Theorem tgcgreq

Description: Congruence and equality. (Contributed by Thierry Arnoux, 27-Aug-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 ) )
	tgcgreq.1	⊢ ( 𝜑 → 𝐴 = 𝐵 )
Assertion	tgcgreq	⊢ ( 𝜑 → 𝐶 = 𝐷 )

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		tgcgreq.1	⊢ ( 𝜑 → 𝐴 = 𝐵 )
11	1 2 3 4 5 6 7 8 9	tgcgreqb	⊢ ( 𝜑 → ( 𝐴 = 𝐵 ↔ 𝐶 = 𝐷 ) )
12	10 11	mpbid	⊢ ( 𝜑 → 𝐶 = 𝐷 )