let L be non empty satisfying_DN_1 ComplLLattStr ; :: thesis: for x, y, z being Element of L holds ((((x + y) `) + (x + z)) `) + y = y
let x, y, z be Element of L; :: thesis: ((((x + y) `) + (x + z)) `) + y = y
(((x + y) `) + ((((x + y) `) + ((x + z) `)) `)) ` = (((x + y) `) + (x + z)) ` by Th45;
hence ((((x + y) `) + (x + z)) `) + y = y by Th40; :: thesis: verum