Theorem elab4g 2990

Description: Membership in a class abstraction, using implicit substitution. (Contributed by NM, 17-Oct-2012.)

Hypotheses

Ref	Expression
elab4g.1	⊢ (x = A → (φ ↔ ψ))
elab4g.2	⊢ B = {x ∣ φ}

Assertion

Ref	Expression
elab4g	⊢ (A ∈ B ↔ (A ∈ V ∧ ψ))

Distinct variable groups:   ψ,x   x,A

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

Proof of Theorem elab4g

Step	Hyp	Ref	Expression
1		elex 2868	. 2 ⊢ (A ∈ B → A ∈ V)
2		elab4g.1	. . 3 ⊢ (x = A → (φ ↔ ψ))
3		elab4g.2	. . 3 ⊢ B = {x ∣ φ}
4	2, 3	elab2g 2988	. 2 ⊢ (A ∈ V → (A ∈ B ↔ ψ))
5	1, 4	biadan2 623	1 ⊢ (A ∈ B ↔ (A ∈ V ∧ ψ))

Syntax hints:   → wi 4   ↔ wb 176   ∧ wa 358   = 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: (None)