let L be non empty satisfying_Sh_1 ShefferStr ; :: thesis: for x, y, z being Element of L holds (x | (y | z)) | ((y | x) | x) = (x | (y | z)) | (x | (y | z))
let x, y, z be Element of L; :: thesis: (x | (y | z)) | ((y | x) | x) = (x | (y | z)) | (x | (y | z))
set X = x | (y | z);
(x | (y | z)) | (y | x) = x by Th25;
hence (x | (y | z)) | ((y | x) | x) = (x | (y | z)) | (x | (y | z)) by Th32; :: thesis: verum