Theorem nmulcomd

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

	Ref	Expression
Hypotheses	nmul.1	⊢ ( 𝜑 → 𝐴 ∈ On )
	nmul.2	⊢ ( 𝜑 → 𝐵 ∈ On )
Assertion	nmulcomd	⊢ ( 𝜑 → ( 𝐴 ·no 𝐵 ) = ( 𝐵 ·no 𝐴 ) )

Proof

Step	Hyp	Ref	Expression
1		nmul.1	⊢ ( 𝜑 → 𝐴 ∈ On )
2		nmul.2	⊢ ( 𝜑 → 𝐵 ∈ On )
3		nmulcom	⊢ ( ( 𝐴 ∈ On ∧ 𝐵 ∈ On ) → ( 𝐴 ·no 𝐵 ) = ( 𝐵 ·no 𝐴 ) )
4	1 2 3	syl2anc	⊢ ( 𝜑 → ( 𝐴 ·no 𝐵 ) = ( 𝐵 ·no 𝐴 ) )