Theorem sbn 882

Description: Negation inside and outside of substitution are equivalent.

Assertion

Ref	Expression
sbn	⊢ ([y / x] ¬ φ ↔ ¬ [y / x]φ)

Proof of Theorem sbn

Step	Hyp	Ref	Expression
1		sbn1 880	. 2 ⊢ ([y / x] ¬ φ → ¬ [y / x]φ)
2		sbn2 881	. 2 ⊢ (¬ [y / x]φ → [y / x] ¬ φ)
3	1, 2	impbi 139	1 ⊢ ([y / x] ¬ φ ↔ ¬ [y / x]φ)

Syntax hints:  ¬ wn 1   ↔ wb 127  [wsb 852

This theorem is referenced by:  sb5y 883  sbi2 885  sbor 887  sban 889  sbea4 894  sb8e 919  sbex 998  sbcn 1459  difab 1693

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

This theorem depends on definitions:  df-bi 128  df-an 198  df-ex 679  df-sb 853

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