Metamath Proof Explorer


Theorem nbbn

Description: Move negation outside of biconditional. Compare Theorem *5.18 of WhiteheadRussell p. 124. (Contributed by NM, 27-Jun-2002) (Proof shortened by Wolf Lammen, 20-Sep-2013)

Ref Expression
Assertion nbbn
|- ( ( -. ph <-> ps ) <-> -. ( ph <-> ps ) )

Proof

Step Hyp Ref Expression
1 xor3
 |-  ( -. ( ph <-> ps ) <-> ( ph <-> -. ps ) )
2 con2bi
 |-  ( ( ph <-> -. ps ) <-> ( ps <-> -. ph ) )
3 bicom
 |-  ( ( ps <-> -. ph ) <-> ( -. ph <-> ps ) )
4 1 2 3 3bitrri
 |-  ( ( -. ph <-> ps ) <-> -. ( ph <-> ps ) )