Metamath Proof Explorer


Theorem hvsubcl

Description: Closure of vector subtraction. (Contributed by NM, 17-Aug-1999) (New usage is discouraged.)

Ref Expression
Assertion hvsubcl ( ( 𝐴 ∈ ℋ ∧ 𝐵 ∈ ℋ ) → ( 𝐴 𝐵 ) ∈ ℋ )

Proof

Step Hyp Ref Expression
1 hvsubval ( ( 𝐴 ∈ ℋ ∧ 𝐵 ∈ ℋ ) → ( 𝐴 𝐵 ) = ( 𝐴 + ( - 1 · 𝐵 ) ) )
2 neg1cn - 1 ∈ ℂ
3 hvmulcl ( ( - 1 ∈ ℂ ∧ 𝐵 ∈ ℋ ) → ( - 1 · 𝐵 ) ∈ ℋ )
4 2 3 mpan ( 𝐵 ∈ ℋ → ( - 1 · 𝐵 ) ∈ ℋ )
5 hvaddcl ( ( 𝐴 ∈ ℋ ∧ ( - 1 · 𝐵 ) ∈ ℋ ) → ( 𝐴 + ( - 1 · 𝐵 ) ) ∈ ℋ )
6 4 5 sylan2 ( ( 𝐴 ∈ ℋ ∧ 𝐵 ∈ ℋ ) → ( 𝐴 + ( - 1 · 𝐵 ) ) ∈ ℋ )
7 1 6 eqeltrd ( ( 𝐴 ∈ ℋ ∧ 𝐵 ∈ ℋ ) → ( 𝐴 𝐵 ) ∈ ℋ )