I'm trying to prove the following theorem:
"Let $~f: \mathbb{N} \times \mathbb{N} \to \mathbb{N}~$ be a function, and let $~c~$ be a natural number.
Show that there exists a unique function $~a: \mathbb{N} \to \mathbb{N}~$ such that $~a(0)=c~$ and $~a(n+1)=f(n,a(n)) ~\forall n \in \mathbb{N}$".
To prove the theorem is it sufficient to show by induction that for every natural number $~N \in \mathbb{N}$ there exists a unique function $~a_N:\{n \in \mathbb{N}: n \leq N\} \to \mathbb{N}~$ such that $~a_N(0)=c~$ and $~a_N(n+1)=f(n,a_N(n)) ~\forall n \in \mathbb{N}~$ such that $~n<N$ ?
Suppose there exists another function $b$ such that $b(0) = c$, and let $m$ be the minimum index such that $a(m) \neq b(m)$
This implies that
$$a(m) = f(m, a(m-1)) \neq b(m) = f(m, b(m-1)$$
But by hypothesis $b(m-1) = a(m-1)$, and this implies
$$f(m, a(m-1)) \neq f(m, a(m-1))$$ which is impossible since $f$ is a function.