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=BaseK
clatglbcl.g G=glbK
Assertion clatglbcl KCLatSBGSB

Proof

Step Hyp Ref Expression
1 clatglbcl.b B=BaseK
2 clatglbcl.g G=glbK
3 eqid lubK=lubK
4 1 3 2 clatlem KCLatSBlubKSBGSB
5 4 simprd KCLatSBGSB