Metamath Proof Explorer


Theorem glbsscl

Description: If a subset of S contains the GLB of S , then the two sets have the same GLB. (Contributed by Zhi Wang, 26-Sep-2024)

Ref Expression
Hypotheses lubsscl.k φKPoset
lubsscl.t φTS
glbsscl.g G=glbK
glbsscl.s φSdomG
glbsscl.x φGST
Assertion glbsscl φTdomGGT=GS

Proof

Step Hyp Ref Expression
1 lubsscl.k φKPoset
2 lubsscl.t φTS
3 glbsscl.g G=glbK
4 glbsscl.s φSdomG
5 glbsscl.x φGST
6 eqid BaseK=BaseK
7 eqid K=K
8 6 7 3 1 4 glbelss φSBaseK
9 2 8 sstrd φTBaseK
10 9 5 sseldd φGSBaseK
11 1 adantr φyTKPoset
12 4 adantr φyTSdomG
13 2 sselda φyTyS
14 6 7 3 11 12 13 glble φyTGSKy
15 14 ralrimiva φyTGSKy
16 breq2 y=GSzKyzKGS
17 simp3 φzBaseKyTzKyyTzKy
18 5 3ad2ant1 φzBaseKyTzKyGST
19 16 17 18 rspcdva φzBaseKyTzKyzKGS
20 19 3expia φzBaseKyTzKyzKGS
21 20 ralrimiva φzBaseKyTzKyzKGS
22 breq1 x=GSxKyGSKy
23 22 ralbidv x=GSyTxKyyTGSKy
24 breq2 x=GSzKxzKGS
25 24 imbi2d x=GSyTzKyzKxyTzKyzKGS
26 25 ralbidv x=GSzBaseKyTzKyzKxzBaseKyTzKyzKGS
27 23 26 anbi12d x=GSyTxKyzBaseKyTzKyzKxyTGSKyzBaseKyTzKyzKGS
28 27 rspcev GSBaseKyTGSKyzBaseKyTzKyzKGSxBaseKyTxKyzBaseKyTzKyzKx
29 10 15 21 28 syl12anc φxBaseKyTxKyzBaseKyTzKyzKx
30 biid yTxKyzBaseKyTzKyzKxyTxKyzBaseKyTzKyzKx
31 6 7 3 30 1 glbeldm2 φTdomGTBaseKxBaseKyTxKyzBaseKyTzKyzKx
32 9 29 31 mpbir2and φTdomG
33 eqidd φBaseK=BaseK
34 3 a1i φG=glbK
35 7 33 34 1 9 10 14 19 posglbdg φGT=GS
36 32 35 jca φTdomGGT=GS