# Why is this predicate logic statement false?

$$\exists x \exists z \forall y(2y-z=4x)$$

Where $$x$$, $$y$$ and $$z$$ are integers.

Rearranging, I got

$$z=2y-4x$$

Since $$y$$ and $$x$$ are integers, $$2y-4x$$ would also be an integer for any given value of $$y$$.

Since $$z$$ is also an integer, shouldn't the predicate logic statement be true as there would always exist an integer $$z$$ which equals the integer $$2y-4x$$?

The explanation on why it was false is here:

False. For every x and z, there is an integer y = 4|x| + |z| + 1, and thus 2y = 8|x| + 2|z| + 2 > 4x + z.

However I don't understand how that shows that it must be false.

• What do you think the difference is between $\exists x,$ $\exists z,$ and $\forall y?$ Do you think the order matters? Oct 28, 2021 at 4:22
• The way that I understood the question was that you either had to prove or disprove that you could alter the numbers x and z so that the expression is true for any value of y. In that case, I don't think that the order matters. Oct 28, 2021 at 4:36
• Then you are wrong. It means “there exists x,z such that (for all y, …).” Order matters. If it started $\forall y\exists x\exists z\dots,$” then you could pick $x,z$ given a specific $y.$ Oct 28, 2021 at 4:52

As user @Thomas Andrews hinted in the comment, order of mixed type quantifiers matters, see a previous post discussing similar subtle FOL issue. Had your $$\forall y$$ moved to any previous position then your reasoning is fine. But now you're clearly given a concrete counterexample construction you cannot avoid. Hope you can see that now...