Disjunction and Negation in Information Based Semantics

Abstract

We investigate an information based generalization of the incompatibility-frame treatment of logics with non-classical negation connectives. Our framework can be viewed as an alternative to the neighbourhood semantics for extensions of lattice logic by various negation connectives, investigated by Hartonas. We set out the basic semantic framework, along with some correspondence results for extensions. We describe three kinds of constructions of canonical models and show that double negation law is not canonical with respect to any of these constructions. We also compare our semantics to Hartonas’.

This paper is an outcome of the project Logical Structure of Information Channels, no. 21-23610M, supported by the Czech Science Foundation and realized at the Institute of Philosophy of the Czech Academy of Sciences.

A Appendix

Proof

(Proposition 5). First, assume that $\bot _t$ is a filter, for every $t \in S$. We show the inductive step for negation, that is, we will assume that $|| \alpha ||$ is a filter, and show that then $|| \lnot \alpha ||$ is a filter as well. It holds that $i \vDash \lnot \alpha$, since for any $t \in ||\alpha ||$, $i \in \bot _t$. Next, assume that $s \vDash \lnot \alpha$ and $s \le t$. Then if $u \vDash \alpha$ then $s \in \bot _u$, and thus $t \in \bot _u$. So, $t \vDash \lnot \alpha$. Assume that $s \vDash \lnot \alpha$ and $t \vDash \lnot \alpha$. Then for any $u \in || \alpha ||$, $s \in \bot _u$ and $t \in \bot _u$, and so $s \circ t \in \bot _u$. Thus $s \circ t \vDash \lnot \alpha$.

Second, assume that for some $t \in S$, $\bot _t$ is not a filter. We will show that there is a valuation V in I, and a formula $\alpha$ such that $|| \alpha ||$ is not a filter in $\left\langle I, V \right\rangle$. We consider three cases: (a) $i \notin \bot _t$; (b) there are $s, u \in S$ such that $s \le u$, $s \in \bot _t$, and $u \notin \bot _t$; (c) there are $s, u \in \bot _t$ such that $s \circ u \notin \bot _t$.

Consider a valuation V such that $V(p)=\{ v \in S~|~t \le v \}$. In the case (a), $t \vDash p$ but $i \notin \bot _t$. So, $i \notin || \lnot p ||$. In the case (b), if $v \vDash p$ then $t \le v$, and thus $s \in \bot _v$. So, $s \vDash \lnot p$. But $u \nvDash \lnot p$, for $t \vDash p$ and $u \notin \bot _t$. Hence, $|| \lnot p ||$ is not a filter. In the case (c) $s \vDash \lnot p$ and $u \vDash \lnot p$ but $s \circ u \nvDash \lnot p$. Hence, again, $|| \lnot p ||$ is not a filter.   $\square$

Proof

(Proposition 9). The proof is by induction on the complexity of $\alpha$, and the only case we will consider is that where $\alpha =\lnot \beta$. The right-to-left direction is immediate, so suppose that $\lnot \beta \notin \varDelta$. Fix $\varGamma =\{\gamma \mid \beta \vdash _{\varLambda }\gamma \}$, and note that $\varGamma \in S_1$ and $\varGamma \vDash \beta$. Suppose that $\varGamma \bot _1 \varDelta$ holds, so that for some $\gamma \in \varGamma$, $\lnot \gamma \in \varDelta$; in that case, $\beta \vdash \gamma$, and so $\lnot \gamma \vdash \lnot \beta$, and thus $\lnot \beta \in \varDelta$, contrary to the assumption. Thus $\varDelta \nvDash \lnot \beta$, as desired.    $\square$

