hvsubdistr1

Description: Scalar multiplication distributive law for subtraction. (Contributed by NM, 19-May-2005) (New usage is discouraged.)

		Ref	Expression
	Assertion	hvsubdistr1	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℋ ∧ 𝐶 ∈ ℋ ) → ( 𝐴 ·_ℎ ( 𝐵 −_ℎ 𝐶 ) ) = ( ( 𝐴 ·_ℎ 𝐵 ) −_ℎ ( 𝐴 ·_ℎ 𝐶 ) ) )

Proof

Step	Hyp	Ref	Expression
1		neg1cn	⊢ - 1 ∈ ℂ
2		hvmulcl	⊢ ( ( - 1 ∈ ℂ ∧ 𝐶 ∈ ℋ ) → ( - 1 ·_ℎ 𝐶 ) ∈ ℋ )
3	1 2	mpan	⊢ ( 𝐶 ∈ ℋ → ( - 1 ·_ℎ 𝐶 ) ∈ ℋ )
4		ax-hvdistr1	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℋ ∧ ( - 1 ·_ℎ 𝐶 ) ∈ ℋ ) → ( 𝐴 ·_ℎ ( 𝐵 +_ℎ ( - 1 ·_ℎ 𝐶 ) ) ) = ( ( 𝐴 ·_ℎ 𝐵 ) +_ℎ ( 𝐴 ·_ℎ ( - 1 ·_ℎ 𝐶 ) ) ) )
5	3 4	syl3an3	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℋ ∧ 𝐶 ∈ ℋ ) → ( 𝐴 ·_ℎ ( 𝐵 +_ℎ ( - 1 ·_ℎ 𝐶 ) ) ) = ( ( 𝐴 ·_ℎ 𝐵 ) +_ℎ ( 𝐴 ·_ℎ ( - 1 ·_ℎ 𝐶 ) ) ) )
6		hvmulcom	⊢ ( ( 𝐴 ∈ ℂ ∧ - 1 ∈ ℂ ∧ 𝐶 ∈ ℋ ) → ( 𝐴 ·_ℎ ( - 1 ·_ℎ 𝐶 ) ) = ( - 1 ·_ℎ ( 𝐴 ·_ℎ 𝐶 ) ) )
7	1 6	mp3an2	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐶 ∈ ℋ ) → ( 𝐴 ·_ℎ ( - 1 ·_ℎ 𝐶 ) ) = ( - 1 ·_ℎ ( 𝐴 ·_ℎ 𝐶 ) ) )
8	7	oveq2d	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐶 ∈ ℋ ) → ( ( 𝐴 ·_ℎ 𝐵 ) +_ℎ ( 𝐴 ·_ℎ ( - 1 ·_ℎ 𝐶 ) ) ) = ( ( 𝐴 ·_ℎ 𝐵 ) +_ℎ ( - 1 ·_ℎ ( 𝐴 ·_ℎ 𝐶 ) ) ) )
9	8	3adant2	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℋ ∧ 𝐶 ∈ ℋ ) → ( ( 𝐴 ·_ℎ 𝐵 ) +_ℎ ( 𝐴 ·_ℎ ( - 1 ·_ℎ 𝐶 ) ) ) = ( ( 𝐴 ·_ℎ 𝐵 ) +_ℎ ( - 1 ·_ℎ ( 𝐴 ·_ℎ 𝐶 ) ) ) )
10	5 9	eqtrd	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℋ ∧ 𝐶 ∈ ℋ ) → ( 𝐴 ·_ℎ ( 𝐵 +_ℎ ( - 1 ·_ℎ 𝐶 ) ) ) = ( ( 𝐴 ·_ℎ 𝐵 ) +_ℎ ( - 1 ·_ℎ ( 𝐴 ·_ℎ 𝐶 ) ) ) )
11		hvsubval	⊢ ( ( 𝐵 ∈ ℋ ∧ 𝐶 ∈ ℋ ) → ( 𝐵 −_ℎ 𝐶 ) = ( 𝐵 +_ℎ ( - 1 ·_ℎ 𝐶 ) ) )
12	11	3adant1	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℋ ∧ 𝐶 ∈ ℋ ) → ( 𝐵 −_ℎ 𝐶 ) = ( 𝐵 +_ℎ ( - 1 ·_ℎ 𝐶 ) ) )
13	12	oveq2d	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℋ ∧ 𝐶 ∈ ℋ ) → ( 𝐴 ·_ℎ ( 𝐵 −_ℎ 𝐶 ) ) = ( 𝐴 ·_ℎ ( 𝐵 +_ℎ ( - 1 ·_ℎ 𝐶 ) ) ) )
14		hvmulcl	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℋ ) → ( 𝐴 ·_ℎ 𝐵 ) ∈ ℋ )
15	14	3adant3	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℋ ∧ 𝐶 ∈ ℋ ) → ( 𝐴 ·_ℎ 𝐵 ) ∈ ℋ )
16		hvmulcl	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐶 ∈ ℋ ) → ( 𝐴 ·_ℎ 𝐶 ) ∈ ℋ )
17	16	3adant2	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℋ ∧ 𝐶 ∈ ℋ ) → ( 𝐴 ·_ℎ 𝐶 ) ∈ ℋ )
18		hvsubval	⊢ ( ( ( 𝐴 ·_ℎ 𝐵 ) ∈ ℋ ∧ ( 𝐴 ·_ℎ 𝐶 ) ∈ ℋ ) → ( ( 𝐴 ·_ℎ 𝐵 ) −_ℎ ( 𝐴 ·_ℎ 𝐶 ) ) = ( ( 𝐴 ·_ℎ 𝐵 ) +_ℎ ( - 1 ·_ℎ ( 𝐴 ·_ℎ 𝐶 ) ) ) )
19	15 17 18	syl2anc	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℋ ∧ 𝐶 ∈ ℋ ) → ( ( 𝐴 ·_ℎ 𝐵 ) −_ℎ ( 𝐴 ·_ℎ 𝐶 ) ) = ( ( 𝐴 ·_ℎ 𝐵 ) +_ℎ ( - 1 ·_ℎ ( 𝐴 ·_ℎ 𝐶 ) ) ) )
20	10 13 19	3eqtr4d	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℋ ∧ 𝐶 ∈ ℋ ) → ( 𝐴 ·_ℎ ( 𝐵 −_ℎ 𝐶 ) ) = ( ( 𝐴 ·_ℎ 𝐵 ) −_ℎ ( 𝐴 ·_ℎ 𝐶 ) ) )

Metamath Proof Explorer

Theorem hvsubdistr1

Proof