Theorem elabg 2987

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 2490	. 2 ⊢ ℲxA
2		nfv 1619	. 2 ⊢ Ⅎxψ
3		elabg.1	. 2 ⊢ (x = A → (φ ↔ ψ))
4	1, 2, 3	elabgf 2984	1 ⊢ (A ∈ V → (A ∈ {x ∣ φ} ↔ ψ))

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

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:  elab2g  2988  intmin3  3955  peano5  4410  findsd  4411  nnadjoin  4521  spfininduct  4541  vfinspclt  4553  elxpi  4801  elimasn  5020  clos1induct  5881  clos1is  5882  elmapg  6013  elce  6176  spacis  6289