Theorem elab3g 2890

Description: Membership in a class abstraction, with a weaker antecedent than elabg 2885. (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 2319	. 2 ⊢ Ⅎ𝑥𝐴
2		nfv 1528	. 2 ⊢ Ⅎ𝑥𝜓
3		elab3g.1	. 2 ⊢ (𝑥 = 𝐴 → (𝜑 ↔ 𝜓))
4	1, 2, 3	elab3gf 2889	1 ⊢ ((𝜓 → 𝐴 ∈ 𝐵) → (𝐴 ∈ {𝑥 ∣ 𝜑} ↔ 𝜓))

Syntax hints:   → wi 4   ↔ wb 105   = wceq 1353   ∈ wcel 2148  {cab 2163

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 709  ax-5 1447  ax-7 1448  ax-gen 1449  ax-ie1 1493  ax-ie2 1494  ax-8 1504  ax-10 1505  ax-11 1506  ax-i12 1507  ax-bndl 1509  ax-4 1510  ax-17 1526  ax-i9 1530  ax-ial 1534  ax-i5r 1535  ax-ext 2159

This theorem depends on definitions:  df-bi 117  df-tru 1356  df-nf 1461  df-sb 1763  df-clab 2164  df-cleq 2170  df-clel 2173  df-nfc 2308  df-v 2741

This theorem is referenced by:  elab3  2891  elssabg  4150  elrnmptg  4881  elrelimasn  4996  fvelrnb  5565  elmapg  6663  lspsnel  13508