(6/6) Incidentally, note now that every A-directed (up-down) edge has one x and one ±y. Good work, Bob!
Now if Bob measures B and C, the resulting state of Alice's A is one of those 4 up-down edges. In particular, it would have one amplitude of x and one amplitude of ±y.
Moreover, the 2-bit outcome Bob sees tells him which of the 4 edges Alice got. So if he informed Alice of these 2 bits, she could easily swap and/or negate some amplitudes to fix A up to x|0> y|1>.
(Quantum teleportation, ta-da.)