theorem Th157:
  for q,x,z,y holds ((x | y) | (((x | (y | (z | (z | z)))) | q) |
  x)) = ((x | y) | (x | (y | (z | (z | z)))))
proof
  let q,x,z,y;
  (x | (y | (z | (z | z)))) | (x | y) = x by Th113;
  hence thesis by Th156;
end;
