Theorem sbcbid 3100

Description: Formula-building deduction rule for class substitution. (Contributed by NM, 29-Dec-2014.)

Hypotheses

Ref	Expression
sbcbid.1	⊢ Ⅎxφ
sbcbid.2	⊢ (φ → (ψ ↔ χ))

Assertion

Ref	Expression
sbcbid	⊢ (φ → ([̣A / x]̣ψ ↔ [̣A / x]̣χ))

Proof of Theorem sbcbid

Step	Hyp	Ref	Expression
1		sbcbid.1	. . . 4 ⊢ Ⅎxφ
2		sbcbid.2	. . . 4 ⊢ (φ → (ψ ↔ χ))
3	1, 2	abbid 2467	. . 3 ⊢ (φ → {x ∣ ψ} = {x ∣ χ})
4	3	eleq2d 2420	. 2 ⊢ (φ → (A ∈ {x ∣ ψ} ↔ A ∈ {x ∣ χ}))
5		df-sbc 3048	. 2 ⊢ ([̣A / x]̣ψ ↔ A ∈ {x ∣ ψ})
6		df-sbc 3048	. 2 ⊢ ([̣A / x]̣χ ↔ A ∈ {x ∣ χ})
7	4, 5, 6	3bitr4g 279	1 ⊢ (φ → ([̣A / x]̣ψ ↔ [̣A / x]̣χ))

This proof depends on syntax axioms:   → wi 4   ↔ wb 176  Ⅎwnf 1544   ∈ wcel 1710  {cab 2339  [̣wsbc 3047

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-7 1734  ax-11 1746  ax-12 1925  ax-ext 2334

This proof depends on definitions:  df-bi 177  df-an 360  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649  df-clab 2340  df-cleq 2346  df-clel 2349  df-sbc 3048

This theorem is used by:  sbcbidv  3101  csbeq2d  3161