Abstract
Many authors have noted that there are types of English modal sentences cannot be formalized in the language of basic firstorder modal logic. Some widely discussed examples include “There could have been things other than there actually are” and “Everyone who is actually rich could have been poor.” In response to this lack of expressive power, many authors have discussed extensions of firstorder modal logic with twodimensional operators. But claims about the relative expressive power of these extensions are often justified only by example rather than by rigorous proof. In this paper, we provide proofs of many of these claims and present a more complete picture of the expressive landscape for such languages.
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Notes
Originally from Hazen (1976, p. 31).
Originally from Cresswell (1990, p. 34).
Hodes (1984c).
Wehmeier (2001).
See Garson (2001) for a tree of such formulations.
Davies and Humberstone (1980).
See Blackburn et al. (2001).
There is a general question as to whether the object quantifier in (R) and (NR) is possibilist or actualist. In general, I assume it should be actualist, in which case the object quantifiers in (4) and (5) should technically be \({\textsf {E}} \)bounded. But to avoid clutter, we assume in the background that nothing can be rich or poor unless it exists (), in which case it does not matter which type of quantifier we use in (4) and (5). None of our models will violate this constraint.
See Blackburn et al. (2001, Chap. 2) for an introduction to bisimulations.
See Blackburn et al. (2001, p. 68) for the proof in the propositional case.
See Goranko and Otto (2006) for a proof in the propositional case. Generalizing to firstorder modal logic is straightforward.
A proof of this was suggested by Hazen (1976, p. 35). He describes his models as follows:
For suppose that [(3)] is false, that the actual world is the only one with infinitely many individuals, and that for every finite set of individuals in the actual world there is a world containing just those individuals, and consider the purely logical sentences true under those suppositions. Now suppose there is added to the system of possible worlds a new world for each old world, containing all the same individuals plus one new individual (the same for each new world) not in any old world. [(3)] will have become true, but no purely logical sentence of the modal language will have changed its truth value.
However, the proof is not correct as stated, since the second model, but not the first, satisfies the formula (there is no world where this new object exists by itself). The natural fix is to add another world to the second model which only contains the new object. The resulting models (which are like ours except the worlds with cofinite domains are removed) satisfy the same formulas, but they are not bisimilar, since they are distinguished by the formula . However, if you restrict the bisimulation game to nsteps, for any finite n, then Eloïse has a winning strategy (that is, the models are nbisimilar for every n); and this suffices to guarantee modal equivalence. Here, we ensure full bisimilarity by including worlds with cofinite domains. Thus, our proof shows that (3) is not even expressible in .
Yanovich (2015, p. 87) claims to have shown that \(\Diamond \Pi {x}\left( @Q(x) \rightarrow Q(x)\right) \) is not expressible in . He proceeds, as we do, by constructing two models that disagree on this sentence, and then argues that they are bisimilar. The first model consists of two worlds w and u, where and where in both w and u, satisfy Q(x) and do not satisfy Q(x). The second model consists of worlds \(w'\), \(u_1'\), and \(u_2'\), where . In \(w'\), satisfy Q(x) while do not. In \(u_1'\), only satisfy Q, and in \(u_2'\), only satisfy Q. He then claims that w and \(w'\) are bisimilar. However, these models are not bisimilar. In fact, they do not even satisfy the same formulas: e.g., \(\Sigma {x}\left( Q(x) \wedge \Diamond \lnot Q(x)\right) \) distinguishes the two models. The claim that such sentences cannot be expressed in is still correct, as can be verified with a bisimulation argument using the models \(\mathcal {N} _1,w,w\) and \(\mathcal {N} _2,w,w\) defined in Fig. 4 below (where \(w = w^\emptyset _\emptyset \)). Another proof that cannot express (R) can be found in Wehmeier (2001).
I claimed in Kocurek (2015, p. 215) that the proof of Proposition 16 extends to immediately. But this needs qualification. We can obtain a quick proof that (4) is not expressible in by restricting the accessibility relations in \(\mathcal {R} _1\) and \(\mathcal {R} _2\); but as Appendix 3 reveals, the proof of UDinexpressibility is more challenging.
I claimed to prove this in Kocurek (2015, pp. 215–216) using models like the ones presented here, except with \(T = \emptyset \) for all worlds. However, my proof was incorrect, since those models are distinguishable by the formula .
The exact relation between , , and relative to \(\mathbf U \) is an open question. In particular, it is unknown whether or even . Fig. 11 in Appendix 4 summarizes the remaining inclusions relative to \(\mathbf U \).
Though our partial order was linear and discrete, this was not crucial to the construction. We could have, for instance, mapped each state in \(\mathcal {M} \) to a rational in the interval [0, 1), and have obtained a dense linear order. Alternatively, via tree unraveling, we could have obtained a branching structure.
Several conjectures have been made about how to construct such formulas. For instance, Needham (1975, pp. 73–74) gives a sentence he claims is not expressible in ; but van Benthem (1977, p. 417) shows it is. van Benthem then gives a genuine example of a temporal formula that is not expressible in . However, it should be noted that even though \(\mathcal {F}\) operators were not the focus of van Benthem (1977), the sentence he gives is expressible in , which is still twodimensional. Forbes (1989, p. 89) also gives a schema that was supposed to show that \((n1)\)dimensional logic is not as expressive as ndimensional logic. But as footnote 29 explains, the example is not correct either. Cresswell (1990, p. 30) suggests one can generate such sentences from (H) since, reading the conditional in (15) as a disjunction, “disjunctions can be extended with no upper limit.”
Forbes (1989, p. 87) gives a purported example of an ndimensional formula not expressible as an \((n+1)\)dimensional formulas. (Following Forbes, we restrict attention to models whose accessibility relations are universal.) Given a model \(\mathcal {M} \), let us say a sequence is an nchain if \(\delta ^\mathcal {M} (w_i) \subset \delta ^\mathcal {M} (w_{i+1})\) for \(1 \le i < n\). Forbes (1989, p. 89) says “it is a very probable conjecture that ‘there is an nchain of worlds’ cannot be expressed in [], so that the hierarchy of modal languages is strictly increasing in expressive power...” Forbes also notes that can express “there is an nchain of worlds”, but claims that it cannot express “there is an \(n+1\)chain of worlds”. However, the conjecture is false. Define . Then the claim that there is a 4chain can be expressed as an formula: \(\Diamond _1\Diamond _2\left( \theta _{1,2} \wedge \Diamond _1\left( \theta _{2,1} \wedge \Diamond _2\theta _{1,2}\right) \right) \). Likewise for \(n > 4\). Also, let and let . Then the claim that there is a 4chain can be expressed as an formula: \(\Diamond _1\Diamond \left( \chi \wedge \Diamond _1\left( \eta \wedge \Diamond \chi \right) \right) \). Again, this generalizes to nchains where \(n > 4\).
A theorem of Yanovich (2015, pp. 85–86) shows that one will not generally be able to use the bisimulation
technique proposed in this paper to establish inexpressibility over the class of models with finite domains. I suspect one could still establish such results, however, by constructing sequences of pairs of finite models that were nbisimilar for each \(n \in \mathbb {N} \). But working out the details must be left for future work.
See Kocurek (2016) for one such syntactic characterization.
See Chang and Keisler (1990, Chap. 4).
See Bell and Slomson (2006, pp. 222–224).
Hodes (1984b, pp. 445–446) claimed to have a proof that . He also constructs two models which he claims satisfy the same formulas, but disagree on the formula . Here are the models \(\mathcal {A} \) and \(\mathcal {B} \) he describes (p. 445, his notation; A(w) and B(w) in Hodes’s notation means \(\delta ^\mathcal {A} (w)\) and \(\delta ^\mathcal {B} (w)\) in ours, and \(\left\langle w,a\right\rangle \in V(P)\) in his notation means \(a \in I(P,w)\) in ours; he also write \(\mathcal {A} ,0 \vDash \varphi \) in place of our \(\mathcal {A} ,0,0 \Vdash \varphi \)):
Let , . Let \(A(0) = B(0) = \omega \), , let V(P) be empty for all \(P \in {\textsf {Pred}} \), \(\mathcal {A} = \left( W,A,V\right) \), \(\mathcal {B} = \left( W',B,V\right) \). Clearly \(\mathcal {A} ,0 \vDash \theta _2\) but \(\mathcal {B} ,0 \nvDash \theta _2\).
However, Hodes’s description of these models is incomplete, since crucially the local domain of 1 in \(\mathcal {B} \) is never specified, and the proof that follows gives no indication of what it might be. Moreover, given the proof requires that \(\mathcal {A},0,0 \Vdash \theta _2\) and \(\mathcal {B},0,0 \nVdash \theta _2\), we can infer that it would have to be that has exactly one member (since is the set of objects that do not exist at 0). But if that is right, then these models are distinguishable by the following formula:
$$\begin{aligned} \Sigma {x}\Sigma {y}\left( x \not \approx y \wedge \lnot {\textsf {E}} (x) \wedge \lnot {\textsf {E}} (y) \wedge \Diamond ({\textsf {E}} (x) \wedge \lnot {\textsf {E}} (y))\right) . \end{aligned}$$The proof of Lemma 35 was inspired by an attempt to fix Hodes’s proof.
Here is the idea. First, every \(@\) in \(\varphi \) is in the immediate scope of a \(\mathcal {F}\), then \(\Vdash _\mathbf U \Box \varphi \leftrightarrow \mathcal {F}@\varphi \). So every \(\Box \) in a formula can be replaced by \(\mathcal {F}@\). Second, note that . So having replaced every \(\Box \) with \(\mathcal {F}@\), every \(\Pi {}\) is in the scope of a \(\mathcal {F}@\), so we can replace every \(\Pi {x}\) with . The result is a \(\mathbf UD \)equivalent formula.
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Acknowledgments
Many thanks to Melissa Fusco, Wes Holliday, and two anonymous reviewers for all their feedback on this paper. Thanks also to those who participated in UC Berkeley’s dissertation seminar in the Spring of 2016 for all their valuable comments and suggestions on an earlier draft of this paper.
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Appendices
Appendix 1: van Benthem’s characterization theorem
In this appendix, we prove Theorem 13. We essentially follow the proof of the corresponding theorem for propositional modal logic in Blackburn et al. (2001, Chap. 2.6). The crucial change is with the definition of a set of formulas being satisfiable.
Definition 19
(Satisfiability) Let be some new variables not in \({\textsf {VAR}} \), and let be the result of extending \(\mathcal {L}\) with . Let be a set of formulas whose only variables not among are \(\overline{x} \). Let \(\mathcal {M} \) be a model, and \(X \subseteq W^2\), and let \(\overline{b} \in D\).

