theorem Th146:
  for z,y,x holds (z | (x | y)) | y = y | ((x | x) | z)
proof
  let z,y,x;
  (((y | ((x | x) | z)) | (y | ((x | x) | z))) | ((x | y) | ((z | z) | y))
  ) = ((z | (x | y)) | y) by Th123;
  hence thesis by Th145;
end;
