Metamath Proof Explorer


Theorem clatglbcl2

Description: Any subset of the base set has a GLB in a complete lattice. (Contributed by NM, 13-Sep-2018)

Ref Expression
Hypotheses clatglbcl.b B = Base K
clatglbcl.g G = glb K
Assertion clatglbcl2 K CLat S B S dom G

Proof

Step Hyp Ref Expression
1 clatglbcl.b B = Base K
2 clatglbcl.g G = glb K
3 1 fvexi B V
4 3 elpw2 S 𝒫 B S B
5 4 biimpri S B S 𝒫 B
6 5 adantl K CLat S B S 𝒫 B
7 eqid lub K = lub K
8 1 7 2 isclat K CLat K Poset dom lub K = 𝒫 B dom G = 𝒫 B
9 simprr K Poset dom lub K = 𝒫 B dom G = 𝒫 B dom G = 𝒫 B
10 8 9 sylbi K CLat dom G = 𝒫 B
11 10 adantr K CLat S B dom G = 𝒫 B
12 6 11 eleqtrrd K CLat S B S dom G