Consider the predicate $P(x, y, z) = “xyz = 1”$, for $x, y, z \in\mathbb{R}, > x, y, z > 0$. What are the truth values of these statements? Justify your answer

  1. $\forall x; \forall y; \exists z; P(x; y; z)$
  2. $\exists x, \forall y,\forall z, P(x, y, z)$

Given Solution:

  1. $\forall x; \forall y; \exists z; P(x; y; z)$ is true.

  2. $\exists x; \forall y; \forall z, P(x, y, z)$ is false: one cannot find a single x such that $\ xyz = 1$, no matter what are $y$ and $z$. Assume such $x$ exists, then for any $y, z = 0$ and $y_1, z_1, xy_1z_1 = 1$ and $x(y_1+1)z_1 = 1$ result in valid solution, hence contradiction.

Could someone please explain to me this line "Assume such $x$ exists, then for any $y , z = 0$ and $y_1, z_1, xy_1z_1 = 1$ and $x(y_1+1)z_1 = 1$ result in valid solution". I have read the notes again and again and reread this solution and it makes no sense to me... Also how can 1. be true but 2. be false when both has the same quantifiers???

  • $\begingroup$ Sorry.. I have no idea how to name the title... how do you name a statement like $\forall x; \forall y; \exists z; P(x; y; z)$ ? Is it a predicate statement or something? $\endgroup$
    – terahertz
    Sep 17, 2019 at 14:48
  • $\begingroup$ "Every man has a woman who loves him," is different from "There is some man who is loved by all women." $\endgroup$
    – saulspatz
    Sep 17, 2019 at 15:11

2 Answers 2


In order to evaluate the truth value of $∃x \ ∀y \ ∀z \ P(x,y,z)$, it is useful to "read" it :

"there is a positive real $x$ such that, for every (positive reals) $y$ and $z$ it is true that $xyz=1$.

The reasoning is : assume that $x > 0$ exists such that .... From $xyz=1$ we get $yz= \dfrac 1 x$ (we can do it because we have $x > 0$) for every $y,z > 0$.

Let $y=z=1$ and we get $1=\dfrac 1 x$. Let $y=1$ and $z=2$ and we get $2 = \dfrac 1 x$. From the first equation we have $x=1$ and form the second we have $x= \dfrac 1 2$.

Conclusion : if we assume that $x$ such that ... exists, we have that $x=1= \dfrac 1 2$.

In general, the order of quantifier matters.

Consider for example $\mathbb N$ and the difference between : $∀n∃m (n < m)$ and $∃m∀n (n < m)$


Statement 1 is true and statement 2 is false, because the order of quantifiers matter.

As the existence quantifier in statement 1 comes after the "for all" quantifers, it claims existence after assignments have been made to the variables named in those.

Contrary to that the existence quantifer in statement 2 comes first, and therefore claims existence of that number no matter what follows.


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