose edges remain in $C$ and whose endpoints remain on the left and right rough boundaries. It crosses every one of the $d$ vertical cuts, so $\mathrm{\ell }:=\mathrm{\mid }P\mathrm{\mid }\ge d$, and self-avoidance bounds $\mathrm{\ell }\le n$.

The path $P$ has zero relative boundary because its internal vertices have even degree and its two endpoints are omitted rough vertices. Hence $R\mathrm{△}P$ has the same syndrome as $R$. Minimum cardinality gives $\mathrm{\mid }R\mathrm{\mid }\le \mathrm{\mid }R\mathrm{△}P\mathrm{\mid }$. Since $P\subseteq E\mathrm{△}R$, each edge of $P$ belongs to exactly one of $E$ and $R$; comparing the two cardinalities and canceling the edges outside $P$ yields $\mathrm{\mid }R\cap P\mathrm{\mid }\le \mathrm{\mid }P\setminus R\mathrm{\mid }=\mathrm{\mid }E\cap P\mathrm{\mid }$. These counts sum to $\mathrm{\ell }$, so $\mathrm{\mid }E\cap P\mathrm{\mid }\ge \lceil \mathrm{\ell }\mathrm{/}2\rceil$.