
theorem Th35:
  for S being non empty RelStr, D1, D2 being Subset of S holds (
  sup_op S).:[:D1,D2:] = D1 "\/" D2
proof
  let S be non empty RelStr, D1, D2 be Subset of S;
  set f = sup_op S;
  reconsider X = [:D1,D2:] as set;
  thus f.:[:D1,D2:] c= D1 "\/" D2
  proof
    let q be object;
    assume
A1: q in f.:[:D1,D2:];
    then reconsider q1 = q as Element of S;
    consider c being Element of [:S,S:] such that
A2: c in [:D1,D2:] and
A3: q1 = f.c by A1,FUNCT_2:65;
    consider x, y being object such that
A4: x in D1 & y in D2 and
A5: c = [x,y] by A2,ZFMISC_1:def 2;
    reconsider x, y as Element of S by A4;
    q = f.(x,y) by A3,A5
      .= x "\/" y by Def5;
    then
    q in { a "\/" b where a, b is Element of S : a in D1 & b in D2 } by A4;
    hence thesis by YELLOW_4:def 3;
  end;
  let q be object;
  assume q in D1 "\/" D2;
  then q in { x "\/" y where x, y is Element of S : x in D1 & y in D2 } by
YELLOW_4:def 3;
  then consider x, y being Element of S such that
A6: q = x "\/" y & x in D1 & y in D2;
  q = f.(x,y) & [x,y] in X by A6,Def5,ZFMISC_1:87;
  hence thesis by FUNCT_2:35;
end;
