let L be non empty satisfying_Sh_1 ShefferStr ; :: thesis: for x, y being Element of L holds x | (y | (x | y)) = x | x
let x, y be Element of L; :: thesis: x | (y | (x | y)) = x | x
(x | y) | (x | ((x | y) | y)) = x by Th15;
hence x | (y | (x | y)) = x | x by Th30; :: thesis: verum