Monotonicity of ordinal $+$ and $\cdot$: strictly increasing and continuous in the right argument, weakly increasing in the left, with left cancellation, and the identities $0+\beta =\beta$ and $1\cdot \beta =\beta$

For ordinals $\mu ,\nu$: $\mu \in \nu$ if and only if ${\mu }^{+}\subseteq \nu$, since $\mu \in \nu$ gives $\mu \subseteq \nu$ by transitivity and $\{\mu \}\subseteq \nu$, while ${\mu }^{+}\subseteq \nu$ gives $\mu \in \nu$; consequently $\mu <\nu ⟺{\mu }^{+}\le \nu$, and $\mu \le \nu$ implies ${\mu }^{+}\le {\nu }^{+}$, because $\mu \le \nu <{\nu }^{+}$ gives $\mu <{\nu }^{+}$.

For a set $A$ of ordinals $\sup A=\bigcup A$ is its least upper bound, so if every member of $A$ is $\le$ some member of $B$ then $\sup A\le \sup B$; for a limit ordinal $\lambda$ one has $\bigcup \lambda =\lambda$, $0\in \lambda$ (because $\mathrm{\varnothing }\subseteq \lambda$ and $\mathrm{\varnothing }\mathrm{\ne }\lambda$), and $\xi \in \lambda \Rightarrow {\xi }^{+}\in \lambda$, so also $1=0^{+}\in \lambda$.

Directly from the clauses: $\beta +0=\beta$; $\beta +1=\beta +0^{+}=(\beta +0{)}^{+}={\beta }^{+}$; $\beta \cdot 0=0$; and $\beta \cdot 1=\beta \cdot 0^{+}=\beta \cdot 0+\beta =0+\beta$.

$0+\beta =\beta$ for every $\beta$, by induction: at $0$ this is $0+0=0$; at ${\delta }^{+}$, $0+{\delta }^{+}=(0+\delta {)}^{+}={\delta }^{+}$; and at a limit $\lambda$, $0+\lambda =\bigcup \{0+\xi :\xi \in \lambda \}=\bigcup \lambda =\lambda$.

$0\cdot \beta =0$ for every $\beta$, by induction: at $0$ this is [L2]; at ${\delta }^{+}$, $0\cdot {\delta }^{+}=0\cdot \delta +0=0+0=0$; and at a limit $\lambda$, $0\cdot \lambda =\bigcup \{0\}=0$.

$1\cdot \beta =\beta$ for every $\beta$, by induction: at $0$ this is [L2]; at ${\delta }^{+}$, $1\cdot {\delta }^{+}=1\cdot \delta +1=\delta +1={\delta }^{+}$ by step 1.3; and at a limit $\lambda$, $1\cdot \lambda =\bigcup \{\xi :\xi \in \lambda \}=\bigcup \lambda =\lambda$.

Claim (b), the inequality: by induction on $\gamma$, for every $\beta \in \gamma$ one has $\alpha +\beta \in \alpha +\gamma$. At $\gamma =0$ there is nothing to prove. At $\gamma ={\delta }^{+}$, $\beta \in {\delta }^{+}$ gives $\beta \le \delta$ by [L3], so $\alpha +\beta \le \alpha +\delta$, using the claim at $\delta$ when $\beta \in \delta$, and $\alpha +\delta \in (\alpha +\delta {)}^{+}=\alpha +{\delta }^{+}$. At $\gamma =\lambda$ a limit, $\beta \in \lambda$ gives ${\beta }^{+}\in \lambda$ by step 1.2, and $\alpha +\beta \in (\alpha +\beta {)}^{+}=\alpha +{\beta }^{+}\subseteq \alpha +\lambda$.

Claim (c), the inequality $\alpha +\gamma \le \beta +\gamma$ for $\alpha \le \beta$: by induction on $\gamma$. At $\gamma =0$ it is $\alpha \le \beta$. At $\gamma ={\delta }^{+}$, the claim at $\delta$ gives $\alpha +\delta \le \beta +\delta$, hence $(\alpha +\delta {)}^{+}\le (\beta +\delta {)}^{+}$ by step 1.1. At $\gamma =\lambda$ a limit, every $\alpha +\xi$ with $\xi \in \lambda$ is $\le \beta +\xi$, so the suprema compare by step 1.2.

$\alpha \cdot 1=\alpha$, since $\alpha \cdot 1=0+\alpha =\alpha$ by step 1.3 and step 2.1; together with step 1.3 and steps 2.1 to 2.3 this proves claim (a).

