Possibilistic Semantics for a Modal KD45 Extension of Gödel Fuzzy Logic

Abstract

In this paper we provide a simplified semantics for the logic $KD45(\mathbf {G})$, i.e. the many-valued Gödel counterpart of the classical modal logic KD45. More precisely, we characterize $KD45(\mathbf {G})$ as the set of valid formulae of the class of possibilistic Gödel Kripke Frames $\langle W, \pi \rangle$, where W is a non-empty set of worlds and $\pi : W \rightarrow [0, 1]$ is a normalized possibility distribution on W.

Appendix

Proof of Lemma 2

Proof

First of all, notice that if $\nu$ satisfies the condition b, then necessarily $\alpha < \beta$. Let $B= \{ \nu (\lambda ) : \lambda \in \varDelta _\varphi , \nu (\lambda )< 1\} \cup \{0 \} = \{ b_0 = 0< b_1 < \ldots b_N \}$. Obviously, $b_N < 1$. For each $0 \le i \le N$, pick $\lambda _i \in \varDelta _\varphi$ such that $\nu (\lambda _i) = b_i$. Define now a continuous strictly function $g:[0,1]\mapsto [0,\delta ) \cup \{1\}$ such that

$g(1)=1$

$g(b_i)= v(\lambda _i)$ for every $0 \le i \le N$

$g[(b_N, 1)]= (\alpha , \delta ).$

Notice that $\alpha = g(b_N)$. In addition, define another continuous strictly increasing function $h:[0,1] \mapsto [\delta , 1]$ such that

$h(0)=\delta$

$h[(0,\beta )] =(\delta , \beta )$

$h(x) = x$, for $x \in [\beta , 1].$

Then we define the valuation $w: Var \cup X \rightarrow [0,1]$ as follows:

$w(p) = \left\{ \begin{array}{ll} g(\nu (p)), &{} \text{ if } \nu (p) < 1,\\ h(u(p)), &{} \text{ if } \nu (p) = 1. \end{array} \right.$

First of all, let us show by induction that this extends to any propositional formula, that is,

$w(\varphi ) = \left\{ \begin{array}{ll} g(\nu (\varphi )), &{} \text{ if } \nu (\varphi ) < 1,\\ h(u(\varphi )), &{} \text{ if } \nu (p) = 1. \end{array} \right.$

Note that, since g and h are strictly increasing mappings, $g\circ \nu$ and $h \circ u$ are valuations as well. So, in the induction steps below we only need to check that everything is fine when both are used at the same time when evaluating a compound formula. The base case holds by definition.

• Assume $\psi = \psi _1 \wedge \psi _2$. We only check the case when $v(\psi ) < 1$ and $v(\psi _1) < 1$ and $v(\psi _2) = 1$. Then $w(\psi ) = \min (w(\psi _1), w(\psi _2)) = \min (g(\nu (\psi _1)), h(u(\psi _2))) = g(\nu (\psi _1))$, since $g(\nu (\psi _1)) < \delta \le h(u(\psi _2)$. But, $g(\nu (\psi _1)) = \min ( g(\nu (\psi _1)), 1) = \min ( g(\nu (\psi _1)), g(\nu (\psi _2))) = g(\nu (\psi _1\wedge \psi _2)) = g(\nu (\psi ))$.

• Assume $\psi = \psi _1 \rightarrow \psi _2$, and consider two subcases:

  (1) $v(\psi _1) < 1$ and $v(\psi _2) = 1$. Then $v(\psi _1 \rightarrow \psi _2) = 1$ and $w(\psi ) = w(\psi _1) \Rightarrow w(\psi _2) = g(\nu (\psi _1)) \Rightarrow h(u(\psi _2)) = 1 = h(u(\psi _1)) \Rightarrow h(u(\psi _2)) = h(u(\psi _1 \rightarrow \psi _2)) = h(u(\psi )).$ (2) $v(\psi _1) = 1$ and $v(\psi _2) < 1$. Then $v(\psi _1 \rightarrow \psi _2) = v(\psi _2) < 1$ and $w(\psi ) = w(\psi _1) \Rightarrow w(\psi _2) = h(u(\psi _1)) \Rightarrow g(\nu (\psi _2)) = g(\nu (\psi _2)) = g(\nu (\psi _1)) \Rightarrow g(\nu (\psi _2)) = g(\nu (\psi _1 \rightarrow \psi _2)) = g(\nu (\psi ))$.

