theorem Th12:
  h * f = g iff f = h" * g
proof
  h * (h" * g) = h * h" * g by Def3
    .= 1_G * g by Def5
    .= g by Def4;
  hence h * f = g implies f = h" * g by Th6;
  assume f = h" * g;
  hence h * f = h * h" * g by Def3
    .= 1_G * g by Def5
    .= g by Def4;
end;
