Theorem sbcbi 1463

Description: Distribution of class substitution over biconditional. (Contributed by Raph Levien, 10-Apr-04.)

Assertion

Ref	Expression
sbcbi	⊢ (A ∈ B → ([A / x](φ ↔ ψ) ↔ ([A / x]φ ↔ [A / x]ψ)))

Proof of Theorem sbcbi

Step	Hyp	Ref	Expression
1		dfsbcq 1442	. 2 ⊢ (y = A → ([y / x](φ ↔ ψ) ↔ [A / x](φ ↔ ψ)))
2		dfsbcq 1442	. . 3 ⊢ (y = A → ([y / x]φ ↔ [A / x]φ))
3		dfsbcq 1442	. . 3 ⊢ (y = A → ([y / x]ψ ↔ [A / x]ψ))
4	2, 3	bibi12d 477	. 2 ⊢ (y = A → (([y / x]φ ↔ [y / x]ψ) ↔ ([A / x]φ ↔ [A / x]ψ)))
5		sbbi 890	. 2 ⊢ ([y / x](φ ↔ ψ) ↔ ([y / x]φ ↔ [y / x]ψ))
6	1, 4, 5	vtoclbg 1384	1 ⊢ (A ∈ B → ([A / x](φ ↔ ψ) ↔ ([A / x]φ ↔ [A / x]ψ)))

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

This theorem is referenced by:  bisbcdv 1468

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-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  df-sbc 1441

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