What are the truth values of these statements? a) ∃!xP(x)→∃xP(x) b) ∀x P(x) → ∃!xP(x) c) ∃!x¬P(x)→¬∀xP(x)

a) The statement ∃!xP(x)→∃xP(x) is always true. This is because the left side of the implication, ∃!xP(x), states that there exists exactly one x such that P(x) is true. The right side of the implication, ∃xP(x), states that there exists at least one x such that P(x) is true. If there exists exactly one x such that P(x) is true, then it is also true that there exists at least one x such that P(x) is true. Therefore, if ∃!xP(x) is true, then ∃xP(x) is also true, making the entire implication true.

b) The statement ∀xP(x)→∃!xP(x) is not always true. The left side of the implication, ∀xP(x), states that for all x, P(x) is true. The right side of the implication, ∃!xP(x), states that there exists exactly one x such that P(x) is true. It is possible for P(x) to be true for all x, but for there to be more than one x for which P(x) is true. In this case, the left side of the implication would be true, but the right side would be false, making the entire implication false.

c) The statement ∃!x¬P(x)→¬∀xP(x) is always true. The left side of the implication, ∃!x¬P(x), states that there exists exactly one x such that P(x) is not true. The right side of the implication, ¬∀xP(x), states that it is not the case that for all x, P(x) is true. If there exists exactly one x such that P(x) is not true, then it is also true that it is not the case that for all x, P(x) is true. Therefore, if ∃!x¬P(x) is true, then ¬∀xP(x) is also true, making the entire implication true.