Properties 1 – 6 in the statement of Lemma 2 now directly follow from the above.

Finally, we prove that $w \in W^v$. By definition of w it is clear that $w \sim _\varphi v$. It remains to check that w validates all the axioms. The axioms of $\mathcal {G}$ are evaluated to 1 by any Gödel valuation. As for the specific axioms of $KD45(\mathbf {G})$, it is an immediate consequence of Property 3 because it implies that if $\nu (\psi \rightarrow \phi )=1$ then $w(\psi \rightarrow \phi )=1$.   $\blacksquare$

Claim 1 from Lemma 3. If $u(\Box \psi )=\alpha <1$, for every $\varepsilon > 0$, there exists a valuation $w\in W$ such that $\pi ^\varphi (w) > w(\psi )$ and $w(\psi ) < \alpha + \varepsilon$, and thus $\pi ^\varphi (w) \Rightarrow w(\psi ) = w(\psi ) < \alpha + \varepsilon$.

Proof

By definition of Gödel’s implication $\Rightarrow$ in [0,1], to grant the required conditions on w it is enough to find $w \in W$ such that $\alpha \le w(\psi )$ and, for any $\theta \in Sub(\varphi )$, $u(\Box \theta )\le w(\theta ) \le u(\Diamond \theta ) \le \alpha$. This is achieved in two stages:

• first producing a valuation $\nu \in W$ satisfying $\nu (\psi )<1$ and preserving the relative ordering conditions the values $w(\theta )$ must satisfy, conditions which may be coded by a theory $\varGamma _{\psi ,u}$;

• and then moving the values $\nu (\theta )$, for $\theta \in Sub(\varphi )$, to the correct valuation w by composing $\nu$ with an increasing bijection of [0,1].

Assume $u(\Box \psi )=\alpha <1$, and define (all formulas involved ranging in $\mathcal {L}_{\square \Diamond }(Var)$)

$$\begin{aligned} \begin{array}{ll} \varGamma _{\psi ,u}= &{} \{\lambda : \lambda \in \varDelta _\varphi \text{ and } u(\lambda ) > \alpha \} \\ &{} \cup \{\lambda \rightarrow \theta : \lambda \in \varDelta _\varphi \text{ and } u(\lambda ) \le u(\Box \theta ) \} \\ &{} \cup \{(\theta \rightarrow \lambda )\rightarrow \lambda : \lambda \in \varDelta _\varphi \text{ and } u(\lambda )< u(\Box \theta )< 1 \} \\ &{} \cup \{\theta \rightarrow \lambda : \lambda \in \varDelta _\varphi \text{ and } u(\Diamond \theta ) \le u(\lambda ) \} \\ &{} \cup \{(\lambda \rightarrow \theta )\rightarrow \theta : \lambda \in \varDelta _\varphi \text{ and } u(\Diamond \theta )< u(\lambda )< 1 \} \\ \end{array} \end{aligned}$$

Then we have $u(\square \xi ) > \alpha$ for each $\xi \in \varGamma _{\psi ,u}$. Indeed, first recall that, by $U_\Box$ and $U_\Diamond$ of Proposition 2, for any $\lambda \in \varDelta _{\varphi }$ we have $u(\lambda ) = u(\Box \lambda ) = u(\Diamond \lambda )$. We analyse case by case. For the first set of formulas, it is clear by construction. For the second set, we have $u(\square (\lambda \rightarrow \theta )) \ge u(\Diamond \lambda \rightarrow \square \theta )= u(\Diamond \lambda ) \Rightarrow u(\Box \theta ) = u(\lambda ) \Rightarrow u(\Box \theta ) = 1$, by FS2. For the third, by FS2 and P, we have $u(\square ((\theta \rightarrow \lambda )\rightarrow \lambda )) \ge u(\Diamond (\theta \rightarrow \lambda ) \rightarrow \Box \lambda ) \ge u((\Box \theta \rightarrow \Diamond \lambda ) \rightarrow \Box \lambda ) = 1$, since $u(\Box \lambda ) = u(\Diamond \lambda ) = u(\square \theta \rightarrow \Diamond \lambda ) < 1$. The fourth and fifth cases are very similar to the second and third ones respectively.

