theorem Th159:
  for z,x,q,y holds (x | y) | (x | (y | q)) = ((x | y) | (x | (y
  | (z | (z | z)))))
proof
  let z,x,q,y;
  ((y | (z | (z | z))) | (y | (z | (z | z)))) = y by Th136;
  hence thesis by Th158;
end;
