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Apr 14 at 14:28 vote accept Miviska
Apr 10 at 15:46 comment added Gro-Tsen (My previous comment should be taken cum grano salis, however, because a Lawvere-Tierney topology (=local operator) $j\colon\Omega\to\Omega$ certainly behaves as a sort of modal operator (whose meaning is something like “locally”), and by definition it makes sense at the truth-value level.)
Apr 10 at 15:42 comment added Gro-Tsen A formal answer to your question has already been given by Simon Henry. But let me make the following more informal remark: even just classically, the whole point of modal operators is that they give rise to a semantic of “possible worlds” (Kripke semantics) where the truth value differs from one world to another: so the whole point of $\Box$ is that the truth value of $\Box p$ doesn't just depend on the truth value of $p$. As such, it is unreasonable to try to make $\Box$ into a function from the object $\Omega$ of truth value to itself.
Apr 10 at 14:31 answer added Simon Henry timeline score: 11
Apr 9 at 22:00 comment added Todd Trimble @SimonHenry Thanks. That sounds right to me.
Apr 9 at 21:34 comment added Simon Henry @ToddTrimble Given the context, my understanding is that p is a variable of type $\Omega$ and this is an internal statement. That is this is the assumption that $(\square,id): \Omega \to \Omega^2$ factor through the order relation.
Apr 9 at 21:19 comment added Simon Henry I feel like the fact mentioned in the cited paragraph that there is no operator with these two properties answer the question. So, are you asking for a proof of that fact? or are you not convinced that this fact answer your question?
Apr 9 at 20:53 comment added Todd Trimble Could you clarify what is $p$, please?
S Apr 9 at 20:45 review First questions
Apr 9 at 21:24
S Apr 9 at 20:45 history asked Miviska CC BY-SA 4.0