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So for example if i have $(\exists y\forall x(P(x) ∨ Q(y)) \implies \forall x\exists y(P(x) \vee Q(y)))$ Is this true or false?

The $\implies$ means logically implies, which is different than implies ($\to$). Logically implies means that for every thing that satisfies the first part, the second part is also satisfied by the same thing.

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    $\begingroup$ the answer to your title is that yes, it matters. However the implication you have written is correct. You can not reverse the arrow though. For example, in plain english "For every rational number $q$ there is an integer $n$ such that $n\cdot q$ is an integer" is true but "There is an integer $n$ such that for every rational number $q$ we have $n\cdot q$ is an integer" is false. $\endgroup$ – Callus Feb 3 '15 at 2:28
  • $\begingroup$ "For every man, there is a woman married to this man" means every man has a wife. "There is a woman who is married to every man" means that a given woman has all the men as husbands. $\endgroup$ – Bernard Massé Feb 3 '15 at 2:34
  • $\begingroup$ Technically, this isn't predicate logic, this is first-order logic. $\endgroup$ – Thomas Andrews Feb 3 '15 at 2:42
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$$\forall x\exists y(x+x=y)$$ means just that we can always find $2x$.

$$\exists y\forall x(x+x=y)$$ means something quite different - that there is some $y$ such that $x+x=y$ for every $x$.

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  • $\begingroup$ Wow this is so simple yet so explanatory. Thank you very much $\endgroup$ – Torched90 Feb 3 '15 at 2:32
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In general $\exists y \forall x\, R(x,y) \implies \forall x \exists y\, R(x,y)$

If there is something related to all things, then everything has something which is related to it.

However, if everything has something which is related to it, it may not be so that there's a same thing related to everything.

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