theorem Th160:
  for x,q,y holds (x | y) | (x | (y | q)) = x
proof
  now
    let q,x,z,y;
    (x | y) | (x | (y | (z | (z | z)))) = x by Th112;
    hence (x | y) | (x | (y | q)) = x by Th159;
  end;
  hence thesis;
end;
