If ⊕ represents XOR Gate in Boolean algebra and A ⊕ B = C, then B ⊕ C equals:

Concept

XOR (⊕) obeys three algebraic laws: it is commutative (X ⊕ Y = Y ⊕ X), associative ((X ⊕ Y) ⊕ Z = X ⊕ (Y ⊕ Z)), and self-inverse, meaning a variable XORed with itself cancels to 0 (X ⊕ X = 0), with 0 acting as the identity (X ⊕ 0 = X). Because these laws hold for every bit pattern, the result is fixed and does not depend on the actual values of A and B.

Application

1. Start from the unknown expression and replace C by its definition C = A ⊕ B, giving B ⊕ C = B ⊕ (A ⊕ B).

2. Use commutativity and associativity to regroup the like terms together: B ⊕ (A ⊕ B) = (B ⊕ B) ⊕ A.

3. Apply the self-inverse law B ⊕ B = 0, so the expression becomes 0 ⊕ A.

4. Apply the identity law 0 ⊕ A = A. Hence B ⊕ C = A.

Cross-check

Test with concrete bits. Let A = 1, B = 0: then C = A ⊕ B = 1, and B ⊕ C = 0 ⊕ 1 = 1 = A. Let A = 1, B = 1: then C = 0, and B ⊕ C = 1 ⊕ 0 = 1 = A. Both trials reproduce A, confirming the identity B ⊕ C = A. The question is fully determined: no specific numeric values of A or B are required, because the answer is the symbol A itself.