In trying to prove the following inequality $\neg(a\leq b)\Longrightarrow b<a$ i could produce the following indirect proof:
Let $\neg(a\leq b)$.
Let $\neg(b<a)$
But $\neg(b<a)\Longrightarrow a\leq b$. By using the trichotomy law.
But since we have assumed $\neg(a\leq b)$. we have a contradiction
Hence $b<a$
However for a direct proof the following proof was suggested to me
"You start with axiom $(a < b)\ V\ (a = b)\ V\ (a > b).$ One of those three possibilities must be true.
$\neg(a \le b) \implies \neg\{(a < b)\ V\ (a = b)\}.$
$\therefore \neg(a \le b) \implies \{\neg(a < b) \wedge \neg(a = b)\}.$ Two are false.
$\therefore \neg(a \le b) \implies a > b.$ The one remaining is true"
Is this direct proof correct??