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Question

The notation \(\exists! x P(x)\) denotes the proposition “there exists a unique \(x\) such that P(\(x\)) is true”.

Give the truth values of the following statements :

I. \(\exists! x P(x) \rightarrow \exists x P(x)\)

II. \(\exists! x \neg P(x) \rightarrow \neg \forall x P(x)\)

This question was previously asked in
UGC NET 2014 Paper 1 Question Paper (28-Dec-2014)
The correct answer is

Both I & II are true.

Statement I: if exactly one \(x\) makes \(P(x)\) true, then at least one such \(x\) exists, so the implication always holds.

Statement II: if exactly one \(x\) makes \(P(x)\) false, then \(P(x)\) is not true for every \(x\), so \(\neg \forall x P(x)\) holds.

Both implications are therefore valid for every domain and every predicate.

Hence, Both I & II are true.

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