# Provide a Kripke model to prove that $\neg(\neg P\lor \neg Q\lor R)\to (P\land Q \land R)$ is false in intuitionistic logic

Could anyone check my working please?

Provide a Kripke model to prove that $$\neg(\neg P\lor \neg Q\lor \neg R)\to (P\land Q \land R)$$ is false in intuitionistic logic.

Here $$k_0$$ forces $$\neg(\neg P\lor \neg Q\lor \neg R)$$: this is equivalent to $$\lnot\lnot P\land\lnot Q\land\lnot\lnot R$$ by deMoran's law. For $$k_0$$ to prove $$\lnot\lnot P$$ either $$k_0$$ or a higher node must force $$P$$ - which $$k_1$$ fufilled. The same goes for $$Q$$ and $$R$$.

$$k_0$$ does not force $$P,Q,R$$. Therefore the antecedent is forced at $$k_0$$, but not the consequent - thus this model falsifies this sentence. However, the analysis is not quite correct. In particular, you say: "For $$k_0$$ to prove $$\lnot \lnot P$$ either $$k_0$$ or a higher node must force $$P$$ - which $$k_1$$ fufilled." In fact, the correct statement is: $$p \Vdash \lnot \lnot P$$ if and only if for all $$q \ge p$$, it is not the case that for all $$r \ge q$$, $$r \not\Vdash P$$. (And if you believe that the underlying universe for the Kripke model satisfies classical logic, or if you're working with a finite Kripke model, then this is equivalent to: $$\forall q \ge p, \exists r \ge q, r \Vdash P$$.) Now, in the given model, it is indeed true that $$k_0 \Vdash \lnot \lnot P$$ -- in either the case $$q = k_0$$ or the case $$q = k_1$$, we can use $$r = k_1$$.
In this case, you could also work directly with the formula as given, instead of going via de Morgan's law. i.e. by definitions, $$k_0 \not\Vdash \lnot P$$ since $$k_1 \ge k_0$$ and $$k_1 \Vdash P$$; similarly, $$k_1 \not\Vdash \lnot P$$; and so on for $$Q$$ and $$R$$. Therefore, $$k_0 \not\Vdash \lnot P \vee \lnot Q \vee \lnot R$$, and similarly $$k_1 \not\Vdash \lnot P \vee \lnot Q \vee \lnot R$$; so again by definitions, $$k_0 \Vdash \lnot(\lnot P \vee \lnot Q \vee \lnot R)$$.
To see the difference between the incorrect statement and the correct statement about $$p \Vdash \lnot \lnot P$$, consider the standard counterexample model to $$P \vee \lnot P$$ where $$k_0 \le k_1, k_0 \le k_2$$ with $$k_1, k_2$$ incomparable, and only $$k_1 \Vdash P$$. Then in this model $$k_0 \not\Vdash \lnot \lnot P$$, despite the fact that $$k_1 \ge k_0$$ and $$k_1 \Vdash P$$. On the other hand, $$k_2 \ge k_0$$, and there is nothing $$\ge k_2$$ which forces $$P$$.