Theorem elabg 2762

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 2229	. 2 ⊢ Ⅎ𝑥𝐴
2		nfv 1467	. 2 ⊢ Ⅎ𝑥𝜓
3		elabg.1	. 2 ⊢ (𝑥 = 𝐴 → (𝜑 ↔ 𝜓))
4	1, 2, 3	elabgf 2759	1 ⊢ (𝐴 ∈ 𝑉 → (𝐴 ∈ {𝑥 ∣ 𝜑} ↔ 𝜓))

Syntax hints:   → wi 4   ↔ wb 104   = wceq 1290   ∈ wcel 1439  {cab 2075

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-io 666  ax-5 1382  ax-7 1383  ax-gen 1384  ax-ie1 1428  ax-ie2 1429  ax-8 1441  ax-10 1442  ax-11 1443  ax-i12 1444  ax-bndl 1445  ax-4 1446  ax-17 1465  ax-i9 1469  ax-ial 1473  ax-i5r 1474  ax-ext 2071

This theorem depends on definitions:  df-bi 116  df-tru 1293  df-nf 1396  df-sb 1694  df-clab 2076  df-cleq 2082  df-clel 2085  df-nfc 2218  df-v 2622

This theorem is referenced by:  elab2g  2763  intmin3  3721  finds  4428  elxpi  4467  ovelrn  5807  indpi  6955  peano5nnnn  7481  peano5nni  8479  eltg  11806  eltg2  11807