infparadox: This isn't about credentials, but da-bacon is the "Bacon" in Bacon-Shor codes.
I think you are misunderstanding the situation or the math. You say "For example, consider a two subsystems entangled with each other, |q1>|q2>." But this system is -not- entangled, it is a separable state by definition, since you wrote it as a tensor product. For the state to be entangled, it needs to be a sum over separable states: sum_i a_i |x_i>|y_i>, with 0<=a_i<1. In this case, you cannot simply factor the state because of the summation. The second system is not in a pure state individually, nor is the first, it is only the joint state of the system which is pure. The reduced state for a single subsystem cannot be pure if it is entangled to other subsystems (due to the need for the summation in order for the state to be entangled: if you can remove the summation, as in the case where [x_0 = 0, x_1 = 1, y_0 = 0, y_1 = 0] then there is no entanglement).
Now, in the case of a separable state |q1>|q2>, you can indeed just consider the state of each subsystem individually as |q1> and |q2>, so you are correct in this regard. However, entangled states are not of this form, and cannot be treated in this way. It's an elementary mistake that many beginners make.
I think I made a misunderstanding by the |q1>|q2> notation. Consider the entangled 2-qubits system a|00>+b|11>. The prob of the 2nd qubit is |0> is |a|^2 and the prob of 2nd qubit is |1> is |b|^2. Now apply I(x)NOT, then the system is a|01>+b|10>. The prob of 2nd qubit to be |0> is now |b|^2 and prob of 2nd qubit is |1> is |a|^2. The same can be obtained by considering the 2nd qubit only as a|0>+b|1> and apply NOT only on the second qubit, without making any operation on the first qubit and without breaking the entanglement.
Yes, this is true for any unitary operation on a different subsystem, and is known as the no-signalling principle. But this doesn't let you remove the second system: you are still stuck with a mixed state on the first system, not a pure state (in your case the state is 1/2 |0><0| + 1/2 |1><1|).
Well, I still don't see any thing wrong with the math. Even the authors rewrite the equations as sum(|A_k>|C_k>) instead of sum(|C_k>) and apply I(x)M_x instead of applying only M_x on |C_k> alone, this will not change the following equations.
I think you are misunderstanding the situation or the math. You say "For example, consider a two subsystems entangled with each other, |q1>|q2>." But this system is -not- entangled, it is a separable state by definition, since you wrote it as a tensor product. For the state to be entangled, it needs to be a sum over separable states: sum_i a_i |x_i>|y_i>, with 0<=a_i<1. In this case, you cannot simply factor the state because of the summation. The second system is not in a pure state individually, nor is the first, it is only the joint state of the system which is pure. The reduced state for a single subsystem cannot be pure if it is entangled to other subsystems (due to the need for the summation in order for the state to be entangled: if you can remove the summation, as in the case where [x_0 = 0, x_1 = 1, y_0 = 0, y_1 = 0] then there is no entanglement).
Now, in the case of a separable state |q1>|q2>, you can indeed just consider the state of each subsystem individually as |q1> and |q2>, so you are correct in this regard. However, entangled states are not of this form, and cannot be treated in this way. It's an elementary mistake that many beginners make.