This implies

$$\begin{aligned} \varGamma _{\psi ,u} \not \vdash _{KD45(\mathbf {G})}\psi , \end{aligned}$$

otherwise there would exist $\xi _{1},\ldots ,\xi _{k}\in \varGamma _{\psi ,u}$ such that $\xi _{1},\ldots ,\xi _{k} \vdash _{KD45(\mathbf {G})} \psi .$ In such a case, we would have $\Box \xi _{1},\ldots ,\Box \xi _{k} \vdash _{KD45(\mathbf {G})}\Box \psi$ by Nec and $K_{\square }$. Then $\Box \xi _{1},\ldots ,\Box \xi _{k},ThKD45(\mathbf {G})\vdash _G\Box \psi$ by Lemma 1 and thus by Proposition 1 (i), and recalling that $u(ThKD45(\mathbf {G}))=1,$

$$\begin{aligned} \alpha < \inf u(\{\Box \xi _{1},\ldots ,\Box \xi _{k}\}\cup ThKD45(\mathbf {G}))\le u(\square \psi )=\alpha , \end{aligned}$$

a contradiction. Therefore, by Proposition 1 (ii) there exists a valuation $\nu :Var\cup X\mapsto [0,1]$ such that $\nu (\varGamma _{\psi ,u} \cup ThKD45(\mathbf {G}))=1$ and $\nu (\psi )<1$. This implies the following relations between u and $\nu ,$ that we list for further use. Given $\lambda \in \varDelta _\varphi , \theta \in \mathcal {L}_{\square \Diamond }(Var)$, we have :

• #1. If $u(\lambda )>\alpha$ then $\nu (\lambda )=1$ (since then $\lambda \in \varGamma _{\psi ,u})$.

• #2. If $u(\lambda )\le u(\square \theta )$ then $\nu (\lambda )\le \nu (\theta )$ (since then $\lambda \rightarrow \theta \in \varGamma _{\psi ,u})$. In particular, if $\lambda _1,\lambda _2 \in \varDelta _\varphi$ and $u(\lambda _1)\le u(\Box \lambda _2) = u(\lambda _2)$ then $\nu (\lambda _1)\le \nu (\lambda _2)$. Furthermore,if $\Box \theta \in \varDelta _\varphi$ then from $u(\Box \theta ) = u(\Box \theta )$ by #2, $\nu (\Box \theta ) \le \nu (\theta )$. That means, taking $\theta = \psi$, $\nu (\Box \psi ) \le \nu (\psi ) < 1$.

• #3. If $u(\lambda )< u(\square \theta )< 1$ then $\nu (\lambda )< \nu (\theta )$ or $\nu (\lambda )=1$ (since then $(\theta \rightarrow \lambda )\rightarrow \lambda )\in \varGamma _{\psi ,u}$). In particular, if $\lambda _1,\lambda _2 \in \varDelta _\varphi$, $u(\lambda _1) < u(\lambda _2)$ and $u(\lambda _2) \le u(\Box \psi )= \alpha$ then $\nu (\lambda _1)< \nu (\psi ) < 1$ and thus $\nu (\lambda _1) < \nu (\lambda _2)$. This means that $\nu$ preserves in a strict sense the order values by u of the formulas $\lambda \in \varDelta _\varphi$ such that $u(\lambda ) \le \alpha$.

• #4. If $u(\Diamond \theta ) \le u(\lambda )$ then $\nu (\theta ) \le \nu (\lambda )$ (because $\theta \rightarrow \lambda \in \varGamma _{\psi ,u}$). In particular, if $\Diamond \theta \in \varDelta _\varphi$ then $\nu (\theta ) \le \nu (\Diamond \theta )$.

• #5. If $u(\Diamond \theta )< u(\lambda )< 1$ then $\nu (\theta ) < \nu (\lambda )$ or $\nu (\theta ) =1$. In particular, if $\lambda _1, \lambda _2 \in \varDelta _\varphi$ and $u(\lambda _1)< u(\lambda _2)\le \alpha = u(\Box \psi )$ then $\nu (\lambda _1)< \nu (\lambda _2)$. Furthermore, if $u(\lambda _2)>0$ then $\nu (\lambda _2)>0$ (making $\lambda _1:=\Diamond \bot$ since $u(\bot )=u(\Diamond \bot )=0$).

