
theorem Th64:
  for L being non empty transitive RelStr for S being non empty
full SubRelStr of L for X being Subset of S st ex_sup_of X,L & "\/"(X,L) in the
  carrier of S holds ex_sup_of X,S & "\/"(X,S) = "\/"(X,L)
proof
  let L be non empty transitive RelStr;
  let S be non empty full SubRelStr of L;
  let X be Subset of S;
  assume that
A1: ex_sup_of X,L and
A2: "\/"(X,L) in the carrier of S;
  reconsider a = "\/"(X,L) as Element of S by A2;
A3: now
    "\/"(X,L) is_>=_than X by A1,Def9;
    hence a is_>=_than X by Th61;
    let b be Element of S;
    reconsider b9 = b as Element of L by Th58;
    assume b is_>=_than X;
    then b9 is_>=_than X by Th62;
    then b9 >= "\/"(X,L) by A1,Def9;
    hence b >= a by Th60;
  end;
  consider a9 being Element of L such that
A4: X is_<=_than a9 and
A5: for b being Element of L st X is_<=_than b holds b >= a9 and
  for c being Element of L st X is_<=_than c & for b being Element of L st
  X is_<=_than b holds b >= c holds c = a9 by A1;
A6: a9 = "\/"(X,L) by A1,A4,A5,Def9;
  thus ex_sup_of X,S
  proof
    take a;
    thus a is_>=_than X & for b being Element of S st b is_>=_than X holds b
    >= a by A3;
    let c be Element of S;
    reconsider c9 = c as Element of L by Th58;
    assume X is_<=_than c;
    then
A7: X is_<=_than c9 by Th62;
    assume for b being Element of S st X is_<=_than b holds b >= c;
    then
A8: a >= c by A3;
    now
      let b be Element of L;
      assume X is_<=_than b;
      then
A9:   b >= a9 by A5;
      a9 >= c9 by A6,A8,Th59;
      hence b >= c9 by A9,ORDERS_2:3;
    end;
    hence thesis by A1,A7,Def9;
  end;
  hence thesis by A3,Def9;
end;
