Orthogonal Frames and Indexed Relations

Abstract

We define and study the notion of an indexed frame. This is a bi-dimensional structure consisting of a Cartesian product equipped with relations which only relate pairs if they coincide in one of their components. We show that these structures are quite ubiquitous in modal logic, showing up in the literature as products of Kripke frames, subset spaces, or temporal frames for STIT logics. We show that indexed frames are completely characterised by their ‘orthogonal’ relations, and we provide their sound and complete logic. Using these ‘orthogonality’ results, we provide necessary and sufficient conditions for an arbitrary Kripke frame to be isomorphic to certain well-known bi-dimensional structures.

Appendix

Proof of Proposition 2. Given an orthogonal $[L_1,L_2]$-frame $(W,R_1,R_2)$, we extend W to the set

$$W' = W \cup \{ x_{wv}: w,v\in W, R^*_2(w)\cap R^*_1(v) = \varnothing \},$$

i.e., we add one element for each pair of connected components which have an empty intersection, and we extend the relations $R_i$ as follows:

• if $F_\bullet \models L_i$, then $R'_i = R_i$;

• if $F_\circ \models L_i$, then $R'_i = R_i \cup \{(x,x):x\in W'\setminus W\}$;

where $F_\bullet$ is the irreflexive singleton frame $(\{*\}, \varnothing )$, and $F_\circ$ is the reflexive singleton frame $(\{*\}, \{(*,*)\})$. (Recall that every logic is satisfied in either $F_\bullet$ or $F_\circ$; this is a consequence of a classical result by Makinson [11].)

Note that, in either case, no elements of W are related to any elements of $W'\setminus W$ and thus $(W,R_1,R_2)$ is a generated subframe of $(W',R'_1, R'_2)$.

We define

Note that $\equiv '_1$ and $\equiv '_2$ satisfy conditions O1 – O3 of Definition 2, and therefore $(W',R'_1,R'_2)$ is a full orthogonal frame. Finally, for $i=1,2$, $(W',R'_i)$ is the disjoint union of the $L_i$-frame $(W,R_i)$ with a family of singleton $L_i$-frames, and thus it is an $L_i$-frame.

Proof of Lemma 2. We leave it to the reader to check that (O3’) implies that $\equiv _1 \circ \equiv _2$ is an equivalence relation. Let $W'$ be an equivalence class of $\equiv _1\circ \equiv _2$. Let $R'_i$ and $\equiv '_i$ be the restrictions of $R_i$ and $\equiv _i$ to $W'$. It is routine to check that (O1) $R'_i\subseteq \equiv '_i$, (O2) $\equiv '_1\cap \equiv '_2 = Id_{W'}$, and (O3) $\equiv '_1 \circ \equiv '_2 = (W')^2$. Each of these is therefore a full orthogonal frame and $(W,R_1,R_2)$ is equal to the disjoint union $\bigcup _{W'\in W/_{\equiv _1\circ \equiv _2}} (W',R'_1,R'_2)$.

Proof (sketch) of Proposition 4. This uses the very standard technique of canonical models; we point the reader to [4, Chapter 4] for full details and we simply offer a sketch here:

Let X be the set of maximal consistent sets of formulas in the language. We define the relations $xR_i y$ iff, for all $\phi$, $\square _i\phi \in x$ implies $\phi \in y$ and $x\equiv _i y$ iff, for all $\phi$, $\blacksquare _i\phi \in x$ implies $\phi \in y$.

The Truth Lemma shows that, given the valuation $V(p) = \{x\in X:p\in x \}$, it is the case that $x\models \phi$ iff $\phi \in x$.

We note that the logic $Log_\dashv ^{L_1 L_2}$ is canonical, for canonicity is preserved by fusions [9, Cor. 6] and the addition of Sahlqvist axioms [4, Chapter 4]. This canonicity ensures that $(X,R_i)\models L_i$; the S5 axioms for the $\blacksquare _i$’s ensure that $\equiv _i$ is an equivalence relation; $\blacksquare _1\blacksquare _2\phi \leftrightarrow \blacksquare _2\blacksquare _1\phi$ ensures that (O3’) is satisfied; finally, the axioms $\blacksquare _i\phi \rightarrow \square _i\phi$ ensure (O1).

Therefore $(X,R_{1,2},\equiv _{1,2})$ is a semistructure and any consistent formula $\phi$ can be satisfied in it.

Proof of Lemma 3. We simply show the existence of a matrix enumeration $f:X\times X\rightarrow X$ whenever X is infinite; we leave further details to the reader. Let $\{X_1, X_2\}$ be a partition of X into two sets which are equipotent to X itself (note that the existence of such partition requires the Axiom of Choice for uncountable cardinalities [16]). Let $f_1:X_1\rightarrow X$ and $f_2:X_2\rightarrow X$ be two surjections. The map

$$f(x,y) = {\left\{ \begin{array}{ll} f_i(x) &{} \text { if } x,y\in X_i \\ f_j(y) &{} \text { if } x\in X_i, y\in X_j, i\ne j \end{array}\right. }$$

is the desired enumeration.

