theorem
  X \ Y [= Z implies X [= Y "\/" Z
proof
  assume X \ Y [= Z;
  then Y "\/" (X "/\" Y`) [= Y "\/" Z by FILTER_0:1;
  then (Y "\/" X) "/\" (Y "\/" Y`) [= Y "\/" Z by LATTICES:11;
  then
A1: (Y "\/" X) "/\" Top L [= Y "\/" Z by LATTICES:21;
  X [= X "\/" Y by LATTICES:5;
  hence thesis by A1,LATTICES:7;
end;
