Metamath Proof Explorer


Theorem isclatd

Description: The predicate "is a complete lattice" (deduction form). (Contributed by Zhi Wang, 29-Sep-2024)

Ref Expression
Hypotheses isclatd.b φB=BaseK
isclatd.u φU=lubK
isclatd.g φG=glbK
isclatd.k φKPoset
isclatd.1 φsBsdomU
isclatd.2 φsBsdomG
Assertion isclatd φKCLat

Proof

Step Hyp Ref Expression
1 isclatd.b φB=BaseK
2 isclatd.u φU=lubK
3 isclatd.g φG=glbK
4 isclatd.k φKPoset
5 isclatd.1 φsBsdomU
6 isclatd.2 φsBsdomG
7 eqid BaseK=BaseK
8 eqid K=K
9 eqid lubK=lubK
10 biid ytyKxzBaseKytyKzxKzytyKxzBaseKytyKzxKz
11 7 8 9 10 4 lubdm φdomlubK=t𝒫BaseK|∃!xBaseKytyKxzBaseKytyKzxKz
12 ssrab2 t𝒫BaseK|∃!xBaseKytyKxzBaseKytyKzxKz𝒫BaseK
13 11 12 eqsstrdi φdomlubK𝒫BaseK
14 elpwi s𝒫BsB
15 14 5 sylan2 φs𝒫BsdomU
16 15 ralrimiva φs𝒫BsdomU
17 dfss3 𝒫BdomUs𝒫BsdomU
18 16 17 sylibr φ𝒫BdomU
19 1 pweqd φ𝒫B=𝒫BaseK
20 2 dmeqd φdomU=domlubK
21 18 19 20 3sstr3d φ𝒫BaseKdomlubK
22 13 21 eqssd φdomlubK=𝒫BaseK
23 eqid glbK=glbK
24 biid ytxKyzBaseKytzKyzKxytxKyzBaseKytzKyzKx
25 7 8 23 24 4 glbdm φdomglbK=t𝒫BaseK|∃!xBaseKytxKyzBaseKytzKyzKx
26 ssrab2 t𝒫BaseK|∃!xBaseKytxKyzBaseKytzKyzKx𝒫BaseK
27 25 26 eqsstrdi φdomglbK𝒫BaseK
28 14 6 sylan2 φs𝒫BsdomG
29 28 ralrimiva φs𝒫BsdomG
30 dfss3 𝒫BdomGs𝒫BsdomG
31 29 30 sylibr φ𝒫BdomG
32 3 dmeqd φdomG=domglbK
33 31 19 32 3sstr3d φ𝒫BaseKdomglbK
34 27 33 eqssd φdomglbK=𝒫BaseK
35 7 9 23 isclat KCLatKPosetdomlubK=𝒫BaseKdomglbK=𝒫BaseK
36 35 biimpri KPosetdomlubK=𝒫BaseKdomglbK=𝒫BaseKKCLat
37 4 22 34 36 syl12anc φKCLat