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 syl6eqr 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 syl6eqr 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 syl6eqr 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