Axiom: 0 is a natural number.
Axiom: For every natural number n, the successor of n, denoted S(n) is a natural number.
Axiom: For all natural numbers m and n, m = n if and only if S(m) = S(n). That is, S is an injection.
Define 1 as S(0)
Define 2 as S(1)
Define addition recursively:
a) 0+n = n
b) If m = S(k), then define m+n as S(k+n). By the third axiom there is at most one k for any m.
Now we are ready for the proof
1+1 = S(0)+1 = S(0+1) = S(1) = 2
