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The fault-tolerant procedures for state preparation and logical operation consider the procedure successful if only a one-qubit error occurs in each encoded block of the output.

So, I thought that the procedure for fault-tolerantly measuring the observable M, as shown in the figure below, should also assume pre-existing errors on the encoded data qubits.

Figure 10.28.

I understand that this circuit is not fault-tolerant, so we have to repeat it three times and take a majority vote to make the error probability $O(p^2)$.

But with a probability $O(p)$, there is a single-qubit pre-existing error, and it may propagate through the controlled-$M'$ gate. And the measurement on the ancilla is likely to be incorrect. The single-qubit pre-existing error will still exist in the second and third measurement trials. So I think the measurement error probability after the majority vote will also be $O(p)$.

What am I misunderstanding? Or do we go through the error recovery process for each of the three measurement procedures for the majority vote?

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