If $S(a) = a \cup \{ a\}$ is the successor function, i want to show that (assuming that the natural numbers are a transitive set, and that each natural number is a transitive set):
if for some $x \in \mathbb{N}$, $S(x) = S(y) \implies x = y ~~~\forall y \in \mathbb{N}$ then $S(S(x)) = S(S(y)) \implies S(x) = S(y)$. (i.e. the successor is unique)
I think the right way to go about this is through finding a contradiction, I assume that $S(S(x)) = S(S(y)) $ and $S(x) \neq S(y)$.
As $x,y \in \mathbb{N}$, then $S(x), S(y) \in \mathbb{N}$ and so we either have that $S(x) < S(y)$ or the other way around.
\begin{align*} S(x) < S(y) &\implies S(x) \in S(y) \\ & \implies S(x) \subsetneq S(y)\\ & \implies S(S(x)) \subsetneq S(S(y)) \\ & \implies S(S(x)) \neq S(S(y)) \end{align*}
The second implication follows by transitivity of the natural numbers and by the assumption that $S(x) \neq S(y)$. This is a contradiction since we started off by assuming that S(S(x)) = S(S(y)). We can do the same in the other direction $S(x) > S(y)$.
I think this is correct, though I have to prove the preliminary result that if $a \subset b$ then $S(a) \subset S(b)$. I'm wondering if this is correct, or if there is an easier way to prove this under this set up
Assume a $\cup$ {a} = b $\cup$ {b} and a /= b.
As a in a $\cup$ {a}, a in b or a in {b}, so a in b.
By transitivity, a subset b. Likewise b subset a.
Whence a = b, a contradiction.