Metamath Proof Explorer

Theorem neeq1i

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 neeq1i 
|- ( A =/= C <-> B =/= C )

Proof

Step Hyp Ref Expression
1 neeq1i.1 
 |-  A = B
2 1 eqeq1i 
 |-  ( A = C <-> B = C )
3 2 necon3bii 
 |-  ( A =/= C <-> B =/= C )