Metamath Proof Explorer


Theorem clatglbcl

Description: Any subset of the base set has a GLB in a complete lattice. (Contributed by NM, 14-Sep-2011)

Ref Expression
Hypotheses clatglbcl.b
|- B = ( Base ` K )
clatglbcl.g
|- G = ( glb ` K )
Assertion clatglbcl
|- ( ( K e. CLat /\ S C_ B ) -> ( G ` S ) e. B )

Proof

Step Hyp Ref Expression
1 clatglbcl.b
 |-  B = ( Base ` K )
2 clatglbcl.g
 |-  G = ( glb ` K )
3 eqid
 |-  ( lub ` K ) = ( lub ` K )
4 1 3 2 clatlem
 |-  ( ( K e. CLat /\ S C_ B ) -> ( ( ( lub ` K ) ` S ) e. B /\ ( G ` S ) e. B ) )
5 4 simprd
 |-  ( ( K e. CLat /\ S C_ B ) -> ( G ` S ) e. B )