let L be non empty satisfying_DN_1 ComplLLattStr ; :: thesis: for x, y being Element of L holds ((((x + y) `) + x) `) + y = y
let x, y be Element of L; :: thesis: ((((x + y) `) + x) `) + y = y
((((x + y) `) + x) `) + y = (y `) ` by Th24;
hence ((((x + y) `) + x) `) + y = y by Th23; :: thesis: verum