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

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+---
+title: "Prefix Normal Form - Implication Exercise"
+date: 2023-02-17T11:05:35-05:00
+draft: false
+tags: []
+math: true
+medium_enabled: false
+---
+
+I recently read through the Wikipedia article on [Prefix Normal Form](https://en.wikipedia.org/wiki/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](https://en.wikipedia.org/wiki/Rewriting#Logic) 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 $\exists x (\phi \implies \psi)$
+$$
+\begin{align*}
+\phi \implies (\exists x \psi) &\iff \neg \phi \vee (\exists x \psi) \tag{0.1}\\\\
+&\iff (\exists x \psi) \vee \neg \phi \tag{symmetry}\\\\
+&\iff \exists x (\psi \vee \neg \phi) \tag{1.2}\\\\
+&\iff \exists x (\neg \phi \vee \psi) \tag{symmetry}\\\\
+&\iff \exists x (\phi \implies \psi) \tag{0.1}
+\end{align*}
+$$