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 CLat S B G S 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 CLat S B lub K S B G S B
5 4 simprd K CLat S B G S B