Metamath Proof Explorer


Theorem disjel

Description: A set can't belong to both members of disjoint classes. (Contributed by NM, 28-Feb-2015)

Ref Expression
Assertion disjel
|- ( ( ( A i^i B ) = (/) /\ C e. A ) -> -. C e. B )

Proof

Step Hyp Ref Expression
1 disj3
 |-  ( ( A i^i B ) = (/) <-> A = ( A \ B ) )
2 eleq2
 |-  ( A = ( A \ B ) -> ( C e. A <-> C e. ( A \ B ) ) )
3 eldifn
 |-  ( C e. ( A \ B ) -> -. C e. B )
4 2 3 syl6bi
 |-  ( A = ( A \ B ) -> ( C e. A -> -. C e. B ) )
5 1 4 sylbi
 |-  ( ( A i^i B ) = (/) -> ( C e. A -> -. C e. B ) )
6 5 imp
 |-  ( ( ( A i^i B ) = (/) /\ C e. A ) -> -. C e. B )