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Theorem elabf 3635
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 2925 . . 3 𝑥𝐴
3 elabf.1 . . 3 𝑥𝜓
4 elabf.3 . . 3 (𝑥 = 𝐴 → (𝜑𝜓))
52, 3, 4elabgf 3634 . 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 1570  wnf 1813  wcel 2143  {cab 2741  Vcvv 3455
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1825  ax-4 1839  ax-5 1940  ax-6 1997  ax-7 2038  ax-8 2145  ax-9 2153  ax-10 2176  ax-11 2192  ax-12 2213  ax-ext 2735
This theorem depends on definitions:  df-bi 210  df-an 401  df-or 861  df-tru 1573  df-ex 1810  df-nf 1814  df-sb 2097  df-clab 2742  df-cleq 2755  df-clel 2838  df-nfc 2912  df-v 3457
This theorem is referenced by:  scottabf  9867  dfon2lem1  36254  sdclem2  38374  sdclem1  38375
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