let L be non empty satisfying_Sh_1 ShefferStr ; :: thesis: for x, y, z being Element of L holds ((x | (y | z)) | z) | x = x | (y | z)
let x, y, z be Element of L; :: thesis: ((x | (y | z)) | z) | x = x | (y | z)
set X = y | z;
(x | (y | z)) | (x | z) = x by Th16;
hence ((x | (y | z)) | z) | x = x | (y | z) by Th15; :: thesis: verum