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 ∧ 𝑆𝐵 ) → 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 𝐵𝑆𝐵 )
5 4 biimpri ( 𝑆𝐵𝑆 ∈ 𝒫 𝐵 )
6 5 adantl ( ( 𝐾 ∈ CLat ∧ 𝑆𝐵 ) → 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 ∧ 𝑆𝐵 ) → dom 𝐺 = 𝒫 𝐵 )
12 6 11 eleqtrrd ( ( 𝐾 ∈ CLat ∧ 𝑆𝐵 ) → 𝑆 ∈ dom 𝐺 )