theorem Th80:
  for p,x,y holds (((y | (x | x)) | (y | (x | x))) | (p | (p | p))
  ) = ((x | x) | y)
proof
  let p,x,y;
  (((x | x) | y) | ((x | x) | y)) = ((y | (x | x)) | (y | (x | x))) by
SHEFFER1:def 15;
  hence thesis by Th71;
end;
