theorem Th118:
  for q,p,x holds (((p | (x | p)) | (p | (x | p))) | (q | (q | q)
  )) = (x | x) | p
proof
  let q,p,x;
  (((x | x) | p) | ((p | p) | p)) = ((p | (x | p)) | (p | (x | p))) by
SHEFFER1:def 15;
  hence thesis by Th79;
end;
