Theorem nadddird

Description: Natural multiplication distributes over natural addition. Deduction form. (Contributed by Scott Fenton, 3-Aug-2026)

	Ref	Expression
Hypotheses	nadddid.1	⊢ ( 𝜑 → 𝐴 ∈ On )
	nadddid.2	⊢ ( 𝜑 → 𝐵 ∈ On )
	nadddid.3	⊢ ( 𝜑 → 𝐶 ∈ On )
Assertion	nadddird	⊢ ( 𝜑 → ( ( 𝐴 +no 𝐵 ) ·no 𝐶 ) = ( ( 𝐴 ·no 𝐶 ) +no ( 𝐵 ·no 𝐶 ) ) )

Proof

Step	Hyp	Ref	Expression
1		nadddid.1	⊢ ( 𝜑 → 𝐴 ∈ On )
2		nadddid.2	⊢ ( 𝜑 → 𝐵 ∈ On )
3		nadddid.3	⊢ ( 𝜑 → 𝐶 ∈ On )
4	3 1 2	nadddid	⊢ ( 𝜑 → ( 𝐶 ·no ( 𝐴 +no 𝐵 ) ) = ( ( 𝐶 ·no 𝐴 ) +no ( 𝐶 ·no 𝐵 ) ) )
5	1 2	naddcld	⊢ ( 𝜑 → ( 𝐴 +no 𝐵 ) ∈ On )
6	5 3	nmulcomd	⊢ ( 𝜑 → ( ( 𝐴 +no 𝐵 ) ·no 𝐶 ) = ( 𝐶 ·no ( 𝐴 +no 𝐵 ) ) )
7	1 3	nmulcomd	⊢ ( 𝜑 → ( 𝐴 ·no 𝐶 ) = ( 𝐶 ·no 𝐴 ) )
8	2 3	nmulcomd	⊢ ( 𝜑 → ( 𝐵 ·no 𝐶 ) = ( 𝐶 ·no 𝐵 ) )
9	7 8	oveq12d	⊢ ( 𝜑 → ( ( 𝐴 ·no 𝐶 ) +no ( 𝐵 ·no 𝐶 ) ) = ( ( 𝐶 ·no 𝐴 ) +no ( 𝐶 ·no 𝐵 ) ) )
10	4 6 9	3eqtr4d	⊢ ( 𝜑 → ( ( 𝐴 +no 𝐵 ) ·no 𝐶 ) = ( ( 𝐴 ·no 𝐶 ) +no ( 𝐵 ·no 𝐶 ) ) )