let L be non empty satisfying_Sh_1 ShefferStr ; :: thesis: for x, y being Element of L holds x | (y | (y | y)) = x | x
let x, y be Element of L; :: thesis: x | (y | (y | y)) = x | x
set Y = y | (x | y);
(y | (x | y)) | (x | y) = y by Th25;
hence x | (y | (y | y)) = x | x by Th58; :: thesis: verum