\(\varGamma \) is satisfiable inXover\(\overline{b} \)(with respect to\(\mathcal {M} \)) if there is a \(\left\langle w,v\right\rangle \in X\) and some \(\overline{a} \in D/\delta (v)/\delta (w)\) such that \(\mathcal {M} ,w,v \Vdash \varGamma [\overline{a},\overline{b} ]\).

\(\varGamma \) is finitelysatisfiable inXover\(\overline{b} \)(with respect to\(\mathcal {M} \)) if every finite subset of \(\varGamma \) is satisfiable in X over \(\overline{b} \) (with respect to \(\mathcal {M} \)).
Note that if the only free variables in \(\varGamma \) are among , then \(\varGamma \) is (finitely) \(\Sigma {}\)satisfiable in X over \(\overline{b} \) iff it is (finitely) satisfiable in X over \(\overline{b} \) iff it is (finitely) satisfiable in X over \(\overline{b} \). We will just use the term “(finitely) satisfiable” when the distinction does not matter.
Definition 20
(Modal saturation) Assume \(\mathcal {F}\), , and \(\Pi {}\) are not among . A model \(\mathcal {M} \) is saturated if for all \(w,v \in W\), all \(\overline{b} \in D\), and all sets of formulas:

(a)
if \(\varGamma \) is finitely satisfiable in over \(\overline{b} \) (with respect to \(\mathcal {M} \)), then it is satisfiable in over \(\overline{b} \);

(b)
if \(\varGamma \) is finitely satisfiable in over \(\overline{b} \) (with respect to \(\mathcal {M} \)), then it is satisfiable in over \(\overline{b} \).
If \(\mathcal {F}\) is among , we just add the following clause:

(c)
if \(\varGamma \) is finitely satisfiable in over \(\overline{b} \) (with respect to \(\mathcal {M} \)), then it is satisfiable in over \(\overline{b} \).
If is among , add the clauses above but with replaced by .
Lemma 21
(Modal saturation implies the Hennessy–Milner property) Suppose \(\mathcal {M} \) and \(\mathcal {N} \) are saturated. Then is an bisimulation between \(\mathcal {M} \) and \(\mathcal {N} \). Hence, if , then it follows that .
Proof
Suppose . Clearly (Atomic) is satisfied (and likewise for (Ex) and (Eq) if \({\textsf {E}} \) or \(\approx \) are among ).

ZigZag.
Let \(u \in R^\mathcal {M} [v]\). Define . Let \(\varDelta \subseteq \varGamma \) be finite and nonempty. Then since \(u \in R^\mathcal {M} [v]\), . Since (and since for each \(\psi \in \varDelta \), we could replace with fresh new variables in \({\textsf {VAR}} \)), it follows that . Hence, \(\varDelta \) is satisfiable in over \(\overline{b} \). By saturation, there is a \(u' \in R^\mathcal {N} [v']\) such that \(\mathcal {M} ,w,u' \Vdash \varGamma [\overline{b} ]\). Thus, . Likewise for the Zag clause. \(\checkmark \)

BackForth.
Let \(a' \in \delta ^\mathcal {M} (v)\). Define . Let \(\varDelta \subseteq \varGamma \) be finite and nonempty. Then . Since , it follows that . Hence, \(\varDelta \) is satisfiable in over \(\overline{b} \) with respect to \(\mathcal {N} \). By saturation, \(\varGamma \) itself is satisfiable in over \(\overline{b} \) with respect to \(\mathcal {N} \). So there is a \(b' \in \delta ^\mathcal {M} (v')\) such that \(\mathcal {N} ,w',v' \Vdash \varGamma [b',\overline{b} ]\). Thus, . Likewise for the Forth clause. \(\checkmark \)
The \(\mathcal {F}\)ZigZagclauses are just like the ZigZag clause above, and the other quantifier BackForth clauses are just like the BackForth clauses above. The Act and Diag clauses are taken care of automatically by the fact that (assuming \(@\)/\({\mathop {\downarrow }}\) is among ). \(\square \)
Definition 22
(Realization) Let be extended with and \(\overline{t} \notin {\textsf {SVAR}} \). Let \(\overline{b} \in D\) and \(\overline{v} \in W\) where and \(\left \overline{v} \right = \left \overline{t} \right \). Let \(\mathcal {M} \) be a model, and let be a set of formulas whose only free variables are among .

\(\varGamma \) is realized over\(\overline{b} \)and\(\overline{v} \)(with respect to\(\mathcal {M} \)) if there are some \(\overline{a} \in D\) and \(\overline{u} \in W\) such that \(\mathcal {M} \vDash \varGamma [\overline{a},\overline{b};\overline{u},\overline{v} ]\). We call \(\overline{b} \) and \(\overline{v} \)parameters.

\(\varGamma \) is finitely realized over\(\overline{b} \)and\(\overline{v} \)(with respect to\(\mathcal {M} \)) if every finite subset of \(\varGamma \) is realized over \(\overline{b} \) and \(\overline{v} \) (with respect to \(\mathcal {M} \)).
Definition 23
(Saturation) We will say \(\mathcal {M} \) is countably saturated if for every finite \(\overline{b} \in D\), every finite \(\overline{v} \in W\), and every set \(\varGamma \) of formulas (where and \(\left \overline{v} \right = \left \overline{t} \right \)) that is finitely realized over \(\overline{b} \) and \(\overline{v} \), \(\varGamma \) is also realized over \(\overline{b} \) and \(\overline{v} \).
Lemma 24
(Countable saturation implies modal saturation) If \(\mathcal {M} \) is countably saturated, then it is \(\mathcal {L}\)saturated.
Proof
Let \(\mathcal {M} \) be a countably saturated model. Suppose a set of formulas is finitely satisfiable in over \(\overline{b} \). Consider the set:
Let \(\varDelta \subseteq \varGamma \) be finite and nonempty. Since \(\varDelta \) is satisfiable in , there is a \(u \in R[v]\) and some \(\overline{a} \in \delta (u)\) such that \(\mathcal {M} ,w,u \Vdash \varDelta [\overline{a},\overline{b} ]\). Let:
Then \(\mathcal {M} \vDash \varDelta ^*[\overline{a},\overline{b};w,v,u]\). But \(\varDelta ^* \subseteq \varGamma ^*\) is finite. So every finite subset of \(\varGamma ^*\) is realized over \(\overline{b} \), w, v, and u. By countable saturation, there are some \(\overline{a} \in D\) and \(u \in W\) such that \(\mathcal {M} \vDash \varGamma ^*[\overline{a},\overline{b};w,v,u]\). But then \(u \in R[v]\), \(\overline{a} \in \delta (u)\), and \(\mathcal {M} ,w,u \Vdash \varGamma [\overline{a},\overline{b} ]\). So \(\varGamma \) is satisfiable in . Likewise for satisfiability in , except you define:
Similarly for the other kinds of satisfiability. \(\square \)
It follows from Lemmas 21 and 24 that:
Corollary 25
(Countable saturation implies Hennessy–Miller property) If \(\mathcal {M} \) and \(\mathcal {N} \) are countably saturated, and it , then it follows that .
Definition 26
(Ultraproducts) Let \(N \ne \emptyset \). An ultrafilter overN is a set \(U \subseteq \wp \left( N\right) \) where U is closed under supersets and finite intersections, \(\emptyset \notin U\), and for all \(S \in \wp \left( N\right) \), either \(S \in U\) or . Let U be an ultrafilter over N. For each \(i \in N\), let \(W_i \ne \emptyset \). Then \(\prod _{i \in N} W_i\) is the set of functions \(f :N \rightarrow \bigcup _{i \in N} W_{i}\) where \(f(i) \in W_i\). We will say \(f \sim _U f'\) if . Define . Finally, we will define the ultraproduct of\(W_i\)moduloU as the set . We will drop the subscript U when the ultrafilter in question is obvious from the context. An ultrapower is an ultraproduct where \(W_i = W\) for all \(i \in N\), which we may write as \(\prod _U W\).
Definition 27
(Ultrapowers of models) Let \(\mathcal {M} = \left\langle W,R,F,D,\delta ,I\right\rangle \) be a model. The ultrapower model of\(\mathcal {M} \)moduloU is the model \(\prod _U \mathcal {M} \) defined as follows:

\(W_U {:}{=}\prod _U W\)

\(R_U(\left[ f_1\right] ,\left[ f_2\right] )\) iff

\(F_U(\left[ f_1\right] ,\left[ f_2\right] )\) iff

\(D_U {:}{=}\prod _U D\)

\(\left[ o\right] \in \delta _U\left( \left[ f\right] \right) \) iff

\(\left\langle \left[ o_1\right] ,\dots ,\left[ o_n\right] \right\rangle \in I_U(P,\left[ f\right] )\) iff .
It is a routine exercise to show that these definitions are welldefined, i.e., they do not depend on the representative of the equivalence class used in their statement.
Theorem 28
(Łoś ’s theorem) The following are equivalent:

(a)
\(\prod _U \mathcal {M} \vDash \alpha [\left[ o_1\right] ,\dots ,\left[ o_n\right] ]\).

(b)
.
This can be proven by induction.^{Footnote 34} Now, define \({f_w}:{i}\mapsto {w}\) and \({o_a}:{i}\mapsto {a}\). If g is a variable assignment over \(\mathcal {M} \), define \({g_U}:{x}\mapsto {o_{g(x)}}\). Let the diagonal map be the map d from \(\mathcal {M} \) to \(\prod _U \mathcal {M} \) such that \(d(w) = f_w\) and \(d(a) = o_a\). Then it is a straightforward corollary of Theorem 28 that the diagonal map is an elementary embedding of \(\mathcal {M} \) into \(\prod _U \mathcal {M} \).
Say that U is countably incomplete if there is a countable subset of U whose intersection is not in U. A standard result from model theory shows that if U is countably incomplete, then \(\prod _U \mathcal {M} \) is countably saturated.^{Footnote 35} The important point, however, is that we can always find a countably saturated elementary extension of \(\mathcal {M} \) (viz., \(\prod _U \mathcal {M} \) where U is a countably incomplete ultrafilter). This is all we will need below.
Proof
(Theorem 13) Let . It suffices to show (by the compactness of ) that \(\varGamma \vDash \alpha \). Suppose \(\mathcal {M} ,g \vDash \varGamma \). Define:
It is easy to show that \(\varDelta \) is satisfiable (again by compactness). Let \(\mathcal {N} ,h \vDash \varDelta \). Then by the way we defined \(\varDelta \). Now, by the results above, there exist elementary extensions \({e}:{\mathcal {M} }\preccurlyeq {\mathcal {M} '}\) and \({f}:{\mathcal {N} }\preccurlyeq {\mathcal {M} '}\) that are countably saturated. Since these are elementary embeddings:
By Corollary 25, since these are countably saturated:
Hence, by invariance under bisimulation, \(\mathcal {M} ',g' \vDash \alpha \) (where \(g'(x) = e(g(x))\). Since \({e}:{\mathcal {M} }\preccurlyeq {\mathcal {M} '}\), it follows that \(\mathcal {M} ,g \vDash \alpha \). \(\square \)
Appendix 2: Bisimulation proofs
In this appendix, we give more formal details regarding the bisimulation proofs from Sect. 4. We start with the proof of Proposition 15. Before reading, recall the definition of a partial isomorphism from Definition 14, and the definition of the models \(\mathcal {E} _1\) and \(\mathcal {E} _2\) (see Fig. 2 for a picture). Note that \(\mathcal {M} ,w,v,\overline{a} \simeq \mathcal {N} ,w',v',\overline{b} \) iff the map \(a_i \mapsto b_i\) is a partial isomorphism between them. In particular, for our models \(\mathcal {E} _1\) and \(\mathcal {E} _2\), if \(a_i \in \delta _1(v_1)\) iff \(b_i \in \delta _2(v_2)\) and \(a_i = a_j\) iff \(b_i = b_j\), then \(\mathcal {E} _1,w_\mathbb {N} ,v_1,\overline{a} \simeq \mathcal {E} _2,w_\mathbb {N} ,v_2,\overline{b} \), since this means \(a_i \mapsto b_i\) is a partial isomorphism. We will make use of this implicitly throughout.
We will define our bisimulation in stages. First, set . Next, suppose we have constructed \(Z_i\) so that for all \(\left\langle w_\mathbb {N} ,v_1,\overline{a};w_\mathbb {N} ,v_2,\overline{b} \right\rangle \in Z_i\), \(\left \delta _1(v_1) \right = \left \delta _2(v_2) \right \) and \(\mathcal {E} _1,w_\mathbb {N} ,v_1,\overline{a} \simeq \mathcal {E} _2,w_\mathbb {N} ,v_2,\overline{b} \). (Clearly this holds for \(Z_0\).) Define the following sets:
Set \(Z_{i+1} {:}{=}Z_i \cup Z_i^\text {ZZfin} \cup Z_i^\text {ZZcofin} \cup Z_i^\text {BF}\).
Lemma 29
(This construction can continue) If \(\left\langle w_\mathbb {N} ,v_1,\overline{a};w_\mathbb {N} ,v_2,\overline{b} \right\rangle \in Z_{i+1}\), then \(\left \delta _1(v_1) \right = \left \delta _2(v_2) \right \) and \(\mathcal {E} _1,w_\mathbb {N} ,v_1,\overline{a} \simeq \mathcal {E} _2,w_\mathbb {N} ,v_2,\overline{b} \).
Proof
Suppose \(\left\langle w_\mathbb {N} ,v_1,\overline{a};w_\mathbb {N} ,v_2,\overline{b} \right\rangle \in Z_{i+1}\). If \(\left\langle w_\mathbb {N} ,v_1,\overline{a};w_\mathbb {N} ,v_2,\overline{b} \right\rangle \in Z_i\), then this is obvious. Otherwise, \(\left\langle w_\mathbb {N} ,v_1,\overline{a};w_\mathbb {N} ,v_2,\overline{b} \right\rangle \) is either in \(Z_i^\text {ZZfin}\), \(Z_i^\text {ZZcofin}\), or \(Z_i^\text {BF}\).

Case 1: \(Z_i^\text {ZZfin}\). So \(v_1 = w_S\) and \(v_2 = w_T\) for some S and T. By the fact that \(\left S \right = \left T \right \), \(\left \delta _1(v_1) \right = \left \delta _2(v_2) \right \). And by construction, \(a_i \in S\) iff \(b_i \in T\), so \(a_i \in \delta _1(v_1)\) iff \(b_i \in \delta _2(v_2)\), and thus \(\mathcal {E} _1,w_\mathbb {N} ,v_1,\overline{a} \simeq \mathcal {E} _2,w_\mathbb {N} ,v_2,\overline{b} \). \(\checkmark \)

Case 2: \(Z_i^\text {ZZcofin}\). Same reasoning, only we know . \(\checkmark \)

Case 3: \(Z_i^\text {BF}\). We already know \(\left \delta _1(v_1) \right = \left \delta _2(v_2) \right \) since this was guaranteed by \(Z_i\). Moreover, both \(a \in \delta _1(v_1)\) and \(b \in \delta _2(v_2)\), so we still meet (Existence). And by the fact that \(a = a_i\) iff \(b = b_i\), we still meet (Equality). \(\checkmark \)
\(\square \)
Lemma 29 guarantees we can continue the construction. Finally, define \(Z \,{:}{=}\, \bigcup _{i \in \omega } Z_i\).
Proof
(Proposition 15) Suppose \(\left\langle w_\mathbb {N} ,v_1,\overline{a};w_\mathbb {N} ,v_2,\overline{b} \right\rangle \in Z\). Then there is some i such that \(\left\langle w_\mathbb {N} ,v_1,\overline{a};w_\mathbb {N} ,v_2,\overline{b} \right\rangle \in Z_i\). By Lemma 29, \(\mathcal {E} _1,w_\mathbb {N} ,v_1,\overline{a} \simeq \mathcal {E} _2,w_\mathbb {N} ,v_2,\overline{b} \), so (Atomic) and (Eq) (as well as (Ex)) are satisfied.

