Metamath Proof Explorer

Theorem necon3abid

Description: Deduction from equality to inequality. (Contributed by NM, 21-Mar-2007)

Ref Expression
Hypothesis necon3abid.1 
|- ( ph -> ( A = B <-> ps ) )
Assertion necon3abid 
|- ( ph -> ( A =/= B <-> -. ps ) )

Proof

Step Hyp Ref Expression
1 necon3abid.1 
 |-  ( ph -> ( A = B <-> ps ) )
2 df-ne 
 |-  ( A =/= B <-> -. A = B )
3 1 notbid 
 |-  ( ph -> ( -. A = B <-> -. ps ) )
4 2 3 syl5bb 
 |-  ( ph -> ( A =/= B <-> -. ps ) )