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Show that it is not the case that if $models lnot (Aland B)$ then $models lnot A$ or $models lnot B$.

Show $models lnot (Aland B)$ is true and neither $models lnot A$ not $models lnot B$ is the case.

Consider the formula $lnot (pland lnot p)$.

$models lnot (pland lnot p)$ then $models lnot p$ or $models lnotlnot p$.

Show $models lnot (pland lnot p)$ is true and neither $models lnot p$ nor $models lnotlnot p$ is the case.

$models$ is for validity, the intutionist logical consequence.

$models lnot p$ if $W = w$, $wRw$, $vw(p) = 0$.

There are several notions of semantics for intuitionistic logic, and hence several meanings of $models$. If you're going to talk about the $models$ relation in intuitionistic logic, you should really make sure that everyone is on the same page by specifying what it means. Based on your question, I'm going to assume you are using Kripke Semantics.

First things first: In intuitionistic logic, $lnot varphi$ can be viewed as an abbreviation for $varphirightarrow bot$. So we have $models lnot varphi$ iff for every intuitionistic Kripke model $M$ and every world $win M$, $wVdash varphirightarrow bot$. By definition, this means that for all $ugeq w$, if $uVdash varphi$, then $uVdash bot$. But we always have $unotVdash bot$. So the condition is equivalent to saying that that for all $ugeq w$, $unotVdash varphi$.

Summing up:

• We have $models lnot varphi$ if and only if for every model $M$ and every world $win M$, $wVdash lnot varphi$.


• If $M$ is a model and $win M$ is a world, then we have $wVdash lnot varphi$ if and only if for all $ugeq w$, $unotVdash varphi$.

Ok, following your question, let's show that $models lnot (pland lnot p)$, but $notmodels lnot p$ and $notmodels lnotlnot p$.

To show that $models lnot (pland lnot p)$, we have to consider an arbitrary model $M$ and an arbitrary world $win M$ and show that for all $ugeq w$, $unotVdash pland lnot p$. By definition, $uVdash pland lnot p$ if and only if $uVdash p$ and $uVdash lnot p$. But if $uVdash lnot p$, we have that for all $u'geq u$, $u'notVdash p$. In particular, taking $u' = u$, we have $unotVdash p$, which contradicts $uVdash p$. So $unotVdash pland lnot p$, as desired.

To show that $notmodels lnot p$, we just need to find a single model $M$ and a single world $w$ such that $wnotVdash lnot p$. Well, consider a model $M$ with a single world $w$ such that $v(p) = w$, i.e. $wVdash p$. Then since $wgeq w$ and $wVdash p$, we have $wnotVdash lnot p$.

To show that $notmodels lnot lnot p$, we just need to find a single model $M$ and a single world $w$ such that $wnotVdash lnotlnot p$. Well, consider a model $M$ with a single world $w$ such that $v(p) = emptyset$, i.e. $wnotVdash p$. Then for all $w'geq w$ ($w' = w$ is the only choice), we have $w'notVdash p$, so $wVdash lnot p$. And since $wgeq w$ and $wVdash lnot p$, we have $wnotVdash lnotlnot p$.