# prenex normal form quantifiers predicates

Are these if and only if equalities always true ?

1 - $\exists x \exists y \;(P(x) \;and \;Q(y))\Leftrightarrow \exists x P(x) \; and \;\exists y \;Q(y)$

2- $\exists x \exists y \;(P(x) \;or \;Q(y))\Leftrightarrow \exists x P(x) \; or \;\exists y \;Q(y)$

3- $\forall x \forall y \; (P(x) \;or \;Q(y)) \Leftrightarrow \forall x P(x) \;or\; \forall y Q(y)$

4- $\forall x \forall y \; (P(x) \;and \;Q(y)) \Leftrightarrow \forall x P(x) \;and\; \forall y Q(y)$

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They all hold. (1): If $P(x_0)$ and $Q(y_0)$ are both true, then certainly $P(x_0)$ and $Q(y_0)$ is. Similarly, if either of those holds, then $P(x_0)$ or $Q(y_0)$ is true.
For the universally quantified sentences, the same argument gets the right-to-left implications. For the left-to-right in 3, suppose I had $x_0, y_0$ so $P(x_0)$ and $Q(y_0)$ were both false. Then the left-hand side would fail as well. So, by the contrapositive, if I get the left-hand side I get at least one disjunct on the right. Think you can extend the argument to (4) now?
For the left hand side you've only made two of the four possible pairings. Under a $\forall$ we need to be able to fix $x$ and then vary $y$ across all its values, and vice versa, so you also need to check $P(x_2)$ or $Q(y_1)$, as well as $P(x_1)$ or $Q(y_2)$. Only three of the four hold. – Kevin Carlson Aug 4 '12 at 13:57
No, I agree with you that the right hand side is false, but I'm saying the left hand side is also false because there's a choice of $x$ and $y$ making $(P(x)$ or $Q(y))$ false. – Kevin Carlson Aug 4 '12 at 14:52