theorem
  X (\/) Y = (X (\+\) Y) (\/) X (/\) Y
proof
  thus X (\/) Y = ((X (\) Y) (\/) X (/\) Y) (\/) Y by Th65
    .= (X (\) Y) (\/) (X (/\) Y (\/) Y) by Th28
    .= (X (\) Y) (\/) Y by Th31
    .= (X (\) Y) (\/) ((Y (\) X) (\/) (Y (/\) X)) by Th65
    .= (X (\+\) Y) (\/) X (/\) Y by Th28;
end;
