[Math] Show the quantified statements are not logically equivalent

Question

∀x(P(x) ⊕ Q(x)) and (∀xP(x)) ⊕ (∀xQ(x))

What I did was simplify the first statement

Ǝx ~(P(x) ↔ Q(x)) Expression for inclusive or

Ǝx ~((P(x) -> Q(x)) ^ (Q(x) -> P(x))) Expression for biconditional

Ǝx ~(~(P(x) V Q(x)) ^ ~(Q(x) V P(x))) Expression for implication

Ǝx ((P(x) V Q(x)) V (Q(x) V P(x))) DeMorgan’s law and double negation

Ǝx (P(x) V Q(x)) Indempotent law

Then I said suppose this statement is true in a certain universe. We can say that P(a) is true or Q(a) is true. If Q(a) is true, then Ǝx P(x) is true, and by the rule of amplification we can say that Ǝx (P(x) V Q(x)) is true. Same thing applies for P(a).

For the second I simplified

~(∀xP(x)) ↔ (∀xQ(x)) Expression for inclusive or

~(((∀xP(x)) -> (∀xQ(x))) ^ ((∀xQ(x)) -> (∀xP(x)))) Expression for bicondtional

~(~((∀xP(x)) V (∀xQ(x))) ^ ~((∀xQ(x)) V (∀xP(x)))) Expression for implication

(((∀xP(x)) V (∀xQ(x))) V ((∀xQ(x)) V (∀xP(x)))) DeMorgan’s + double negation

∀xP(x) V ∀xQ(x) Indempotent law

But I'm not really sure how to explain this in words. I'm also not completely sure if I'm doing this whole question the correct way.

Answer 1

You can disprove logical equivalence through a counterexample. Think about Natural numbers as a domain, $q(x)=1$ if x even and $p(x)=1$ if x is odd .

What happens to the left side? What happens to the right side ?