
theorem Th32:
  for L being satisfying_Sheffer_1 satisfying_Sheffer_2
  satisfying_Sheffer_3 properly_defined non empty ShefferOrthoLattStr, x, y
  being Element of L holds x | (x | x) = y | (y | y)
proof
  let L be satisfying_Sheffer_1 satisfying_Sheffer_2 satisfying_Sheffer_3
  properly_defined non empty ShefferOrthoLattStr;
  let x, y be Element of L;
  x | (x | x) = ((x | x") | (x | x"))" by Def13
    .= ((x | x") | (y | y"))" by Def14
    .= ((y | y") | (x | (x | x)))" by Th31
    .= ((y | y")")" by Def14
    .= y | (y | y) by Def13;
  hence thesis;
end;
