Theorem naddcomd

Description: Natural addition commutes. Deduction form. (Contributed by Scott Fenton, 30-Jul-2026)

	Ref	Expression
Hypotheses	nadd.1	|- ( ph -> A e. On )
	nadd.2	|- ( ph -> B e. On )
Assertion	naddcomd	|- ( ph -> ( A +no B ) = ( B +no A ) )

Proof

Step	Hyp	Ref	Expression
1		nadd.1	|- ( ph -> A e. On )
2		nadd.2	|- ( ph -> B e. On )
3		naddcom	|- ( ( A e. On /\ B e. On ) -> ( A +no B ) = ( B +no A ) )
4	1 2 3	syl2anc	|- ( ph -> ( A +no B ) = ( B +no A ) )