Theorem elab2 1419

Description: Membership in a class abstraction, using implicit substitution.

Hypotheses

Ref	Expression
elab2.1	⊢ A ∈ V
elab2.2	⊢ (x = A → (φ ↔ ψ))
elab2.3	⊢ B = {x∣φ}

Assertion

Ref	Expression
elab2	⊢ (A ∈ B ↔ ψ)

DiEtinct variable group(s):   ψ,x   x,A

Proof of Theorem elab2

Step	Hyp	Ref	Expression
1		elab2.1	. 2 ⊢ A ∈ V
2		elab2.2	. . 3 ⊢ (x = A → (φ ↔ ψ))
3		elab2.3	. . 3 ⊢ B = {x∣φ}
4	2, 3	elab2g 1418	. 2 ⊢ (A ∈ V → (A ∈ B ↔ ψ))
5	1, 4	ax-mp 6	1 ⊢ (A ∈ B ↔ ψ)

Syntax hints:   → wi 2   ↔ wb 127  {cab 1090   = wceq 1091   ∈ wcel 1092  Vcvv 1348

This theorem is referenced by:  elint 1971  elom 2375  eldm 2527  elrn 2562  elima 2606  elec 3216  elqs 3227  aceq3lem 3555  aceq5lem4 3561  kmlem8 3587  1pr 3911  ltexprlem3 3938  ltexprlem4 3939  reclem2pr 3951  suppsr 4016  suppsr3 4018  supsrlem4 4022  supre 4054  infxpidmlem2 4934

This theorem was proved from axioms:  ax-1 3  ax-2 4  ax-3 5  ax-mp 6  ax-4 673  ax-5 674  ax-6 675  ax-7 676  ax-gen 677  ax-8 798  ax-9 799  ax-10 800  ax-11 801  ax-12 802  ax-16 922  ax-17 925  ax-ext 1074

This theorem depends on definitions:  df-bi 128  df-an 198  df-ex 679  df-sb 853  df-clab 1093  df-cleq 1097  df-clel 1099  df-v 1349

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