
theorem
  for S,T being non empty RelStr
  for f being Function of S,T, g being Function of S,T opp st
  f = g holds (f is monotone iff g is antitone) &
  (f is antitone iff g is monotone)
proof
  let S,T be non empty RelStr;
  let f be Function of S,T, g be Function of S,T~ such that
A1: f = g;
  thus f is monotone implies g is antitone
  proof
    assume
A2: for x,y being Element of S st x <= y holds f.x <= f.y;
    let x,y be Element of S;
    assume x <= y;
    then
A3: f.x <= f.y by A2;
    (f.x)~ = f.x & (f.y)~ = f.y;
    hence thesis by A1,A3,LATTICE3:9;
  end;
  thus g is antitone implies f is monotone
  proof
    assume
A4: for x,y being Element of S st x <= y for a,b being Element of T
    opp st a = g.x & b = g.y holds a >= b;
    let x,y be Element of S;
    assume x <= y;
    then
A5: g.y <= g.x by A4;
    ~(g.x) = g.x & ~(g.y) = g.y;
    hence thesis by A1,A5,Th1;
  end;
  thus f is antitone implies g is monotone
  proof
    assume
A6: for x,y being Element of S st x <= y for a,b being Element of T
    st a = f.x & b = f.y holds a >= b;
    let x,y be Element of S;
    assume x <= y;
    then
A7: f.y <= f.x by A6;
    (f.x)~ = f.x & (f.y)~ = f.y;
    hence thesis by A1,A7,LATTICE3:9;
  end;
  thus g is monotone implies f is antitone
  proof
    assume
A8: for x,y being Element of S st x <= y holds g.x <= g.y;
    let x,y be Element of S;
    assume x <= y;
    then
A9: g.y >= g.x by A8;
    ~(g.x) = g.x & ~(g.y) = g.y;
    hence thesis by A1,A9,Th1;
  end;
end;
