Metamath Proof Explorer


Theorem hvaddcan2

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

Ref Expression
Assertion hvaddcan2 ( ( 𝐴 ∈ ℋ ∧ 𝐵 ∈ ℋ ∧ 𝐶 ∈ ℋ ) → ( ( 𝐴 + 𝐶 ) = ( 𝐵 + 𝐶 ) ↔ 𝐴 = 𝐵 ) )

Proof

Step Hyp Ref Expression
1 ax-hvcom ( ( 𝐶 ∈ ℋ ∧ 𝐴 ∈ ℋ ) → ( 𝐶 + 𝐴 ) = ( 𝐴 + 𝐶 ) )
2 1 3adant3 ( ( 𝐶 ∈ ℋ ∧ 𝐴 ∈ ℋ ∧ 𝐵 ∈ ℋ ) → ( 𝐶 + 𝐴 ) = ( 𝐴 + 𝐶 ) )
3 ax-hvcom ( ( 𝐶 ∈ ℋ ∧ 𝐵 ∈ ℋ ) → ( 𝐶 + 𝐵 ) = ( 𝐵 + 𝐶 ) )
4 3 3adant2 ( ( 𝐶 ∈ ℋ ∧ 𝐴 ∈ ℋ ∧ 𝐵 ∈ ℋ ) → ( 𝐶 + 𝐵 ) = ( 𝐵 + 𝐶 ) )
5 2 4 eqeq12d ( ( 𝐶 ∈ ℋ ∧ 𝐴 ∈ ℋ ∧ 𝐵 ∈ ℋ ) → ( ( 𝐶 + 𝐴 ) = ( 𝐶 + 𝐵 ) ↔ ( 𝐴 + 𝐶 ) = ( 𝐵 + 𝐶 ) ) )
6 hvaddcan ( ( 𝐶 ∈ ℋ ∧ 𝐴 ∈ ℋ ∧ 𝐵 ∈ ℋ ) → ( ( 𝐶 + 𝐴 ) = ( 𝐶 + 𝐵 ) ↔ 𝐴 = 𝐵 ) )
7 5 6 bitr3d ( ( 𝐶 ∈ ℋ ∧ 𝐴 ∈ ℋ ∧ 𝐵 ∈ ℋ ) → ( ( 𝐴 + 𝐶 ) = ( 𝐵 + 𝐶 ) ↔ 𝐴 = 𝐵 ) )
8 7 3coml ( ( 𝐴 ∈ ℋ ∧ 𝐵 ∈ ℋ ∧ 𝐶 ∈ ℋ ) → ( ( 𝐴 + 𝐶 ) = ( 𝐵 + 𝐶 ) ↔ 𝐴 = 𝐵 ) )