 Solution

The PDF places the extra $X$ on $q_2$ after that qubit has interacted with $a_1$ but before it interacts with $a_2$. Its timing is essential. With no incoming error, the first extracted parity is zero. The fault then changes the data toThe remaining parity measurement gives $a_2=1$ on both branches. Thus the measured error syndrome is $01$, which the rule in the preceding part interprets as an error on $q_1$. Applying $X_1$ leavesThis is outside the original encoded subspace: neither of its supporting basis strings is $000$ or $111$. The incorrect syndrome causes a second data error rather than restoring the input. This timed bit flip during repetition-code syndrome extraction demonstrates why a simple parity circuit need not tolerate its own faults.

The same failure can be checked for every incoming channel error. Let $f\in \{0,1\}$ indicate the internal fault and let $c(s)$ be the correction bit pattern from the table. The observed syndrome and the final pre-correction data pattern areBecause $s$ is linear over binary addition and $s(c(s_{\mathrm{obs}}))=s_{\mathrm{obs}}$, the residual pattern $r=d\oplus c(s_{\mathrm{obs}})$ satisfiesWhenever the fault occurs, the residual syndrome is $10$, so recovery never returns the state to the encoded subspace, even if a channel error also occurred. The diagram therefore must not be treated as an extra independent pre-extraction data flip that would always be diagnosed normally.