let L be non empty satisfying_Sh_1 ShefferStr ; :: thesis: for x, y, z being Element of L holds (x | ((y | z) | x)) | (y | y) = (y | z) | (y | y)
let x, y, z be Element of L; :: thesis: (x | ((y | z) | x)) | (y | y) = (y | z) | (y | y)
set Y = y | z;
set X = x | ((y | z) | x);
((y | z) | (y | z)) | ((x | ((y | z) | x)) | (y | z)) = y | z by Th11;
hence (x | ((y | z) | x)) | (y | y) = (y | z) | (y | y) by Th50; :: thesis: verum