# $\forall x (\forall y \exists z (y = xz) \implies (x \neq 0))$ is True but $\forall x \forall y \exists z ((y = xz) \implies (x \neq 0))$ is False

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.

• I think you might have transcribed it incorrectly, since there are identical as written... – Don Thousand Jan 20 at 18:25
• @DonThousand I have added a screenshot of the prof's course notes to show how it is originally written. – IntegrateThis Jan 20 at 18:26
• 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. – 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.
• 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. – player3236 Jan 20 at 18:30
• 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 $x=0$. – player3236 Jan 20 at 18:51