Validity of Well formed Fromulas

Question

How to prove whether this well formed formula is a tautology or not?
∃x,P(x)∧∃x,Q(x)⇒∃x,(P(x)∧Q(x))
So far my approach has been assuming P(x) to be a statement with a truth value(say,"x is odd") and Q(x) to be (say ,"x is even"). But its too confusing for me.
Can anybody suggest a mechanical way of solving this problem(say using truth table)?

Answer 1

Let's take your example : - $P(x)$ is "$x$ is odd" - $Q(x)$ is "$x$ is even"

So, $\exists x, P(x) \land \exists x, Q(x)$ means "There is a $x$ odd, and there is a $x$ even", while $\exists x, P(x)\land Q(x)$ means "There is a $x$ which is both odd and even".

So, the LHS is true while the RHS is false. Thus, the implication is false.

Answer 2

To prove that :

$∃x P(x) ∧ ∃x Q(x) \to ∃x(P(x)∧Q(x))$

is not valid, you have to "manufacture" a suitable counter-example.

To prove that e.g.

$∃x (P(x)∧Q(x)) \to ∃xP(x)∧∃xQ(x)$

is valid, you have to argument "semantically" (for simple cases) or derive it with some proof system, like Natural Deduction or Tableau Method.