# Public announcements and knowledge(necessity)

In the context of public announcement logic and epistemic logic(or more general general modal logic), the general semantic definition for the public announcement operator is:

$(M,w)\models[\phi]\psi$ if $(M,w)\models\phi$ implies $(M|\phi,w)\models\psi$

I.e. the formula $\psi$ is still valid in the $\phi$-restricted model(where all worlds are cut out which deny $\phi$).

For recalling purposes, the semantic definition for knowledge(necessity) is:

$(M,w)\models K_a\phi$ if $\forall v\in M:(w,v)\in R_a$ implies $(M,v)\models\phi$

I just enforced the multi-agent case here(it doesn't make much difference for the upcoming question). For the interplay between announcements and knowledge or necessity, I'm curious about the following:

In the reference material I'm reading, the formula $[\phi]K_a\psi$ is obviously interpreted as

$(M,w)\models[\phi]K_a\psi$ if $(M,w)\models\phi$ implies $\forall v\in M|\phi:(w,v)\in R_a'$ implies $(M|\phi,v)\models\psi$

$R_a'$ is the accessibility relation of the agent $a$ in the restricted model. But couldn't someone just enforce conditional semantics? I.e. would this not be equivalent to

$(M,w)\models\phi$ implies $\forall v\in M:(w,v)\in R_a$ and $(M,v)\models\phi$ implies $(M,v)\models\psi$

Edit: Additionally, would it then be valid to introduce a conditionalized operator $K_a(\phi,\psi)$ with

$(M,w)\models K_a(\phi,\psi)$ if $\forall v\in W: (w,v)\in R_a$ and $(M,v)\models\phi$ implies $(M,v)\models\psi$. If so, this would then imply

$[\phi]K_a\psi\leftrightarrow \phi\rightarrow K_a(\phi,\psi)$

There is a classic counterexample: a Moore's sentence. Let's consider a formula $[p \wedge \neg K_a p] K_a \neg (p \wedge \neg K_a p)$, and a model $M$ with two states $w$ and $v$, and universal relation for $a$. Also, let $(M,w) \models p$ and $(M,w) \not \models p$. Now, by the semantics, we have $$(M,w) \models [p \wedge \neg K_a p] K_a \neg (p \wedge \neg K_a p) \quad \textrm{iff} \quad (M,w) \models p \wedge \neg K_a p \ \textrm{implies} \ \forall v \in W^{p \wedge \neg K_a p}: (M\mid(p \wedge \neg K_a p),v) \models \neg (p \wedge \neg K_a p).$$ In other words, after announcement of $p \wedge \neg K_a p$ we 'delete' all the incompatible states (i.e. $v$ with $\neg p$), and in the end it holds that $\neg(p \wedge \neg K_a p)$ in the remaining single state $w$.
Now, it seems that you want to imitate somehow the following axiom of PAL: $[\varphi]K_a\psi \leftrightarrow (\varphi \rightarrow K_a [\varphi]\psi)$. It seems that your variant works only if $\psi$ is a propositional variable. Returning to the counterexample, your hypothesis suggests that there is a state with $(p \wedge \neg K_a p) \rightarrow \neg (p \wedge \neg K_a p)$ (an an agent knows it), which is a contradiction (also, you use 'and', suggesting that every state is reachable for a given agent, which is not generally true).
As for the conditionalised operator, the same counterexample holds (and again you claim that for an agent $a$ every state is reachable).