Is there any difference between a quantum- and classically-controlled gate if I know my basis?

Question

Consider an unrealistic 2-qubit plus 1 ancilla bit-flip error correction code (images generated by quirk), where I know by some means or other that an error may have happen on qubit0 (represented by the octagon)

First, qubit1 is entangled with q0, then after the potential error, calculate the parity on q2 (the ancilla), and then apply a CNOT to q0.

Now I know that my ancilla MUST be either |0> or |1> because of how it has been prepared, independent of the initial state of q0. Does that mean I can perform the parity measurement classically, and get the same result? i.e.:

Are these two circuits completely equivalent? Or is there some subtlety of collapsing the entangled state that I am missing? Does the initial state of q0 matter? What if q0 is already entangled with some other qubit (not shown)?

Answer 1

Be cautious using logic like "it must be A or B, both of which imply C, therefore C" when dealing with quantum circuits. It can fail if A, B, or C don't commute (e.g. Hardy's paradox).

Personally, I would say the reason the circuit works is because of the no communication theorem and the deferred measurement principle. By the no communication theorem, operations you apply to the third qubit at the end of the circuit cannot affect measurement statistic over other qubits (note that I'm assuming you're not going to do any other multiqubit operations with the third qubit that you're not showing). So you can introduce a measurement operation on the third qubit at the end of the circuit. Then, by the deferred measurement principle, this measurement commutes with controls, so you can measure before the controlled operation instead of after.