let L be non empty satisfying_Sh_1 ShefferStr ; :: thesis: for x, y being Element of L holds (x | x) | (x | (y | x)) = x
let x, y be Element of L; :: thesis: (x | x) | (x | (y | x)) = x
x | ((x | x) | x) = x | x by Th5;
hence (x | x) | (x | (y | x)) = x by Def1; :: thesis: verum