Preserving addition of one in inductive step of modulo I am trying to prove the following: $$\forall m\ n : \mathbb{N}, (m + 1)\ mod\ n \neq 0 \rightarrow m\ mod\ n + 1 = (m + 1)\ mod\ n$$
So far I've done the following.
Assume arbitrary $m$ and $n$. Now assume $(m + 1)\ mod\ n \neq 0$. Proceed by cases formed by the trichotomy $m < n \vee m = n \vee m > n$. The first two disjuncts are trivial and follow from basic facts about modulo.
Now let us assume $m > n$. By induction on $m$, we have inductive base where $m = 0$. Then $0 > n$ which is a contradiction. In the inductive step, let us show $(m + 1)\ mod\ n + 1 = (m + 2)\ mod\ n$ assuming $(m + 1)\ mod\ n \neq 0 \rightarrow n < m \rightarrow m\ mod\ n + 1 = (m + 1)\ mod\ n$.
Sadly I am not sure how to continue here. I know $m + 1 > n$ which is not particularly helpful with the inductive hypothesis. Any ideas?

Assume $m=qn+r$ with $0\le r<n$. Then $m+1=qn+(r+1)$. Either $r+1<n$ and we are done, or $r+1=n$ and that means $m+1\equiv 0\pmod n$.
