Metamath Proof Explorer

Theorem necon3bbid

Description: Deduction from equality to inequality. (Contributed by NM, 2-Jun-2007)

Ref Expression
Hypothesis necon3bbid.1 
|- ( ph -> ( ps <-> A = B ) )
Assertion necon3bbid 
|- ( ph -> ( -. ps <-> A =/= B ) )

Proof

Step Hyp Ref Expression
1 necon3bbid.1 
 |-  ( ph -> ( ps <-> A = B ) )
2 1 bicomd 
 |-  ( ph -> ( A = B <-> ps ) )
3 2 necon3abid 
 |-  ( ph -> ( A =/= B <-> -. ps ) )
4 3 bicomd 
 |-  ( ph -> ( -. ps <-> A =/= B ) )