let L be non empty satisfying_DN_1 ComplLLattStr ; :: thesis: for x, y being Element of L holds (x + ((x + (y `)) `)) ` = (x + y) `
let x, y be Element of L; :: thesis: (x + ((x + (y `)) `)) ` = (x + y) `
(x + ((((x + y) `) + x) `)) ` = (x + y) ` by Th8;
hence (x + ((x + (y `)) `)) ` = (x + y) ` by Th44; :: thesis: verum