theorem Th63:
  (X (\) Y) (/\) Y = EmptyMS I
proof
A1: Y (/\) X c= Y by Th15;
  thus (X (\) Y) (/\) Y = (Y (/\) X) (\) Y by Th62
    .= EmptyMS I by A1,Th52;
  thus thesis;
end;
