reserve L for satisfying_Sh_1 non empty ShefferStr;

theorem Th1:
  for x, y, z, u being Element of L holds ((x | (y | z)) | (x | (x
| (y | z)))) | ((z | ((x | z) | z)) | (u | (x | (y | z)))) = z | ((x | z) | z)
proof
  let x, y, z, u be Element of L;
  set Y = z | ((x | z) | z);
  set X = x | (y | z);
  (X | ((Y | X) | X)) | (Y | (u | X)) = Y by Def1;
  hence thesis by Def1;
end;
