The domain of $x$ and $y$ is the same, real numbers. I was wondering if there was a $P(x)$ for which these two would be non equivalent.
$P(x, y)$ for which $\exists x\forall yP(x, y)$ and $\exists y\forall xP(x, y)$ are not equivalent
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$\begingroup$
logic
predicate-logic
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1... or rather $P(x,y)\equiv x\le y^2$. Or simply $P(x,y)\equiv x=42$. – 2017-02-12
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0Natural numbers, $P=\le$. – 2017-02-12
2 Answers
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You could let $P(x,y)$ mean $x=0$. (Yes, $y$ deliberately doesn't appear there).
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You can also consider:
$$P(x,y)="y\times\lfloor x^2+1\rfloor\text{ is an integer}".$$
Then
$$\exists x,\quad\forall y,\quad P(x,y)$$
is false since the domain you consider is the set of real numbers.
But
$$\exists y,\quad\forall x,\quad P(x,y)$$
is true (you can take $y=0$ or $y=k\in \mathbb Z$ for instance).
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0@HenningMakholm Oops, the "$y$" disappeared mysteriously. I edited. – 2017-02-12