website/content/blog/implications-prefix-normal-form.md

title	date	draft	tags	math	medium_enabled
Prefix Normal Form - Implication Exercise	2023-02-17T11:05:35-05:00	false		true	false

I recently read through the Wikipedia article on Prefix 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 $\exists x (\phi \implies \psi)$ $$ \begin{align*} (\forall x \phi) \implies \psi &\iff \neg (\forall x \phi) \vee \psi \tag{0.1} \\ &\iff (\exists x \neg \phi) \vee \psi \tag{2.2}\\ &\iff \exists x (\neg \phi \vee \psi) \tag{2.1}\\ &\iff \exists x (\neg \phi \implies \psi) \tag{0.1} \end{align*} $$ 2. 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*}