let L be non empty satisfying_DN_1 ComplLLattStr ; :: thesis: for x, y, z being Element of L holds (((((x + (y `)) + z) `) + y) `) ` = y
let x, y, z be Element of L; :: thesis: (((((x + (y `)) + z) `) + y) `) ` = y
(((((x + (y `)) + z) `) + y) `) + (((y `) + y) `) = ((((x + (y `)) + z) `) + y) ` by Th32;
hence (((((x + (y `)) + z) `) + y) `) ` = y by Th36; :: thesis: verum