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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 −ℎ 𝐵 ) = ( 𝐴 −ℎ 𝐶 ) ↔ 𝐵 = 𝐶 ) )

Proof

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