Theorem tgcgreq 28505

Description: Congruence and equality. (Contributed by Thierry Arnoux, 27-Aug-2019.)

Hypotheses

Ref	Expression
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

Ref	Expression
tgcgreq	⊢ (𝜑 → 𝐶 = 𝐷)

Proof of Theorem tgcgreq

Step	Hyp	Ref	Expression
1		tgcgreq.1	. 2 ⊢ (𝜑 → 𝐴 = 𝐵)
2		tkgeom.p	. . 3 ⊢ 𝑃 = (Base‘𝐺)
3		tkgeom.d	. . 3 ⊢ − = (dist‘𝐺)
4		tkgeom.i	. . 3 ⊢ 𝐼 = (Itv‘𝐺)
5		tkgeom.g	. . 3 ⊢ (𝜑 → 𝐺 ∈ TarskiG)
6		tgcgrcomlr.a	. . 3 ⊢ (𝜑 → 𝐴 ∈ 𝑃)
7		tgcgrcomlr.b	. . 3 ⊢ (𝜑 → 𝐵 ∈ 𝑃)
8		tgcgrcomlr.c	. . 3 ⊢ (𝜑 → 𝐶 ∈ 𝑃)
9		tgcgrcomlr.d	. . 3 ⊢ (𝜑 → 𝐷 ∈ 𝑃)
10		tgcgrcomlr.6	. . 3 ⊢ (𝜑 → (𝐴 − 𝐵) = (𝐶 − 𝐷))
11	2, 3, 4, 5, 6, 7, 8, 9, 10	tgcgreqb 28504	. 2 ⊢ (𝜑 → (𝐴 = 𝐵 ↔ 𝐶 = 𝐷))
12	1, 11	mpbid 232	1 ⊢ (𝜑 → 𝐶 = 𝐷)

Syntax hints:   → wi 4   = wceq 1537   ∈ wcel 2106  ‘cfv 6563  (class class class)co 7431  Basecbs 17245  distcds 17307  TarskiGcstrkg 28450  Itvcitv 28456

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1792  ax-4 1806  ax-5 1908  ax-6 1965  ax-7 2005  ax-8 2108  ax-9 2116  ax-ext 2706  ax-nul 5312

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-3an 1088  df-tru 1540  df-fal 1550  df-ex 1777  df-sb 2063  df-clab 2713  df-cleq 2727  df-clel 2814  df-ne 2939  df-ral 3060  df-rab 3434  df-v 3480  df-sbc 3792  df-dif 3966  df-un 3968  df-in 3970  df-ss 3980  df-nul 4340  df-if 4532  df-sn 4632  df-pr 4634  df-op 4638  df-uni 4913  df-br 5149  df-iota 6516  df-fv 6571  df-ov 7434  df-trkgc 28471  df-trkg 28476

This theorem is referenced by:  tgcgrextend  28508  tgidinside  28594  tgbtwnconn1lem3  28597  krippenlem  28713  ragcgr  28730  lmiisolem  28819  cgrg3col4  28876