Metamath Proof Explorer

Theorem necon3abii

Description: Deduction from equality to inequality. (Contributed by NM, 9-Nov-2007)

Ref Expression
Hypothesis necon3abii.1 
|- ( A = B <-> ph )
Assertion necon3abii 
|- ( A =/= B <-> -. ph )

Proof

Step Hyp Ref Expression
1 necon3abii.1 
 |-  ( A = B <-> ph )
2 df-ne 
 |-  ( A =/= B <-> -. A = B )
3 2 1 xchbinx 
 |-  ( A =/= B <-> -. ph )