How to Prove (∀x P(x) → A) → ∃x(P(x) → A)

Assume the intended statement is $((\forall x\,P(x))\to A)\to\exists x\,(P(x)\to A)$, where $A$ does not depend on $x$. In classical first-order logic, this is valid provided the domain is nonempty. The proof is short once you split into the cases $A$ and $\neg A$.

The assumptions matter: classical logic allows the case split, and a nonempty domain provides an element when $A$ is true. The result is not valid for an empty domain, and it is not generally derivable in intuitionistic logic.

A classical proof

Let $H$ denote the premise:

$$H: (\forall x\,P(x))\to A.$$

We need to prove $\exists x\,(P(x)\to A)$. By the law of excluded middle, either $A$ is true or $\neg A$ is true.

Case 1: $A$ is true. Since the domain is nonempty, choose an element $d$. The implication $P(d)\to A$ is true because its conclusion, $A$, is true. Therefore $d$ witnesses the existential:

$$\exists x\,(P(x)\to A).$$

Case 2: $\neg A$. We show that not every element satisfies $P$. If $\forall x\,P(x)$ held, then $H$ would give $A$, contradicting $\neg A$. Thus $\neg\forall x\,P(x)$. Classical quantifier negation gives

$$\exists x\,\neg P(x).$$

Choose a witness $c$ with $\neg P(c)$. Then $P(c)\to A$ holds: if $P(c)$ were true, it would contradict $\neg P(c)$, so the implication is true. Hence $c$ witnesses $\exists x\,(P(x)\to A)$.

Both cases establish the desired conclusion, so the full implication is true.

Why the domain must be nonempty

The first case uses an element of the domain as a witness. If the domain is empty, the universal statement $\forall x\,P(x)$ is vacuously true and the existential statement $\exists x\,(P(x)\to A)$ is false. For example, take $A$ to be true. Then the premise $(\forall x\,P(x))\to A$ is true, while the conclusion is false. So the displayed implication fails in an empty-domain interpretation.

Many introductory logic courses build nonempty domains into their definition of first-order structures. Under that convention, the caveat is automatic. If empty domains are allowed, the nonempty-domain condition must be stated.

Why classical logic matters

The proof uses the law of excluded middle to split into $A$ and $\neg A$. It also uses classical quantifier negation to pass from $\neg\forall x\,P(x)$ to $\exists x\,\neg P(x)$. Those steps are not generally available in intuitionistic logic, so the formula is not an intuitionistic theorem for arbitrary predicates and domains.

The distinction is not about whether the proof is written formally. It is about which logical rules the proof system permits. If a course is using classical predicate logic, the case proof above is standard. If it is using intuitionistic logic, you need additional assumptions or a different target conclusion.

Common mistakes

• Forgetting to choose a witness. From $A$ alone, you cannot introduce an existential over an empty domain. In the usual proof, use the nonempty-domain assumption to choose $d$.
• Treating the implication as a biconditional. The premise only says that if every object satisfies $P$, then $A$ follows. It does not say that $A$ implies $\forall x\,P(x)$.
• Using $\neg\forall xP(x)\Rightarrow\exists x\neg P(x)$ without checking the logic. This is a classical quantifier-negation rule; it is not generally intuitionistically valid.
• Letting $A$ depend on $x$. The argument assumes $x$ is not free in $A$. If $A$ contains $x$, the formula has a different meaning and this proof does not automatically apply.

The key idea is to use $A$ itself as one case. When $A$ fails, the premise rules out the possibility that $P$ holds everywhere, leaving an object that makes the conditional true.