Is $\exists x [(P(x) \vee Q(x))\rightarrow R(x)]$ logically equivalent to $\exists x [(P(x) \rightarrow R(x)) \vee (Q(x)\rightarrow R(x))]$?

Question

Is $\exists x [(P(x) \vee Q(x))\rightarrow R(x)]$ logically equivalent to $\exists x [(P(x) \rightarrow R(x)) \vee (Q(x)\rightarrow R(x))]$?

What about if I replace $\exists$ with $\forall$?

Answer 1

No. In fact, $(P(x) ∨ Q(x)) ⇒ R(x)$ is equivalent to $(P(x) ⇒ R(x)) \color{red}{∧} (Q(x) ⇒ R(x))$.

Proof:

• Suppose $(P(x) ∨ Q(x)) ⇒ R(x)$. If $P(x)$, then $P(x) ∨ Q(x)$, and thus $R(x)$, which proves $P(x) ⇒ R(x)$. Likewise, $Q(x) ⇒ R(x)$. So, $(P(x) ⇒ R(x)) ∧ (Q(x) ⇒ R(x))$.
• In the other direction, suppose $(P(x) ⇒ R(x)) ∧ (Q(x) ⇒ R(x))$. If $P(x) ∨ Q(x)$, then in the case $P(x)$ we get $R(x)$ by $P(x) ⇒ R(x)$, and in the case $Q(x)$, we get $R(x)$ by $Q(x) ⇒ R(x)$. Thus, $(P(x) ∨ Q(x)) ⇒ R(x)$.

Now you should be able to easily find counterexamples to the equivalence under an $∃$ or $∀$ quantifier.

Answer 2

To supplement the answer given by Jean Abou Samra: If $P(x)$ and $R(x)$ are identically false, while $Q(x)$ is identically true, then $(P(x) \lor Q(x)) \to R(x)$ is false, but $(P(x) \to R(x)) \lor (Q(x) \to R(x))$ is true. As these predicates do not depend on $x$, it does not matter whether you quantify them existentially or universally.