let L be non empty satisfying_Sheffer_1 satisfying_Sheffer_2 satisfying_Sheffer_3 ShefferStr ; :: thesis: for x, y being Element of L holds x = (x | x) | (y | x)
let x, y be Element of L; :: thesis: x = (x | x) | (y | x)
(y | x) | ((y | y) | x) = x by Th92;
hence x = (x | x) | (y | x) by Th120; :: thesis: verum