Zig.
Let \(u_1 \in W_1\). Suppose that \(u_1 = w_S\) for some finite nonempty \(S \subseteq \mathbb {N} \). We want to show that there is a \(T \subseteq \mathbb {N} ^\infty \) such that \(\left\langle w_\mathbb {N} ,w_S,\overline{a};w_\mathbb {N} ,w_T,\overline{b} \right\rangle \in Z_i^\text {ZZfin}\). List the elements . Let be nmany distinct elements from , and set . Then \(T \subseteq \mathbb {N} ^\infty \) is also finite and nonempty, \(\left S \right = \left T \right \), and \(a_i \in S\) iff \(b_i \in T\). So \(\left\langle w_\mathbb {N} ,w_S,\overline{a};w_\mathbb {N} ,w_T,\overline{b} \right\rangle \in Z_i^\text {ZZfin}\). The case where \(u_1 = w_{\mathbb {N}  S}\) is essentially the same, except one does not need \(\left S \right = \left T \right \). \(\checkmark \)

Zag.
As above. \(\checkmark \)

Forth.
Let \(a \in \delta _1(v_1)\). If \(a = a_i\), then we can just pick \(b = b_i\), and note that \(\left\langle w_\mathbb {N} ,v_1,\overline{a},a_i;w_\mathbb {N} ,v_2,\overline{b},b_i\right\rangle \in Z_i^\text {BF}\). So suppose a is not among \(\overline{a} \). Since \(\left \delta _1(v_1) \right = \left \delta _2(v_2) \right \), we have that . And the former is not empty since . So pick any . Then we have that \(\left\langle w_\mathbb {N} ,v_1,\overline{a},a;w_\mathbb {N} ,v_2,\overline{b},b\right\rangle \in Z_i^\text {BF}\). \(\checkmark \)

Back.
As above. \(\checkmark \)
\(\square \)
Now for the proof of Proposition 16. As before, set . Now suppose we have constructed \(Z_i\) and for all \(\left\langle w,u_1,\overline{a};w,u_2,\overline{b} \right\rangle \in Z_i\), \(\mathcal {R} _1,w,u_1,\overline{a} \simeq \mathcal {R} _2,w,\overline{u} _2,\overline{b} \) and \(u_1 = w\) iff \(u_2 = w\). Define the following sets:
Then set: \(Z_{i+1} = Z_i \cup Z_i^\text {Act} \cup Z_i^\text {ZZ} \cup Z_i^{\mathrm{ZZ}\emptyset } \cup Z_i^\text {BF}\).
Lemma 30
(This construction can continue too) If \(\left\langle w,u_1,\overline{a};w,u_2,\overline{b} \right\rangle \in Z_{i+1}\), then \(\mathcal {R} _1,w,u_1,\overline{a} \simeq \mathcal {R} _2,w,\overline{u} _2,\overline{b} \) and \(u_1 = w\) iff \(u_2 = w\).
Proof
Suppose \(\left\langle w,u_1,\overline{a};w,u_2,\overline{b} \right\rangle \in Z_{i+1}\). It is easy to verify that \(u_1 = w\) iff \(u_2 = w\) by looking at the constructions above. If \(\left\langle w,u_1,\overline{a};w,u_2,\overline{b} \right\rangle \in Z_i\), then we are done. So suppose \(\left\langle w,u_1,\overline{a};w,u_2,\overline{b} \right\rangle \notin Z_i\).
First, (Predicate). If \(\left\langle w,u_1,\overline{a};w,u_2,\overline{b} \right\rangle \in Z_i^\text {Act} \cup Z_i^\text {ZZ} \cup Z_i^{\mathrm{ZZ}\emptyset }\), then we know . So \(\mathcal {R} _1,w,u_1',\overline{a} \simeq \mathcal {R} _2,w,u_2',\overline{b} \). But since \(u_1' = w\) iff \(u_2' = w\), that means \(a_i \in \mathbb {N} \) iff \(b_i \in \mathbb {N} \). So since \(u_1 = w\) iff \(u_2 = w\), \(\mathcal {R} _1,w,u_1,\overline{a} \) and \(\mathcal {R} _2,w,u_2,\overline{b} \) satisfy (Predicate). If instead \(\left\langle w,u_1,\overline{a};w,u_2,\overline{b} \right\rangle = \left\langle w,u_1,\overline{c},c;w,u_2,\overline{d},d\right\rangle \in Z_i^\text {BF}\), then \(\left\langle w,u_1,\overline{c};w,u_2,\overline{d} \right\rangle \in Z_i\), which (by the same reasoning as above) means \(c_i \in \mathbb {N} \) iff \(d_i \in \mathbb {N} \). And by construction of \(Z_i^\text {BF}\), \(c \in \mathbb {N} \) iff \(d \in \mathbb {N} \). So since \(u_1 = w\) iff \(u_2 = w\), again \(\mathcal {R} _1,w,u_1,\overline{a} \) and \(\mathcal {R} _2,w,u_2,\overline{b} \) satisfy (Predicate).
Next, (Existence). This is trivial if \(\left\langle w,u_1,\overline{a};w,u_2,\overline{b} \right\rangle \in Z_i^\text {Act}\), since \(\delta _1(w) = \delta _2(w) = \mathbb {Z} \). It is guaranteed by construction in all other cases.
Finally, (Equality). This is trivial in every case, except \(Z_i^\text {BF}\), in which case it is guaranteed by construction. \(\square \)
As before, define \(Z \,{:}{=}\, \bigcup _{i \in \omega } Z_i\).
Proof
(Proposition 16) Suppose \(\left\langle w,u_1,\overline{a};w,u_2,\overline{b} \right\rangle \in Z\). Then there is an i such that \(\left\langle w,u_1,\overline{a};w,u_2,\overline{b} \right\rangle \in Z_i\). By Lemma 30, (Atomic) and (Eq) (as well as (Ex)) are satisfied.

Act.
By construction of \(Z_i^\text {Act}\), \(\left\langle w,w,\overline{a};w,w,\overline{b} \right\rangle \in Z_{i+1}\). \(\checkmark \)

Zig.
Let \(u_1' \in W_1\). If \(u_1' = w\), then this is covered by the above case. So let \(u_1' = v_S\) instead. Define where is arbitrary. Then T is finite and nonempty, and \(a_i \in S\) iff \(b_i \in T\). So \(\left\langle w,v_S,\overline{a};w,v_T,\overline{b} \right\rangle \in Z_i^\text {ZZ}\). \(\checkmark \)

Zag.
As above, except in the case where we pick \(v_\emptyset \in W_2\). In that case, let \(S \subseteq \mathbb {N} \) be any finite nonempty set such that . Then \(\left\langle w,v_S,\overline{a};w,v_\emptyset ,\overline{b} \right\rangle \in Z_i^{\mathrm{ZZ}\emptyset }\). \(\checkmark \)

Back.
Let \(a \in \delta _1(u_1)\) (we assume without loss of generality that ). If \(a \in \mathbb {N} ^\), then pick any new \(b \in \mathbb {N} ^\). If instead \(a \in \mathbb {N} \), then since \(\delta _2(u_2) \cap \mathbb {N} \) is infinite, pick any new \(b \in \delta _2(u_2) \cap \mathbb {N} \). Either way, \(\left\langle w,u_1,\overline{a},a;w,u_2,\overline{b},b\right\rangle \in Z_i^\text {BF}\). \(\checkmark \)

Forth.
As above. \(\checkmark \)
\(\square \)
Other inexpressibility proofs are straightforward once the winning strategy for Eloïse is worked out.
Appendix 3: Inexpressibility in
In this appendix, we will prove that (4) is not expressible as an formula. Recall (4):
First, we define our models \(\mathcal {R} _3 = \left\langle W_3,R_3,F_3,D_3,\delta _3,I_3\right\rangle \) and \(\mathcal {R} _4 = \left\langle W_4,R_4,F_4,D_4,\delta _4,I_4\right\rangle \). Our global domains will contain \(\mathbb {Z} \) plus a disjoint copy of \(\mathbb {N} \), which we will call . So \(D_3 = D_4 = \mathbb {Z} \cup \mathbb {N} _\infty \). All the accessibility relations are universal in their respective models. If \(T \subseteq \mathbb {N} \), let . Please note: throughout this section, when we write , we mean \(\mathbb {N} _\infty  T\); when we write where \(S \subseteq \mathbb {N} \), we mean \(\mathbb {N}  S\).
For each finite nonempty\(S \subseteq \mathbb {N} \), and each finite \(T \subseteq \mathbb {N} \), \(W_1\) will contain a world \(w^T_S\) and a world \(v^T_S\). Intuitively, \(w^T_S\) is a world where (i) every negative integer is poor, (ii) every integer of \((\mathbb {N}  S)\) is rich, (iii) every object of \(T_\infty \) is rich, and (iv) nothing in exists. \(v^T_S\) is like \(w^T_S\) except the rich and the poor are flipped. In addition, for any finite \(T \subseteq \mathbb {N} \), there will be a world of the form \(w^T_\emptyset \) in \(W_1\) (our actual world will be \(w {:}{=}w^\emptyset _\emptyset \)). \(W_2\) is like \(W_1\) except it also contains worlds of the form \(v^T_\emptyset \). See Fig. 7 for a picture.
Observe \(\mathcal {R} _3 \nvDash \) (4)[w] while \(\mathcal {R} _4 \vDash \) (4)[w]. Furthermore, recall that the reason \(\mathcal {R} _1\) and \(\mathcal {R} _2\) could not be used to show that (4) is not expressible as an formula was because they disagreed on the following formulas at w:
Observe that this is no longer the case: w does not satisfy either (8) or (9) in \(\mathcal {R} _3\) or \(\mathcal {R} _4\).
Proposition 31
(Strengthened inexpressibility of (R)) \(\mathcal {R} _3,w,w \leftrightarrows _{\approx ,@,{\mathop {\downarrow }},\mathcal {F}} \mathcal {R} _4,w,w\).
Proof
Clearly \(\mathcal {R} _3,w,w \simeq \mathcal {R} _4,w,w\). So suppose that \(\mathcal {R} _3,s^{T_1}_{S_1},t^{T_2}_{S_2},\overline{a} \simeq \mathcal {R} _4,s^{T_1'}_{S_1'}, t^{T_2'}_{S_2'},\overline{b} \), where:
Notice in particular that \(\mathcal {R} _3,w,w \simeq \mathcal {R} _4,w,w\) meets all of these constraints vacuously. We will show using (I)–(V) that, regardless of Abelard’s move, Eloïse can continue the game in a way that preserves (I)–(V). Note throughout that if I use the same letter, say u, for \(u^T_S\) and \(u^{T'}_{S'}\), I mean for \(u^T_S\) to be a wworld iff \(u^{T'}_{S'}\) is a wworld.
First, suppose Abelard decides to pick an object \(a \in \delta _3(t^{T_2}_{S_2})\) (the case where he picks a \(b \in \delta _4(t^{T_2'}_{S_2'})\) is symmetric). If he does this, then obviously (I) and (V) are met regardless of what Eloïse picks. So she just needs to ensure (II)–(IV) are met. If \(a \in \mathbb {N} ^\), then Eloïse can pick an arbitrary \(b \in \mathbb {N} ^\) that has not already been picked. Otherwise, since \(a \in \delta _3(t^{T_2}_{S_2})\), that means . There are two cases to consider:

