Proof verification for $a > b \implies -a < -b$ I am aware that this question asks  for the verification of a proof of (almost) the same problem but my proof is different and in my opinion, a bit simpler and more intuitive. Here's how it goes :
Let us assume that $a > b$.

We can write this inequality in the form of an equation as follows :
$$a = b + x \text{, where } x > 0$$
On multiplying the LHS and RHS by $-1$, we obtain :
$$-a = -(b+x) = -b-x \implies -b = -a+x$$
We have already mentioned that $x > 0$. From this, we can say that $-b$ is obtained when we add a positive number ($x$) to $-a$. Hence, $-b > -a \implies -a < -b$
Thanks!

If you are allowed to use $a>b\iff a+c>b+c$, then the theorem can be reduced to
$$a>0\iff -a<0,$$ which seems easier.

Update:
Using the above lemma, you could as well write the whole proof as
"WLOG $b=0$ (translate as required) and $a>0\iff -a<0$".
