Theorem elab4g 3668

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 3487	. 2 ⊢ (𝐴 ∈ 𝐵 → 𝐴 ∈ V)
2		elab4g.1	. . 3 ⊢ (𝑥 = 𝐴 → (𝜑 ↔ 𝜓))
3		elab4g.2	. . 3 ⊢ 𝐵 = {𝑥 ∣ 𝜑}
4	2, 3	elab2g 3665	. 2 ⊢ (𝐴 ∈ V → (𝐴 ∈ 𝐵 ↔ 𝜓))
5	1, 4	biadanii 819	1 ⊢ (𝐴 ∈ 𝐵 ↔ (𝐴 ∈ V ∧ 𝜓))

Syntax hints:   → wi 4   ↔ wb 205   ∧ wa 395   = wceq 1533   ∈ wcel 2098  {cab 2703  Vcvv 3468

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1789  ax-4 1803  ax-5 1905  ax-6 1963  ax-7 2003  ax-8 2100  ax-9 2108  ax-ext 2697

This theorem depends on definitions:  df-bi 206  df-an 396  df-tru 1536  df-ex 1774  df-sb 2060  df-clab 2704  df-cleq 2718  df-clel 2804  df-v 3470

This theorem is referenced by:  isprs  18260  ispos  18277  istrkgc  28209  istrkgb  28210  istrkgcb  28211  istrkge  28212  istrkgl  28213  eulerpartlemt0  33898  istrkg2d  34207