Proof

(Proposition 13). Again the proof is by induction on the complexity of $\alpha$. The inductive step for negation amounts to Lemma 3.13 in [14]. Since our notation is quite different, let us reconstruct this step. For the left-to-right direction assume that $\varDelta \vDash \lnot \beta$. Then for any $\varGamma \in S_2$, if $\varGamma \vDash \beta$, i.e. $\beta \in \varGamma$, then $\varGamma ^* \subseteq \varDelta$. Let us fix $\varGamma =\{\gamma \mid \beta \vdash _{\varLambda }\gamma \}$. Then $\gamma \in \varGamma ^*$ iff for some $\delta$, $\delta \vdash _{\varLambda } \beta$ and $\lnot \delta \vdash _{\varLambda } \gamma$, which is equivalent to $\lnot \beta \vdash _{\varLambda } \gamma$. Hence, $\varGamma ^*= \{\gamma \mid \lnot \beta \vdash _{\varLambda }\gamma \}$ and thus $\lnot \beta \in \varDelta$. For the right-to-left direction assume that $\lnot \beta \in \varDelta$. Take any $\varGamma$  such that $\varGamma \vDash \beta$  and thus $\beta \in \varGamma$. Let $\gamma \in \varGamma ^*$. Then there is $\delta$ such that $\delta \vdash _{\varLambda }\beta$ and $\lnot \delta \vdash _{\varLambda } \gamma$. Hence, $\lnot \beta \vdash _{\varLambda } \gamma$ and $\gamma \in \varDelta$. We have shown that $\varGamma ^* \subseteq \varDelta$, i.e. $\varGamma \bot _2 \varDelta$. It follows that $\varDelta \vDash \lnot \beta$.    $\square$

Proof

(Proposition 15). The inductive step for negation can be proved as follows: $[\alpha ] \vDash \lnot \beta$ iff for all $\gamma$, if $[\gamma ] \vDash \beta$ then $[\gamma ] \bot _3 [\alpha ]$ iff for all $\gamma$, if $\gamma \vdash _{\varLambda } \beta$ then $\alpha \vdash _{\varLambda } \lnot \gamma$ iff $\alpha \vdash _{\varLambda } \lnot \beta$.    $\square$

Proof

(Proposition 16). (b) First, take any IIF I in which the condition is satisfied, i.e. for every filter F and every state $s \notin F$ there is a state t such that $s \notin \bot _t$ and $F \subseteq \bot ^t$. Assume $s \nvDash p$, i.e. $s \notin V(p)$. Thus, there is t such that $s \notin \bot _t$ and $V(p) \subseteq \bot ^t$. Then $t \vDash \lnot p$ (any state supporting p is incompatible with t) but since it is not the case that $t \bot s$ we obtain $s \nvDash \lnot \lnot p$. We have shown that $\lnot \lnot p \vdash p$ is valid in I. Second, take an IIF I in which the condition is not satisfied, i.e. there is a filter F and a state $s \notin F$ such that for every state t, if $F \subseteq \bot ^t$ then $s \in \bot _t$. Consider a valuation V such that $V(p)=F$. Then $s \nvDash p$. Moreover, if $t \vDash \lnot p$, then $V(p) \subseteq \bot ^t$, and, consequently $t \bot s$. It follows that $s \vDash \lnot \lnot p$. We have shown that $\lnot \lnot p \vdash p$ is not valid in I.

(c) First, note that if $\bot$ is symmetric, $\bot _t = \bot ^t$ for any state t. Hence, it follows from (a) and (b) that $\{ p \vdash \lnot \lnot p, \lnot \lnot p \vdash p \}$ characterizes the class of IIFs where (i) the compatibility relation is symmetric and (ii) for any filter F and any state $s \notin F$ there is an incompatibility filter G such that $F \subseteq G$ and $s \notin G$. Now it suffices to show that (ii) is equivalent to the condition stating that every filter is the intersection of a set of incompatibility filters. Let $UpF=\{G~|~G\text { is an incompatibility filter such that }F \subseteq G \}$. It is obvious that $F \subseteq \bigcap UpF$. It holds that there is F that is not the intersection of any set of incompatibility filters iff there is F that is a proper subset of $\bigcap UpF$ iff there is F and $s \notin F$ such that $s \in \bigcap UpF$ iff (ii) does not hold.    $\square$

