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