(Addition and ordering compatibility) You have \(n≤ m\) if and only if \(n+k≤ m+k\).
(Multiplication and ordering compatibility) When \(k≠ 0\) you have \(n≤ m\) if and only if \(n× k≤ m× k\).
In particular (remembering [28M]) the map \(n\mapsto n× h\) is strictly increasing (and hence injective).
We will use some properties left for exercise.
If \(n≤ m\), by definition \(m=n+h\), then \(n+k≤ m+k\) because \(m+k=n+h+k\) (note that we are using associativity). If \(n+k≤ m+k\) let then \(j\) the only natural number such that \(n+k+j= m+k\) but then \(n+j= m\) by cancellation [27V].
If \(n≤ m\) then \(m=n+h\) therefore \(m× k=n× k+h× k\) so \(n× k≤ m× k\). Vice versa let \(k≠ 0\) and \(n× k≤ m× k\) i.e. \(n× k + j= m× k\): divide \(j\) by \(k\) using the division [28J], we write \(j=q× k + r\) therefore for associativity \((n+q)× k + r= m× k\) but for the uniqueness of the division \(r=0\); eventually collecting \((n+q)× k = m× k\) and using [28M] we conclude that \((n+q) = m\).