Case 1: . That means \(a \in (\mathbb {N}  (S_1 \cup S_2)) \cup (T_{1\infty } \cup T_{2\infty })\). But since \(S_1'\), \(S_2'\), \(T_1'\), and \(T_2'\) are all finite, there will be infinitely many \(b \in (\mathbb {N}  (S_1' \cup S_2')) \cup (T_{1\infty }' \cup T_{2\infty }')\) that have not been picked yet. So Eloïse can just pick an arbitrary one of those, in which case and . \(\checkmark \)

Case 2: . That means we need to ensure that while also ensuring that . That means we need:
But since \(\left (S_1  S_2) \cup (T_{2\infty }  T_{1\infty }) \right = \left (S_1'  S_2') \cup (T_{2\infty }'  T_{1\infty }') \right \), and since (II)–(IV) hold for \(\overline{a} \) and \(\overline{b} \), it is easy to show that:
Hence, there must be some \(b \in (S_1'  S_2') \cup (T_{2\infty }'  T_{1\infty }')\) that has not been picked yet. So Eloïse can just pick an arbitrary one of those. \(\checkmark \)
Next, suppose Abelard decides to relocate the game. If he uses the \(@\) or \({\mathop {\downarrow }}\) moves, then the constraints will all be vacuously satisfied. So suppose he decides to relocate the game in \(\mathcal {R} _3\) to \(\left\langle s^{T_1}_{S_1},u^{T_3}_{S_3}\right\rangle \). If \(T_3 = T_1\) and \(S_3 = S_1\), then Eloïse should pick \(u^{T_1'}_{S_1'}\). If \(T_3 = T_2\) and \(S_3 = S_2\), then she should pick \(u^{T_2'}_{S_2'}\). Otherwise, Eloïse can pick a \(T_3'\) and \(S_3'\) using a different method as follows. Define the following sets:
One can verify that if \(S_3' \subseteq \mathbb {N} \) and \(T_{3\infty }' \subseteq \mathbb {N} _\infty \) such that and , then iff . We will now show the following:
Claim
There are \(S_3' \subseteq \mathbb {N} \) and \(T_{3\infty }' \subseteq \mathbb {N} _\infty \) such that:
 1.

