Theorem elab3g 2693

Description: Membership in a class abstraction, with a weaker antecedent than elabg 2688. (Contributed by NM, 29-Aug-2006.)

Hypothesis

Ref	Expression
elab3g.1	⊢ (𝑥 = 𝐴 → (𝜑 ↔ 𝜓))

Assertion

Ref	Expression
elab3g	⊢ ((𝜓 → 𝐴 ∈ 𝐵) → (𝐴 ∈ {𝑥 ∣ 𝜑} ↔ 𝜓))

Distinct variable groups:   𝜓,𝑥   𝑥,𝐴

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

Proof of Theorem elab3g

Step	Hyp	Ref	Expression
1		nfcv 2178	. 2 ⊢ Ⅎ𝑥𝐴
2		nfv 1421	. 2 ⊢ Ⅎ𝑥𝜓
3		elab3g.1	. 2 ⊢ (𝑥 = 𝐴 → (𝜑 ↔ 𝜓))
4	1, 2, 3	elab3gf 2692	1 ⊢ ((𝜓 → 𝐴 ∈ 𝐵) → (𝐴 ∈ {𝑥 ∣ 𝜑} ↔ 𝜓))

Syntax hints:   → wi 4   ↔ wb 98   = wceq 1243   ∈ wcel 1393  {cab 2026

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-io 630  ax-5 1336  ax-7 1337  ax-gen 1338  ax-ie1 1382  ax-ie2 1383  ax-8 1395  ax-10 1396  ax-11 1397  ax-i12 1398  ax-bndl 1399  ax-4 1400  ax-17 1419  ax-i9 1423  ax-ial 1427  ax-i5r 1428  ax-ext 2022

This theorem depends on definitions:  df-bi 110  df-tru 1246  df-nf 1350  df-sb 1646  df-clab 2027  df-cleq 2033  df-clel 2036  df-nfc 2167  df-v 2559

This theorem is referenced by:  elab3  2694  elssabg  3902  elrnmptg  4586  elreimasng  4691  fvelrnb  5221