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

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

Rearranging, I got


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.

  • 2
    $\begingroup$ What do you think the difference is between $\exists x,$ $\exists z,$ and $\forall y?$ Do you think the order matters? $\endgroup$ Oct 28, 2021 at 4:22
  • $\begingroup$ 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. $\endgroup$
    – FarmerZee
    Oct 28, 2021 at 4:36
  • 1
    $\begingroup$ 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 $y.$ $\endgroup$ Oct 28, 2021 at 4:52

1 Answer 1


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...


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