I'm struggling with the proof of the above proposition. I could prove that indeed every natural number is either $0$ or a successor, but I'm struggling with uniqueness. In the lecture notes it says:
To see uniqueness, suppose $n = l^{+} = m^{+}$. Then $ l \cup \{l\} = m \cup \{m\}$. Then $l \leq m$ and $m \leq l$, so that $l =m$.
I don't really see how we get $l \leq m$ and $m \leq l$ from the previous statement. Could someone help me out? It seems so easy...
NB: We defined $ R_{\leq} = \{ (n,m) \in \mathbb{N} \times \mathbb{N} : (n \in m) \vee (n=m)\}$, and we write $n\leq m$ for $(n,m) \in R_{\leq}$.
If $\ell \cup \{\ell\} = m \cup \{m\}$, then $\ell \in \{ \ell \} \subseteq \ell \cup \{\ell\} = m \cup \{m\}$ so either $\ell \in m$ or $\ell \in \{m\}$. That is either $\ell \in m$ or $\ell=m$. Therefore, by your definition, $\ell \leq m$.