The truth value of the statements: ∃!x P(x) → ∃x P(x) and ∃!x ¬P(x) → ¬∀x…
2013
The truth value of the statements:
∃!x P(x) → ∃x P(x)
and
∃!x ¬P(x) → ¬∀x P(x),
(where the notation ∃!x P(x) denotes the proposition “There exists a unique x such that P(x) is true”) are:
Answer: D. True and True — Statement 1: ∃!x P(x) → ∃x P(x). The unique-existence quantifier ∃!x P(x) asserts that exactly one element satisfies P, which in particular means at least one…
- A.
True and False
- B.
False and True
- C.
False and False
- D.
True and True
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Correct answer: D
Statement 1: ∃!x P(x) → ∃x P(x). The unique-existence quantifier ∃!x P(x) asserts that exactly one element satisfies P, which in particular means at least one element does. Hence ∃x P(x) holds whenever the antecedent does, and the implication is a tautology — True.
Statement 2: ∃!x ¬P(x) → ¬∀x P(x). The antecedent ∃!x ¬P(x) says exactly one element makes P false. Whenever it is true, there is some element for which P fails, so ∀x P(x) is false and therefore ¬∀x P(x) is true. The implication can never be true → false, so it too is a tautology — True.
Both statements are true, so the answer is True and True.
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