Theorem elab4g 3642

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 3476	. 2 ⊢ (𝐴 ∈ 𝐵 → 𝐴 ∈ V)
2		elab4g.1	. . 3 ⊢ (𝑥 = 𝐴 → (𝜑 ↔ 𝜓))
3		elab4g.2	. . 3 ⊢ 𝐵 = {𝑥 ∣ 𝜑}
4	2, 3	elab2g 3639	. 2 ⊢ (𝐴 ∈ V → (𝐴 ∈ 𝐵 ↔ 𝜓))
5	1, 4	biadanii 833	1 ⊢ (𝐴 ∈ 𝐵 ↔ (𝐴 ∈ V ∧ 𝜓))

Syntax hints:   → wi 4   ↔ wb 209   ∧ wa 400   = wceq 1570   ∈ wcel 2143  {cab 2741  Vcvv 3455

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1825  ax-4 1839  ax-5 1940  ax-6 1997  ax-7 2038  ax-8 2145  ax-9 2153  ax-ext 2735

This theorem depends on definitions:  df-bi 210  df-an 401  df-tru 1573  df-ex 1810  df-sb 2097  df-clab 2742  df-cleq 2755  df-clel 2838  df-v 3457

This theorem is referenced by:  isprs  18347  ispos  18365  istrkgc  28723  istrkgb  28724  istrkgcb  28725  istrkge  28726  istrkgl  28727  eulerpartlemt0  34759  istrkg2d  35053