2.
, and

3.
\(\left (S_1  S_3) \cup (T_{3\infty }  T_{1\infty }) \right = \left (S_1'  S_3') \cup (T_{3\infty }'  T_{1\infty }') \right \).
Suppose not. That is, suppose that every \(S_3' \subseteq \mathbb {N} \) and \(T_{3\infty }' \subseteq \mathbb {N} _\infty \) that satisfy (i) and (ii) fail to satisfy (iii). We will show that from this assumption, we can derive a contradiction.
First, suppose there is an \(S_3' \subseteq \mathbb {N} \) and \(T_{3\infty }' \subseteq \mathbb {N} _\infty \) satisfying (i) and (ii) such that \(\left (S_1  S_3) \cup (T_{3\infty }  T_{1\infty }) \right > \left (S_1'  S_3') \cup (T_{3\infty }'  T_{1\infty }') \right \). Let n be such that \(\left (S_1  S_3) \cup (T_{3\infty }  T_{1\infty }) \right = \left (S_1'  S_3') \cup (T_{3\infty }'  T_{1\infty }') \right + n\) (both sets are finite after all). Since \(T_{1\infty }'\) and \(T_{3\infty }'\) are finite, we can pick n arbitrary objects and set . But then \(\left (S_1  S_3) \cup (T_{3\infty }  T_{1\infty }) \right = \left (S_1'  S_3') \cup (T_{3\infty }''  T_{1\infty }') \right \), and (ii) is still met replacing \(T_{3\infty }'\) with \(T_{3\infty }''\).
Hence, it must be that for every \(S_3' \subseteq \mathbb {N} \) and \(T_{3\infty }' \subseteq \mathbb {N} _\infty \) satisfying (i) and (ii), \(\left (S_1  S_3) \cup (T_{3\infty }  T_{1\infty }) \right < \left (S_1'  S_3') \cup (T_{3\infty }'  T_{1\infty }') \right \). Now, we can assume without loss of generality that and . Here is why. Suppose . Then pick a and set . Then \(\left S_1'  S_3'' \right < \left S_1'  S_3' \right \), so \(\left (S_1'  S_3'') \cup (T_{3\infty }'  T_{1\infty }') \right < \left (S_1'  S_3') \cup (T_{3\infty }'  T_{1\infty }') \right \). \(S_3''\) still satisfies (i), so by hypothesis, it still must be that \(\left (S_1  S_3) \cup (T_{3\infty }  T_{1\infty }) \right < \left (S_1'  S_3'') \cup (T_{3\infty }'  T_{1\infty }') \right \). So we can just keep adding objects from to \(S_3'\) in this way until . Likewise, we can keep removing objects in \(T_{3\infty }'\) from until .
Thus, we may assume that and . It follows that . But if \(b_i \in (S_1'  S_3') \cup (T_{3\infty }'  T_{1\infty }')\), then \(a_i \in (S_1  S_3) \cup (T_{3\infty }  T_{1\infty })\) by (III) and by the fact that (i) and (ii) imply that iff . This gives rise to an injection from to , which means . This completes our proof of the claim above.
Thus, Eloïse can just pick any such \(S_3'\) and \(T_{3\infty }'\), and it will have the desired properties. If instead Abelard decides to relocate the game in \(\mathcal {R} _4\), the strategy is the same: the reasoning above did not rely on Abelard’s \(S_3\) being nonempty. Finally, the case where Abelard decides to relocate the game in \(\mathcal {R} _3\) to \(\left\langle u^{T_3}_{S_3},t^{T_2}_{S_1}\right\rangle \) is symmetric. \(\square \)
Appendix 4: Mapping the expressive hierarchy
In this appendix, we map out in more detail the relative expressive power for various fragments of . We will start by showing that, ignoring \({\textsf {E}} \), \(\approx \), and \(\Pi {}\), the relative expressive power for the remaining languages is accurately diagrammed by Fig. 8. This includes the strict inclusions and incomparabilities the diagram suggests. Note that for any class of models \(\mathbf C \), \(\le _\mathbf C \) is a preorder.
Lemma 32
(Adding only \({\mathop {\downarrow }}\) does nothing) .
Proof
First, note that, by induction, for any formula \(\varphi \), \(\mathcal {M} ,w,v,g \Vdash \varphi \) iff \(\mathcal {M} ,w',v,g \Vdash \varphi \). Thus, where \(\varphi \) is an formula, let \(\varphi ^\) be the result of removing every instance of \({\mathop {\downarrow }}\) from \(\varphi \). Then it is easy to show by induction (using this fact for the \({\mathop {\downarrow }}\)case) that \(\Vdash \varphi \leftrightarrow \varphi ^\). Hence, , and therefore . \(\square \)
Throughout, when I say “\(I^\mathcal {M} \) is empty” or “\(I^\mathcal {M} = \emptyset \)”, what I mean is that for all \(w \in W^\mathcal {M} \) and all predicates P, \(I^\mathcal {M} (P,w) = \emptyset \). Also, if \(\overline{a} \) is clear from context, I will use “\(a_i\)” to stand for an arbitrary element of \(\overline{a} \).
Lemma 33
(Adding \(\mathcal {F}\)) .
Proof
Let \(\mathcal {M} _1 = \left\langle W_1,R_1,F_1,D_1,\delta _1,I_1\right\rangle \), where , , , and \(I_1 = \emptyset \). Let \(\mathcal {M} _2\) be just like \(\mathcal {M} _1\) except \(F_2 = \emptyset \). Then \(\mathcal {M} _1,w,w \leftrightarrows _{\approx ,@,{\mathop {\downarrow }},\Pi {}} \mathcal {M} _2,w,w\), but \(\mathcal {M} _1,w,w \nVdash \mathcal {F}\bot \) while \(\mathcal {M} _2,w,w \Vdash \mathcal {F}\bot \). \(\square \)
Where \(\mathcal {M} \) is a model, let \(\mathcal {M} ^{{\textsf {E}} \rightarrow P}\) be the model just like \(\mathcal {M} \) except \(I^{\mathcal {M} ^{{\textsf {E}} \rightarrow P}}(P,w) = \delta ^\mathcal {M} (w)\) for all \(w \in W^\mathcal {M} \). That is, \(\mathcal {M} ^{{\textsf {E}} \rightarrow P}\) effectively makes P an existence predicate. Define \(\mathcal {M} ^{\approx \rightarrow P}\) likewise. It will be useful to note the following:
Lemma 34
(Replacing \({\textsf {E}} \)) If \(\mathcal {M} ,w,v,\overline{a} \leftrightarrows _{\mathcal {L}({\textsf {E}})} \mathcal {N} ,w',v',\overline{b} \), then \(\mathcal {M} ^{{\textsf {E}} \rightarrow P},w,v,\overline{a} \leftrightarrows _{\mathcal {L}({\textsf {E}})} \mathcal {N} ^{{\textsf {E}} \rightarrow P},w',v',\overline{b} \). In addition, if \(I^\mathcal {M} (P,u) = \emptyset = I^\mathcal {N} (P,u')\) for all \(u \in W^\mathcal {M} \) and all \(u' \in W^\mathcal {N} \), then the converse holds as well. Likewise for \(\approx \) in place of \({\textsf {E}} \).
We can use this trick to bootstrap off of previous inexpressibility results which used \({\textsf {E}} \) or \(\approx \) for languages without \({\textsf {E}} \) or \(\approx \). For instance, it is relatively easy to show . Take models \(\mathcal {E} _1\) and \(\mathcal {E} _2\) from Fig. 2. Since \(\mathcal {E} _1,w_\mathbb {N} ,w_\mathbb {N} \leftrightarrows _{\approx } \mathcal {E} _2,w_\mathbb {N} ,w_\mathbb {N} \), we can use Lemma 34 to conclude \(\mathcal {E} _1^{{\textsf {E}} \rightarrow P},w_\mathbb {N} ,w_\mathbb {N} \leftrightarrows _{\approx } \mathcal {E} _2^{{\textsf {E}} \rightarrow P},w_\mathbb {N} ,w_\mathbb {N} \), though they disagree on .
However, this proof does not show that , since the models are distinguishable by \(\Sigma {x}\lnot P(x)\). Lemma 35 strengthens this result to include \(\Pi {}\), again using Lemma 34:^{Footnote 36}
Lemma 35
(Adding \(@\)) .
Proof
It is easy to show that if \(\varphi \) is an formula, then \(\Vdash _\mathbf U \mathcal {F}\varphi \leftrightarrow \varphi \) (just use the observation from the proof of Lemma 32). So every formula is \(\mathbf U \)equivalent to an formula. But if \(\psi \) is an formula, then \(\Vdash _\mathbf U {\mathop {\downarrow }}\psi \leftrightarrow \psi \). Putting these together, every formula is \(\mathbf U \)equivalent to an formula. So it suffices to find two bisimilar models in UD that disagree on some formula.
Let \(\mathcal {M} _1 = \left\langle W_1,R_1,F_1,D_1,\delta _1,I_1\right\rangle \), where:
\(R_1\) and \(F_1\) are universal, \(D_1 = \mathbb {Z} \), \(\delta _1(w) = \mathbb {N} \), \(\delta _1(v_S) = (\mathbb {N}  S) \cup T\), and \(I_1 = \emptyset \). Let \(\mathcal {M} _2\) be like \(\mathcal {M} _1\) except that , and . Observe:
However, we will show that \(\mathcal {M} _1,w,w \leftrightarrows _{\approx ,\Pi {}} \mathcal {M} _2,w,w\). Clearly \(w,w \simeq w,w\). Suppose \(w,u_1,\overline{a} \simeq w,u_2,\overline{b} \). Since \(\delta _1(u_1)\) and are infinite, if Abelard picks an \(a' \in D_1\), then Eloïse can find a \(b' \in D_2\) so that \(w,u_1,\overline{a},a' \simeq w,u_2,\overline{b},b'\) by ensuring that \(a' \in \delta _1(u_1)\) iff \(b' \in \delta _2(u_2)\). Likewise if Abelard picks a \(b' \in D_2\). Now, suppose Abelard picks an \(u_1' \in W_1\). Define and . (If \(\left S \right \le 1\), add a couple of elements from to S. If \(\left T \right \le 1\), add a couple of elements from to T.) Then \(a_i \in \delta _1(u_1)\) iff \(b_i \in \delta _2(v_S^T)\). So \(w,v_S^T,\overline{a} \simeq w,v_{S'}^{T'},\overline{b} \). Likewise if Abelard chooses a \(u_2' \in W_2\), even if \(u_2' = v\). Thus, using Lemma 34, . \(\square \)
Lemma 36
(Adding two operators) .
Proof
First, . Let \(\mathcal {M} _1 = \left\langle W_1,R_1,F_1,D_1,\delta _1,I_1\right\rangle \), where , \(R_1 = W_1 \times W_1\), \(F_1 = \emptyset \), , and \(I_1 = \emptyset \). Let \(\mathcal {M} _2\) be like \(\mathcal {M} _1\) except . Then \(\mathcal {M} _1,w,w \leftrightarrows _{\approx ,@,\mathcal {F},\Pi {}} \mathcal {M} _2,w,w\) (since \(F_1[w] = F_2[w] = \emptyset \)), but \(\mathcal {M} _1,w,w \Vdash \Box {\mathop {\downarrow }}\mathcal {F}\bot \), while \(\mathcal {M} _2,w,w \nVdash \Box {\mathop {\downarrow }}\mathcal {F}\bot \). So .
Next, . Consider the models \(\mathcal {N} _1\) and \(\mathcal {N} _2\) from Fig. 4. Modify them so that \(F_1 = F_2 = \emptyset \), and call the resulting models \(\mathcal {N} _1'\) and \(\mathcal {N} _2'\). Then \(\mathcal {N} _1',z,z \leftrightarrows _{\approx ,@,\mathcal {F},\Pi {}} \mathcal {N} _2',z,z\), but they disagree on . Hence, . \(\square \)
It is tedious, but straightforward, to show the following using the lemmas above:
Theorem 37
(Completeness of Fig. 8) Figure 8 is a complete diagram of the expressive power of the languages presented in that diagram.
Now we turn to extensions with \({\textsf {E}} \), \(\approx \), and \(\Pi {}\). It will help to define backandforth games for and some of its fragments. First, an formula is almost\({\textsf {E}} \)free if it can be built from atomics other than those of the form \({\textsf {E}} (x;s)\) using negation, conjunction, object quantification, state quantification, and \({\textsf {E}} \)bounded object quantification. (Thus, \({\textsf {E}} \) only occurs as the bounds of object quantifiers.) Let be the \(\approx \)free fragment, be the \(\approx \)free and almost \({\textsf {E}} \)free fragment, and be the \({\textsf {E}} \)bounded fragment of .
Definition 38
(Backandforth system) Let \(\mathcal {M} \) and \(\mathcal {N} \) be models. A backandforth system between\(\mathcal {M} \)and\(\mathcal {N} \) is a nonempty variably polyadic relation Z such that whenever \(Z(\overline{w},\overline{a};\overline{v},\overline{b})\), and , and if \(Z(\overline{w},\overline{a};\overline{v},\overline{b})\), then:

(\({\textsf {TS}} \) Atomic) :

(\({\textsf {TS}} \) Eq) \(\forall {n,m \le \left \overline{a} \right :} a_n = a_m\) iff \(b_n = b_m\)

(\({\textsf {TS}} \) StEq) \(w_k = w_l\) iff \(v_k = v_l\)

(\({\textsf {TS}} \) Ex) iff \(b_n \in \delta ^\mathcal {N} (v_k)\)

(\({\textsf {TS}} \) Acc) \(R(w_k,w_l)\) iff \(R(v_k,v_l)\) and \(F(w_k,w_l)\) iff \(F(v_k,v_l)\)

(\({\textsf {TS}} \) Zig)

(\({\textsf {TS}} \) Zag)

(\({\textsf {TS}} \) Forth)

(\({\textsf {TS}} \) Back) .
We may write \(\mathcal {M} ,\overline{w},\overline{a} \leftrightarrows _{{\textsf {TS}}} \mathcal {N} ,\overline{v},\overline{b} \) to indicate that \(\mathcal {M} ,\overline{w},\overline{a} \) and \(\mathcal {N} ,\overline{v},\overline{b} \) are backandforth equivalent. If we drop (\({\textsf {TS}} \) Eq) and (\({\textsf {TS}} \) StEq), we get a notion of a backandforth system for . We get a notion of a backandforth system for if we drop (\({\textsf {TS}} \) Eq), (\({\textsf {TS}} \) StEq), and (\({\textsf {TS}} \) Ex) and we add:

