theorem Th84:
  for x,p holds (p | x) | (p | ((x | x) | (x | 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 SHEFFER1:def 15;
end;
