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 Th22;
hence ((((x + y) ` ) + x) ` ) + y = (y ` ) ` by Th23; :: thesis: verum