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Theorem elabf 3620
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 2902 . . 3 𝑥𝐴
3 elabf.1 . . 3 𝑥𝜓
4 elabf.3 . . 3 (𝑥 = 𝐴 → (𝜑𝜓))
52, 3, 4elabgf 3619 . 2 (𝐴 ∈ V → (𝐴 ∈ {𝑥𝜑} ↔ 𝜓))
61, 5ax-mp 5 1 (𝐴 ∈ {𝑥𝜑} ↔ 𝜓)
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 207   = wceq 1547  wnf 1790  wcel 2119  {cab 2718  Vcvv 3432
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1802  ax-4 1816  ax-5 1917  ax-6 1974  ax-7 2015  ax-8 2121  ax-9 2129  ax-10 2152  ax-11 2168  ax-12 2189  ax-ext 2712
This theorem depends on definitions:  df-bi 208  df-an 397  df-or 854  df-tru 1550  df-ex 1787  df-nf 1791  df-sb 2074  df-clab 2719  df-cleq 2732  df-clel 2815  df-nfc 2889  df-v 3434
This theorem is referenced by:  dfon2lem1  36016  sdclem2  38116  sdclem1  38117  scottabf  44691
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