Proof

(Prop. 17). We will just consider the case of $\lnot p\wedge \lnot q\vdash \lnot (p\vee q)$ – the others are similar, and this is the axiom that distinguishes our basic setting from that employing a standard treatment of disjunction (and all ‘prime’ states). For canonicity$_1$, suppose that $\varGamma ,\varSigma ,\varDelta \in S_1$, $\varGamma \bot _1\varDelta$, and $\varSigma \bot _1\varDelta$. It follows that there are $\alpha \in \varGamma$ and $\beta \in \varSigma$ such that $\lnot \alpha ,\lnot \beta \in \varDelta$. Thus $\lnot \alpha \wedge \lnot \beta \in \varDelta$, and thus $\lnot (\alpha \vee \beta )\in \varDelta$. It is immediate that $\alpha \vee \beta \in \varGamma \circ _1\varSigma =\varGamma \cap \varSigma$, and thus $\varGamma \circ _1\varSigma \bot _1\varDelta$, as desired. For canonicity$_3$, we’ll work in the more complex setting, not assuming that there is an explosive formula. To that end, suppose that $s,t,u\in S_3$, $s\bot _3 u$, and $t\bot _3 u$. If any of s, t, u is $i_3$ then we’re done, as $i_3\bot _3 v$, $v\bot _3i_3$, and $v\circ _3i_3=i_3\circ _3v=i_3$ all hold for every v, so suppose that none is: i.e., that there are $\alpha ,\beta ,\gamma \in Prop_\varLambda$ such that $s=[\alpha ]$, $t=[\beta ]$, and $u=[\gamma ]$. By the supposition, then, $\gamma \vdash \lnot \alpha$ and $\gamma \vdash \lnot \beta$, and thus $\gamma \vdash \lnot \alpha \wedge \lnot \beta$. It follows that $\gamma \vdash \lnot (\alpha \vee \beta )$, and thus, since $s\circ _3t=[\alpha \vee \beta ]$, $s\circ _3\bot _3u$, as desired.    $\square$

Proof

(Proposition 18). First, we will show that $\{ p \vdash \lnot \lnot p, \lnot \lnot p \vdash p \}$ is not canonical$_1$. Let $\varLambda$ be the distributive lattice logic enriched with (N) and both $\lnot \lnot p \vdash p$ and $p \vdash \lnot \lnot p$. We will show the following: There is a filter of non-empty $\varLambda$-theories F (i.e. a set of non-empty $\varLambda$-theories that is closed under intersection and stronger $\varLambda$-theories) and there is a non-empty $\varLambda$-theory $\varDelta \notin F$ such that $\varDelta \bot _1 \varOmega$ for all $\varOmega$ such that $\varGamma \bot _1 \varOmega$  for each $\varGamma \in F$. This will imply that F  cannot be expressed as an intersection of incompatibility filters. Let us construct F and $\varDelta$ satisfying the desired property. Take for each $i \in \mathbb {N}$ an infinite set of atomic formulas $X_i=\{p^i_1, p^i_2, \ldots \}$, assuming that if $i \ne j$  then $X_i$  and $X_j$ are disjoint. Let $\varGamma _i$ be the $\varLambda$-theory generated by $X_i$. Let F be the filter generated by the set of $\varLambda$-theories $\{\varGamma _1, \varGamma _2, \ldots \}$. Now we can construct $\varDelta$. Let Y be the set of all disjunctions of atomic formulas from $\bigcup _{i \in \mathbb {N}} X_i$ satisfying the following:

• $p^{i_1}_{j_1} \vee \ldots \vee p^{i_n}_{j_n} \in Y$ iff all $i_1, \ldots , i_n$ are distinct and it is not the case that there is k such that $j_1 = \ldots = j_n = k$ and $i_1, \ldots , i_n \le k$.

This definition can be illustrated with the following table. Y contains all disjunctions of atomic formulas from different columns except those that are connected by a path built from the horizontal lines.

Let $\varDelta$  be the $\varLambda$-theory generated by Y. Let $\varGamma \in F$. Then there are $\varGamma _{i_1}, \ldots , \varGamma _{i_n}$ such that $\varGamma _{i_1} \cap \ldots \cap \varGamma _{i_n} \subseteq \varGamma$. Take $k=max\{i_1, \ldots , i_n\}$. Then $p^1_k \vee \ldots \vee p^k_k \in \varGamma$ but $p^1_k \vee \ldots \vee p^k_k \notin \varDelta$, and thus $\varDelta \ne \varGamma$. We have shown that $\varDelta \notin F$.

