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	|- ( ph -> G e. TarskiG )
	tgcgrcomlr.a	|- ( ph -> A e. P )
	tgcgrcomlr.b	|- ( ph -> B e. P )
	tgcgrcomlr.c	|- ( ph -> C e. P )
	tgcgrcomlr.d	|- ( ph -> D e. P )
	tgcgrcomlr.6	|- ( ph -> ( A .- B ) = ( C .- D ) )
Assertion	tgcgrcomlr	|- ( ph -> ( 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	|- ( ph -> G e. TarskiG )
5		tgcgrcomlr.a	|- ( ph -> A e. P )
6		tgcgrcomlr.b	|- ( ph -> B e. P )
7		tgcgrcomlr.c	|- ( ph -> C e. P )
8		tgcgrcomlr.d	|- ( ph -> D e. P )
9		tgcgrcomlr.6	|- ( ph -> ( A .- B ) = ( C .- D ) )
10	1 2 3 4 5 6	axtgcgrrflx	|- ( ph -> ( A .- B ) = ( B .- A ) )
11	1 2 3 4 7 8	axtgcgrrflx	|- ( ph -> ( C .- D ) = ( D .- C ) )
12	9 10 11	3eqtr3d	|- ( ph -> ( B .- A ) = ( D .- C ) )