The following is a theorem from the book The Real Numbers and Real Analysis by Bloch which I am currently self-studying. I am pretty sure my proof for uniqueness is correct, but I am wondering is there any other way to prove uniqueness for this theorem. The following is the axiom used:
Axiom: There exists a set $\mathbb{N}$ with an element $1 \in\mathbb{N}$ and a function $s: \mathbb{N}\rightarrow\mathbb{N}$ that satisfy the following three properties.
There is no n $\in\mathbb{N}$ such that $s(n)=1$.
The function $s$ is injective.
Let $G\subseteq\mathbb{N}$ be a set. Suppose $1 \in G$, and that if g $\in G$ then $s(g)\in G$. Then $G=\mathbb{N}$.
The following is the theorem I want to prove uniqueness for.
Theorem: There is a unique binary operation $\circ:\mathbb{N}\times\mathbb{N}\rightarrow\mathbb{N}$ that satisfies the following two properties for all $n,m\in\mathbb{N}$.
$n\circ 1=n$
$n\circ s(m)=n\circ m +n$
Proof: Suppose there are two binary operations, $\circ$ and $\bullet$, that follow the two properties. Let $G=\{x\in\mathbb{N}:\forall n\in\mathbb{N} (n\circ x=n\bullet x)\}$. It is easy to see $G\subseteq\mathbb{N}$. When $x=1$, $n\circ 1=n=n\bullet 1$ by part 1 of the theorem, so $1\in G$. Suppose $p\in G$, then $n\circ s(p)=n\circ p +n=n\bullet p +n=n\bullet s(p)$ using the 2nd part of the theorem and the induction hypothesis that $n\circ p=n\bullet p$. Since $n\circ s(p)=n\bullet s(p)$, $s(p)\in\mathbb{N}$ Thus, by part 3 of the axiom $G=\mathbb{N}$ which means the two binary operations are the same.
I saw here that uniqueness for a different theorem was proven using the same process (assuming the object is not unique) but also with a different, less redundant process (using inverses). I am wondering if there is a known second way to prove uniqueness for the theorem in this question.
