Which of the following expressions is a tautology?[Fill in the blank]

Question

BlackTom AI · Accepted Answer

P⇔Q asserts P and Q have the same truth value, so when combined with ¬Q it forces ¬P; therefore ((P⇔Q) ∧ ¬Q) ⇒ ¬P is always true.

Relevant concepts:
- Tautology: a formula true for every truth assignment to its atomic propositions.
- Biconditional P⇔Q: true exactly when P and Q are both true or both false (equivalently (P∧Q) ∨ (¬P∧¬Q)).
- Material implication A ⇒ B is true whenever A is false or B is true.

Step-by-step reasoning:
1. Assume the antecedent (P⇔Q) ∧ ¬Q is true. That means both P⇔Q is true and Q is false.
2. From P⇔Q being true we know P and Q share the same truth value. With Q false, the shared value must be false, so P is false.
3. Therefore ¬P is true.
4. Since whenever the antecedent holds the consequent ¬P also holds, the conditional ((P⇔Q) ∧ ¬Q) ⇒ ¬P is true under every possible assignment. Thus it is a tautology.

Algebraic viewpoint:
- P⇔Q ≡ (P∧Q) ∨ (¬P∧¬Q). If ¬Q holds, the only satisfied disjunct is (¬P∧¬Q), which directly yields ¬P. Hence antecedent ⇒ ¬P is always satisfied.

Why the other options fail:
- (a) ((P⇔Q) ∨ P) ⇒ Q is not a tautology. Counterexample: P = T, Q = F. Then (P⇔Q) is F, but (F ∨ P) = T, so antecedent true while Q is false, making the implication false.
- (b) (P ⇒ Q) ⇔ (¬Q ∨ P) is not a tautology because P ⇒ Q is equivalent to (¬P ∨ Q), not (¬Q ∨ P). Counterexample: P = T, Q = F gives left false, right true, so the biconditional is false.
- (c) (P ⇒ Q) ⇔ (¬P ∧ Q) is false: take P = T, Q = T. Then P ⇒ Q is true but ¬P ∧ Q is false, so the biconditional fails.

Conclusion: only statement (d) ((P⇔Q) ∧ ¬Q) ⇒ ¬P is a tautology, which matches the provided correct answer.

Which of the following expressions is a tautology?[Fill in the blank]单项选择题

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