Scott Garrabrant comments on Committing, Assuming, Externalizing, and Internalizing

[Scott Garrabrant]

This is not correct. It is true, however that $V$ is observable in ${\text{External}}_V\left({\text{Internal}}_V\right(C\left)\right)$.

A counter example is the 2 by 2 matrix where $A$ chooses whether to carry and umbrella and $E$ chooses whether or not it rains. Externalizing whether or not it rains has no effect on the frame, but the agent still cannot observe the rain.

[Rohin Shah]

Hmm, I’m not seeing it. Taking your example, let’s say that $A=\{u,n\}$, $E=\{r,s\}$, and $W=\{ur,us,nr,ns\}$, all in the obvious way.

Whether or not it rains would be formalized by the partition $V=\left\{\right\{ur,nr\},\{us,ns\left\}\right\}$.

Plugging this in to the definition from worlds, I get that $B=\left\{\right\{u\},\{n\left\}\right\}$.

Plugging this in to the definition of a quotient, I get that $A/B=\left\{\text{id}\right\}$ (the singleton containing the identity function).

Since ${\text{External}}^B\left(C\right)=(A/B,B\times E,\star )$, we get out a Cartesian frame whose agent has only one option, for which all properties are trivially observable.

[Scott Garrabrant]

I think $B=\left\{\right\{u,n\left\}\right\}$.

[Rohin Shah]

Oh yup I was misinterpreting how B was defined, and that would also mess up my proof. Thanks!