$$(p\to q)\Leftrightarrow (\neg p \vee q)$$.
I consider only $$(p\to q)\to(\neg p \vee q)$$
$w_0\vdash$
$w_1\vdash p$
$w_2\vdash q$
Then we can also say that :
$w_0\vdash \not \neg p$ in other words $w_0\not\vdash \neg p$
$w_1\vdash \not \neg q$ in other words $w_1\not\vdash \neg q$
Moreover, I should add:
$w_1\vdash \not q$
$w_2\vdash \not p$
Now, lets conclude:
$w_0\not\vdash q,p\neg q, \neg p$
$w_1\not\vdash (\neg p\vee q), (p\to q)$
$w_2\vdash (\neg p\vee q), (p\to q)$
Because $w_0\not\vdash (\neg p\vee q)$ and $w_1\vdash (p\to q)$ so $w_0\not\vdash (p\to q)\to(\neg p \vee q)$.
- Is it ok ?
- Is it true that $w\vdash \not p$ is the same as $w\not\vdash p$ ?
- Am I right that determining for each state which variable (here $p,q$) one of the possibility: $p,\not p, q,\not q$ is in my hands ( I try to find counterexample) ?. Of course choice of these possibility give us following additional information:
For example if I decide that $w_0\vdash p$ then I can say that $w_0\not \vdash\neg p$. Similary, If I decide that $w\vdash \neg p$ then I can say that $w\not\vdash p$. Generally, play in deciding about $p,q$ is aimed to find a counterexample.
Yeah ?
Long comment : it seems to me that there are some typos in your proof.
Assuming that : $w_0 \le w_1,w_2$, with the model :
we have :
and also : $w_1 \nVdash (p→q)$ because it is not true that for every $w \ge w_1$ if $w \Vdash p$, then $w \Vdash q$ : consider $w_1$ itself.
But : $w_2 \Vdash (p→q),(¬p∨q)$; why $w_2 \Vdash (p→q)$ ? Because $w_2 \nVdash p$ and thus the conditional holds vacuously.
If so, also $w_0 \nVdash p$ and thus also $w_0 \Vdash (p→q)$.
So, it seems to me, the counter-example works, because of :
Note. I'm not very skilled in the art of conjuring Kripke countermodels...
If the above counterexample works, then it seems to me that a single-node model $w_0$ that forces neither $p$ nor $q$ (and thus $w_0 \nVdash (\lnot p \lor q)$) will suffice, becuse vacuously : $w_0 \Vdash (p \to q)$.