theorem Th119:
  for p,x holds p | (x | p) = (x | x) | p
proof
  now
    let q,p,x;
    (((p | (x | p)) | (p | (x | p))) | (q | (q | q))) = (p | (x | p)) by Th71;
    hence p | (x | p) = (x | x) | p by Th118;
  end;
  hence thesis;
end;
