theorem Th6:
  h * g = h * f or g * h = f * h implies g = f
proof
  assume h * g = h * f or g * h = f * h;
  then h" * (h * g) = h" * h * f or g * h * h" = f * (h * h") by Def3;
  then h" * (h * g) = 1_G * f or g * (h * h") = f * (h * h") by Def3,Def5;
  then h" * (h * g) = f or g * 1_G = f * (h * h") by Def4,Def5;
  then h" * h * g = f or g = f * (h * h") by Def3,Def4;
  then h" * h * g = f or g = f * 1_G by Def5;
  then 1_G * g = f or g = f by Def4,Def5;
  hence thesis by Def4;
end;
