Theorem elab4g 3640

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 3475	. 2 ⊢ (𝐴 ∈ 𝐵 → 𝐴 ∈ V)
2		elab4g.1	. . 3 ⊢ (𝑥 = 𝐴 → (𝜑 ↔ 𝜓))
3		elab4g.2	. . 3 ⊢ 𝐵 = {𝑥 ∣ 𝜑}
4	2, 3	elab2g 3637	. 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 2147  {cab 2741  Vcvv 3454

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 1945  ax-6 2002  ax-7 2043  ax-8 2149  ax-9 2157  ax-ext 2735

This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1573  df-ex 1813  df-sb 2102  df-clab 2742  df-cleq 2755  df-clel 2839  df-v 3456

This theorem is used by:  isprs  18520  ispos  18538  istrkgc  28973  istrkgb  28974  istrkgcb  28975  istrkge  28976  istrkgl  28977  eulerpartlemt0  35112  istrkg2d  35406