Theorem elabg 1417

Description: Membership in a class abstraction with implicit substitution. Compare Theorem 6.13 of [Quine] p. 44.

Hypothesis

Ref	Expression
elabg.1	⊢ (x = A → (φ ↔ ψ))

Assertion

Ref	Expression
elabg	⊢ (A ∈ B → (A ∈ {x∣φ} ↔ ψ))

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

Proof of Theorem elabg

Step	Hyp	Ref	Expression
1		ax-17 925	. 2 ⊢ (y ∈ A → ∀x y ∈ A)
2		ax-17 925	. 2 ⊢ (ψ → ∀xψ)
3		elabg.1	. 2 ⊢ (x = A → (φ ↔ ψ))
4	1, 2, 3	elabgf 1416	1 ⊢ (A ∈ B → (A ∈ {x∣φ} ↔ ψ))

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

This theorem is referenced by:  elab2g 1418  elab3g 1420  finds 2397  scott0 3542  nnind 4434

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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