
theorem
  for L being non empty Poset, R being auxiliary(i) auxiliary(ii) (
  Relation of L), C being Subset of L st (for c being Element of L holds
  ex_sup_of SetBelow (R,C,c),L) holds SupBelow (R,C) is sup-closed
proof
  let L be non empty Poset, R be auxiliary(i) auxiliary(ii) (Relation of L), C
  be Subset of L;
  assume
A1: for c being Element of L holds ex_sup_of SetBelow (R,C,c),L;
  let X be Subset of SupBelow (R,C);
  set s = "\/"(X,L);
  assume
A2: ex_sup_of X,L;
A3: ex_sup_of SetBelow (R,C,s),L by A1;
  X is_<=_than sup SetBelow (R,C,s)
  proof
    let x be Element of L;
A4: ex_sup_of SetBelow (R,C,x),L by A1;
    assume
A5: x in X;
    then
A6: x = sup SetBelow (R,C,x) by Def10;
    SetBelow (R,C,x) c= SetBelow (R,C,s) by A2,A5,Th17,YELLOW_4:1;
    hence x <= sup SetBelow (R,C,s) by A3,A6,A4,YELLOW_0:34;
  end;
  then
A7: s <= sup SetBelow (R,C,s) by A2,YELLOW_0:def 9;
A8: the carrier of subrelstr SupBelow (R,C) = SupBelow (R,C) by YELLOW_0:def 15
;
  SetBelow (R,C,s) is_<=_than s by Th16;
  then sup SetBelow (R,C,s) <= s by A3,YELLOW_0:def 9;
  then s = sup SetBelow (R,C,s) by A7,ORDERS_2:2;
  then s in SupBelow (R,C) by Def10;
  hence thesis by A8,A2,YELLOW_0:64;
end;
