Lemma: Decoder conjugation certificate

Proof. The two controlled-$Z$ gates map $Z_1{X}_2{Z}_3$ to $X_2$ and leave $Y_1{Z}_2{Y}_3$ unchanged. On positions 1 and 3, $H{S}^{†}$ maps $Y$ to $Z$. The two CNOTs then map $Z_1{Z}_2{Z}_3$ to $Z_2$, leaving $X_2$ unchanged. The final $Y_2$ reverses the signs of both $X_2$ and $Z_2$. This cancels the two minus signs in (eq:reps).