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