Which one of the following propositional logic formulas is TRUE when exactly…

Correct formula: (¬(p ↔ q) ∧ r) ∨ (p ∧ q ∧ ¬r)

Reasoning: The formula is the disjunction of two cases that together capture all and only the assignments where exactly two variables are true.

• Case 1: p and q are both true and r is false. Then p ∧ q ∧ ¬r is true, so the whole formula is true. (assignment: p = T, q = T, r = F)

• Case 2: p and q have different truth values and r is true. Then ¬(p ↔ q) ∧ r is true, so the whole formula is true. (assignments: p = T, q = F, r = T or p = F, q = T, r = T)

These two cases correspond exactly to the three assignments with two true variables: (T,T,F), (T,F,T), and (F,T,T).

Why the other formulas fail:

• ((p ↔ q) ∧ r) ∨ (p ∧ q ∧ ¬r) is true when p = q = r = true because (p ↔ q) ∧ r holds, so it accepts the case with all three true.

• ((p → q) ∧ r) ∨ (p ∧ q ∧ ¬r) can be true when only one variable is true (for example p = F, q = F, r = T makes p → q true and r true), so it accepts one-true assignments.

• (¬(p ↔ q) ∧ r) ∧ (p ∧ q ∧ ¬r) conjuncts contradictory requirements for r (both r and ¬r), so it is always false and cannot represent exactly-two-true.