Theorem elabf 2985

Description: Membership in a class abstraction, using implicit substitution. (Contributed by NM, 1-Aug-1994.) (Revised by Mario Carneiro, 12-Oct-2016.)

Hypotheses

Ref	Expression
elabf.1	⊢ Ⅎxψ
elabf.2	⊢ A ∈ V
elabf.3	⊢ (x = A → (φ ↔ ψ))

Assertion

Ref	Expression
elabf	⊢ (A ∈ {x ∣ φ} ↔ ψ)

Distinct variable group:   x,A

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

Proof of Theorem elabf

Step	Hyp	Ref	Expression
1		elabf.2	. 2 ⊢ A ∈ V
2		nfcv 2490	. . 3 ⊢ ℲxA
3		elabf.1	. . 3 ⊢ Ⅎxψ
4		elabf.3	. . 3 ⊢ (x = A → (φ ↔ ψ))
5	2, 3, 4	elabgf 2984	. 2 ⊢ (A ∈ V → (A ∈ {x ∣ φ} ↔ ψ))
6	1, 5	ax-mp 5	1 ⊢ (A ∈ {x ∣ φ} ↔ ψ)

Syntax hints:   → wi 4   ↔ wb 176  Ⅎwnf 1544   = wceq 1642   ∈ wcel 1710  {cab 2339  Vcvv 2860

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-7 1734  ax-11 1746  ax-12 1925  ax-ext 2334

This theorem depends on definitions:  df-bi 177  df-or 359  df-an 360  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649  df-clab 2340  df-cleq 2346  df-clel 2349  df-nfc 2479  df-v 2862

This theorem is referenced by:  elab  2986