Is this assertion: $¬[∃x\ ∃y\ ∃z \ (x ≤ y ≤ z)]$ equivalent to this: $∀x\ ∀y\ ∀z\ (x>y>z)$?

Question

Is this assertion: $¬[∃x\ ∃y\ ∃z \ (x ≤ y ≤ z)]$ equivalent to this: $∀x\ ∀y\ ∀z\ (x>y>z)$?

Answer 1

No. The negation of $x \le y \le z$ is not $ x > y > z$:

Note first that $x \le y \le z$ is a shorthand for $(x \le y) \land (y \le z)$. So we have to apply de Morgan to the negation and we thus get

$\lnot(x \le y) \lor \lnot(y \le z)$ which is (in a linear order!) $(x > y) \lor (y > z)$, while $x > y > z$ is shorthand for $(x > y) \land (y > z)$.

Answer 2

No it is not .

In the context of all real numbers your $$∀x\ ∀y\ ∀z\ (x>y>z)$$

should have been $$∀x\ ∀y\ ∀z\ (x>y \text { or } y>z)$$

Answer 3

Note that \begin{align} \lnot \big [(\exists x)(\exists y)(\exists z)(x\leq y\leq z)\big] \longleftrightarrow & \lnot \big [(\exists x)(\exists y)\big(\exists z)\big((x\leq y) \mbox{ and } (y\leq z)\big)\big] \\ \longleftrightarrow & (\forall x)(\forall y)(\forall z)\lnot\big((x\leq y) \mbox{ and } (y\leq z)\big) \\ \longleftrightarrow & (\forall x)(\forall y)(\forall z)\big(\lnot(x\leq y) \mbox{ or } \lnot(y\leq z)\big) \end{align}

As Godel observed in the case of an exercise in logic, we do not know whether $\leq$ is a relation of total order. But if we know that $\leq $ is a relation of total order then we can affirm that $\lnot (x\leq y )\longleftrightarrow $y

$$ \lnot \big [(\exists x)(\exists y)(\exists z)(x\leq y\leq z)\big] \longleftrightarrow (\forall x)(\forall y)(\forall z)\big((x> y) \mbox{ or } (y> z)\big) $$

Answer 4

Why not think about what these sentences say in words?

The first: "You can't find three numbers weakly increasing in order".

The second: "Any arrangement of three numbers puts them in strictly decreasing order".

Common sense can help you avoid errors in formal manipulation of symbols.

Answer 5

Correct me if wrong:

The negation should read:

For all $x$, for all $y$, for all $z$, the statement $x \le y \le z $ is not true.

Statement $x \le y \le z$ is equivalent to:

$x \le y$ AND $y \le z$.

Negation of this statement:

$x \gt y$ OR $y \gt z$.

The complete negation reads:

For all $x$, for all $y$, for all $z$,
$x\gt y$ or $y \gt z$.

P.S.As user Goedel and others point out we are dealing with a total order.

Answer 6

The negation of $x\leq y$ is not $x>y$ in all context.

Consider $A:=\{a,b,c\}$ and $\leq:=\{(a,a);(b,b);(c,c);(a,b);(a,c)\}$.

Here you have that $\neg(b\leq c)$ is true but this does not imply that $(c<b)$.