theorem Th56:
  (i => k)"/\"(j => k) [= (i"\/"j) => k
proof
A1: (i"/\"((i=>k)"/\"(j=>k)))"\/"(j"/\"((i=>k)"/\"(j=>k))) = (i"\/"j)"/\" ((
  i=>k)"/\"(j=>k)) by LATTICES:def 11;
A2: j"/\"((j=>k)"/\"(i=>k)) = j"/\"(j=>k) "/\"(i=>k) by LATTICES:def 7;
  j"/\"(j=>k) [= k by FILTER_0:def 7;
  then
A3: j"/\"(j=>k)"/\"(i=>k) [= k by FILTER_0:2;
  i"/\"(i=>k) [= k by FILTER_0:def 7;
  then
A4: i"/\"(i=>k)"/\"(j=>k) [= k by FILTER_0:2;
  i"/\"((i=>k)"/\"(j=>k)) = i"/\"(i=>k)"/\"(j=>k) by LATTICES:def 7;
  then (i"\/"j)"/\"((i=>k)"/\"(j=>k)) [= k by A4,A3,A1,A2,FILTER_0:6;
  hence thesis by FILTER_0:def 7;
end;
