let L be non empty satisfying_Sh_1 ShefferStr ; :: thesis: for x, y being Element of holds x | (y | (x | x)) = x | x
let x, y be Element of ; :: thesis: x | (y | (x | x)) = x | x
set Y = x | x;
(x | x) | (x | x) = x by Th11;
hence x | (y | (x | x)) = x | x by Th11; :: thesis: verum