Propositional Intuitionistic Logic Question

Question

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

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

Consider the formula $\lnot (p\land \lnot p)$.

$\models \lnot (p\land \lnot p)$ then $\models \lnot p$ or $\models \lnot\lnot p$.

Show $\models \lnot (p\land \lnot p)$ is true and neither $\models \lnot p$ nor $\models \lnot\lnot p$ is the case.

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

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

Answer 1

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 $\varphi\rightarrow \bot$. So we have $\models \lnot \varphi$ iff for every intuitionistic Kripke model $M$ and every world $w\in M$, $w\Vdash \varphi\rightarrow \bot$. By definition, this means that for all $u\geq w$, if $u\Vdash \varphi$, then $u\Vdash \bot$. But we always have $u\not\Vdash \bot$. So the condition is equivalent to saying that that for all $u\geq w$, $u\not\Vdash \varphi$.

Summing up:

• We have $\models \lnot \varphi$ if and only if for every model $M$ and every world $w\in M$, $w\Vdash \lnot \varphi$.
• If $M$ is a model and $w\in M$ is a world, then we have $w\Vdash \lnot \varphi$ if and only if for all $u\geq w$, $u\not\Vdash \varphi$.

Ok, following your question, let's show that $\models \lnot (p\land \lnot p)$, but $\not\models \lnot p$ and $\not\models \lnot\lnot p$.

To show that $\models \lnot (p\land \lnot p)$, we have to consider an arbitrary model $M$ and an arbitrary world $w\in M$ and show that for all $u\geq w$, $u\not\Vdash p\land \lnot p$. By definition, $u\Vdash p\land \lnot p$ if and only if $u\Vdash p$ and $u\Vdash \lnot p$. But if $u\Vdash \lnot p$, we have that for all $u'\geq u$, $u'\not\Vdash p$. In particular, taking $u' = u$, we have $u\not\Vdash p$, which contradicts $u\Vdash p$. So $u\not\Vdash p\land \lnot p$, as desired.

To show that $\not\models \lnot p$, we just need to find a single model $M$ and a single world $w$ such that $w\not\Vdash \lnot p$. Well, consider a model $M$ with a single world $w$ such that $v(p) = \{w\}$, i.e. $w\Vdash p$. Then since $w\geq w$ and $w\Vdash p$, we have $w\not\Vdash \lnot p$.

To show that $\not\models \lnot \lnot p$, we just need to find a single model $M$ and a single world $w$ such that $w\not\Vdash \lnot\lnot p$. Well, consider a model $M$ with a single world $w$ such that $v(p) = \emptyset$, i.e. $w\not\Vdash p$. Then for all $w'\geq w$ ($w' = w$ is the only choice), we have $w'\not\Vdash p$, so $w\Vdash \lnot p$. And since $w\geq w$ and $w\Vdash \lnot p$, we have $w\not\Vdash \lnot\lnot p$.