Proof for the cancellation law of addition for natural numbers I'm trying to prove this by induction but I'm afraid I didn't quite make good use of the induction hypothesis. Here's my attempt where I assume $0\in \mathbb{N}$ (that's what I mean by $\mathbb{N_{0}}$):
Proof. We are trying to show that $\forall m,n,p\in \mathbb{N_{0}}:m+p=n+p\implies m=n$.
Base case: Fix $m,n\in \mathbb{N_{0}}$. If $p=0$ and $m+p=n+p$. Then,
\begin{align*}m&:=m+0=m+p=n+p=n+0:=n\end{align*}
So, $p=0\implies (m+p=n+p\implies m=n)$.
Induction step: Suppose $m+q=n+q\implies m=n$ for an arbitrary but fixed $q\in \mathbb{N_{0}}$. We may further assume\begin{align*}m+(q+1)=n+(q+1)\tag{$1$}\end{align*}
From $(1)$ we know that $(m+q)+1=(n+q)+1$. Now, our induction hypothesis abstractly tells us (my interpretation): $a+b=c+b\implies a=c$. Since $1=1$ and $a+1=c+1$ where $a=m+q$ and $c=n+q$ then $a=c$,that is $m+q=n+q$ but, again, from the induction hypothesis we can conclude now $m=n$ what finishes our induction on $p$.
 A: You are trying to use the $p=1$ case of the claim, which you can't at that point. Instead, the needed $$x+1=y+1\implies x=y$$ is immediately one of the Peano axioms (perhaps written in terms of the successor function intead of "$+1$")
A: The explanation in the inductive step of your proof is a little weird ('abstractly tells us'?) ...
Also, while the cancellation theorem indeed says that if $a + c = b + c$, then $a = b$, you are in the middle of proving exactly that. So, there is a real danger of circular reasoning here.
But why even bring up $a$, $b$, and $c$?
Like you say, you can go from
$m+(q+1)=n+(q+1)$
to
$(m+q)+1=(n+q)+1$
to
$m+q=n+q$
to
$m = n$
Of course, for the first step you need to assume Association (can you assume that? Have you proven that? It can be proven from the Peano axioms using induction) and for the second step you need to already have shown that $m+1 = n + 1 \implies m +n$ ... or simply regard that as an 'obvious' mathematical truth (as Hagen points out, this is actually an elementary Peano axiom)
