Theorem elab2g 1418

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

Hypotheses

Ref	Expression
elab2g.1	⊢ (x = A → (φ ↔ ψ))
elab2g.2	⊢ B = {x∣φ}

Assertion

Ref	Expression
elab2g	⊢ (A ∈ C → (A ∈ B ↔ ψ))

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

Proof of Theorem elab2g

Step	Hyp	Ref	Expression
1		elab2g.1	. . 3 ⊢ (x = A → (φ ↔ ψ))
2	1	elabg 1417	. 2 ⊢ (A ∈ C → (A ∈ {x∣φ} ↔ ψ))
3		elab2g.2	. . 3 ⊢ B = {x∣φ}
4	3	eleq2i 1153	. 2 ⊢ (A ∈ B ↔ A ∈ {x∣φ})
5	2, 4	syl5bb 410	1 ⊢ (A ∈ C → (A ∈ B ↔ ψ))

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

This theorem is referenced by:  elab2 1419  eldif 1496  elun 1601  elin 1635  elprg 1822  eluni 1922  eliun 1998  eliin 1999  elong 2207  tfrlem12 2960  elnp 3886  hcauchy 5103  sh 5116  closedsub 5128  ch2 5149  stelt 5671

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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