Metamath Proof Explorer


Theorem hvsubcan

Description: Cancellation law for vector addition. (Contributed by NM, 18-May-2005) (New usage is discouraged.)

Ref Expression
Assertion hvsubcan ( ( 𝐴 ∈ ℋ ∧ 𝐵 ∈ ℋ ∧ 𝐶 ∈ ℋ ) → ( ( 𝐴 𝐵 ) = ( 𝐴 𝐶 ) ↔ 𝐵 = 𝐶 ) )

Proof

Step Hyp Ref Expression
1 hvsubval ( ( 𝐴 ∈ ℋ ∧ 𝐵 ∈ ℋ ) → ( 𝐴 𝐵 ) = ( 𝐴 + ( - 1 · 𝐵 ) ) )
2 1 3adant3 ( ( 𝐴 ∈ ℋ ∧ 𝐵 ∈ ℋ ∧ 𝐶 ∈ ℋ ) → ( 𝐴 𝐵 ) = ( 𝐴 + ( - 1 · 𝐵 ) ) )
3 hvsubval ( ( 𝐴 ∈ ℋ ∧ 𝐶 ∈ ℋ ) → ( 𝐴 𝐶 ) = ( 𝐴 + ( - 1 · 𝐶 ) ) )
4 3 3adant2 ( ( 𝐴 ∈ ℋ ∧ 𝐵 ∈ ℋ ∧ 𝐶 ∈ ℋ ) → ( 𝐴 𝐶 ) = ( 𝐴 + ( - 1 · 𝐶 ) ) )
5 2 4 eqeq12d ( ( 𝐴 ∈ ℋ ∧ 𝐵 ∈ ℋ ∧ 𝐶 ∈ ℋ ) → ( ( 𝐴 𝐵 ) = ( 𝐴 𝐶 ) ↔ ( 𝐴 + ( - 1 · 𝐵 ) ) = ( 𝐴 + ( - 1 · 𝐶 ) ) ) )
6 neg1cn - 1 ∈ ℂ
7 hvmulcl ( ( - 1 ∈ ℂ ∧ 𝐵 ∈ ℋ ) → ( - 1 · 𝐵 ) ∈ ℋ )
8 6 7 mpan ( 𝐵 ∈ ℋ → ( - 1 · 𝐵 ) ∈ ℋ )
9 hvmulcl ( ( - 1 ∈ ℂ ∧ 𝐶 ∈ ℋ ) → ( - 1 · 𝐶 ) ∈ ℋ )
10 6 9 mpan ( 𝐶 ∈ ℋ → ( - 1 · 𝐶 ) ∈ ℋ )
11 hvaddcan ( ( 𝐴 ∈ ℋ ∧ ( - 1 · 𝐵 ) ∈ ℋ ∧ ( - 1 · 𝐶 ) ∈ ℋ ) → ( ( 𝐴 + ( - 1 · 𝐵 ) ) = ( 𝐴 + ( - 1 · 𝐶 ) ) ↔ ( - 1 · 𝐵 ) = ( - 1 · 𝐶 ) ) )
12 10 11 syl3an3 ( ( 𝐴 ∈ ℋ ∧ ( - 1 · 𝐵 ) ∈ ℋ ∧ 𝐶 ∈ ℋ ) → ( ( 𝐴 + ( - 1 · 𝐵 ) ) = ( 𝐴 + ( - 1 · 𝐶 ) ) ↔ ( - 1 · 𝐵 ) = ( - 1 · 𝐶 ) ) )
13 8 12 syl3an2 ( ( 𝐴 ∈ ℋ ∧ 𝐵 ∈ ℋ ∧ 𝐶 ∈ ℋ ) → ( ( 𝐴 + ( - 1 · 𝐵 ) ) = ( 𝐴 + ( - 1 · 𝐶 ) ) ↔ ( - 1 · 𝐵 ) = ( - 1 · 𝐶 ) ) )
14 neg1ne0 - 1 ≠ 0
15 6 14 pm3.2i ( - 1 ∈ ℂ ∧ - 1 ≠ 0 )
16 hvmulcan ( ( ( - 1 ∈ ℂ ∧ - 1 ≠ 0 ) ∧ 𝐵 ∈ ℋ ∧ 𝐶 ∈ ℋ ) → ( ( - 1 · 𝐵 ) = ( - 1 · 𝐶 ) ↔ 𝐵 = 𝐶 ) )
17 15 16 mp3an1 ( ( 𝐵 ∈ ℋ ∧ 𝐶 ∈ ℋ ) → ( ( - 1 · 𝐵 ) = ( - 1 · 𝐶 ) ↔ 𝐵 = 𝐶 ) )
18 17 3adant1 ( ( 𝐴 ∈ ℋ ∧ 𝐵 ∈ ℋ ∧ 𝐶 ∈ ℋ ) → ( ( - 1 · 𝐵 ) = ( - 1 · 𝐶 ) ↔ 𝐵 = 𝐶 ) )
19 5 13 18 3bitrd ( ( 𝐴 ∈ ℋ ∧ 𝐵 ∈ ℋ ∧ 𝐶 ∈ ℋ ) → ( ( 𝐴 𝐵 ) = ( 𝐴 𝐶 ) ↔ 𝐵 = 𝐶 ) )