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say that ∀x∃y in all possible integers (negative integers, 0 and positive integers) is x*y = x is it safe to say that ∃y∀x is also true. If not can someone explain why its not true. The way I'm thinking of it is that for every value of x, there is a y such that x times y is equal to x. For the second statement, Im thinking of it as there is a y such that for every value of x, x times y is x.

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    $\begingroup$ Both statements are true, but generally the latter statement is stronger. $\forall_{x}\exists_{y}\phi(x,y)$ means that for each $x$ there's a $y$ (possibly a different one for every $x$) that makes $\phi(x,y)$ true. On the other hand, $\exists_{y}\forall_{x}\phi(x,y)$ means that there's a particular $y$ that makes $\phi(x,y)$ simultaneously true for all $x$. The second statement always implies the first, but not vice versa. (In your case, though, there is a particular $y$ such that $x*y=x$ for every $x$: it's $1$.) $\endgroup$
    – mjqxxxx
    Jan 27, 2015 at 4:03
  • $\begingroup$ Thank you so much. I wasn't 100% sure but I'm glad I was correct. $\endgroup$
    – Torched90
    Jan 27, 2015 at 4:04
  • $\begingroup$ @mjqxxxx Could you please transform your comment into an answer? $\endgroup$
    – Workaholic
    Feb 3, 2016 at 13:02

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Both statements are true in this case; in general the second statement is strictly stronger than the first. "$ \forall{x}\exists{y}\varphi(x,y) $" means that for each $x$ there's a $y$ (possibly a different one for every $x$) that makes $\varphi(x,y)$ true. On the other hand, "$\exists{y}\forall{x}\varphi(x,y)$" means that there's a particular $y$ that makes $\varphi(x,y)$ simultaneously true for all $x$. The second statement always implies the first, but not vice versa. (In your case, though, there is a particular $y$ such that $x*y=x$ for every $x$: it's $y=1$.)

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