let L be non empty reflexive antisymmetric up-complete RelStr ; :: thesis: for D being non empty directed Subset of [:L,L:] holds sup D = [(sup ()),(sup ())]
let D be non empty directed Subset of [:L,L:]; :: thesis: sup D = [(sup ()),(sup ())]
reconsider D1 = proj1 D, D2 = proj2 D as non empty directed Subset of L by ;
reconsider C = the carrier of L as non empty set ;
reconsider D9 = D as non empty Subset of [:C,C:] by YELLOW_3:def 2;
A1: ex_sup_of D1,L by WAYBEL_0:75;
the carrier of [:L,L:] = [:C,C:] by YELLOW_3:def 2;
then consider d1, d2 being object such that
A2: ( d1 in C & d2 in C ) and
A3: sup D = [d1,d2] by ZFMISC_1:def 2;
A4: ex_sup_of D2,L by WAYBEL_0:75;
reconsider d1 = d1, d2 = d2 as Element of L by A2;
A5: ex_sup_of D,[:L,L:] by WAYBEL_0:75;
D2 is_<=_than d2
proof
let b be Element of L; :: according to LATTICE3:def 9 :: thesis: ( not b in D2 or b <= d2 )
assume b in D2 ; :: thesis: b <= d2
then consider x being object such that
A6: [x,b] in D by XTUPLE_0:def 13;
reconsider x = x as Element of D1 by ;
D is_<=_than [d1,d2] by ;
then [x,b] <= [d1,d2] by A6;
hence b <= d2 by YELLOW_3:11; :: thesis: verum
end;
then A7: sup D2 <= d2 by ;
D1 is_<=_than d1
proof
let b be Element of L; :: according to LATTICE3:def 9 :: thesis: ( not b in D1 or b <= d1 )
assume b in D1 ; :: thesis: b <= d1
then consider x being object such that
A8: [b,x] in D by XTUPLE_0:def 12;
reconsider x = x as Element of D2 by ;
D is_<=_than [d1,d2] by ;
then [b,x] <= [d1,d2] by A8;
hence b <= d1 by YELLOW_3:11; :: thesis: verum
end;
then sup D1 <= d1 by ;
then A9: [(sup D1),(sup D2)] <= sup D by ;
A10: ex_sup_of [:D1,D2:],[:L,L:] by WAYBEL_0:75;
reconsider D1 = D1, D2 = D2 as non empty Subset of L ;
D9 c= [:D1,D2:] by YELLOW_3:1;
then sup D <= sup [:D1,D2:] by ;
then sup D <= [(sup ()),(sup ())] by ;
hence sup D = [(sup ()),(sup ())] by ; :: thesis: verum