Metamath Proof Explorer


Theorem hlcomb

Description: The half-line relation is symmetric. Theorem 6.6 of Schwabhauser p. 44. (Contributed by Thierry Arnoux, 21-Feb-2020)

Ref Expression
Hypotheses ishlg.p ⊢ 𝑃 = ( Base ‘ 𝐺 )
ishlg.i ⊢ 𝐼 = ( Itv ‘ 𝐺 )
ishlg.k ⊢ 𝐾 = ( hlG ‘ 𝐺 )
ishlg.a ⊢ ( 𝜑 → 𝐴 ∈ 𝑃 )
ishlg.b ⊢ ( 𝜑 → 𝐵 ∈ 𝑃 )
ishlg.c ⊢ ( 𝜑 → 𝐶 ∈ 𝑃 )
ishlg.g ⊢ ( 𝜑 → 𝐺 ∈ 𝑉 )
Assertion hlcomb ( 𝜑 → ( 𝐴 ( 𝐾 ‘ 𝐶 ) 𝐵 ↔ 𝐵 ( 𝐾 ‘ 𝐶 ) 𝐴 ) )

Proof

Step Hyp Ref Expression
1 ishlg.p ⊢ 𝑃 = ( Base ‘ 𝐺 )
2 ishlg.i ⊢ 𝐼 = ( Itv ‘ 𝐺 )
3 ishlg.k ⊢ 𝐾 = ( hlG ‘ 𝐺 )
4 ishlg.a ⊢ ( 𝜑 → 𝐴 ∈ 𝑃 )
5 ishlg.b ⊢ ( 𝜑 → 𝐵 ∈ 𝑃 )
6 ishlg.c ⊢ ( 𝜑 → 𝐶 ∈ 𝑃 )
7 ishlg.g ⊢ ( 𝜑 → 𝐺 ∈ 𝑉 )
8 3ancoma ⊢ ( ( 𝐴 ≠ 𝐶 ∧ 𝐵 ≠ 𝐶 ∧ ( 𝐴 ∈ ( 𝐶 𝐼 𝐵 ) ∨ 𝐵 ∈ ( 𝐶 𝐼 𝐴 ) ) ) ↔ ( 𝐵 ≠ 𝐶 ∧ 𝐴 ≠ 𝐶 ∧ ( 𝐴 ∈ ( 𝐶 𝐼 𝐵 ) ∨ 𝐵 ∈ ( 𝐶 𝐼 𝐴 ) ) ) )
9 orcom ⊢ ( ( 𝐴 ∈ ( 𝐶 𝐼 𝐵 ) ∨ 𝐵 ∈ ( 𝐶 𝐼 𝐴 ) ) ↔ ( 𝐵 ∈ ( 𝐶 𝐼 𝐴 ) ∨ 𝐴 ∈ ( 𝐶 𝐼 𝐵 ) ) )
10 9 a1i ⊢ ( 𝜑 → ( ( 𝐴 ∈ ( 𝐶 𝐼 𝐵 ) ∨ 𝐵 ∈ ( 𝐶 𝐼 𝐴 ) ) ↔ ( 𝐵 ∈ ( 𝐶 𝐼 𝐴 ) ∨ 𝐴 ∈ ( 𝐶 𝐼 𝐵 ) ) ) )
11 10 3anbi3d ⊢ ( 𝜑 → ( ( 𝐵 ≠ 𝐶 ∧ 𝐴 ≠ 𝐶 ∧ ( 𝐴 ∈ ( 𝐶 𝐼 𝐵 ) ∨ 𝐵 ∈ ( 𝐶 𝐼 𝐴 ) ) ) ↔ ( 𝐵 ≠ 𝐶 ∧ 𝐴 ≠ 𝐶 ∧ ( 𝐵 ∈ ( 𝐶 𝐼 𝐴 ) ∨ 𝐴 ∈ ( 𝐶 𝐼 𝐵 ) ) ) ) )
12 8 11 bitrid ⊢ ( 𝜑 → ( ( 𝐴 ≠ 𝐶 ∧ 𝐵 ≠ 𝐶 ∧ ( 𝐴 ∈ ( 𝐶 𝐼 𝐵 ) ∨ 𝐵 ∈ ( 𝐶 𝐼 𝐴 ) ) ) ↔ ( 𝐵 ≠ 𝐶 ∧ 𝐴 ≠ 𝐶 ∧ ( 𝐵 ∈ ( 𝐶 𝐼 𝐴 ) ∨ 𝐴 ∈ ( 𝐶 𝐼 𝐵 ) ) ) ) )
13 1 2 3 4 5 6 7 ishlg ⊢ ( 𝜑 → ( 𝐴 ( 𝐾 ‘ 𝐶 ) 𝐵 ↔ ( 𝐴 ≠ 𝐶 ∧ 𝐵 ≠ 𝐶 ∧ ( 𝐴 ∈ ( 𝐶 𝐼 𝐵 ) ∨ 𝐵 ∈ ( 𝐶 𝐼 𝐴 ) ) ) ) )
14 1 2 3 5 4 6 7 ishlg ⊢ ( 𝜑 → ( 𝐵 ( 𝐾 ‘ 𝐶 ) 𝐴 ↔ ( 𝐵 ≠ 𝐶 ∧ 𝐴 ≠ 𝐶 ∧ ( 𝐵 ∈ ( 𝐶 𝐼 𝐴 ) ∨ 𝐴 ∈ ( 𝐶 𝐼 𝐵 ) ) ) ) )
15 12 13 14 3bitr4d ⊢ ( 𝜑 → ( 𝐴 ( 𝐾 ‘ 𝐶 ) 𝐵 ↔ 𝐵 ( 𝐾 ‘ 𝐶 ) 𝐴 ) )