Now assume $\varGamma \bot _1 \varOmega$ for each $\varGamma \in F$. Then $p^1_{j_1} \wedge \ldots \wedge p^1_{j_n} \vdash _{\varLambda } \alpha$ for some $p^1_{j_1}, \ldots , p^1_{j_n} \in X_1$ and some $\alpha$ such that $\lnot \alpha \in \varOmega$. Then $\lnot (p^1_{j_1} \wedge \ldots \wedge p^1_{j_n}) \in \varOmega$. Take $k=max\{j_1, \ldots , j_n \}+1$. There must be some $p^k_{i_1}, \ldots , p^k_{i_m} \in X_k$ such that $\lnot (p^k_{i_1} \wedge \ldots \wedge p^k_{i_m}) \in \varOmega$. Hence $\lnot ((p^1_{j_1} \wedge \ldots \wedge p^1_{j_n}) \vee (p^k_{i_1} \wedge \ldots \wedge p^k_{i_m})) \in \varOmega$. Moreover, k was selected in such a way that it also holds that $p^1_j \vee p^k_i \in \varDelta$, for each $j \in \{j_1, \ldots , j_n \}$  and $i \in \{i_1, \ldots , i_m \}$. Then $(p^1_{j_1} \wedge \ldots \wedge p^1_{j_n}) \vee (p^k_{i_1} \wedge \ldots \wedge p^k_{i_m}) \in \varDelta$, by distributivity, and thus $\varDelta \bot _1 \varOmega$  as desired.

Second, we will show that $\{ p \vdash \lnot \lnot p, \lnot \lnot p \vdash p \}$ is not canonical$_2$. Let $\varLambda$ be the lattice logic enriched with (N) and both $\lnot \lnot p \vdash p$ and $p \vdash \lnot \lnot p$. We will show that $\bot _2$ is not symmetric. Let $\varGamma$ be the $\varLambda$-theory generated by $\{ p \}$ and $\varDelta$ the $\varLambda$-theory generated by an infinite set of atomic formulas $X=\{ q_1, q_2 \ldots \}$ assuming that $p \notin X$. Then $\varGamma ^*$ is the $\varLambda$-theory generated by $\{ \lnot p \}$ and $\varDelta ^*=\emptyset$ (for there is no $\beta$ such that $\beta \vdash _{\varLambda } \delta$  for each $\delta \in \varDelta$). Hence, $\varDelta \bot _2 \varGamma$  but not $\varGamma \bot _2 \varDelta$.

Finally, we will show that $\{ p \vdash \lnot \lnot p, \lnot \lnot p \vdash p \}$ is not canonical$_3$. Let $\varLambda$ be again the lattice logic enriched with (N) and with $\lnot \lnot p \vdash p$ and $p \vdash \lnot \lnot p$. Let $X=\{q_1, q_2, q_3, \ldots \}$ be an infinite set of atomic formulas such that $p \notin X$. Let F be the filter in $I^{\varLambda }_3$ generated by $\{[q_1], [q_2], [q_3], \ldots \}$. Recall that $I^{\varLambda }_3$ is given by the Lindenbaum-Tarski algebra turned upside down and enriched with a new top element $i_3$. It holds that $[\alpha ] \in F$ iff there are $q_{i_1}, \ldots , q_{i_n}$ such that $\alpha \vdash _{\varLambda } q_{i_1} \vee \ldots \vee q_{i_n}$. It follows that $[p] \notin F$. In order to show that $\{ p \vdash \lnot \lnot p, \lnot \lnot p \vdash p \}$ is not canonical$_3$ it is sufficient to observe that for any $t \in S_3$, if $F \subseteq \bot ^{t}_3$ then $[p] \bot _3 t$. Note that there is no $\beta$ such that $\beta \vdash _{\varLambda } \lnot \alpha$ for all $\alpha \in F$. This follows for example from the fact that $\varLambda$ has the variable sharing property and so for any $\beta$ we can take any $q_i$ not occurring in $\beta$ and then $\beta \nvdash _{\varLambda } \lnot q_i$. It follows that $F \subseteq \bot ^{t}_3$ only if $t=i_3$. But it holds $[p] \bot i_3$ which finishes the proof.    $\square$

Proof

(Proposition 19). First, we show that $\nu (x)=\{ y \in X~|~x \le _{\nu } y \}$. By the definition of $\varGamma _{\nu }$ as $\lambda \rho$ it generally holds that if $y \in \varGamma _{\nu } (\{x \})$ then $\varGamma _{\nu }(\{y\}) \subseteq \varGamma _{\nu } (\{x \})$. By (a) and (d) of Definition 11 we obtain: if $y \in \nu (x)$ then $x \le _{\nu } y$, i.e. $\nu (x) \subseteq \{ y \in X~|~x \le _{\nu } y \}$. On the other side, if $x \le _{\nu } y$ then $\nu (y) \subseteq \nu (x)$, and since $y \in \nu (y)$ (because in general $y \in \lambda \rho (\{y\})$), it follows that $y \in \nu (x)$, and thus $\{ y \in X~|~x \le _{\nu } y \} \subseteq \nu (x)$. Since any $\nu$-stable set is of the form $\nu (x)$, for some $x \in X$, $\nu$-stable sets are exactly the principle upsets w.r.t. $\le _{\nu }$. Since $\le _{\nu }$ is a partial order, $\nu$ is a bijection between X and the lattice of $\nu$-stable sets.

