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Metamath Proof Explorer

Theorem hvaddeq0

Description: If the sum of two vectors is zero, one is the negative of the other. (Contributed by NM, 10-Jun-2006) (New usage is discouraged.)

Ref Expression
Assertion hvaddeq0 ( ( 𝐴 ∈ ℋ ∧ 𝐵 ∈ ℋ ) → ( ( 𝐴 +ℎ 𝐵 ) = 0ℎ ↔ 𝐴 = ( - 1 ·ℎ 𝐵 ) ) )

Proof

Step Hyp Ref Expression
1 hvaddsubval ⊢ ( ( 𝐴 ∈ ℋ ∧ 𝐵 ∈ ℋ ) → ( 𝐴 +ℎ 𝐵 ) = ( 𝐴 −ℎ ( - 1 ·ℎ 𝐵 ) ) )
2 1 eqeq1d ⊢ ( ( 𝐴 ∈ ℋ ∧ 𝐵 ∈ ℋ ) → ( ( 𝐴 +ℎ 𝐵 ) = 0ℎ ↔ ( 𝐴 −ℎ ( - 1 ·ℎ 𝐵 ) ) = 0ℎ ) )
3 neg1cn ⊢ - 1 ∈ ℂ
4 hvmulcl ⊢ ( ( - 1 ∈ ℂ ∧ 𝐵 ∈ ℋ ) → ( - 1 ·ℎ 𝐵 ) ∈ ℋ )
5 3 4 mpan ⊢ ( 𝐵 ∈ ℋ → ( - 1 ·ℎ 𝐵 ) ∈ ℋ )
6 hvsubeq0 ⊢ ( ( 𝐴 ∈ ℋ ∧ ( - 1 ·ℎ 𝐵 ) ∈ ℋ ) → ( ( 𝐴 −ℎ ( - 1 ·ℎ 𝐵 ) ) = 0ℎ ↔ 𝐴 = ( - 1 ·ℎ 𝐵 ) ) )
7 5 6 sylan2 ⊢ ( ( 𝐴 ∈ ℋ ∧ 𝐵 ∈ ℋ ) → ( ( 𝐴 −ℎ ( - 1 ·ℎ 𝐵 ) ) = 0ℎ ↔ 𝐴 = ( - 1 ·ℎ 𝐵 ) ) )
8 2 7 bitrd ⊢ ( ( 𝐴 ∈ ℋ ∧ 𝐵 ∈ ℋ ) → ( ( 𝐴 +ℎ 𝐵 ) = 0ℎ ↔ 𝐴 = ( - 1 ·ℎ 𝐵 ) ) )