theorem Th90:
  for q,p holds ((p | p) | (p | ((q | q) | q))) = ((((q | q) | (q
  | q)) | p) | ((q | q) | p))
proof
  let q,p;
  p | ((q | q) | q) = p | p by Th70;
  hence thesis by SHEFFER1:def 15;
end;