Proof of Proposition 6. Soundness is routine. For completeness, given a formula $\phi \notin Log_\dashv$, it suffices to use Theorem 2 to find a standard orthogonal structure $(W,R_{1,2},\equiv _{1,2})$ that refutes $\phi$, construct the indexed frame $(W/_{\equiv _2}\times W/_{\equiv _1}, \mathsf {R}_1, \mathsf {R}_2)$ isomorphic to $(W,R_1,R_2)$ via Proposition 1 and note that the equivalence relation $([w]_2,[v]_1) \cong _i ([w']_2,[v']_1)$ iff $x_{wv}\equiv _i x_{w'v'}$ relates two pairs if and only if their j-th coordinate coincides, for $j\ne i$.

Proof (sketch) of Proposition 7. This involves a rather standard filtration argument. (See [4, Chapter 2] for details on this technique).

Given a consistent formula $\phi$, we let $(W,R_{1,2},\equiv _{1,2})$ be a semistructure satisfying $\phi$ at a point $w_0$, and $\varGamma$ be a finite set of formulas closed under subformulas such that $\phi \in \varGamma$, and we define an equivalence relation $w\sim _\varGamma v$ iff for all $\psi \in \varGamma$, ($w\models \psi$ iff $v\models \psi$). We define relations in the quotient set $W/_{\sim _\varGamma }$ as follows: for $i=1,2$,

• $[w]_\varGamma \equiv '_i [v]_\varGamma$ iff, for all $\blacksquare _i\psi \in \varGamma$, ($w\models \blacksquare _i\psi$ iff $v\models \blacksquare _i\psi$), and

• $[w]_\varGamma R'_i [v]_\varGamma$ iff $[w]_\varGamma \equiv '_i [v]_\varGamma$ and for all $\square _i\psi \in \varGamma$ ($w\models \square _i\psi$ implies $v\models \psi$).

We leave it to the reader to check that the resulting tuple is a semistructure and a filtration and therefore that $[w_0]_\varGamma \models \phi$. We can then use Proposition 5 and Lemma 2 to obtain an indexed frame satisfying $\phi$.

Proof of Proposition 9. For the left-to-right direction, given a subset space frame we consider the relations $(x,U) R_K (y,V)$ iff $U=V$, $(x,U) R_\square (y,V)$ iff $x=y$ and $U\supseteq V$, and $(x,U)\equiv _\square (y,V)$ iff $x=y$. We note that $(R_K\circ \equiv _\square )(x,U) = \{ (x',U')\in \mathcal {O}_X: x'\in U \}$, and we leave it to the reader to check that this satisfies all the properties in Proposition 9.

Let us now consider a frame with these properties. We let $[.]_\square$ and $[.]_K$ denote the equivalence classes of $\equiv _\square$ and $R_K$. Let us define the subset space

Note that $[w]_\square \in U_v$ if and only if $[w]_\square \cap [v]_K \ne \varnothing$.

By (O2), an intersection $[w]_\square \cap [v]_K$ of two equivalence classes is at most a singleton. Let us map and element $([w]_\square , U_v)$ in the graph of $(X,\mathcal {O})$ to the unique element in $[w]_\square \cap [v]_K$. This is a bijection whose inverse maps each $w\in W$ to $([w]_\square , U_w)$. We define relations $\equiv _K$ and $\ge _\square$ on this graph as in Example 2 and, to show that this map is an isomorphism, it suffices to show that

We start with the second item. From left to right, if $w R_\square v$, then $[w]_\square = [v]_\square$ by (O1), and let us see that $U_w \supseteq U_v$. If $[y]_\square \in U_v$, then there is a unique element $x\in [y]_\square \cap [v]_K$. But since $wR_\square v R_K x$, it follows by (SS3) that there must exist some $x'$ such that $wR_K x'R_\square x$. Since $x'\equiv _\square x$, by (O1), and $x\equiv _\square y$, it follows that $x'\in [w]_K \cap [y]_\square$, and thus $[y]_\square \in U_w$. From right to left, it suffices to see that $U_w\supseteq U_v$ and $w\equiv _\square v$ implies $wR_\square v$. But this follows directly from (SS4), noting that $U_w \supseteq U_v$ implies $[R_k \circ \equiv _\square ](w) \supseteq [R_k \circ \equiv _\square ](v)$.

For the first item it suffices to show that $wR_K v$ iff $U_w = U_v$. The left-to-right direction is immediate from the definition of $U_w$, whereas the right-to-left direction follows from (SS5).