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Theorem elabf 3611
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 2955 . . 3 𝑥𝐴
3 elabf.1 . . 3 𝑥𝜓
4 elabf.3 . . 3 (𝑥 = 𝐴 → (𝜑𝜓))
52, 3, 4elabgf 3610 . 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 209   = wceq 1538  wnf 1785  wcel 2111  {cab 2776  Vcvv 3441
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1911  ax-6 1970  ax-7 2015  ax-8 2113  ax-9 2121  ax-10 2142  ax-11 2158  ax-12 2175  ax-ext 2770
This theorem depends on definitions:  df-bi 210  df-an 400  df-or 845  df-tru 1541  df-ex 1782  df-nf 1786  df-sb 2070  df-clab 2777  df-cleq 2791  df-clel 2870  df-nfc 2938  df-v 3443
This theorem is referenced by:  elab  3615  dfon2lem1  33141  sdclem2  35180  sdclem1  35181  scottabf  40948
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