Theorem tgcgreq 25793

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 25792	. 2 ⊢ (𝜑 → (𝐴 = 𝐵 ↔ 𝐶 = 𝐷))
12	1, 11	mpbid 224	1 ⊢ (𝜑 → 𝐶 = 𝐷)

Syntax hints:   → wi 4   = wceq 1658   ∈ wcel 2166  ‘cfv 6122  (class class class)co 6904  Basecbs 16221  distcds 16313  TarskiGcstrkg 25741  Itvcitv 25747

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1896  ax-4 1910  ax-5 2011  ax-6 2077  ax-7 2114  ax-9 2175  ax-10 2194  ax-11 2209  ax-12 2222  ax-13 2390  ax-ext 2802  ax-nul 5012

This theorem depends on definitions:  df-bi 199  df-an 387  df-or 881  df-3an 1115  df-tru 1662  df-ex 1881  df-nf 1885  df-sb 2070  df-mo 2604  df-eu 2639  df-clab 2811  df-cleq 2817  df-clel 2820  df-nfc 2957  df-ral 3121  df-rex 3122  df-rab 3125  df-v 3415  df-sbc 3662  df-dif 3800  df-un 3802  df-in 3804  df-ss 3811  df-nul 4144  df-if 4306  df-sn 4397  df-pr 4399  df-op 4403  df-uni 4658  df-br 4873  df-iota 6085  df-fv 6130  df-ov 6907  df-trkgc 25759  df-trkg 25764

This theorem is referenced by:  tgcgrextend  25796  tgidinside  25882  tgbtwnconn1lem3  25885  krippenlem  26001  ragcgr  26018  lmiisolem  26104  cgrg3col4  26151