Metamath Proof Explorer

Theorem pssne

Description: Two classes in a proper subclass relationship are not equal. (Contributed by NM, 16-Feb-2015)

Ref Expression
Assertion pssne 
|- ( A C. B -> A =/= B )

Proof

Step Hyp Ref Expression
1 df-pss 
 |-  ( A C. B <-> ( A C_ B /\ A =/= B ) )
2 1 simprbi 
 |-  ( A C. B -> A =/= B )