P+K does not include S, but can output any arbitrary S. That is the whole reason for introducing P, you need a way of getting K(S) into your program without "ha

P+K does not include S, but can output any arbitrary S. That is the whole reason for introducing P, you need a way of getting K(S) into your program without "hard coding" it directly and thus including the complexity of S. So, if we assume K is computable, then both S and P+K can output S, however, for an arbitrarily large S, K(P+K) < K(S). This is the proof by contradiction. Specifically, the value K(S) is supposed to be the the shortest program that can output S, yet P+K, which can also output S, is shorter.