The notation ∃!xP(x) denotes the proposition “there exists a unique x such…
2014
The notation ∃!xP(x) denotes the proposition “there exists a unique x such that P(x) is true”. Give the truth values of the following statements :
I. ∃!xP(x) → ∃xP(x)
II. ∃!x ¬ P(x) → ¬∀xP(x)
- A.
Both I & II are true.
- B.
Both I & II are false.
- C.
I – false, II – true
- D.
I – true, II – false
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Correct answer: A
Final answer: Both statements are true.
I. The statement ∃!x P(x) means there exists an x with P(x) and that x is unique. Formally, ∃!x P(x) ⇔ (∃x P(x)) ∧ (uniqueness condition).
Therefore, from ∃!x P(x) we can directly conclude ∃x P(x). So the implication holds.
II. The statement ∃!x ¬P(x) asserts that there exists exactly one x for which P(x) is false. In particular, there exists an x with ¬P(x).
Because at least one element fails P, it cannot be true that every element satisfies P. Hence ¬∀x P(x) follows, so the implication holds.
Summary: Uniqueness always includes existence, and existence of a counterexample to a universal claim negates that universal claim.
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