Proposition 1 (« addition of inequalities »)

— In an ordered monoid M, let $(x_i)$ and $(y_i)$ ($1 \leq i \leq n$) be two sequences of n elements such that $x_i \leq y_i$ for all $i$; then one has

If moreover all the elements $x_i, y_i$ are cancellable (I, p. 15, Def. 5) (in particular if M is a group), and if $x_i < y_i$ for some $i$, then

The general case reduces by induction to the case $n = 2$, using the fact that a sum of cancellable elements is cancellable for the second assertion (I, p. 15, Prop. 2). The first assertion follows from the relations

which are consequences of the hypotheses and of (OM). This being the case, the relation

would imply

whence $x_2 = y_2$ and $x_1 = y_1$ if $x$, and $y_2$ are cancellable, which proves the second assertion.