by Patrick LaVictoire 1035 days ago | link | parent I can prove the property that for each hypothesis $$A()=a$$ there is at most one $$u$$ such that $$U()=u$$ has a high valuation (for sufficiently high PA+N), with the following caveat: it can sometimes take many steps to prove that $$u\neq u'$$ in PA+N, so we’ll need to include the length of that proof in our bound. In what follows, we will take all subscripts of $$d$$ and $$\nu$$ to be $$PA+N, A()=a$$ for $$N$$ large. For any $$\phi$$, $$d(\bot) - d(\neg\phi)\leq d(\phi)\leq d(\bot)$$ and thus $1 - \frac{d(\phi)}{d(\bot)} \leq \nu(\phi) \leq \frac{d(\bot)}{d(\phi)+d(\bot)}.$ Also, $$d(U()=u)+d(U()=u')+d(u\neq u')\geq d(\bot)$$. This implies $$\max\{d(U()=u),d(U()=u')\}\geq \frac12(d(\bot)-d(u\neq u))$$, which implies $\min\{\nu(U()=u),\nu(U()=u')\}\leq \min\{\frac{d(\bot)}{d(U()=u)+d(\bot)},\frac{d(\bot)}{d(U()=u')+d(\bot)}\} \leq \frac{2d(\bot)}{3d(\bot)-d(u\neq u')}.$ So we see that $$\nu(U()=u)$$ and $$\nu(U()=u')$$ cannot both be significantly larger than 2/3 if there is a short proof that $$u\neq u'$$.

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