theorem Th145:
  for y,z,x holds ((y | ((x | x) | z)) | (y | ((x | x) | z))) | (
  (x | y) | ((z | z) | y)) = y | ((x | x) | z)
proof
  let y,z,x;
  (y | ((x | x) | z)) | (y | ((x | x) | z)) = (x | y) | ((z | z) | y) by Th74;
  hence thesis by Th121;
end;
