In the next section we outline a $\text{ZF}$ construction of the natural numbers under addition using a 'duality' argument (the choice of the word duality is subjective and has no formal meaning).
Although I am not providing all the details, I am looking for a foundational proof verification of the step-by-step logic exposition.
If there is an error please respond by explaining it in an answer.
An expansion of the proof detail can also be provided in an answer.
Also, if someone wants me to expand on the sketched theory please make a comment - I will then attempt to provide the argument details in an answer.
Given:
$\;$ The set $\omega$ of finite ordinals.
$\;$ The successor mapping $S: \omega \to \omega$ given by
$\quad S: \alpha \mapsto \alpha \cup \{\alpha\}$
satisfies
$\quad S \text{ is injective}$
$\quad \text{The image of } S \text{ is equal to } \omega \setminus \{\emptyset\}$
$\quad \text{If } \emptyset \in \nu \subset \omega \text{ and } S(\nu) \subset \nu \text{ then } \nu = \omega$
(c.f. the Peano axioms)
Let $\mathcal A$ denote the monoid of all injective endomorphisms on the set $\omega$.
Let
$\quad \mathcal G = \{T \in \mathcal A \mid T \circ S = S \circ T\}$
Proposition 1: If $T, U \in \mathcal G$ then $T \circ U \in \mathcal G$ and $\mathcal G$ is a monoid.
Using recursion we define a function $\tau: \omega \to \mathcal G$ as follows:
$\quad \tau(\emptyset) = id_{\omega}$
$\quad \tau(S(n)) = S \circ \tau(n)$
The principle of induction is used to prove many of the following statements but we'll usually gloss over it, perhaps just pointing the way to nailing down the induction step.
Proposition 2: For every $n \in \omega$ there exists one and only one $T \in \mathcal G$ such that $T(\emptyset) = n$.
Proof sketch:
Besides induction the proof requires building the endomophism using recursion. $\quad \blacksquare$
We agree to denote $\tau(n)$ by $S^n$; we have
Theorem 3: The function $\tau$ is a bijective mapping and $S^n$ is defined by
$\quad \emptyset \mapsto n$
Proof sketch:
Write as true $S^n(\emptyset) = n$ and $S^{S(n)} = S \circ S^n$. $\quad \blacksquare$
Theorem 4: Composition is a commutative binary operation for the monoid $\mathcal G$.
Proof sketch:
Write as true $S^n \circ S^m = S^m \circ S^n$. Then
$\quad \displaystyle S^n \circ S^{S(m)} = S^n \circ (S \circ S^m) = (S^n \circ S) \circ S^m = (S \circ S^n) \circ S^m = S \circ (S^n \circ S^m) = S^{S(m)} \circ S^n $
$\blacksquare$
Theorem 5: The cancellation property holds true for the monoid $\mathcal G$.
Proof:
Suppose $S^k \circ S^m = S^k \circ S^n$. Since the injective mapping $S^k$ has a left inverse we conclude that $S^m = S^n$.
$\blacksquare$
Theorem 6: Let $S^m, S^n \in \mathcal G$ with $m \ne n$. Then $m \subsetneq n \text{ XOR } n \subsetneq m$.
Moreover, if say $m \subsetneq n$ then there exists $k \in N$ such $S^m \circ S^k = S^n$.
Proof sketch:
Using recursion we construct the sequence
$\quad \bigr(\; (S^m, S^{\emptyset}),(S^m, S^{\{\emptyset\}}),\dots,(S^m, S^k),\dots \;\bigr)$
and thereby equate (using induction)
$\quad \bigr(\; S^m \circ S^{\emptyset}, S^m \circ S^{\{\emptyset\}}, \dots,S^m \circ S^k, \dots \;\bigr)$
with
$\quad \bigr(\; S^m, S^{S(m)}, \dots, S^n,\dots \;\bigr)$
which completes the proof. $\quad \blacksquare$
We have constructed the monoid $(\mathcal G,\circ)$ and derived the properties making it isomorphic to the natural numbers under addition. We can also bijectively bring composition of functions in $\mathcal G$ to the set $\omega$ calling it addition; for $m,n \in \omega$,
$\quad \displaystyle m + n = \tau^{-1}(S^m \circ S^n) = S^m(S^n(\emptyset))$
so that we can write
$\quad \displaystyle S^m \circ S^n = S^{m+n}$
My work and motivation
I made another attempt at this before (see this), but wanted to try again. I also wanted to write this up as a response to Andrés E. Caicedo answer, suggesting that there is a fourth way of defining addition.
