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Theorem elqsg 6156
Description: Closed form of elqs 6157. (Contributed by Rodolfo Medina, 12-Oct-2010.)
Assertion
Ref Expression
elqsg (𝐵𝑉 → (𝐵 ∈ (𝐴 / 𝑅) ↔ ∃𝑥𝐴 𝐵 = [𝑥]𝑅))
Distinct variable groups:   𝑥,𝐴   𝑥,𝐵   𝑥,𝑅
Allowed substitution hint:   𝑉(𝑥)

Proof of Theorem elqsg
Dummy variable 𝑦 is distinct from all other variables.
StepHypRef Expression
1 eqeq1 2046 . . 3 (𝑦 = 𝐵 → (𝑦 = [𝑥]𝑅𝐵 = [𝑥]𝑅))
21rexbidv 2327 . 2 (𝑦 = 𝐵 → (∃𝑥𝐴 𝑦 = [𝑥]𝑅 ↔ ∃𝑥𝐴 𝐵 = [𝑥]𝑅))
3 df-qs 6112 . 2 (𝐴 / 𝑅) = {𝑦 ∣ ∃𝑥𝐴 𝑦 = [𝑥]𝑅}
42, 3elab2g 2689 1 (𝐵𝑉 → (𝐵 ∈ (𝐴 / 𝑅) ↔ ∃𝑥𝐴 𝐵 = [𝑥]𝑅))
Colors of variables: wff set class
Syntax hints:  wi 4  wb 98   = wceq 1243  wcel 1393  wrex 2307  [cec 6104   / cqs 6105
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-io 630  ax-5 1336  ax-7 1337  ax-gen 1338  ax-ie1 1382  ax-ie2 1383  ax-8 1395  ax-10 1396  ax-11 1397  ax-i12 1398  ax-bndl 1399  ax-4 1400  ax-17 1419  ax-i9 1423  ax-ial 1427  ax-i5r 1428  ax-ext 2022
This theorem depends on definitions:  df-bi 110  df-tru 1246  df-nf 1350  df-sb 1646  df-clab 2027  df-cleq 2033  df-clel 2036  df-nfc 2167  df-rex 2312  df-v 2559  df-qs 6112
This theorem is referenced by:  elqs  6157  elqsi  6158  ecelqsg  6159
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