Theorem sbcal 1464

Description: Move universal quantifier in and out of class substitution.

Assertion

Ref	Expression
sbcal	⊢ (A ∈ B → ([A / y]∀xφ ↔ ∀x[A / y]φ))

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

Proof of Theorem sbcal

Step	Hyp	Ref	Expression
1		dfsbcq 1442	. 2 ⊢ (z = A → ([z / y]∀xφ ↔ [A / y]∀xφ))
2		dfsbcq 1442	. . 3 ⊢ (z = A → ([z / y]φ ↔ [A / y]φ))
3	2	bialdv 935	. 2 ⊢ (z = A → (∀x[z / y]φ ↔ ∀x[A / y]φ))
4		sbal 997	. 2 ⊢ ([z / y]∀xφ ↔ ∀x[z / y]φ)
5	1, 3, 4	vtoclbg 1384	1 ⊢ (A ∈ B → ([A / y]∀xφ ↔ ∀x[A / y]φ))

Syntax hints:   → wi 2   ↔ wb 127  ∀wal 672  [wsb 852   = wceq 1091   ∈ wcel 1092  [wsbc 1440

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-or 197  df-an 198  df-ex 679  df-sb 853  df-clab 1093  df-cleq 1097  df-clel 1099  df-v 1349  df-sbc 1441

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