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Theorem elabg 2710
Description: Membership in a class abstraction, using implicit substitution. Compare Theorem 6.13 of [Quine] p. 44. (Contributed by NM, 14-Apr-1995.)
Hypothesis
Ref Expression
elabg.1 (𝑥 = 𝐴 → (𝜑𝜓))
Assertion
Ref Expression
elabg (𝐴𝑉 → (𝐴 ∈ {𝑥𝜑} ↔ 𝜓))
Distinct variable groups:   𝜓,𝑥   𝑥,𝐴
Allowed substitution hints:   𝜑(𝑥)   𝑉(𝑥)

Proof of Theorem elabg
StepHypRef Expression
1 nfcv 2194 . 2 𝑥𝐴
2 nfv 1437 . 2 𝑥𝜓
3 elabg.1 . 2 (𝑥 = 𝐴 → (𝜑𝜓))
41, 2, 3elabgf 2707 1 (𝐴𝑉 → (𝐴 ∈ {𝑥𝜑} ↔ 𝜓))
Colors of variables: wff set class
Syntax hints:  wi 4  wb 102   = wceq 1259  wcel 1409  {cab 2042
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 103  ax-ia2 104  ax-ia3 105  ax-io 640  ax-5 1352  ax-7 1353  ax-gen 1354  ax-ie1 1398  ax-ie2 1399  ax-8 1411  ax-10 1412  ax-11 1413  ax-i12 1414  ax-bndl 1415  ax-4 1416  ax-17 1435  ax-i9 1439  ax-ial 1443  ax-i5r 1444  ax-ext 2038
This theorem depends on definitions:  df-bi 114  df-tru 1262  df-nf 1366  df-sb 1662  df-clab 2043  df-cleq 2049  df-clel 2052  df-nfc 2183  df-v 2576
This theorem is referenced by:  elab2g  2711  intmin3  3669  finds  4350  elxpi  4388  ovelrn  5676  indpi  6497  peano5nnnn  7023  peano5nni  7992  peano5setOLD  10438
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