I am not expert in the field, I just took a course, but I know that a mixed state is not the same as an entangled state. In their case it is entangled not mixed and so there is no need to take the partial trace. For example, consider a two subsystems entangled with each other, |q1>|q2>. I we want to apply an operator U on the second subsytem we apply I(x)U or we can simply consider the second subsystem and apply U|q2>. This is valid as long as the entanglement holds and Identity gates are assumed to be applied on the first subsystem.

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.