Each ordered semigroup is cancellative: reference?

Question

It is easy enough to show that $a+b < a+c\Rightarrow b < c$ holds in totally ordered semigroups. Indeed this must be very well known. Can anyone please provide a reference for this result? A textbook will do!

Proof: Suppose $c \ge b$. Then $a+c \ge a+b$ as + respects the order. We are done by contraposition.

Answer 1

EDIT. This is an answer to the initial question of the OP, which has been changed later on.

This is not true. Actually, the fact that $+$ respects order just means that $a \leqslant b$ implies $a+c \leqslant b+c$.

For a counterexample to your claim, take $M = \{0, 1\}$ under the usual multiplication and the usual order $0 < 1$.

EDIT. The answer to the new question is "yes", but the proof should be

Suppose that $c \leqslant b$. Then $a+c \leqslant a+b$ as + respects the order. The claim follows by contraposition.