Theorem bisbcdv 1468

Description: Formula-building deduction rule for class substitution.

Hypothesis

Ref	Expression
bisbcdv.1	⊢ (φ → (ψ ↔ χ))

Assertion

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

Distinct variable group(s):   φ,x

Proof of Theorem bisbcdv

Step	Hyp	Ref	Expression
1		a4sbc 1444	. . . 4 ⊢ (A ∈ B → (∀x(ψ ↔ χ) → [A / x](ψ ↔ χ)))
2		bisbcdv.1	. . . . 5 ⊢ (φ → (ψ ↔ χ))
3	2	19.21aiv 943	. . . 4 ⊢ (φ → ∀x(ψ ↔ χ))
4	1, 3	syl5 22	. . 3 ⊢ (A ∈ B → (φ → [A / x](ψ ↔ χ)))
5		sbcbi 1463	. . 3 ⊢ (A ∈ B → ([A / x](ψ ↔ χ) ↔ ([A / x]ψ ↔ [A / x]χ)))
6	4, 5	sylibd 177	. 2 ⊢ (A ∈ B → (φ → ([A / x]ψ ↔ [A / x]χ)))
7	6	imp 277	1 ⊢ ((A ∈ B ∧ φ) → ([A / x]ψ ↔ [A / x]χ))

Syntax hints:   → wi 2   ↔ wb 127   ∧ wa 196  ∀wal 672   ∈ wcel 1092  [wsbc 1440

This theorem is referenced by:  sbcgf 1469

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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