Metamath Proof Explorer

Theorem neeq12i

Description: Inference for inequality. (Contributed by NM, 24-Jul-2012) (Proof shortened by Wolf Lammen, 25-Nov-2019)

Ref Expression
Hypotheses neeq1i.1 
|- A = B
neeq12i.2 
|- C = D
Assertion neeq12i 
|- ( A =/= C <-> B =/= D )

Proof

Step Hyp Ref Expression
1 neeq1i.1 
 |-  A = B
2 neeq12i.2 
 |-  C = D
3 1 2 eqeq12i 
 |-  ( A = C <-> B = D )
4 3 necon3bii 
 |-  ( A =/= C <-> B =/= D )