Theorem tgcgreq 28953

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 28952	. 2 ⊢ (𝜑 → (𝐴 = 𝐵 ↔ 𝐶 = 𝐷))
12	1, 11	mpbid 235	1 ⊢ (𝜑 → 𝐶 = 𝐷)

This proof depends on syntax axioms:   → wi 4   = wceq 1569   ∈ wcel 2144  ‘cfv 6543  (class class class)co 7424  Basecbs 17396  distcds 17446  TarskiGcstrkg 28898  Itvcitv 28904

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040  ax-8 2146  ax-9 2154  ax-ext 2732  ax-nul 5259

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1572  df-fal 1582  df-ex 1812  df-sb 2099  df-clab 2739  df-cleq 2752  df-clel 2835  df-ne 2956  df-ral 3077  df-rab 3413  df-v 3452  df-sbc 3739  df-ss 3897  df-dif 4046  df-in 4090  df-un 4150  df-nul 4278  df-if 4482  df-sn 4585  df-pr 4587  df-op 4835  df-uni 4867  df-br 5103  df-iota 6497  df-fv 6551  df-ov 7427  df-trkgc 28919  df-trkg 28924

This theorem is used by:  tgcgrextend  28956  tgidinside  29043  tgbtwnconn1lem3  29046  krippenlem  29171  symquadprlnglem  29174  mirleqb  29175  ragcgr  29191  lmiisolem  29310  tgaaddcpbllem1  29358  cgrg3col4  29381  tgextalem1inag  33079