let L be non empty satisfying_Sh_1 ShefferStr ; :: thesis: for x, y being Element of L holds x | (y | x) = x | (y | y)
let x, y be Element of L; :: thesis: x | (y | x) = x | (y | y)
set Y = x | (y | y);
(x | (y | y)) | (x | y) = x by Th16;
hence x | (y | x) = x | (y | y) by Th38; :: thesis: verum