Theorem om2

Description: Two ways to double an ordinal. (Contributed by RP, 3-Jan-2025)

	Ref	Expression
Assertion	om2	$$⊢A\in \mathrm{On}\to A{+}_{𝑜}A=A{\cdot }_{𝑜}{2}_{𝑜}$$

Proof

Step	Hyp	Ref	Expression
1		df-2o	$$⊢2_{𝑜}=\mathrm{suc}{1}_{𝑜}$$
2	1	oveq2i	$$⊢A{\cdot }_{𝑜}{2}_{𝑜}=A{\cdot }_{𝑜}\mathrm{suc}{1}_{𝑜}$$
3		1on	$$⊢1_{𝑜}\in \mathrm{On}$$
4		omsuc	$$⊢\left(A\in \mathrm{On}\wedge 1_{𝑜}\in \mathrm{On}\right)\to A{\cdot }_{𝑜}\mathrm{suc}{1}_{𝑜}=\left(A{\cdot }_{𝑜}{1}_{𝑜}\right){+}_{𝑜}A$$
5	3 4	mpan2	$$⊢A\in \mathrm{On}\to A{\cdot }_{𝑜}\mathrm{suc}{1}_{𝑜}=\left(A{\cdot }_{𝑜}{1}_{𝑜}\right){+}_{𝑜}A$$
6		om1	$$⊢A\in \mathrm{On}\to A{\cdot }_{𝑜}{1}_{𝑜}=A$$
7	6	oveq1d	$$⊢A\in \mathrm{On}\to \left(A{\cdot }_{𝑜}{1}_{𝑜}\right){+}_{𝑜}A=A{+}_{𝑜}A$$
8	5 7	eqtrd	$$⊢A\in \mathrm{On}\to A{\cdot }_{𝑜}\mathrm{suc}{1}_{𝑜}=A{+}_{𝑜}A$$
9	2 8	eqtr2id	$$⊢A\in \mathrm{On}\to A{+}_{𝑜}A=A{\cdot }_{𝑜}{2}_{𝑜}$$