hvsubdistr2

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

		Ref	Expression
	Assertion	hvsubdistr2	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℋ ) → ( ( 𝐴 − 𝐵 ) ·_ℎ 𝐶 ) = ( ( 𝐴 ·_ℎ 𝐶 ) −_ℎ ( 𝐵 ·_ℎ 𝐶 ) ) )

Proof

Step	Hyp	Ref	Expression
1		hvmulcl	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐶 ∈ ℋ ) → ( 𝐴 ·_ℎ 𝐶 ) ∈ ℋ )
2	1	3adant2	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℋ ) → ( 𝐴 ·_ℎ 𝐶 ) ∈ ℋ )
3		hvmulcl	⊢ ( ( 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℋ ) → ( 𝐵 ·_ℎ 𝐶 ) ∈ ℋ )
4	3	3adant1	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℋ ) → ( 𝐵 ·_ℎ 𝐶 ) ∈ ℋ )
5		hvsubval	⊢ ( ( ( 𝐴 ·_ℎ 𝐶 ) ∈ ℋ ∧ ( 𝐵 ·_ℎ 𝐶 ) ∈ ℋ ) → ( ( 𝐴 ·_ℎ 𝐶 ) −_ℎ ( 𝐵 ·_ℎ 𝐶 ) ) = ( ( 𝐴 ·_ℎ 𝐶 ) +_ℎ ( - 1 ·_ℎ ( 𝐵 ·_ℎ 𝐶 ) ) ) )
6	2 4 5	syl2anc	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℋ ) → ( ( 𝐴 ·_ℎ 𝐶 ) −_ℎ ( 𝐵 ·_ℎ 𝐶 ) ) = ( ( 𝐴 ·_ℎ 𝐶 ) +_ℎ ( - 1 ·_ℎ ( 𝐵 ·_ℎ 𝐶 ) ) ) )
7		mulm1	⊢ ( 𝐵 ∈ ℂ → ( - 1 · 𝐵 ) = - 𝐵 )
8	7	oveq1d	⊢ ( 𝐵 ∈ ℂ → ( ( - 1 · 𝐵 ) ·_ℎ 𝐶 ) = ( - 𝐵 ·_ℎ 𝐶 ) )
9	8	adantr	⊢ ( ( 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℋ ) → ( ( - 1 · 𝐵 ) ·_ℎ 𝐶 ) = ( - 𝐵 ·_ℎ 𝐶 ) )
10		neg1cn	⊢ - 1 ∈ ℂ
11		ax-hvmulass	⊢ ( ( - 1 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℋ ) → ( ( - 1 · 𝐵 ) ·_ℎ 𝐶 ) = ( - 1 ·_ℎ ( 𝐵 ·_ℎ 𝐶 ) ) )
12	10 11	mp3an1	⊢ ( ( 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℋ ) → ( ( - 1 · 𝐵 ) ·_ℎ 𝐶 ) = ( - 1 ·_ℎ ( 𝐵 ·_ℎ 𝐶 ) ) )
13	9 12	eqtr3d	⊢ ( ( 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℋ ) → ( - 𝐵 ·_ℎ 𝐶 ) = ( - 1 ·_ℎ ( 𝐵 ·_ℎ 𝐶 ) ) )
14	13	3adant1	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℋ ) → ( - 𝐵 ·_ℎ 𝐶 ) = ( - 1 ·_ℎ ( 𝐵 ·_ℎ 𝐶 ) ) )
15	14	oveq2d	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℋ ) → ( ( 𝐴 ·_ℎ 𝐶 ) +_ℎ ( - 𝐵 ·_ℎ 𝐶 ) ) = ( ( 𝐴 ·_ℎ 𝐶 ) +_ℎ ( - 1 ·_ℎ ( 𝐵 ·_ℎ 𝐶 ) ) ) )
16		negcl	⊢ ( 𝐵 ∈ ℂ → - 𝐵 ∈ ℂ )
17		ax-hvdistr2	⊢ ( ( 𝐴 ∈ ℂ ∧ - 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℋ ) → ( ( 𝐴 + - 𝐵 ) ·_ℎ 𝐶 ) = ( ( 𝐴 ·_ℎ 𝐶 ) +_ℎ ( - 𝐵 ·_ℎ 𝐶 ) ) )
18	16 17	syl3an2	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℋ ) → ( ( 𝐴 + - 𝐵 ) ·_ℎ 𝐶 ) = ( ( 𝐴 ·_ℎ 𝐶 ) +_ℎ ( - 𝐵 ·_ℎ 𝐶 ) ) )
19		negsub	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( 𝐴 + - 𝐵 ) = ( 𝐴 − 𝐵 ) )
20	19	3adant3	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℋ ) → ( 𝐴 + - 𝐵 ) = ( 𝐴 − 𝐵 ) )
21	20	oveq1d	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℋ ) → ( ( 𝐴 + - 𝐵 ) ·_ℎ 𝐶 ) = ( ( 𝐴 − 𝐵 ) ·_ℎ 𝐶 ) )
22	18 21	eqtr3d	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℋ ) → ( ( 𝐴 ·_ℎ 𝐶 ) +_ℎ ( - 𝐵 ·_ℎ 𝐶 ) ) = ( ( 𝐴 − 𝐵 ) ·_ℎ 𝐶 ) )
23	6 15 22	3eqtr2rd	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℋ ) → ( ( 𝐴 − 𝐵 ) ·_ℎ 𝐶 ) = ( ( 𝐴 ·_ℎ 𝐶 ) −_ℎ ( 𝐵 ·_ℎ 𝐶 ) ) )

Metamath Proof Explorer

Theorem hvsubdistr2

Proof