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Consider the predicate $$P(x,y,z) = xyz = 1",$$ for $$ x,y, z \in R,$$ $$x; y; z > 0.$$

$1 - \forall x; \forall y; \exists z; P(x; y; z). $

$2 - \exists x; \forall y; \forall z; P(x; y; z). $

Ain't the above 2 statement truth value the same? But my solution says other wise.

$\forall x; \forall y; \exists z; P(x; y; z)$ is true: choose any $x$ and any $y$, then there exists a $z$, namely $z = \frac{1}{xy}$ such that $xyz = 1$.

$\exists x; \forall y; \forall z; P(x; y; z). $ is false: one cannot find a single $x$ such that $xyz = 1$ no matter what are $y$ and $z$. This is because once $yz$ are chosen, then $x$ is completely determined, so $x$ changes whenever $yz$ does.

Why can't statement 2 be justified as true using the same logic for statement 1? That is choose any $y$ and any $z$, then there exists a $x$, namely $x = \frac{1}{yz}$

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  • $\begingroup$ You've got me convinced. The first statement is true and the second is false, for the reasons you gave. Who says the truth values are the same? $\endgroup$ – bof Nov 21 '14 at 5:16
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    $\begingroup$ The line you added in your latest edit is an argument for the truth of the statement $3-\forall y\forall z\exists x P(x,y,z)$. You know that the order of the quantifiers matters, don't you? $\endgroup$ – bof Nov 21 '14 at 5:19

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