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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?

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3 is false I think – mehdi Aug 4 '12 at 13:42
P(x2) true Q(y2) false, P(x1) false Q(y1) true, then right hand side is false and left is true – mehdi Aug 4 '12 at 13:45
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
Sure, but I dont think there is a need to check for both cases if you have one false example – mehdi Aug 4 '12 at 14:19
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

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