# Equivalency of Two First-Order Formulae

I was reading a paper, and I encountered a definition of some concept. The definition was of the form:

(+) $\ \qquad\qquad (\exists x) \phi(x) \quad \Rightarrow \quad (\exists y) \psi(y)$

where $\phi$ and $\psi$ are two formulae.

I was wondering if the author's could write the above formula in the form:

(++) $\qquad\qquad (\forall x)(\exists y) \quad \phi(x) \Rightarrow \psi(y)$

That is, if (+) and (++) are equivalent.

Please prove the equivalency, or give a counterexample.

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Yes the two formulas say the same. $((\exists x)\phi(x))\to((\exists y)\psi(y))$ is equivalent to $(\lnot(\exists x)\phi(x))\lor((\exists y)\psi(y))$ which is equivalent to $((\forall x)\lnot\phi(x))\lor((\exists y)\psi(y))$ which is equivalent to $(\forall x)(\exists y)(\lnot\phi(x)\lor\psi(y))$ which is what you are looking for, namely $(\forall x)(\exists y)(\phi(x)\to\psi(y))$.
Could you please elaborate on how $((\forall x)\lnot\phi(x))\lor((\exists y)\psi(y))$ implies $(\forall x)(\exists y)(\lnot\phi(x)\lor\psi(y))$? –  Sadeq Dousti Jan 27 '11 at 22:41
@Sadeq Dousti: Essentially this is because $x$ doesn't appear in $\psi(y)$ (and $y$ doesn't appear in $\phi(x)$). Formally you can see the semantic equivalence through the T-schema: If a sentence is not dependent on $x$, saying that this sentence is true and saying that for every $x$ the sentence is true is the same. –  Apostolos Jan 27 '11 at 23:02
(+) implies (++): suppose first that there is an $x_0$ for which $\varphi(x_0)$ holds, and that (+) holds. Let $y_0$ be a witness for (+) (must exist, since the antecedent holds); then $y_0$ is a witness for (++) for every $x$. If, on the other hand, $\forall(x)(\neg\varphi(x))$ holds, then (++) holds by vacuity (this includes the case in which the domain is empty).
(++) implies (+): if the domain is empty, then (+) holds by vacuity. If the domain is not empty, and $\forall x(\neg\phi(x))$ holds, then (+) is true by vacuity again. And if there exists $x_0$ such that $\phi(x_0)$ holds, then by (++) there exists $y_0$ such that $\phi(x_0)\Rightarrow \psi(y_0)$ holds, and since $\phi(x_0)$ is true then $\psi(y_0)$ holds, hence (+) holds because the consequent is true.