Theorem sbcng 3087

Description: Move negation in and out of class substitution. (Contributed by NM, 16-Jan-2004.)

Assertion

Ref	Expression
sbcng	⊢ (A ∈ V → ([̣A / x]̣ ¬ φ ↔ ¬ [̣A / x]̣φ))

Proof of Theorem sbcng

Dummy variable y is distinct from all other variables.

Step	Hyp	Ref	Expression
1		dfsbcq2 3050	. 2 ⊢ (y = A → ([y / x] ¬ φ ↔ [̣A / x]̣ ¬ φ))
2		dfsbcq2 3050	. . 3 ⊢ (y = A → ([y / x]φ ↔ [̣A / x]̣φ))
3	2	notbid 285	. 2 ⊢ (y = A → (¬ [y / x]φ ↔ ¬ [̣A / x]̣φ))
4		sbn 2062	. 2 ⊢ ([y / x] ¬ φ ↔ ¬ [y / x]φ)
5	1, 3, 4	vtoclbg 2916	1 ⊢ (A ∈ V → ([̣A / x]̣ ¬ φ ↔ ¬ [̣A / x]̣φ))

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 176   = wceq 1642  [wsb 1648   ∈ wcel 1710  [̣wsbc 3047

This theorem was proved from 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 theorem depends on definitions:  df-bi 177  df-or 359  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-nfc 2479  df-v 2862  df-sbc 3048

This theorem is referenced by:  sbcrext  3120  sbcnel12g  3154  sbcne12g  3155