Theorem elabg 2895

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 2329	. 2 ⊢ Ⅎ𝑥𝐴
2		nfv 1538	. 2 ⊢ Ⅎ𝑥𝜓
3		elabg.1	. 2 ⊢ (𝑥 = 𝐴 → (𝜑 ↔ 𝜓))
4	1, 2, 3	elabgf 2891	1 ⊢ (𝐴 ∈ 𝑉 → (𝐴 ∈ {𝑥 ∣ 𝜑} ↔ 𝜓))

Syntax hints:   → wi 4   ↔ wb 105   = wceq 1363   ∈ wcel 2158  {cab 2173

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 710  ax-5 1457  ax-7 1458  ax-gen 1459  ax-ie1 1503  ax-ie2 1504  ax-8 1514  ax-10 1515  ax-11 1516  ax-i12 1517  ax-bndl 1519  ax-4 1520  ax-17 1536  ax-i9 1540  ax-ial 1544  ax-i5r 1545  ax-ext 2169

This theorem depends on definitions:  df-bi 117  df-tru 1366  df-nf 1471  df-sb 1773  df-clab 2174  df-cleq 2180  df-clel 2183  df-nfc 2318  df-v 2751

This theorem is referenced by:  elab2g  2896  intmin3  3883  finds  4611  elxpi  4654  elabrexg  5772  ovelrn  6037  elfi  6984  indpi  7355  peano5nnnn  7905  peano5nni  8936  lss1d  13572  lspsn  13605  eltg  13848  eltg2  13849