This question is about the proof of Theorem 3.13 b) in Goldreis' "Classic Set Theory":
For all natural numbers $n,m,a$, if $a>0$ and $n>m$ then $a\cdot n>a\cdot m \quad(1.)$
It has previously been established that if $m<n$ then $m+a<n+a \quad (2.)$.
$n^+$ denotes the "successor" $S(n)$ of any natural number $n$ (n+1 if you will).
Goldrei writes that for all $n \leq m$ the statement $(1.)$ holds vacuously, so "...that the smallest $n$ for which there are anything significant to prove, namely because $m<n$, is $m^+$.". He proceeds with proving $(1.)$ when $n=m^+$.
This is where my first question comes from, because I don't think proving this "base case" is necessary, precisely because $(1.)$ holds vacuously for all $n\leq m$. Even if these cases are true vacuously they are still true, right? So we could just use $n=0$ as the base case.
Then he aims to prove $(1.)$ by induction with the induction hypothesis that $(1.)$ holds for $n > m$ in the following way: $a\cdot n^+ = (a\cdot n) + a > (a\cdot n) + 0\quad(2.) = a\cdot n > a\cdot m$ (by induction hypothesis)
My issue here is that the last step using the induction hypothesis assumes that $n > m$ but the only thing we know from $n^+ > m$ is that $n\geq m$. Of course this is not a big problem since it is sufficient that $a\cdot n = a \cdot m$ for $a\cdot n^+ > a \cdot n = a \cdot m$
So the big question for me is was the base case with $n=m^+$ even necessary?