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I have this propositional function:

$p(x,y):y-x=y+x^2$

and I have to find truth value for:

  • $\forall{x}\exists{y}$ $p(x,y)$
  • $\exists{y}\forall{x}$ $p(x,y)$

Set of all numbers is integer numbers $Z$. I am not sure if this is the same or not. My solution for both is $0$. Is this two propositions the same?

Johny
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  • Have you written this correctly? As it stands, the $y's$ cancel and you just get $x=0,-1$ regardless of the value of $y$. – lulu Nov 16 '15 at 12:35
  • yes i written it correctly. – Johny Nov 16 '15 at 12:36
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    Then, presuming this is meant to hold over $\mathbb Z$ or $\mathbb Q$ or $\mathbb R$ or something like that, then both statements are false. No value of $y$ makes the statement $p(1,y)$ true, for example. – lulu Nov 16 '15 at 12:41
  • sorry, i forgot to write that we are in set $Z$, I correct it – Johny Nov 16 '15 at 12:42
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    To be clear: a priori, the statements are inequivalent. The second one is stronger. For example, take the proposition $q(x,y):x+y=0$. Then $\forall x \exists y;q(x,y)$ is true but $\exists y \forall x ;q(x,y)$ is false. – lulu Nov 16 '15 at 13:08
  • So, for my function are the same and my solution is correct? – Johny Nov 16 '15 at 13:40
  • I don't know what "my solution for both is $0$" means in this context. Both statements are false. I would not say that they are logically equivalent, unless you want to say that all false statements are logically equivalent. – lulu Nov 16 '15 at 15:32

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