[math: ∀x(¬P(x)→Q(x))]

Question

BlackTom AI · Accepted Answer

An implication with a universal quantifier is not automatically true; it fails exactly when there exists some x for which the antecedent is true and the consequent is false. Here the inner formula is (¬P(x) → Q(x)). Recall the truth table for implication: T → T = T, T → F = F, F → T = T, F → F = T. So (¬P(x) → Q(x)) is false precisely when ¬P(x) is true and Q(x) is false.

Step-by-step reasoning and a concrete counterexample:
1. Pick a simple domain with one element, say D = {a}.
2. Define P and Q on this domain by P(a) = false and Q(a) = false.
3. Evaluate the antecedent: ¬P(a) = ¬false = true.
4. Evaluate the consequent: Q(a) = false.
5. Evaluate the implication at a: (¬P(a) → Q(a)) = (true → false) = false (by the implication truth table).
6. Since the universal quantifier requires the inner formula to be true for every element of the domain, and it is false for a, the universally quantified statement ∀x(¬P(x) → Q(x)) is false in this interpretation.

Why the correct answer is False: The question asks whether the formula holds in all interpretations (i.e., whether it is valid/always true). The counterexample above demonstrates an interpretation that makes the formula false, so it is not valid. Note: the statement can be true in some interpretations (for example if P(x) is true for every x, the implication is vacuously true), but that does not make it universally valid.

Why "True" is incorrect: Claiming "True" would mean the formula is true under every possible interpretation; the single counterexample shows this is not the case.

Conclusion: The correct answer is False, which matches the provided answer.

[math: ∀x(¬P(x)→Q(x))]True/False

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