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Theorem elabf 3640
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 2974 . . 3 𝑥𝐴
3 elabf.1 . . 3 𝑥𝜓
4 elabf.3 . . 3 (𝑥 = 𝐴 → (𝜑𝜓))
52, 3, 4elabgf 3639 . 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 2115  {cab 2799  Vcvv 3471
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 1912  ax-6 1971  ax-7 2016  ax-8 2117  ax-9 2125  ax-10 2146  ax-11 2162  ax-12 2178  ax-ext 2793
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 2071  df-clab 2800  df-cleq 2814  df-clel 2892  df-nfc 2960  df-v 3473
This theorem is referenced by:  elab  3644  dfon2lem1  33035  sdclem2  35060  sdclem1  35061  scottabf  40733
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