let L be non empty satisfying_DN_1 ComplLLattStr ; :: thesis: for x, y, z being Element of L holds x + ((((y + z) `) + ((y + x) `)) `) = y + x
let x, y, z be Element of L; :: thesis: x + ((((y + z) `) + ((y + x) `)) `) = y + x
x + ((((y + z) `) + ((y + x) `)) `) = ((y + x) `) ` by Th26;
hence x + ((((y + z) `) + ((y + x) `)) `) = y + x by Th23; :: thesis: verum