Theorem elabg 2761

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 → (𝜑 ↔ 𝜓))

Assertion

Ref	Expression
elabg	⊢ (𝐴 ∈ 𝑉 → (𝐴 ∈ {𝑥 ∣ 𝜑} ↔ 𝜓))

Distinct variable groups:   𝜓,𝑥   𝑥,𝐴

Allowed substitution hints:   𝜑(𝑥)   𝑉(𝑥)

Proof of Theorem elabg

Step	Hyp	Ref	Expression
1		nfcv 2228	. 2 ⊢ Ⅎ𝑥𝐴
2		nfv 1466	. 2 ⊢ Ⅎ𝑥𝜓
3		elabg.1	. 2 ⊢ (𝑥 = 𝐴 → (𝜑 ↔ 𝜓))
4	1, 2, 3	elabgf 2758	1 ⊢ (𝐴 ∈ 𝑉 → (𝐴 ∈ {𝑥 ∣ 𝜑} ↔ 𝜓))

Syntax hints:   → wi 4   ↔ wb 103   = wceq 1289   ∈ wcel 1438  {cab 2074

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-io 665  ax-5 1381  ax-7 1382  ax-gen 1383  ax-ie1 1427  ax-ie2 1428  ax-8 1440  ax-10 1441  ax-11 1442  ax-i12 1443  ax-bndl 1444  ax-4 1445  ax-17 1464  ax-i9 1468  ax-ial 1472  ax-i5r 1473  ax-ext 2070

This theorem depends on definitions:  df-bi 115  df-tru 1292  df-nf 1395  df-sb 1693  df-clab 2075  df-cleq 2081  df-clel 2084  df-nfc 2217  df-v 2621

This theorem is referenced by:  elab2g  2762  intmin3  3713  finds  4413  elxpi  4452  ovelrn  5785  indpi  6891  peano5nnnn  7417  peano5nni  8415