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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 +ℎ 𝐶 ) = ( 𝐵 +ℎ 𝐶 ) ↔ 𝐴 = 𝐵 ) )

Proof

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