Theorem elab4g 3644

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 3478	. 2 ⊢ (𝐴 ∈ 𝐵 → 𝐴 ∈ V)
2		elab4g.1	. . 3 ⊢ (𝑥 = 𝐴 → (𝜑 ↔ 𝜓))
3		elab4g.2	. . 3 ⊢ 𝐵 = {𝑥 ∣ 𝜑}
4	2, 3	elab2g 3641	. 2 ⊢ (𝐴 ∈ V → (𝐴 ∈ 𝐵 ↔ 𝜓))
5	1, 4	biadanii 834	1 ⊢ (𝐴 ∈ 𝐵 ↔ (𝐴 ∈ V ∧ 𝜓))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401   = wceq 1570   ∈ wcel 2146  {cab 2743  Vcvv 3457

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2148  ax-9 2156  ax-ext 2737

This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1573  df-ex 1813  df-sb 2100  df-clab 2744  df-cleq 2757  df-clel 2840  df-v 3459

This theorem is used by:  isprs  18374  ispos  18392  istrkgc  28774  istrkgb  28775  istrkgcb  28776  istrkge  28777  istrkgl  28778  eulerpartlemt0  34824  istrkg2d  35118