Theorem tgcgreq 28831

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 28830	. 2 ⊢ (𝜑 → (𝐴 = 𝐵 ↔ 𝐶 = 𝐷))
12	1, 11	mpbid 235	1 ⊢ (𝜑 → 𝐶 = 𝐷)

This proof depends on syntax axioms:   → wi 4   = wceq 1570   ∈ wcel 2145  ‘cfv 6537  (class class class)co 7417  Basecbs 17307  distcds 17357  TarskiGcstrkg 28776  Itvcitv 28782

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2147  ax-9 2155  ax-ext 2734  ax-nul 5267

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1573  df-fal 1583  df-ex 1813  df-sb 2100  df-clab 2741  df-cleq 2754  df-clel 2837  df-ne 2958  df-ral 3079  df-rab 3415  df-v 3455  df-sbc 3743  df-dif 3905  df-un 3907  df-in 3909  df-ss 3919  df-nul 4283  df-if 4486  df-sn 4588  df-pr 4590  df-op 4594  df-uni 4871  df-br 5108  df-iota 6493  df-fv 6545  df-ov 7420  df-trkgc 28797  df-trkg 28802

This theorem is used by:  tgcgrextend  28834  tgidinside  28921  tgbtwnconn1lem3  28924  krippenlem  29049  symquadprlnglem  29052  mirleqb  29053  ragcgr  29069  lmiisolem  29188  tgaaddcpbllem1  29236  cgrg3col4  29259