Left cancellation for $+$: if $\beta \mathrm{\ne }\gamma$ then $\beta \in \gamma$ or $\gamma \in \beta$ by [L4], so $\alpha +\beta \mathrm{\ne }\alpha +\gamma$ by step 2.4 and [L3]; and $\alpha =\alpha +0\le \alpha +\beta$ with equality exactly when $\beta =0$, again by step 2.4. This completes claim (b).

$\beta \le \alpha +\beta$: since $0\le \alpha$, step 2.5 gives $0+\beta \le \alpha +\beta$, and $0+\beta =\beta$ by step 2.1. This completes claim (c).

Claim (d), the inequality: let $\alpha >0$; by induction on $\gamma$, for every $\beta \in \gamma$ one has $\alpha \cdot \beta \in \alpha \cdot \gamma$. At $\gamma =0$ there is nothing to prove. At $\gamma ={\delta }^{+}$, $\beta \le \delta$ gives $\alpha \cdot \beta \le \alpha \cdot \delta$ using the claim at $\delta$, and $\alpha \cdot \delta =\alpha \cdot \delta +0\in \alpha \cdot \delta +\alpha =\alpha \cdot {\delta }^{+}$ by step 2.4 applied to $0\in \alpha$. At $\gamma =\lambda$ a limit, ${\beta }^{+}\in \lambda$ by step 1.2 and $\alpha \cdot \beta \in \alpha \cdot {\beta }^{+}\subseteq \alpha \cdot \lambda$.

Claim (e): let $\alpha \le \beta$; by induction on $\gamma$. At $\gamma =0$ both sides are $0$. At $\gamma ={\delta }^{+}$, the claim at $\delta$ gives $\alpha \cdot \delta \le \beta \cdot \delta$, so $\alpha \cdot {\delta }^{+}=\alpha \cdot \delta +\alpha \le \beta \cdot \delta +\alpha \le \beta \cdot \delta +\beta =\beta \cdot {\delta }^{+}$, the first inequality by step 2.5 and the second by step 2.4. At $\gamma =\lambda$ a limit, the suprema compare by step 1.2.

The rest of claim (d): for $\alpha >0$, $\beta \mathrm{\ne }\gamma$ gives $\alpha \cdot \beta \mathrm{\ne }\alpha \cdot \gamma$ by step 3.4 and [L4], which is cancellation; $\alpha =\alpha \cdot 1\le \alpha \cdot \beta$ for $1\le \beta$ by step 3.4 and step 3.1; and $\alpha \cdot \beta =0$ forces $\alpha =0$ or $\beta =0$, since $\alpha >0$ and $\beta >0$ give $\alpha \cdot \beta \ge \alpha \cdot 1=\alpha >0$, while $\alpha =0$ or $\beta =0$ each give $0$ by step 2.2 and [L2].

Claim (f): the first two identities are [L1] and [L2] with $\sup =\bigcup$. For the refinement, let $D\subseteq \lambda$ be nonempty with $\sup D=\lambda$; then $\{\alpha +\eta :\eta \in D\}\subseteq \{\alpha +\xi :\xi \in \lambda \}$ gives $\le$, and conversely each $\xi \in \lambda =\bigcup D$ lies in some $\eta \in D$, so $\alpha +\xi <\alpha +\eta$ by step 2.4 and the suprema compare by step 1.2; the same argument with step 3.4 in place of step 2.4 gives the multiplicative half when $\alpha >0$.

Claim (g): $\alpha +\lambda \mathrm{\ne }0$, because $1\in \lambda$ by step 1.2 and so ${\alpha }^{+}=\alpha +1\le \alpha +\lambda$ by step 2.4 and step 1.3; and $\alpha +\lambda$ is not a successor, since $\alpha +\lambda ={\mu }^{+}$ would put $\mu \in \bigcup \{\alpha +\xi :\xi \in \lambda \}$, hence $\mu \in \alpha +\xi$ for some $\xi \in \lambda$, whence ${\mu }^{+}\le \alpha +\xi <\alpha +{\xi }^{+}\le \alpha +\lambda ={\mu }^{+}$ by step 1.1 and step 2.4, which [L3] forbids; the same argument with step 3.4 in place of step 2.4, and $\alpha \cdot 1=\alpha >0$ in place of $\alpha +1$, shows $\alpha \cdot \lambda$ is a limit ordinal when $\alpha >0$.

Claims (a) to (g) are established.