This is an inofficial mirror of http://metamath.tirix.org for personal testing of a visualizer extension only.

Metamath Proof Explorer

Theorem neeq1d

Description: Deduction for inequality. (Contributed by NM, 25-Oct-1999) (Proof shortened by Wolf Lammen, 19-Nov-2019)

Ref Expression
Hypothesis neeq1d.1 
|- ( ph -> A = B )
Assertion neeq1d 
|- ( ph -> ( A =/= C <-> B =/= C ) )

Proof

Step Hyp Ref Expression
1 neeq1d.1 
 |-  ( ph -> A = B )
2 1 eqeq1d 
 |-  ( ph -> ( A = C <-> B = C ) )
3 2 necon3bid 
 |-  ( ph -> ( A =/= C <-> B =/= C ) )