let L be non empty satisfying_Sh_1 ShefferStr ; :: thesis: for x, y being Element of L holds (x | x) | (y | x) = x
let x, y be Element of L; :: thesis: (x | x) | (y | x) = x
set Y = y | x;
x | ((((y | x) | (y | x)) | x) | x) = y | x by Th10;
hence (x | x) | (y | x) = x by Th7; :: thesis: verum