$\displaystyle a \not\equiv b \ \mbox{(mod m)}$, then $\displaystyle b \not\equiv a \ \mbox{(mod m)}$.
Is this all there is to this one:
$\displaystyle m\nmid (a-b)\rightarrow mx\neq (-1)(b-a)\rightarrow m(-x)\neq (b-a)\rightarrow m\nmid (b-a)$
$\displaystyle a \not\equiv b \ \mbox{(mod m)}$, then $\displaystyle b \not\equiv a \ \mbox{(mod m)}$.
Is this all there is to this one:
$\displaystyle m\nmid (a-b)\rightarrow mx\neq (-1)(b-a)\rightarrow m(-x)\neq (b-a)\rightarrow m\nmid (b-a)$
Oh, ok. Different authors define congruence differently, I guess.
If you rephrase this, the problem is: If $\displaystyle m $ does not divide $\displaystyle a-b $, then also $\displaystyle m $ does not divide $\displaystyle b-a $.
Now, $\displaystyle a-b $ and $\displaystyle b-a $ have the same divisors. So if $\displaystyle m $ is not a divisor of $\displaystyle a-b $, it can't be a divisor of $\displaystyle b-a $.
This is essentially the same explantion "chiph588@" gave.