Proving whether a predicate proposistion is valid under all interpretations?

Question

Is the sentence:

$(∃xQ(x)∧∃xR(x)) → ∃x(Q(x)∧R(x))$

valid?

(Where $Q(x)$ and $R(x)$ are predicates and $x$ is the set of real numbers)

If it is explain why. If it isn't, give an interpretation under which it is false.

How do I know which interpretations could be false for this sentence?

Answer 1

Suppose your domain of interpretation is the set $\mathbb{R}$ of reals. Now, interpret $Q(x)$ as "$x \geq 0$", and $R(x)$ as "$x < 0$". What can you conclude?

Clearly, $\exists x Q(x) \land \exists xR(x) \, $ is true under this interpretation, because there exists a non-negative real number (e.g. $1$) and there exists a negative real number (e.g. $-1$).

On the contrary, $\exists x (Q(x) \land R(x)) \, $ is false under the above interpretation, because there is no real number that is at the same time non-negative and negative.

Therefore, the sentence $(\exists x Q(x) \land \exists x R(x)) \to \exists x (Q(x) \land R(x)) \, $ is false under the above interpretation, since the antecedent (i.e. the left-hand side) of the implication is true but the consequent (i.e. the right-hand side) is false.

Answer 2

Interpret $Q(x)$ as '$x$ is positive' and $R(x)$ as '$x$ is negative'