Consider the following statements: \(S_1 : \forall x P(x) \vee \forall x…

Answer: Both statements S1 and S2 are not logically equivalent. Below are concise counterexamples.

• S1: (∀x P(x)) ∨ (∀x Q(x)) is not equivalent to ∀x (P(x) ∨ Q(x)). Example: let the domain be {1,2}, let P hold only at 1 and Q hold only at 2. Then for each element x, P(x) ∨ Q(x) is true, so ∀x (P(x) ∨ Q(x)) is true. But neither ∀x P(x) nor ∀x Q(x) holds, so (∀x P(x)) ∨ (∀x Q(x)) is false. Thus the two formulas differ in truth value.

• S2: (∃x P(x)) ∧ (∃x Q(x)) is not equivalent to ∃x (P(x) ∧ Q(x)). Example: with the same domain {1,2}, let P hold only at 1 and Q hold only at 2. Then ∃x P(x) and ∃x Q(x) are both true, so the conjunction is true, but there is no single element that satisfies both P and Q, so ∃x (P(x) ∧ Q(x)) is false. Hence the two formulas are not equivalent.

Remark: In each case one direction does hold in general (for instance, (∀x P(x)) ∨ (∀x Q(x)) implies ∀x (P(x) ∨ Q(x)), and ∃x (P(x) ∧ Q(x)) implies (∃x P(x)) ∧ (∃x Q(x))), but the converses fail, which is why the pairs are not logically equivalent.