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

logic

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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Best Answer

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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