Answer to Question #300002 in Discrete Mathematics for shahana

Question

 Show that ∃xP(x) ∧ ∃xQ(x) and ∃x(P(x) ∧ Q(x)) are not logically equivalent.

Answer 1

Let us show that $∃xP(x) ∧ ∃xQ(x)$ and $∃x(P(x) ∧ Q(x))$ are not logically equivalent.

Let "P(x)=``x>0"" and "Q(x)=``x<0"", where the domain of $x$ is the set of real numbers.

Then $∃xP(x)$ is true and $∃xQ(x)$ is true, and hence $∃xP(x) ∧ ∃xQ(x)$ is true.

On the other hand the statement $∃x(P(x) ∧ Q(x)) =∃x(x>0 ∧ x<0)$ is false.

Therefore, $∃xP(x) ∧ ∃xQ(x)$ and $∃x(P(x) ∧ Q(x))$ are not logically equivalent.