I recently read through the Wikipedia article on Prenex Normal Form. It first describes the two equivalences for conjunction/disjunction. $$ (\forall x \phi) \vee \psi \iff \forall x(\phi \vee \psi) \tag{1.1} $$

$$ (\exists x \phi) \vee \psi \iff \exists x (\phi \vee \psi) \tag{1.2} $$

They show these rules similarly for conjunction. In the next section, they describe the rules for negation: $$ \neg \exists x \phi \iff \forall x \neg \phi \tag{2.1} $$

$$ \neg \forall x \phi \iff \exists x \neg \phi \tag{2.2} $$

In the third section, they describe the rules related to implication. With it comes the following quote:

These rules can be derived by rewriting the implication $\phi \implies \psi$ as $\neg \phi \vee \psi$ and applying the rules for disjunction above.

This sounds like “we leave this as an exercise to the reader”, and a reader I am! Let’s label the rule in the quote as $0.1$.

1. Show that $(\forall x \phi) \implies \psi$ is equivalent to $\exists x (\phi \implies \psi)$ $$ \begin{align*} (\forall x \phi) \implies \psi &\iff \neg (\forall x \phi) \vee \psi \tag{0.1} \\ &\iff (\exists x \neg \phi) \vee \psi \tag{2.2}\\ &\iff \exists x (\neg \phi \vee \psi) \tag{1.2}\\ &\iff \exists x (\phi \implies \psi) \tag{0.1} \end{align*} $$ 2. Show that $(\exists x \phi) \implies \psi$ is equivalent to $\forall x (\phi \implies \psi)$ $$ \begin{align*} (\exists x \phi) \implies \psi &\iff \neg(\exists x \phi) \vee \psi \tag{0.1}\\ &\iff \forall x (\neg \phi) \vee \psi \tag{2.1}\\ &\iff \forall x (\neg \phi \vee \psi) \tag{1.1}\\ &\iff \forall x (\phi \implies \psi) \tag{0.1} \end{align*} $$ 3. Show that $\phi \implies (\exists x \psi)$ is equivalent to $\exists x (\phi \implies \psi)$ $$ \begin{align*} \phi \implies (\exists x \psi) &\iff \neg \phi \vee (\exists x \psi) \tag{0.1}\\ &\iff (\exists x \psi) \vee \neg \phi \tag{symmetry}\\ &\iff \exists x (\psi \vee \neg \phi) \tag{1.2}\\ &\iff \exists x (\neg \phi \vee \psi) \tag{symmetry}\\ &\iff \exists x (\phi \implies \psi) \tag{0.1} \end{align*} $$ 4. Show that $\phi \implies (\forall x \psi)$ is equivalent to $\forall x (\phi \implies \psi)$ $$ \begin{align*} \phi \implies (\forall x \psi) &\iff \neg \phi \vee (\forall x \psi) \tag{0.1}\\ &\iff \forall x(\psi) \vee \neg \phi \tag{symmetry} \\ &\iff \forall x (\psi \vee \neg \phi) \tag{1.1}\\ &\iff \forall x (\neg \phi \vee \psi) \tag{symmetry} \\ &\iff \forall x (\phi \implies \psi) \tag{0.1} \end{align*} $$