Metamath Proof Explorer

Theorem sgrp0

Description: Any set with an empty base set and any group operation is a semigroup. (Contributed by AV, 28-Aug-2021)

Ref Expression
Assertion sgrp0 
|- ( ( M e. V /\ ( Base ` M ) = (/) ) -> M e. Smgrp )

Proof

Step Hyp Ref Expression
1 mgm0 
 |-  ( ( M e. V /\ ( Base ` M ) = (/) ) -> M e. Mgm )
2 rzal 
 |-  ( ( Base ` M ) = (/) -> A. x e. ( Base ` M ) A. y e. ( Base ` M ) A. z e. ( Base ` M ) ( ( x ( +g ` M ) y ) ( +g ` M ) z ) = ( x ( +g ` M ) ( y ( +g ` M ) z ) ) )
3 2 adantl 
 |-  ( ( M e. V /\ ( Base ` M ) = (/) ) -> A. x e. ( Base ` M ) A. y e. ( Base ` M ) A. z e. ( Base ` M ) ( ( x ( +g ` M ) y ) ( +g ` M ) z ) = ( x ( +g ` M ) ( y ( +g ` M ) z ) ) )
4 eqid 
 |-  ( Base ` M ) = ( Base ` M )
5 eqid 
 |-  ( +g ` M ) = ( +g ` M )
6 4 5 issgrp 
 |-  ( M e. Smgrp <-> ( M e. Mgm /\ A. x e. ( Base ` M ) A. y e. ( Base ` M ) A. z e. ( Base ` M ) ( ( x ( +g ` M ) y ) ( +g ` M ) z ) = ( x ( +g ` M ) ( y ( +g ` M ) z ) ) ) )
7 1 3 6 sylanbrc 
 |-  ( ( M e. V /\ ( Base ` M ) = (/) ) -> M e. Smgrp )