Metamath Proof Explorer

Theorem neeq2i

Description: Inference for inequality. (Contributed by NM, 29-Apr-2005) (Proof shortened by Wolf Lammen, 19-Nov-2019)

Ref Expression
Hypothesis neeq1i.1 
|- A = B
Assertion neeq2i 
|- ( C =/= A <-> C =/= B )

Proof

Step Hyp Ref Expression
1 neeq1i.1 
 |-  A = B
2 1 eqeq2i 
 |-  ( C = A <-> C = B )
3 2 necon3bii 
 |-  ( C =/= A <-> C =/= B )