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Theorem elqs 3227
Description: Membership in a quotient set.
Hypothesis
Ref Expression
elqs.1 BV
Assertion
Ref Expression
elqs (B ∈ (A / R) ↔ ∃x(xAB = [x]R))
Distinct variable group(s):   x,A   x,B   x,R

Proof of Theorem elqs
StepHypRef Expression
1 elqs.1 . . 3 BV
2 cleq1 1107 . . . 4 (y = B → (y = [x]RB = [x]R))
32birexdv 1220 . . 3 (y = B → (∃xA y = [x]R ↔ ∃xA B = [x]R))
4 df-qs 3205 . . 3 (A / R) = {y∣∃xA y = [x]R}
51, 3, 4elab2 1419 . 2 (B ∈ (A / R) ↔ ∃xA B = [x]R)
6 df-rex 1206 . 2 (∃xA B = [x]R ↔ ∃x(xAB = [x]R))
75, 6bitr 151 1 (B ∈ (A / R) ↔ ∃x(xAB = [x]R))
Colors of variables: wff set class
Syntax hints:   ↔ wb 127   ∧ wa 196  ∃wex 678   = wceq 1091   ∈ wcel 1092  ∃wrex 1202  Vcvv 1348  [cec 3198   / cqs 3199
This theorem is referenced by:  elqsi 3228  ecelqsi 3229
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-16 922  ax-17 925  ax-ext 1074
This theorem depends on definitions:  df-bi 128  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-qs 3205
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