theorem Th123:
  for x,z,y holds (((x | ((y | y) | z)) | (x | ((y | y) | z))) |
  ((y | x) | ((z | z) | x))) = (z | (y | x)) | x
proof
  let x,z,y;
  (x | x) | (y | x) = x by Th121;
  hence thesis by Th116;
end;
