Theorem ablcomd

Description: An abelian group operation is commutative, deduction version. (Contributed by Thierry Arnoux, 15-Feb-2026)

	Ref	Expression
Hypotheses	ablcomd.1	$$⊢B={\mathrm{Base}}_G$$
	ablcomd.2	$$⊢\underset{\dot{}}{+}={+}_G$$
	ablcomd.3	$$⊢\phi \to G\in \mathrm{Abel}$$
	ablcomd.4	$$⊢\phi \to X\in B$$
	ablcomd.5	$$⊢\phi \to Y\in B$$
Assertion	ablcomd	$$⊢\phi \to X\underset{\dot{}}{+}Y=Y\underset{\dot{}}{+}X$$

Proof

Step	Hyp	Ref	Expression
1		ablcomd.1	$$⊢B={\mathrm{Base}}_G$$
2		ablcomd.2	$$⊢\underset{\dot{}}{+}={+}_G$$
3		ablcomd.3	$$⊢\phi \to G\in \mathrm{Abel}$$
4		ablcomd.4	$$⊢\phi \to X\in B$$
5		ablcomd.5	$$⊢\phi \to Y\in B$$
6	1 2	ablcom	$$⊢\left(G\in \mathrm{Abel}\wedge X\in B\wedge Y\in B\right)\to X\underset{\dot{}}{+}Y=Y\underset{\dot{}}{+}X$$
7	3 4 5 6	syl3anc	$$⊢\phi \to X\underset{\dot{}}{+}Y=Y\underset{\dot{}}{+}X$$