let L be non empty satisfying_DN_1 ComplLLattStr ; :: thesis: for x, y being Element of L holds ((x + (y `)) + y) ` = ((y `) + y) `
let x, y be Element of L; :: thesis: ((x + (y `)) + y) ` = ((y `) + y) `
((((x + (y `)) `) `) + y) ` = ((y `) + y) ` by Th34;
hence ((x + (y `)) + y) ` = ((y `) + y) ` by Th23; :: thesis: verum