Thanks! I agree it’s hard to imagine a lower-semicomputable M that would dominate both F and -F when S is algorithmically random, but a more solid proof would be nice because about half the steps in your argument aren’t intuitively obvious to me.
Thanks! I agree it’s hard to imagine a lower-semicomputable M that would dominate both F and -F when S is algorithmically random, but a more solid proof would be nice because about half the steps in your argument aren’t intuitively obvious to me.