Proving a stronger version of transitivity law

Question

Given $$(x<y):=\lnot(x\ge y).$$ and the transitivity law of $\le$, i.e. $$ \mbox{If } (x\le y) \mbox{ AND } (y\le z),\mbox{ then }( x \le z). $$

I wish to prove the stronger version of transitivity law of $<$, i.e. $$ \mbox{If } (x< y) \mbox{ AND } (y< z),\mbox{ then }( x < z). $$

My attempt: $$ (x< y) \mbox{ AND } (y< z) $$ $$ \lnot(x \ge y) \mbox{ AND } \lnot(y\ge z) $$

$$ \lnot [(x \ge y) \mbox{ OR } (y\ge z)] \tag{*} $$ (Perhaps some arguments are needed here about the OR relation ??) $$ \lnot [(x \ge y) \mbox{ AND } (y\ge z)] $$ Apply the transitivity law of weaker version: $$ \lnot (x \ge z) $$ $$ x<z $$

My Question is: How to argue the intermediate step after the equation (*)?

Answer 1

You can go from $\neg (P \lor Q)$ to $\neg (P \land Q)$ using a proof by contradiction:

Assume $P \land Q$

Then $P$

Hence $P \lor Q$

But this contradicts with the given $\neg (P \lor Q)$

So, $\neg (P \land Q)$