Metamath Proof Explorer


Theorem hvsub0

Description: Subtraction of a zero vector. (Contributed by NM, 2-Apr-2000) (New usage is discouraged.)

Ref Expression
Assertion hvsub0 ( 𝐴 ∈ ℋ → ( 𝐴 0 ) = 𝐴 )

Proof

Step Hyp Ref Expression
1 ax-hv0cl 0 ∈ ℋ
2 hvsubval ( ( 𝐴 ∈ ℋ ∧ 0 ∈ ℋ ) → ( 𝐴 0 ) = ( 𝐴 + ( - 1 · 0 ) ) )
3 1 2 mpan2 ( 𝐴 ∈ ℋ → ( 𝐴 0 ) = ( 𝐴 + ( - 1 · 0 ) ) )
4 neg1cn - 1 ∈ ℂ
5 hvmul0 ( - 1 ∈ ℂ → ( - 1 · 0 ) = 0 )
6 4 5 ax-mp ( - 1 · 0 ) = 0
7 6 oveq2i ( 𝐴 + ( - 1 · 0 ) ) = ( 𝐴 + 0 )
8 3 7 eqtrdi ( 𝐴 ∈ ℋ → ( 𝐴 0 ) = ( 𝐴 + 0 ) )
9 ax-hvaddid ( 𝐴 ∈ ℋ → ( 𝐴 + 0 ) = 𝐴 )
10 8 9 eqtrd ( 𝐴 ∈ ℋ → ( 𝐴 0 ) = 𝐴 )