Metamath Proof Explorer


Theorem necon3bid

Description: Deduction from equality to inequality. (Contributed by NM, 23-Feb-2005) (Proof shortened by Andrew Salmon, 25-May-2011)

Ref Expression
Hypothesis necon3bid.1
|- ( ph -> ( A = B <-> C = D ) )
Assertion necon3bid
|- ( ph -> ( A =/= B <-> C =/= D ) )

Proof

Step Hyp Ref Expression
1 necon3bid.1
 |-  ( ph -> ( A = B <-> C = D ) )
2 df-ne
 |-  ( A =/= B <-> -. A = B )
3 1 necon3bbid
 |-  ( ph -> ( -. A = B <-> C =/= D ) )
4 2 3 bitrid
 |-  ( ph -> ( A =/= B <-> C =/= D ) )