theorem Th85:
  for x,p holds (p | x) | (p | x) = (((x | x) | (x | x)) | p) | ((
  (x | x) | (x | x)) | p)
proof
  let x,p;
  (x | x) | (x | x) = x by SHEFFER1:def 13;
  hence thesis by Th84;
end;