(\({\textsf {TS}} \)\({\textsf {E}} \)Forth)

(\({\textsf {TS}} \)\({\textsf {E}} \)Back) .
If we replace (\({\textsf {TS}} \) Forth) and (\({\textsf {TS}} \) Back) with (\({\textsf {TS}} \)\({\textsf {E}} \)Forth) and (\({\textsf {TS}} \)\({\textsf {E}} \)Back), we get a notion of a backandforth system for .
Definition 39
(Equivalence) \(\mathcal {M} ,\overline{w},\overline{a} \) and \(\mathcal {N} ,\overline{v},\overline{b} \) are equivalent if for all formulas \(\alpha (\overline{x};\overline{s})\) (where \(\left \overline{x} \right \le \left \overline{a} \right \) and \(\left \overline{s} \right \le \left \overline{w} \right \)), \(\mathcal {M} \vDash \alpha [\overline{a};\overline{w} ]\) iff \(\mathcal {N} \vDash \alpha [\overline{b};\overline{v} ]\). We may write “\(\mathcal {M} ,\overline{w},\overline{a} \equiv _{\textsf {TS}} \mathcal {N} ,\overline{v},\overline{b} \)” to indicate that \(\mathcal {M} ,\overline{w},\overline{a} \) and \(\mathcal {N} ,\overline{v},\overline{b} \) are equivalent. Likewise for the various fragments of .
It is easy to check that \(\mathcal {M} ,\overline{w},\overline{a} \leftrightarrows _{\textsf {TS}} \mathcal {N} ,\overline{v},\overline{b} \) implies \(\mathcal {M} ,\overline{w},\overline{a} \equiv _{\textsf {TS}} \mathcal {N} ,\overline{v},\overline{b} \), and likewise for the various fragments of . Now, say \(\mathcal {L}_1 \le ^* \mathcal {L}_2\) if every \(\mathcal {L}_1\)formula is equivalent to some \(\mathcal {L}_2\)formula. This is more stringent than \(\le \), since some \(\mathcal {L}_1\)formula might only be expressible as a set of \(\mathcal {L}_2\)formulas. Observe by Definition 6 that , that , and that .
Lemma 40
(Adding \({\textsf {E}} \) and \(\approx \)) If , then \(\mathcal {L}< \mathcal {L}({\textsf {E}}) < \mathcal {L}(\approx )\). Likewise if relativize to \(\mathbf D \), \(\mathbf U \), or UD.
Proof
Recall the models \(\mathcal {E} \) and \(\mathcal {E} '\) from Fig. 1. It is easy to check that via our original bisimulation, \(\mathcal {E},w,w \leftrightarrows _{{\textsf {TS}}\approx ,({\textsf {E}})} \mathcal {E} ',w,w\) (remember, you do not need to satisfy (\({\textsf {TS}} \) Eq) or (\({\textsf {TS}} \) Ex) in this backandforth game!). But these models are distinguishable by the formula . So . Suppose now for reductio that \(\mathcal {L}({\textsf {E}}) \le \mathcal {L}\). Since (easily verified by induction), , . So \(\mathcal {L}({\textsf {E}}) \nleq \mathcal {L}\), and thus \(\mathcal {L}< \mathcal {L}({\textsf {E}})\).
As for \(\mathcal {L}({\textsf {E}}) < \mathcal {L}(\approx )\), revise \(\mathcal {E} \) and \(\mathcal {E} '\) by deleting the world w from the models. Call the resulting models \(\mathcal {E} _{}\) and \(\mathcal {E} _{}'\). Then \(\mathcal {E} _{},v \leftrightarrows _{{\textsf {TS}}\approx } \mathcal {E} _{}',v\), but they disagree on . So . But \(\mathcal {L}({\textsf {E}}) \le \mathcal {L}(\approx )\), so reasoning as before (noting that ), we have that \(\mathcal {L}({\textsf {E}}) < \mathcal {L}(\approx )\). \(\square \)
Now, where \(\mathcal {L}_1\) and \(\mathcal {L}_2\) were languages in Fig. 8 such that \(\mathcal {L}_1 < \mathcal {L}_2\), we can show that the inclusions involving their extensions with \({\textsf {E}} \) or \(\approx \) can be diagrammed as in Fig. 9. First, the arrows that are present are immediate by Lemma 40 and by the fact that if \(\mathcal {L}_1 < \mathcal {L}_2\) in Fig. 8, then we already have \(\mathcal {L}_1 <^* \mathcal {L}_2\). Next, \(\mathcal {L}_1({\textsf {E}}) \nleq \mathcal {L}_2\), since if it were, we would have , contrary to Lemma 40. Likewise, \(\mathcal {L}_1(\approx ) \nleq \mathcal {L}_2({\textsf {E}})\). Finally, observe that in the results used above to show that \(\mathcal {L}_2 \nleq \mathcal {L}_1\), we already showed that \(\mathcal {L}_2 \nleq \mathcal {L}_1(\approx ,\Pi {})\). Thus, \(\mathcal {L}_2 \nleq \mathcal {L}_1(\approx )\).
Lemma 41
(Adding \(\Pi {}\)) If , then \(\mathcal {L}< \mathcal {L}(\Pi {})\). Likewise if we relativize to \(\mathbf U \). Also, if , then \(\mathcal {L}<_\mathbf D \mathcal {L}(\Pi {})\).
Proof
Let \(\mathcal {M} _1 = \left\langle W_1,R_1,F_1,D_1,\delta _1,I_1\right\rangle \), where , , , , . Let \(\mathcal {M} _2\) be just like \(\mathcal {M} _1\) except . Then \(\mathcal {M} _1,w \leftrightarrows _{{\textsf {TS}} \upharpoonright {\textsf {E}}} \mathcal {M} _2,w\), but they disagree on the formula \(\Sigma {x}\lnot P(x)\). So . Reasoning as before, \(\mathcal {L}< \mathcal {L}(\Pi {})\).
Suppose now we restrict to \(\mathbf D \). Let \(\mathcal {M} _1 = \left\langle W_1,R_1,F_1,D_1,\delta _1,I_1\right\rangle \), where , , \(\delta _1(w) = \mathbb {N} \), , \(I(P,w) = \emptyset \), and for each \(n \in \mathbb {N} \), . Let \(\mathcal {M} _2\) be just like \(\mathcal {M} _1\) except , where \(u \notin W_1\), , \(F_2 = F_1\), \(\delta _2(u) = \mathbb {N} \), and \(I(P,u) = \emptyset \). One can show that \(\mathcal {M} _1,w,w \leftrightarrows _{\approx ,@,{\mathop {\downarrow }},\mathcal {F}} \mathcal {M} _2,w,w\). But, they disagree on the formula \(\Box \Sigma {x} P(x)\). So . So \(\mathcal {L}<_\mathbf D \mathcal {L}(\Pi {})\). \(\square \)
So once again, using Lemma 41 and the results above, we can verify that if \(\mathcal {L}_1\) and \(\mathcal {L}_2\) are in Fig. 8 and \(\mathcal {L}_1 < \mathcal {L}_2\), then their extensions involving \(\Pi {}\) can be represented in Fig. 10. This holds even if we add \({\textsf {E}} \) or \(\approx \). Thus, Fig. 5 from Sect. 4 is correct. Moreover, it is still correct even relative to \(\mathbf D \).
We now turn to asking to what extent these results hold relative to \(\mathbf U \) and UD. We only give a partial answer here. First, set aside \({\textsf {E}} \), \(\approx \), and \(\Pi {}\), and focus just on \(\mathbf U \). Then the diagram of expressive power looks something like Fig. 11 (whether we should include the dashed arrows has yet to be determined).
First, if \(\varphi \) is \(@\)free, then \(\Vdash _\mathbf U (\mathcal {F}\varphi \leftrightarrow \varphi )\) and \(\Vdash _\mathbf U ({\mathop {\downarrow }}\varphi \leftrightarrow \varphi )\). So . But still by Lemma 35. And the remarks on page 4 (together with Lemma 34) show that and that . As for the lack of inclusion from to :
Lemma 42
(\(\mathcal {F},@\) Not included in \(@,{\mathop {\downarrow }}\)) .
Proof
Let \(\mathcal {M} _1 = \left\langle W_1,R_1,F_1,D_1,\delta _1,I_1\right\rangle \) where:
\(R_1 = F_1 = W_1 \times W_1\), \(D_1 = \mathbb {Z} \), \(\delta _1(v^T_S) = (\mathbb {N}  S) \cup T\), and \(I_1 = \emptyset \). Let \(\mathcal {M} _2\) be just like \(\mathcal {M} _1\), except we allow \(\left S \right = 1\). Let \(w = v^\emptyset _\emptyset \). Observe that:
However, we will show \(\mathcal {M} _1,w,w \leftrightarrows _{\approx ,@,{\mathop {\downarrow }}} \mathcal {M} _2,w,w\). Clearly \(w,w \simeq w,w\). Suppose throughout that \(u_1,u_2,\overline{a} \simeq u_1',u_2',\overline{b} \) and that the following hold:

