  Steel and Welch have shown that K is \Sigma^1_3 correct if the reals are
closed under sharps but 0^\pistol doesn't exist. We'll give a simple and purely
combinatorial proof of the following: K is \Sigma^1_3 correct if the reals are
closed under sharps, there is no inner model with a Woodin cardinal, K exists,
and * holds. Here, * is an assertion which can easily be verified if 0^\pistol
doesn't exist. We conjecture that * holds outright. (* is denoted by \clubsuit
in the paper.)
