Metamath Proof Explorer


Theorem isclat

Description: The predicate "is a complete lattice". (Contributed by NM, 18-Oct-2012) (Revised by NM, 12-Sep-2018)

Ref Expression
Hypotheses isclat.b ⊢ B = Base K
isclat.u ⊢ U = lub ⁡ K
isclat.g ⊢ G = glb ⁡ K
Assertion isclat ⊢ K ∈ CLat ↔ K ∈ Poset ∧ dom ⁡ U = 𝒫 B ∧ dom ⁡ G = 𝒫 B

Proof

Step Hyp Ref Expression
1 isclat.b ⊢ B = Base K
2 isclat.u ⊢ U = lub ⁡ K
3 isclat.g ⊢ G = glb ⁡ K
4 fveq2 ⊢ l = K → lub ⁡ l = lub ⁡ K
5 4 2 eqtr4di ⊢ l = K → lub ⁡ l = U
6 5 dmeqd ⊢ l = K → dom ⁡ lub ⁡ l = dom ⁡ U
7 fveq2 ⊢ l = K → Base l = Base K
8 7 1 eqtr4di ⊢ l = K → Base l = B
9 8 pweqd ⊢ l = K → 𝒫 Base l = 𝒫 B
10 6 9 eqeq12d ⊢ l = K → dom ⁡ lub ⁡ l = 𝒫 Base l ↔ dom ⁡ U = 𝒫 B
11 fveq2 ⊢ l = K → glb ⁡ l = glb ⁡ K
12 11 3 eqtr4di ⊢ l = K → glb ⁡ l = G
13 12 dmeqd ⊢ l = K → dom ⁡ glb ⁡ l = dom ⁡ G
14 13 9 eqeq12d ⊢ l = K → dom ⁡ glb ⁡ l = 𝒫 Base l ↔ dom ⁡ G = 𝒫 B
15 10 14 anbi12d ⊢ l = K → dom ⁡ lub ⁡ l = 𝒫 Base l ∧ dom ⁡ glb ⁡ l = 𝒫 Base l ↔ dom ⁡ U = 𝒫 B ∧ dom ⁡ G = 𝒫 B
16 df-clat ⊢ CLat = l ∈ Poset | dom ⁡ lub ⁡ l = 𝒫 Base l ∧ dom ⁡ glb ⁡ l = 𝒫 Base l
17 15 16 elrab2 ⊢ K ∈ CLat ↔ K ∈ Poset ∧ dom ⁡ U = 𝒫 B ∧ dom ⁡ G = 𝒫 B