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 + z)
let x, y, z be Element of L; :: thesis: ((x `) `) + (y + z) = (y + x) + (y + z)
(x `) + (((y + x) + (y + z)) `) = x ` by Th43;
hence ((x `) `) + (y + z) = (y + x) + (y + z) by Th51; :: thesis: verum