Proof Critique: For any $b\in \mathbf{N}$, $b++$ is unique.

Question

As I was trying to answer this question (using only the peano axioms), I came up with the following "proof". On a second look, I noticed a flaw. But I'm struggling to articulate why or what the flaw is. Could somebody take a look, and let me know if I'm on the right track? Thanks!

$\textbf{Incorrect Proof}$: Take any arbitrary $b \in \mathbf{N}$. Suppose for the sake of contradiction that there is some $a,a' \in \mathbf{N}$ such that $b++=a$ and $b++ = a'$. Then we can write $a = b++ = a'$ yielding $a=a'$ as required.

$\textbf{Self-critique}$: I am assuming that $b++$ doesn't have two different successors. What if we're working with some number system where it does? The right way to show this is induction on $b$ that uses the axiom "if $n\neq m$, then $n++ \neq m++$."

Answer 1

You write

Suppose for the sake of contradiction that there is some $a,a' \in \mathbf{N}$ such that $b++=a$ and $b++ = a'$. Then we can write $a = b++ = a'$ yielding $a=a'$ as required.

We cannot conclude this, as we are assuming that $b\!+\!\!+$ is not unique.