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Theorem elabf 2663
Description: Membership in a class abstraction, using implicit substitution. (Contributed by NM, 1-Aug-1994.) (Revised by Mario Carneiro, 12-Oct-2016.)
Hypotheses
Ref Expression
elabf.1 xψ
elabf.2 A V
elabf.3 (x = A → (φψ))
Assertion
Ref Expression
elabf (A {xφ} ↔ ψ)
Distinct variable group:   x,A
Allowed substitution hints:   φ(x)   ψ(x)

Proof of Theorem elabf
StepHypRef Expression
1 elabf.2 . 2 A V
2 nfcv 2160 . . 3 xA
3 elabf.1 . . 3 xψ
4 elabf.3 . . 3 (x = A → (φψ))
52, 3, 4elabgf 2662 . 2 (A V → (A {xφ} ↔ ψ))
61, 5ax-mp 7 1 (A {xφ} ↔ ψ)
Colors of variables: wff set class
Syntax hints:  wi 4  wb 98   = wceq 1228  wnf 1329   wcel 1374  {cab 2008  Vcvv 2535
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 617  ax-5 1316  ax-7 1317  ax-gen 1318  ax-ie1 1363  ax-ie2 1364  ax-8 1376  ax-10 1377  ax-11 1378  ax-i12 1379  ax-bnd 1380  ax-4 1381  ax-17 1400  ax-i9 1404  ax-ial 1409  ax-i5r 1410  ax-ext 2004
This theorem depends on definitions:  df-bi 110  df-tru 1231  df-nf 1330  df-sb 1628  df-clab 2009  df-cleq 2015  df-clel 2018  df-nfc 2149  df-v 2537
This theorem is referenced by:  elab  2664
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