Given the Boolean expression (A ⊕ B) ∧ (B → C), which of the following rows in…

Concept: An exclusive-OR X ⊕ Y is 1 exactly when its two operands hold different truth values, and 0 whenever they agree. An implication X → Y is false in one situation alone — a true antecedent with a false consequent (X = 1, Y = 0) — and is 1 in every other situation, so any row with X = 0 makes it 1. A conjunction P ∧ Q is 1 when P and Q are both 1, and 0 otherwise.

Application: The given expression is a conjunction of A ⊕ B and B → C, so both parts must come out 1 at the same time. Testing the row A = 1, B = 0, C = 1:

1. A = 1 and B = 0 hold different truth values, so A ⊕ B = 1.

2. B = 0 makes the antecedent of B → C false, so B → C = 1 whatever C may be; here C = 1.

3. Both parts are 1, so (A ⊕ B) ∧ (B → C) = 1 ∧ 1 = 1.

The row A = 1, B = 0, C = 1 therefore produces an output of 1 (True).

Cross-check: evaluating each of the four given rows part by part.

Contrast:

• Across the complete eight-row truth table three assignments give an output of 1 — (A, B, C) = (1, 0, 0), (1, 0, 1) and (0, 1, 1) — but of these, just A = 1, B = 0, C = 1 is offered among the rows listed here.

• Common pitfall: treating ⊕ as an ordinary OR. An OR would let A = 1, B = 1, C = 1 give an output of 1 as well, whereas exclusive-OR is 0 whenever A and B agree — that is precisely what separates those two rows.