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 ∈ TarskiG )
	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 ∈ TarskiG )
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 ) )