reserve x,y,z for Element of REAL+;

theorem
  y <=' x implies x - (y + z) = x -' y - z
proof
  assume
A1: y <=' x;
  per cases;
  suppose
A2: y + z <=' x;
    then z <=' x -' y by A1,Lm8;
    then
A3: x -' y - z = x -' y -' z by Def2;
    x - (y + z) = x -' (y + z) by A2,Def2;
    hence thesis by A3,Lm9;
  end;
  suppose that
A4: not y + z <=' x and
A5: x <=' y;
A6: not z <=' x -' y by A1,A4,Lm8;
A7: (x + z) -' x = z by Lm5
      .= z -' (x -' x) by Lm3,Lm4;
    x = y by A1,A5,Th4;
    hence x - (y + z) = [{},z -' (x -' y)] by A4,A7,Def2
      .= x -' y - z by A6,Def2;
  end;
  suppose that
A8: not y + z <=' x and
A9: not x <=' y;
A10: not z <=' x -' y by A1,A8,Lm8;
    y + z -' x = z -' (x -' y) by A9,Lm11;
    hence x - (y + z) = [{},z -' (x -' y)] by A8,Def2
      .= x -' y - z by A10,Def2;
  end;
end;
