Question

 Show that ∃x(P(x) ∨ Q(x)) and ∃xP(x) ∨ ∃xQ(x) are logically equivalent.

Answer 1

Let us show that $∃x(P(x) ∨ Q(x))$ and $∃xP(x) ∨ ∃xQ(x)$ are logically equivalent. Denote by $D$ the domain of $x.$

Let the value of $∃x(P(x) ∨ Q(x))$ is true. Then $P(x_0) ∨ Q(x_0)$ is true for some $x_0\in D.$ Therefore, $P(x_0)$ is true or $Q(x_0)$ is true. It follows that $∃xP(x)$ is true or $∃xQ(x)$ is true. We conclude that $∃xP(x) ∨ ∃xQ(x)$ is also true.

Let the value of $∃xP(x) ∨ ∃xQ(x)$ is true. It follows that $∃xP(x)$ is true or $∃xQ(x)$ is true. Therefore, $P(x_0)$ is true for some $x_0\in D$ or $Q(y_0)$ is true for some $y_0\in D.$ Then $P(x_0) ∨ Q(x_0)$ is true or $P(y_0) ∨ Q(y_0)$ is true. We conclude that $∃x(P(x) ∨ Q(x))$ is also true.

Consequntly, $∃x(P(x) ∨ Q(x))$ is true if and only if $∃xP(x) ∨ ∃xQ(x)$ is true, and hence $∃x(P(x) ∨ Q(x))$ and $∃xP(x) ∨ ∃xQ(x)$ are logically equivalent.