Why is $\forall x, \forall y, \exists z$ true but $\exists x, \forall y, \forall z$ is False??

logicpredicate-logicquantifiers

Question

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???

Best Answer

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 $∃m∀n (n < m)$

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