Prenex normal form in the sunshine

Question

Consider the following formula that is to transformed into prenex normal form: $$(\exists x_1. p(x_1)) \lor (\forall x_2. q(x_2)).$$

When consulting the rules for transformations of quantors (e.g., from enter link description here), the one option would be the following steps $$\exists x_1. (p(x_1)) \lor \forall x_2. q(x_2))$$ $$\exists x_1. ( \forall x_2.(p(x_1) \lor q(x_2))).$$

But another option would be a different order of applying the same rules $$ \forall x_2.((\exists x_1.p(x_1)) \lor q(x_2)))$$ $$ \forall x_2.(\exists x_1.(p(x_1) \lor q(x_2))),$$ which is different formula (this could also have been obtained, by using commutative of $\lor$ in the first formulat).

What is the implicit convention, that tells which order of rules has to be used? And, as a consequence, which of these two options is correct?

Answer 1

To distribute recall that the Existential Quantifier is a Disjunction over the Domain, while the Universal is a Conjunction.

$(\exists x~P(x))~\vee~(\forall y~Q(y))$

In the universe of $\{a,b\}$ that is

$$\begin{align}&~\big(P(a)\vee P(b)\big)~\vee~ \big(Q(a)\wedge Q(b)\big) \\\equiv&~ \big(P(a)\vee P(b)\vee Q(a)\big)~\wedge~\big(P(a)\vee P(b)\vee Q(b)\big) \\\equiv&~ \Big(\big(P(a)\vee Q(a)\big)\vee \big(P(b)\vee Q(a)\big)\Big)~\wedge~\Big(\big(P(a)\vee Q(b)\big)\vee \big(P(b)\vee Q(b)\big)\Big)\end{align}$$

but also that

$$\begin{align}&~\big(P(a)\vee P(b)\big)~\vee~ \big(Q(a)\wedge Q(b)\big) \\\equiv&~ \Big(P(a)\vee \big(Q(a)\wedge Q(b)\big)\Big)~\vee~\Big(P(b)\vee \big(Q(a)\wedge Q(b)\big)\Big) \\\equiv&~ \Big(\big(P(a)\vee Q(a)\big)~\vee~\big(P(a)\vee Q(b)\big)\Big)~\wedge~\Big(\big(P(b)\vee Q(b)\big)\wedge\big(P(b)\vee Q(b)\big)\Big)\end{align}$$

And generally: $$\begin{align}&~\exists x~P(x)~\vee~\forall y~Q(y)\\\equiv&~\forall y~\Big(\big(\exists x~P(x)\big)~\vee~Q(y)\Big)\\\equiv&~\forall y~\exists x~\big(P(x)\vee Q(y)\big)\end{align}$$

And $$\begin{align}&~\exists x~P(x)~\vee~\forall y~Q(y)\\\equiv&~\exists x ~\Big(P(x)~\vee~\big(\forall y~Q(y)\big)\Big)\\\equiv&~\exists x~\forall y~\big(P(x)\vee Q(y)\big)\end{align}$$

In short: As long as $x$ is free in $\forall y~Q(y)$ and $y$ is free in $\forall x~P(x)$ then $\forall y~\exists x~(P(x)\vee Q(y))\iff \exists x~\forall y~(P(x)\vee Q(y))$