Theorem snec 3232

Description: The singleton of an equivalence class.

Hypothesis

Ref	Expression
snec.1	⊢ A ∈ V

Assertion

Ref	Expression
snec	⊢ {[A]R} = ({A} / R)

Proof of Theorem snec

Step	Hyp	Ref	Expression
1		df-rex 1206	. . . . 5 ⊢ (∃x ∈ {A}y = [x]R ↔ ∃x(x ∈ {A} ∧ y = [x]R))
2		elsn 1820	. . . . . . 7 ⊢ (x ∈ {A} ↔ x = A)
3	2	anbi1i 368	. . . . . 6 ⊢ ((x ∈ {A} ∧ y = [x]R) ↔ (x = A ∧ y = [x]R))
4	3	biex 733	. . . . 5 ⊢ (∃x(x ∈ {A} ∧ y = [x]R) ↔ ∃x(x = A ∧ y = [x]R))
5		snec.1	. . . . . 6 ⊢ A ∈ V
6		eceq2 3215	. . . . . . 7 ⊢ (x = A → [x]R = [A]R)
7	6	cleq2d 1112	. . . . . 6 ⊢ (x = A → (y = [x]R ↔ y = [A]R))
8	5, 7	ceqsexv 1371	. . . . 5 ⊢ (∃x(x = A ∧ y = [x]R) ↔ y = [A]R)
9	1, 4, 8	3bitr 155	. . . 4 ⊢ (∃x ∈ {A}y = [x]R ↔ y = [A]R)
10	9	bicomi 150	. . 3 ⊢ (y = [A]R ↔ ∃x ∈ {A}y = [x]R)
11	10	biabi 1181	. 2 ⊢ {y∣y = [A]R} = {y∣∃x ∈ {A}y = [x]R}
12		df-sn 1811	. 2 ⊢ {[A]R} = {y∣y = [A]R}
13		df-qs 3205	. 2 ⊢ ({A} / R) = {y∣∃x ∈ {A}y = [x]R}
14	11, 12, 13	3eqtr4 1126	1 ⊢ {[A]R} = ({A} / R)

Syntax hints:   ∧ wa 196  ∃wex 678  {cab 1090   = wceq 1091   ∈ wcel 1092  ∃wrex 1202  Vcvv 1348  {csn 1808  [cec 3198   / cqs 3199

This theorem is referenced by:  ecqs 3233

This theorem was proved from axioms:  ax-1 3  ax-2 4  ax-3 5  ax-mp 6  ax-4 673  ax-5 674  ax-6 675  ax-7 676  ax-gen 677  ax-8 798  ax-9 799  ax-10 800  ax-11 801  ax-12 802  ax-13 804  ax-14 805  ax-16 922  ax-17 925  ax-ext 1074  ax-rep 1075  ax-pow 1077

This theorem depends on definitions:  df-bi 128  df-or 197  df-an 198  df-ex 679  df-sb 853  df-clab 1093  df-cleq 1097  df-clel 1099  df-rex 1206  df-v 1349  df-dif 1489  df-un 1490  df-in 1491  df-ss 1492  df-nul 1708  df-pw 1799  df-sn 1811  df-pr 1812  df-op 1815  df-br 2063  df-opab 2098  df-xp 2424  df-cnv 2426  df-dm 2428  df-rn 2429  df-res 2430  df-ima 2431  df-ec 3202  df-qs 3205

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