theorem Th18:
  g * h = h * g iff g" * h" = h" * g"
proof
  thus g * h = h * g implies g" * h" = h" * g"
  proof
    assume
A1: g * h = h * g;
    hence g" * h" = (g * h)" by Th16
      .= h" * g" by A1,Th17;
  end;
  assume
A2: g" * h" = h" * g";
  thus g * h = (g * h)"" .= (h" * g")" by Th16
    .= h"" * g"" by A2,Th16
    .= h * g;
end;
