Consider two logical statements for real numbers $x,y,z$

  1. $\forall x (\forall y \exists z (y = xz) \implies (x \neq 0))$

  2. $\forall x \forall y \exists z ((y = xz) \implies (x \neq 0))$

In some course notes on logic I am instructed that 1) is true and 2 is false

For 1) I believe there is two cases. Suppose $x=0$, then the statement $\forall y \exists z (y=xz)$ is false for $y=10$, and so we get False implies True, which is True. If $x \neq 0$ then we get True implies True, which is True.

For 2) I'm not sure why I can't simply use the same reasoning, or what the real difference is between 1 and 2)

Any insights appreciated.

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  • $\begingroup$ I think you might have transcribed it incorrectly, since there are identical as written... $\endgroup$ – Don Thousand Jan 20 at 18:25
  • $\begingroup$ @DonThousand I have added a screenshot of the prof's course notes to show how it is originally written. $\endgroup$ – IntegrateThis Jan 20 at 18:26
  • $\begingroup$ Ahh, I see. I would be more careful with parentheses than the prof was, but the proposition was probably $$\forall x(\forall y\exists z,y=xz)\to(x\neq0)$$That makes more sense. $\endgroup$ – Don Thousand Jan 20 at 18:28

For 2, the statement is: for all $x, y$, there exists $z$ such that [$y = xz \implies x \ne 0$ is true].

Consider when $x= y= 0$. We can choose $z=0$, then $y=xz$ is true while $x\ne 0$ is false, and [true implies false] is false. This shows that 2 is false.

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    $\begingroup$ And so this wouldn't work for statement 1 since if we set x to be 0, then the statement for every y there exists a z such that y=xz is false? $\endgroup$ – IntegrateThis Jan 20 at 18:29
  • $\begingroup$ Indeed. By choosing a 'bad' $y$, we can force $\forall y \exists z (y =xz)$ to be false, and get away with [false implies false] being true. $\endgroup$ – player3236 Jan 20 at 18:30
  • $\begingroup$ For statement 1, if I choose x=0, then in the inner statement for every y there exists z if i let y=0, then z can be anything and y=xz is true, but isn't the overall statement false? I'm still a bit confused tbh what the difference is between the two. $\endgroup$ – IntegrateThis Jan 20 at 18:44
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    $\begingroup$ When not explicitly parenthesized, we should consider $\forall y \exists z (y=xz)$ as one entity [molecule]. When $x=0$, $y=0$ will make $\exists z(y=xz)$ true, but not other values of $y$. Hence $\forall y \exists z (y=xz)$ is false for $x=0$. $\endgroup$ – player3236 Jan 20 at 18:51

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