orthonormal comments on Proof Length and Logical Counterfactuals Revisited

[orthonormal]

I can prove the property that for each hypothesis $A\left(\right)=a$ there is at most one $u$ such that $U\left(\right)=u$ has a high valuation (for sufficiently high PA+N), with the following caveat: it can sometimes take many steps to prove that $u\ne u^{\prime }$ 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 $PA+N,A\left(\right)=a$ for $N$ large.

For any $\varphi$, $d\left(\perp \right)-d\left(\neg \varphi \right)\le d\left(\varphi \right)\le d\left(\perp \right)$ and thus $$1-\frac{d\left(\varphi \right)}{d\left(\perp \right)}\le \nu \left(\varphi \right)\le \frac{d\left(\perp \right)}{d\left(\varphi \right)+d\left(\perp \right)}.$$

Also, $d\left(U\right()=u)+d\left(U\right()=u^{\prime })+d(u\ne u^{\prime })\ge d\left(\perp \right)$. This implies $max\left\{d\right(U\left(\right)=u),d(U\left(\right)=u^{\prime }\left)\right\}\ge \frac{1}{2}\left(d\right(\perp )-d(u\ne u\left)\right)$, which implies $$min\left\{\nu \right(U\left(\right)=u),\nu (U\left(\right)=u^{\prime }\left)\right\}\le min\{\frac{d\left(\perp \right)}{d\left(U\right()=u)+d\left(\perp \right)},\frac{d\left(\perp \right)}{d\left(U\right()=u^{\prime })+d\left(\perp \right)}\}\le \frac{2d\left(\perp \right)}{3d\left(\perp \right)-d(u\ne u^{\prime })}.$$

So we see that $\nu \left(U\right()=u)$ and $\nu \left(U\right()=u^{\prime })$ cannot both be significantly larger than ²⁄₃ if there is a short proof that $u\ne u^{\prime }$.