theorem
  (X (\+\) Y) (\) Z = (X (\) (Y (\/) Z)) (\/) (Y (\) (X (\/) Z))
proof
  thus (X (\+\) Y) (\) Z = (X (\) Y (\) Z) (\/) (Y (\) X (\) Z) by Th72
    .= (X (\) (Y (\/) Z)) (\/) (Y (\) X (\) Z) by Th73
    .= (X (\) (Y (\/) Z)) (\/) (Y (\) (X (\/) Z)) by Th73;
end;
