Theorem tgcgreq 26843

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 26842	. 2 ⊢ (𝜑 → (𝐴 = 𝐵 ↔ 𝐶 = 𝐷))
12	1, 11	mpbid 231	1 ⊢ (𝜑 → 𝐶 = 𝐷)

Syntax hints:   → wi 4   = wceq 1539   ∈ wcel 2106  ‘cfv 6433  (class class class)co 7275  Basecbs 16912  distcds 16971  TarskiGcstrkg 26788  Itvcitv 26794

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1798  ax-4 1812  ax-5 1913  ax-6 1971  ax-7 2011  ax-8 2108  ax-9 2116  ax-10 2137  ax-11 2154  ax-12 2171  ax-ext 2709  ax-nul 5230

This theorem depends on definitions:  df-bi 206  df-an 397  df-or 845  df-3an 1088  df-tru 1542  df-fal 1552  df-ex 1783  df-nf 1787  df-sb 2068  df-mo 2540  df-eu 2569  df-clab 2716  df-cleq 2730  df-clel 2816  df-ral 3069  df-rex 3070  df-rab 3073  df-v 3434  df-sbc 3717  df-dif 3890  df-un 3892  df-in 3894  df-ss 3904  df-nul 4257  df-if 4460  df-sn 4562  df-pr 4564  df-op 4568  df-uni 4840  df-br 5075  df-iota 6391  df-fv 6441  df-ov 7278  df-trkgc 26809  df-trkg 26814

This theorem is referenced by:  tgcgrextend  26846  tgidinside  26932  tgbtwnconn1lem3  26935  krippenlem  27051  ragcgr  27068  lmiisolem  27157  cgrg3col4  27214