According to the properties #1, #2 and #3, it is clear that $\nu$ satisfies the conditions of Lemma 2. Consequently, for all $\epsilon > 0$ (such that $\alpha +\varepsilon <\beta$), taking $\delta = \alpha + \varepsilon$ in Lemma 2, there exists a valuation $w \in W^{v}$ such that $w(\psi ) < \alpha +\varepsilon = \delta$. Then in order to finish our proof, it remains to show that:

$$\begin{aligned} \pi ^\varphi (w)= \inf _{\lambda \in sub(\varphi )} \min (v(\Box \lambda ) \Rightarrow w(\lambda ), w(\lambda ) \Rightarrow v(\Diamond \lambda )) >w(\varphi ) \end{aligned}$$

(3)

To do so, we will prove that, for any $\lambda \in sub(\varphi )$, both implications in (3) are greater than $\delta$.² First we prove it for the first implication by cases:

• - If $v(\Box \lambda ) \le \alpha < 1$ then $\min (v(\Box \lambda ) \Rightarrow w(\lambda ), w(\lambda ) \Rightarrow v(\Diamond \lambda )) = 1$. Indeed, first of all, by #2, from $u (\Box \lambda ) = v(\Box \lambda ) \le \alpha = u(\Box \psi )$ it follows $\nu (\Box \lambda ) \le \nu (\psi ) < 1$. Now, since $u(\Box \lambda ) \le u(\Box \lambda )$, by #2, we have $1 \ne \nu (\Box \lambda ) \le \nu (\lambda )$, and by 3 of Lemma 2 we have $v(\Box \lambda ) = w(\Box \lambda ) \le w(\lambda )$. Then $v(\Box \lambda ) \Rightarrow w(\lambda ) =1$.

• - If $v(\Box \lambda ) > \alpha$ then by #1 and #2, $1=\nu (\Box \lambda )\le \nu (\lambda )$. Therefore, by 1 of Lemma 2, $w(\lambda )>\delta$ which implies $v(\Box \lambda ) \Rightarrow w(\lambda ) > \delta$.

For the second implication we also consider two cases:

• - If $v(\Diamond \lambda ) = u(\Diamond \lambda ) > \delta$ then it is obvious that $w(\lambda ) \Rightarrow v(\Diamond \lambda ) > \delta$.

• - If $u(\Diamond \lambda ) < \delta$, by definition of $\delta$ and taking into account that $\Diamond \lambda \in \varDelta _\varphi$, then $u(\Diamond \lambda ) < \alpha$. Now from $u(\Diamond \lambda ) = u(\Diamond \lambda )$ we obtain by #4, that $\nu (\lambda ) \le \nu (\Diamond \lambda ) < 1$. Then by Lemma 2 we have $w(\lambda ) \le w(\Diamond \lambda ) = v(\Diamond \lambda )$ and thus $w(\lambda ) \Rightarrow v(\Diamond \lambda ) = 1$. $\blacksquare$

Claim 2 from Lemma 3. If $u(\Diamond \psi )=\alpha >0$ then, for any $\varepsilon >0,$ there exists a valuation $w'\in W$ such that $w'(\psi )=1$ and $\pi ^\varphi (w')\ge \alpha -\varepsilon$, and thus $\min (w'(\psi ), \pi ^\varphi (w')) \ge \alpha -\varepsilon$.

Proof

Assume $u(\Diamond \psi )=\alpha >0$ and define $\varGamma _{\psi ,u}$ in the same way that it was defined in the proof of Claim 1. Then we consider two cases:

• - If $u(\Diamond \psi )=1$, let $U_{\psi ,u}^{{}} = \{ \lambda : \ \lambda \in \varDelta _\varphi \text{ and } u(\lambda ) < 1 \}$. We claim that

$$\begin{aligned} \psi , \varGamma _{\psi ,u} \not \vdash _{KD45(\mathbf {G})} \bigvee U_{\psi ,u} \ , \end{aligned}$$

otherwise we would have $\theta _1, \ldots , \theta _n \in \varGamma _{\psi ,u}$ such that $\vdash _{KD45(\mathbf {G})} \psi \rightarrow ((\theta _1\wedge \ldots \wedge \theta _n) \rightarrow \bigvee U_{\psi ,u})$, and then we would also have $\vdash _{KD45(\mathbf {G})} \Diamond \psi \rightarrow \Diamond ((\theta _1\wedge \ldots \wedge \theta _n) \rightarrow \bigvee U_{\psi ,u})$, that would imply in turn that $\vdash _{KD45(\mathbf {G})} \Diamond \psi \rightarrow ((\Box \theta _1\wedge \ldots \wedge \Box \theta _n) \rightarrow \Diamond \bigvee U_{\psi ,u})$. In that case, taking the evaluation u it would yield: $1= u(\Diamond \psi ) \le u(\Box \theta _1\wedge \ldots \wedge \Box \theta _n ) \Rightarrow u( \Diamond \bigvee U_{\psi ,u})$, a contradiction, since $u(\Box \theta _1\wedge \ldots \wedge \Box \theta _n) = 1$ and $u( \Diamond \bigvee U_{\psi ,u})) < 1$.³

