Metamath Proof Explorer


Theorem bj-disjcsn

Description: A class is disjoint from its singleton. A consequence of regularity. Shorter proof than bnj521 and does not depend on df-ne . (Contributed by BJ, 4-Apr-2019)

Ref Expression
Assertion bj-disjcsn
|- ( A i^i { A } ) = (/)

Proof

Step Hyp Ref Expression
1 elirr
 |-  -. A e. A
2 disjsn
 |-  ( ( A i^i { A } ) = (/) <-> -. A e. A )
3 1 2 mpbir
 |-  ( A i^i { A } ) = (/)