Is this true? $\forall x,\forall y, \forall z, [(x\leqslant y)\land(y\leqslant z)\Rightarrow(x\leqslant z)]$ I think the following expression is false for the order relation on the real line.
I do know  the transitive relation exists on the real line. I don't see how the expression states this.
$$\forall x,\forall y, \forall z, [(x\leqslant y)\land(y\leqslant z)\Rightarrow(x\leqslant z)]$$
I read this as, firstly  "take any x ,take any y, for all values of x,y and apply the "less then or equal to", but there must be values for which this isn't true?
The first bracketed expression is "and" so then must be false?
The only way I see the expression as correct is as an implication statement. "If  A then B " in which case, the implication forces the correct values for x,y,z, then expression makes sense?
thanks,
ralph
 A: It is in fact an implication. In plain words, it reads:

For all $x$, $y$, and $z$, if $x$ is not greater than $y$ and $y$ is not greater than $z$, then $x$ is not greater than $z$.

If $x$, $y$, and $z$ are real numbers then it is indeed true.
A: There is a logical equivalence of implication: $$P(x)\implies Q(x)\equiv\forall x\in\mathbb D\neg P(x)\lor Q(x)$$ which is, "if $P(x)$, then $Q(x)$" is logically equivalent to "not $P(x)$ or $Q(x)$ for all $x$ in domain".
This logical equivalence can be proven using truth table. Note that 1 means True and 0 means False.




$p$
$q$
$p\to q$
$\neg p$
$\neg p\lor q$




0
0
1
1
1


0
1
1
1
1


1
0
0
0
0


1
1
1
0
1




Indeed, there exists some 3 real numbers such that $p$ is false. However, $\neg p$ does not contradict with your implication at all.
A: Does equivalently rewriting it as $$\forall x\;\forall y\;\forall z\; \Big((x\leqslant y \;\land\; y\leqslant z)\Rightarrow x\leqslant z\Big) $$ help?
Plainly put: for each triple $(x,y,z),\,$ if $x\leq y\leq z,\,$ then $x\leq z.$
