Metamath Proof Explorer

Theorem disj1

Description: Two ways of saying that two classes are disjoint (have no members in common). (Contributed by NM, 19-Aug-1993)

Ref Expression
Assertion disj1 
|- ( ( A i^i B ) = (/) <-> A. x ( x e. A -> -. x e. B ) )

Proof

Step Hyp Ref Expression
1 disj 
 |-  ( ( A i^i B ) = (/) <-> A. x e. A -. x e. B )
2 df-ral 
 |-  ( A. x e. A -. x e. B <-> A. x ( x e. A -> -. x e. B ) )
3 1 2 bitri 
 |-  ( ( A i^i B ) = (/) <-> A. x ( x e. A -> -. x e. B ) )