Use the principle of mathematical induction to show that$ \forall n \in N: \forall m \in N: n<m \Rightarrow \exists r \in N: n+r=m$.
My attempt:
Let $ B=\lbrace n \in N \mid \forall m \in N: n<m \Rightarrow \exists r \in N: n+r=m \rbrace $.
We will prove that B is inductive the set.
a) $ 1 \in B$ since if $n=1, 1<m$ and this implies that $m\neq 1$ and since $ m \in N $, there exists $r \in N$ such that $1+r =m$.
b) Now we will prove that for all $n \in R, n \in B \Rightarrow n+1 \in B$. That is, we will prove that
$ \forall m \in N: n+1<m \Rightarrow \exists r \in N: (n+1)+r=m$. Here I got stuck.