let L be non empty satisfying_DN_1 ComplLLattStr ; :: thesis: for x, y being Element of L holds (((((x + y) `) + x) `) + ((x + y) `)) ` = x
let x, y be Element of L; :: thesis: (((((x + y) `) + x) `) + ((x + y) `)) ` = x
set X = (x + y) ` ;
(((((x + y) `) + x) `) + ((((x + ((x + y) `)) `) + ((((((x + (x `)) `) + x) `) + ((x + y) `)) `)) `)) ` = x by Th4;
hence (((((x + y) `) + x) `) + ((x + y) `)) ` = x by Th5; :: thesis: verum