theorem Th152:
  for z,p,x holds z | (x | p) = z | (p | x)
proof
  now
    let y,z,p,x;
    (x | (y | (y | y))) | (x | (y | (y | y))) = x by Th136;
    hence z | (x | p) = z | (p | x) by Th151;
  end;
  hence thesis;
end;