Therefore, there is a Gödel valuation $\nu '$ (not necessarily in W) such that $\nu '(\psi )=\nu '(\varGamma _{\psi ,u})=\nu '(ThKD45(\mathbf {G}))=1$ and $\nu '(\bigvee U_{\psi ,u} )<1$. By #2 and #3, it follows that for any $\lambda _{1},\lambda _{2} \in \varDelta _\varphi$ such that $u(\lambda _{1}), u(\lambda _{2}) \le \alpha$, we have $u(\lambda _{1})< u(\lambda _{2}) \le \alpha \ \text{ iff } \ \nu '(\lambda _{1}) < \nu '(\lambda _{2}).$ Thus, $\nu '$ satisfies the conditions of Lemma 2 because it is strictly increasing in $\varDelta _\varphi$ (i.e. it satisfies condition b of Lemma 2), and $\nu '(ThKD45(\mathbf {G}))=1$. Therefore, there exists a valuation $w' \in W$ such that $w'(\psi )=1$.

It remains to show that $\pi ^\varphi (w') = 1$. Indeed, by construction, it holds that $u(\Box \theta ) \le w'(\theta )\le u(\Diamond \theta )$, and hence $\min (u(\Box \theta ) \Rightarrow w'(\theta ) , w'(\theta )\Rightarrow u(\Diamond \theta )) =1$.

• - If $1> u(\Diamond \psi )=\alpha >0$, then we let $U_{\psi ,u}^{{}} = \ (\Diamond \psi \rightarrow \psi ) \rightarrow \psi .$ We claim that

$$\begin{aligned} \Box \top , \varGamma _{\psi ,u} \not \vdash _{KD45(\mathbf {G})} U_{\psi ,u} \ , \end{aligned}$$

otherwise there would exist $\theta _1, \ldots , \theta _n \in \varGamma _{\psi ,u}$ such that $\vdash _{KD45(\mathbf {G})} \Box \top \rightarrow ((\theta _1\wedge \ldots \wedge \theta _n) \rightarrow U_{\psi ,u})$, and then we would have $\vdash _{KD45(\mathbf {G})} \Diamond \Box \top \rightarrow \Diamond ((\theta _1\wedge \ldots \wedge \theta _n) \rightarrow U_{\psi ,u})$, which would imply $\vdash _{KD45(\mathbf {G})} \Box \top \rightarrow ((\Box \theta _1\wedge \ldots \wedge \Box \theta _n) \rightarrow \Diamond U_{\psi ,u})$. In that case, evaluating with u it would yield $1= u(\Box \top ) \le u(\Box \theta _1\wedge \ldots \wedge \Box \theta _n ) \Rightarrow u( \Diamond U_{\psi ,u}))$, contradiction, since $u(\Box \theta _1\wedge \ldots \wedge \Box \theta _n) > \alpha$ and $u( \Diamond U_{\psi ,u})) \le \alpha$ (because $u(\Diamond ((\Diamond \psi \rightarrow \psi ) \rightarrow \psi ))\le u(\Box (\Diamond \psi \rightarrow \psi ) \rightarrow \Diamond \psi )\le u(\Diamond \psi )\le \alpha$).

Therefore, there is an evaluation $\nu '$ such that $\nu '(ThKD45(\mathbf {G}) )=\nu '(\varGamma _{\psi ,u})=1$ and $\nu '(\bigvee U_{\psi ,u} )<1$. Hence, we can conclude that the three pre-conditions $\mathbf a$, $\mathbf b$ and $\mathbf c$ required in Lemma 2 are satisfied. In addition, the following condition is also satisfied:

• d. $\nu '(\Diamond \psi ) = \nu '(\psi )$.

At this point, we can now do a proof dual to the one for Claim 1. Again, by Lemma 2 for $\delta = \frac{\beta - \alpha }{2}$, we obtain from $\nu '$ an evaluation $w' \in W^{v}$ such that $w'(\psi )=\alpha$. It only remains then to show that $\pi ^\varphi (w)> \alpha$. But in this case, the proof is the same than the one given for Eq. (3) using $w'$ instead of w. This finishes the proof.   $\blacksquare$