By (c) of Definition 11 we obtain $\{ \iota \} \subseteq \nu (x)$, for every $x \in X$, which means (by (d)) that $\iota$ the top element w.r.t. $\le _{\nu }$. It remains to be shown that $\nu ^{-1}(\varGamma _{\nu } (\nu (x) \cup \nu (y)))$ is the greatest lower bound of x and y w.r.t. $\le _{\nu }$. It holds that $\nu (x) \subseteq \varGamma _{\nu } (\nu (x) \cup \nu (y))$ and thus $\nu ^{-1}(\varGamma _{\nu } (\nu (x) \cup \nu (y))) \le _{\nu } x$. By the same reasoning $\nu ^{-1}(\varGamma _{\nu } (\nu (x) \cup \nu (y))) \le _{\nu } y$, so $\nu ^{-1}(\varGamma _{\nu } (\nu (x) \cup \nu (y)))$ is a lower bound of $\{x , y \}$. Assume that $z \le _{\nu } x$ and $z \le _{\nu } y$. Then $\nu (x) \subseteq \nu (z)$ and $\nu (y) \subseteq \nu (z)$. So, $\nu (x) \cup \nu (y) \subseteq \nu (z)$ and since $\nu (z)$ is $\nu$-stable we obtain $\varGamma _{\nu } (\nu (x) \cup \nu (y)) \subseteq \nu (z)$. It follows that $z \le _{\nu } \nu ^{-1}(\varGamma _{\nu } (\nu (x) \cup \nu (y)))$ which is what we needed to show.    $\square$

Proof

(Proposition 21). Let $N=\langle X, Y, \nu , \iota , \bot \rangle$ be a NHNF. Proposition 19 shows that $\langle X, \circ ^{\nu }, \iota \rangle$ is an IF. Moreover, the conditions (a) and (b) from Definition 12 guarantee that $I=\langle X, \circ ^{\nu }, \iota , \bot \rangle$ is an IIF. Since $\nu (x)$ is always a (principal) filter, the valuations in N are always the valuations in I. Let V be a valuation in N (in the sense of Definition 11). We will show by induction that for any $\alpha \in L^{\lnot }$ it holds that $x \Vdash \alpha$ in $\langle N, V \rangle$ iff $x \vDash \alpha$ in $\langle I, V \rangle$. The case of atomic formulas and the inductive stapes for conjunction and negation are immediate. Let us show the inductive step for disjunction. Assume that the claim holds for some $L^{\lnot }$-formulas $\alpha , \beta$. It follows from Propositions 19 and 20 that there are states $z_{\alpha }, z_{\beta }$ such that $\nu (z_{\alpha }) = \{ x \in X~|~x \Vdash \alpha \}$ and $\nu (z_{\beta }) = \{ x \in X~|~x \Vdash \beta \}$. Now we can observe that the following claims are equivalent:

• $x \vDash \alpha \vee \beta$,

• there are $u, v \in X$ such that $u \vDash \alpha$, $v \vDash \beta$ and $u \circ ^{\nu } v \le _{\nu } x$,

• there are $u, v \in X$ such that $u \Vdash \alpha$, $v \Vdash \beta$ and $\nu ^{-1}(\varGamma _{\nu } (\nu (u) \cup \nu (v))) \le _{\nu } x$,

• there are $u, v \in X$ such that $u \Vdash \alpha$, $v \Vdash \beta$ and $\nu (x) \subseteq \varGamma _{\nu } (\nu (u) \cup \nu (v))$,

• $\nu (x) \subseteq \varGamma _{\nu } (\nu (z_{\alpha }) \cup \nu (z_{\beta }))$,

• $x \in \varGamma _{\nu } (\nu (z_{\alpha }) \cup \nu (z_{\beta }))$,

• $x \Vdash \alpha \vee \beta$.

It follows that $\nu (Y)$ is closed under the required operation and thus $N'=\langle I, \nu (Y) \rangle$ is a GIIF such that the valuations in N coincide with the valuations in $N'$, which finishes the proof.    $\square$