Metamath Proof Explorer


Theorem tgcgrcomlr

Description: Congruence commutes on both sides. (Contributed by Thierry Arnoux, 23-Mar-2019)

Ref Expression
Hypotheses tkgeom.p ⊢ P = Base G
tkgeom.d ⊢ - ˙ = dist ⁡ G
tkgeom.i ⊢ I = Itv ⁡ G
tkgeom.g ⊢ φ → G ∈ 𝒢 Tarski
tgcgrcomlr.a ⊢ φ → A ∈ P
tgcgrcomlr.b ⊢ φ → B ∈ P
tgcgrcomlr.c ⊢ φ → C ∈ P
tgcgrcomlr.d ⊢ φ → D ∈ P
tgcgrcomlr.6 ⊢ φ → A - ˙ B = C - ˙ D
Assertion tgcgrcomlr ⊢ φ → B - ˙ A = D - ˙ C

Proof

Step Hyp Ref Expression
1 tkgeom.p ⊢ P = Base G
2 tkgeom.d ⊢ - ˙ = dist ⁡ G
3 tkgeom.i ⊢ I = Itv ⁡ G
4 tkgeom.g ⊢ φ → G ∈ 𝒢 Tarski
5 tgcgrcomlr.a ⊢ φ → A ∈ P
6 tgcgrcomlr.b ⊢ φ → B ∈ P
7 tgcgrcomlr.c ⊢ φ → C ∈ P
8 tgcgrcomlr.d ⊢ φ → D ∈ P
9 tgcgrcomlr.6 ⊢ φ → A - ˙ B = C - ˙ D
10 1 2 3 4 5 6 axtgcgrrflx ⊢ φ → A - ˙ B = B - ˙ A
11 1 2 3 4 7 8 axtgcgrrflx ⊢ φ → C - ˙ D = D - ˙ C
12 9 10 11 3eqtr3d ⊢ φ → B - ˙ A = D - ˙ C