let L be non empty satisfying_DN_1 ComplLLattStr ; :: thesis: for x, y, z being Element of L holds (x ` ) + (((y + x) + z) ` ) = x `
let x, y, z be Element of L; :: thesis: (x ` ) + (((y + x) + z) ` ) = x `
set X = x ` ;
(x ` ) + (((y + ((x ` ) ` )) + z) ` ) = x ` by Th42;
hence (x ` ) + (((y + x) + z) ` ) = x ` by Th23; :: thesis: verum