Is this statement logically true? If so How?

Question

Q) Is the statement (∃xQ(x)∧∃xR(x))↔∃x(Q(x)∧R(x)) logically true? If it is, explain why. If it is not give an interpretation under which it is false.

I have asked this previously but did not get an answer that I could understand. Usually I do this sort of questions by drawing a truth table but now the quantifiers ∃ is confusing me a little.

This is how would start this question

[(∃xQ(x)∧∃xR(x))→∃x(Q(x)∧R(x))]∧[∃x(Q(x)∧R(x))→(∃xQ(x)∧∃xR(x))]

Now do I take the possible T or F values for Q(x) and R(x) or ∃Q(x) and ∃R(x)?

From my understanding I should take the T or F values for Q(x) and R(x). So from now I find out the possible truth values of the statement by taking a variety of truth value combinations of Q(x) and R(x).

So Q(x), R(x): (T,T),(T,F),(F,T),(F,F)

This is where my problem is, if there was no quantifier symbol "∃" over here I could easily solve this but I do not understand the question with the quantifiers involved. Any help would be appreciated, Thanks!

Answer 1

Notice that the scope of the quantified variables is different. Indeed, an equivalent statement is: $$ (\exists x ~ Q(x) \land \exists y ~ R(y)) \leftrightarrow \exists z ~ (Q(z) \land R(z)) $$

It should now be easier to see why this statement might be false. For example, suppose that our universe is the set of all integers and suppose that we define the following predicates for all $n \in \mathbb Z$: $$ Q(n): n \text{ is even} \\ R(n): n \text{ is odd} $$ Then $\exists x ~ Q(x)$ is true (take $x = 2$) and $\exists y ~ R(y)$ is true (take $y = 3$), but $\exists z ~ (Q(z) \land R(z))$ is false (no integer is both even and odd at the same time).

Answer 2

It is true that $ \exists x (Q(x) \wedge R(x))\rightarrow (\exists xQ(x) \wedge \exists xR(x))$. That is because $$ Q(x) \wedge R(x) \rightarrow Q(x) $$ $$ Q(x) \wedge R(x) \rightarrow R(x) $$ That means that $$ \exists x( Q(x) \wedge R(x)) \rightarrow \exists x Q(x) $$ $$ \exists x (Q(x) \wedge R(x)) \rightarrow \exists x R(x) $$ Therefore $$ \exists x (Q(x) \wedge R(x))\rightarrow (\exists xQ(x) \wedge \exists xR(x))$$

However $ (\exists xQ(x) \wedge \exists xR(x)) \not\rightarrow \exists x (Q(x) \wedge R(x))$. For example you can have $R(x) = \neg Q(x)$ where Q(x) is a sentence that is true for some $x$ but is not true for some other.