Metamath Proof Explorer


Theorem minel

Description: A minimum element of a class has no elements in common with the class. (Contributed by NM, 22-Jun-1994) (Proof shortened by JJ, 14-Jul-2021)

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

Proof

Step Hyp Ref Expression
1 inelcm
 |-  ( ( A e. C /\ A e. B ) -> ( C i^i B ) =/= (/) )
2 1 expcom
 |-  ( A e. B -> ( A e. C -> ( C i^i B ) =/= (/) ) )
3 2 necon2bd
 |-  ( A e. B -> ( ( C i^i B ) = (/) -> -. A e. C ) )
4 3 imp
 |-  ( ( A e. B /\ ( C i^i B ) = (/) ) -> -. A e. C )