Theorem elab4g 3685

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 3498	. 2 ⊢ (𝐴 ∈ 𝐵 → 𝐴 ∈ V)
2		elab4g.1	. . 3 ⊢ (𝑥 = 𝐴 → (𝜑 ↔ 𝜓))
3		elab4g.2	. . 3 ⊢ 𝐵 = {𝑥 ∣ 𝜑}
4	2, 3	elab2g 3682	. 2 ⊢ (𝐴 ∈ V → (𝐴 ∈ 𝐵 ↔ 𝜓))
5	1, 4	biadanii 822	1 ⊢ (𝐴 ∈ 𝐵 ↔ (𝐴 ∈ V ∧ 𝜓))

Syntax hints:   → wi 4   ↔ wb 206   ∧ wa 395   = wceq 1536   ∈ wcel 2105  {cab 2711  Vcvv 3477

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1791  ax-4 1805  ax-5 1907  ax-6 1964  ax-7 2004  ax-8 2107  ax-9 2115  ax-ext 2705

This theorem depends on definitions:  df-bi 207  df-an 396  df-tru 1539  df-ex 1776  df-sb 2062  df-clab 2712  df-cleq 2726  df-clel 2813  df-v 3479

This theorem is referenced by:  isprs  18353  ispos  18371  istrkgc  28476  istrkgb  28477  istrkgcb  28478  istrkge  28479  istrkgl  28480  eulerpartlemt0  34350  istrkg2d  34659