(I)
\(a_i \in \delta _1(u_1)\) iff \(b_i \in \delta _1(u_1)\)

(II)
\(u_1 = u_2\) iff \(u_1' = u_2'\)

(III)
.
Observe that no matter what \(u_1\) and \(u_2\) are, is infinite, and is finite. Likewise for \(u_1'\) and \(u_2'\).
First, suppose Abelard picks a new \(a \in \delta _1(u_2)\). If \(a \in \delta _1(u_1)\), then since is infinite, Eloïse will always be able to match a with a . If instead \(a \notin \delta _1(u_1)\), then by (III), we can find a to match a with. A symmetric argument applies if Abelard instead picks a \(b \in \delta _2(u_2')\).
Next, suppose Abelard decides to relocate the game. If he invokes (Act) or (Diag), then clearly (I)–(III) hold. So suppose he decides to pick a \(u_3 \in W_3\) to relocate to. Eloïse’s choice is obvious if \(u_3 = u_1\), so suppose \(u_3 \ne u_1\). We will construct a \(T_3'\) and a \(S_3'\) of the appropriate sort and show they meet (I)–(III). First, pick two elements (note that is infinite since each world has cofinitely many positive integers) and define:
Note that where \(u_1' = v^{T_1'}_{S_1'}\), is finite, so \(S_3'\) is finite and . Second, define . Observe that:
Now, where , pick kmany elements (notice that is infinite, since each world only has finitely many negative integers). Define . We will show that if Eloïse chooses \(u_3' = v^{T_3'}_{S_3'}\), then all the necessary constraints are met.
We first need to show \(u_1,u_3,\overline{a} \simeq u_1',v^{T_3'}_{S_3'},\overline{b} \)—in particular, \(a_i \in \delta _1(u_3)\) iff \(b_i \in \delta _2(u_3')\). Suppose \(a_i \in \delta _1(u_3)\). Either \(b_i \in \mathbb {N} \) or \(b_i \in \mathbb {N} ^\). If \(b_i \in \mathbb {N} \), then \(b_i \notin S_3'\), so \(b_i \in \delta _2(u_3')\). If \(b_i \in \mathbb {N} ^\), then \(b_i \in T_{3,0}' \subseteq T_3'\), so \(b_i \in \delta _2(u_3')\). Suppose instead \(a_i \notin \delta _1(u_3)\). Again, either \(b_i \in \mathbb {N} \) or \(b_i \in \mathbb {N} ^\). If \(b_i \in \mathbb {N} \), then \(b_i \in S_3'\), so \(b_i \notin \delta _2(u_3')\). If \(b_i \in \mathbb {N} ^\), then \(b_i \notin T_3'\), so \(b_i \notin \delta _2(u_3')\). No matter what, \(a_i \in \delta _1(u_3)\) iff \(b_i \in \delta _2(u_3')\).
Next, we need to show (I)–(III). (I) holds by default. Now, we assumed above \(u_3 \ne u_1\), so we need \(u_3' \ne u_1'\). But recall that we picked c, d so that \(c,d \in \delta _2(u_1')\). But \(c,d \in S_3'\), so \(c,d \notin \delta _2(u_3')\). Thus, \(u_3' \ne u_1'\). So (II) holds. Finally, using the calculations above, since , we find that:
where . So (III) holds. Thus, if Abelard relocates to \(u_3\), Eloïse can choose to relocate to \(u_3'\). And since \(\left S_3' \right > 1\), a symmetric argument applies if Abelard decides to relocate the game in \(\mathcal {M} _2\). The proof is completed with one application of Lemma 34. \(\square \)
Now, because the proof of Lemma 40 only uses models in \(\mathbf U \) (in fact, in UD), we can still safely say that adding \({\textsf {E}} \) or \(\approx \) can be diagrammed as in Fig. 9. Adding \(\Pi {}\) makes things more complicated. Recall that relative to the class of all models, we could simply say that if , then \(\mathcal {L}_i < \mathcal {L}_i(\Pi {})\), \(\mathcal {L}_1(\Pi {}) < \mathcal {L}_2(\Pi {})\), and \(\mathcal {L}_1(\Pi {})\) and \(\mathcal {L}_2\) were incomparable. However, showing that \(\mathcal {L}_2 \nleq \mathcal {L}_1(\Pi {})\) crucially relied on the fact that all of our inexpressibility proofs for showing \(\mathcal {L}_2 \nleq \mathcal {L}_1\) already showed that \(\mathcal {L}_2 \nleq \mathcal {L}_1(\Pi {})\). But because Lemma 42 left out \(\Pi {}\) (which is crucial, as we will see below), we cannot conclude that \(\mathcal {L}_2 \nleq _\mathbf U \mathcal {L}_1(\Pi {})\).
We can still verify by hand that in some cases, \(\mathcal {L}_2 \nleq \mathcal {L}_1(\Pi {})\). For one thing, by Lemma 35. We also have that by Proposition 17. Likewise, . But importantly, some of these languages without \({\textsf {E}} \) and \(\Pi {}\) that were distinct collapse when you add \({\textsf {E}} \) and \(\Pi {}\):
Lemma 43
(Collapse) . Likewise if we add \(\approx \) to these languages.
Proof
Throughout, let . Note that the following are all \(\mathbf U \)valid (where \(\alpha \) is an atomic formula):
Likewise, all of these are \(\mathbf U \)valid:
Using these rules, we can push each \(@\) and each \({\mathop {\downarrow }}\) inwards as much as possible until \(@\) only occurs right before a \(\mathcal {F}\) or an atomic, and \({\mathop {\downarrow }}\) only occurs right before a \(\Box \). Moreover, we can delete any \(\mathcal {F}\) and \({\mathop {\downarrow }}\) if it does not scope over an \(@\), and repeat. After this entire process, the resulting formula will be \(\mathbf U \)equivalent to our original. So assume without loss of generality that our formula has already gone through this preprocessing.
Now, say that an \(\mathcal {L}^*\)formula is in normal form if it is either a nonmodal formula, or if it is of the form:
where (the quantifier block may be empty), \({\textsf {BC}} \) is some boolean combination of its components, \(\overline{\psi } \) are all nonmodal, each , and \(\overline{\theta } \) are all in normal form. By induction, one can convert every \(\mathcal {L}^*\)formula into one of normal form (essentially by preprocessing as above, and then replacing bound variables and pulling out quantifiers). Thus, we may assume without loss of generality that our formula is already in normal form.
Finally, suppose an \(\mathcal {L}^*\)formula has been preprocessed and is in the form:
where \(\overline{\varphi } \) are all nonmodal, and \(\overline{\psi } \), \(\overline{\theta } \), \(\overline{\chi } \), and \(\overline{\xi } \) are all in normal form (notice that since we preprocessed, each \(\psi \) is either an atomic or of the form \(\mathcal {F}\psi '\)). Then the following are \(\mathbf U \)valid:
Thus, in our original formula, we can replace any \(\mathcal {F}\) with \({\mathop {\downarrow }}\Box \) or vice versa. \(\square \)
To sum up, the following questions have yet to be answered about the relative \(\mathbf U \)expressive power of these languages:

Is ?

Is ?

Is or ?

Is or ?
Answering these questions would settle the rest.
We now finally turn to UD. Excluding \({\textsf {E}} \), \(\approx \), and \(\Pi {}\), the diagram in Fig. 11 is still correct (again, the dashed arrows have not been determined). And again, Fig. 9 is still correct when adding either \({\textsf {E}} \) or \(\approx \). But adding \(\Pi {}\) is even trickier than before, since we can no longer appeal to Lemma 41. We still have the lack of inclusions mentioned above Lemma 43. We also have the following lack of inclusions:
Lemma 44
(Inexpressibility of \(@\) with \(\Pi {}\)) .
Proof
Recall that \(\mathcal {R} _1,w,w \leftrightarrows _{\approx ,@} \mathcal {R} _2,w,w\) (Fig. 3). But the models are distinguished by . \(\square \)
Lemma 45
(Inexpressibility of \(@,\Pi {}\) with \(@,{\mathop {\downarrow }},\mathcal {F}\)) .
Proof
This immediately follows from Proposition 31. \(\square \)
However, we now have more inclusions. For example, (just set ).^{Footnote 37} Likewise, , thoughthe proof is a bit more roundabout.^{Footnote 38} These inclusions are strict by Lemma 35. The questions mentioned above for \(\mathbf U \)expressive power are still unanswered for UDexpressive power. And again, answering these questions suffices to settle the remaining inclusions.
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Kocurek, A.W. On the expressive power of firstorder modal logic with twodimensional operators. Synthese 195, 4373–4417 (2018). https://doi.org/10.1007/s1122901611723
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DOI: https://doi.org/10.1007/s1122901611723
Keywords
 Firstorder modal logic
 Expressive power
 Twodimensional operators
 Actually
 Fixedly
 Vlach Operators
 Bisimulation