A complete classification of the expressiveness of interval logics of Allen’s relations: the general and the dense cases

Abstract

Interval temporal logics take time intervals, instead of time points, as their primitive temporal entities. One of the most studied interval temporal logics is Halpern and Shoham’s modal logic of time intervals HS, which associates a modal operator with each binary relation between intervals over a linear order (the so-called Allen’s interval relations). In this paper, we compare and classify the expressiveness of all fragments of HS on the class of all linear orders and on the subclass of all dense linear orders. For each of these classes, we identify a complete set of definabilities between HS modalities, valid in that class, thus obtaining a complete classification of the family of all 4096 fragments of HS with respect to their expressiveness. We show that on the class of all linear orders there are exactly 1347 expressively different fragments of HS, while on the class of dense linear orders there are exactly 966 such expressively different fragments.

Appendices

Appendix 1: Complete proof of Lemma 2

Lemma 2 (Soundness for $\langle L \rangle$ over $\mathsf {Den}$) The following definabilities hold relative to the class $\mathsf {Den}$:

• $\langle L \rangle$  is definable in $\mathsf {D O}$,

• $\langle L \rangle$  is definable in $\overline{\mathsf {B}}\mathsf {D}$,

• $\langle L \rangle$  is definable in $\mathsf {E O}$,

• $\langle L \rangle$  is definable in $\mathsf {B O}$ and

• $\langle L \rangle$  is definable in $\overline{\mathsf {L}}\mathsf {O}$.

Proof

The proof of the first claim is on page 11 in the main body of the paper. Consider now the definability equation $\langle L \rangle p \equiv \langle \overline{B} \rangle [ D ] \langle \overline{B} \rangle \langle D \rangle \langle \overline{B} \rangle p$. Suppose that $M,[a,b] \Vdash \langle L \rangle p$ for an interval [a, b] in a model M. Thus, as before, there exists an interval [c, d] in M such that $b < c$ and $M,[c,d] \Vdash p$. By definition of $R_{\overline{B}}$, it holds that $[a,b] R_{\overline{B}} [a,c]$. We now show that [a, c] satisfies $[ D ] \langle \overline{B} \rangle \langle D \rangle \langle \overline{B} \rangle p$. First, every interval [e, f], with $[a,c] R_D [e,f]$, is such that $e < c$. We claim that $M,[e,f] \Vdash \langle \overline{B} \rangle \langle D \rangle \langle \overline{B} \rangle p$. To see this, let us consider the interval [e, d]. We observe that $[e,f] R_{\overline{B}} [e,d]$ holds. Moreover, by the density of M, there exists a point $d'$, with $c < d' < d$, such that $[e,d] R_{D} [c,d']$ holds and $[c,d']$ satisfies $\langle \overline{B} \rangle p$, because p holds over [c, d] and $[c,d'] R_{\overline{B}} [c,d]$. Thus, $M,[e,f] \Vdash \langle \overline{B} \rangle \langle D \rangle \langle \overline{B} \rangle p$, as claimed.

As for the opposite direction, suppose that $M,[a,b] \Vdash \langle \overline{B} \rangle [ D ] \langle \overline{B} \rangle \langle D \rangle \langle \overline{B} \rangle p$ for an interval [a, b] in a model M. That means that there exists a point $c > b$ such that the interval [a, c] satisfies $[ D ] \langle \overline{B} \rangle \langle D \rangle \langle \overline{B} \rangle p$. As a particular instance of the latter formula, every interval [e, f] such that $b < e < f < c$ (the existence of such an interval [e, f] is guaranteed by the density of M) must satisfy $\langle \overline{B} \rangle \langle D \rangle \langle \overline{B} \rangle p$ which means that there exists a point $g > f$ such that $M,[e,g] \Vdash \langle D \rangle \langle \overline{B} \rangle p$, which implies, in turn, the existence of two points h, i, with $e < h < i$, such that $M,[h,i] \Vdash p$. Since $h>b$, we have that $M,[a,b] \Vdash \langle L \rangle p$.

Next, let us focus on $\langle L \rangle p \equiv \langle O \rangle [ E ] \langle O \rangle \langle O \rangle p$. Suppose that $M,[a,b] \Vdash \langle L \rangle p$ for an interval [a, b] in a model M. Once again, this means that there exists an interval [c, d] in M such that $b < c$ and $M,[c,d] \Vdash p$. Consider an interval $[a',c]$, with $a < a' < b$ (the existence of such a point $a'$ is guaranteed by the density of M). It holds that $[a,b] R_O [a',c]$. We prove that $M,[a',c] \Vdash [ E ] \langle O \rangle \langle O \rangle p$. Indeed, for every interval [e, c], with $[a',c] R_E [e,c]$, by the density of M, there exist a point f, with $e < f < c$, and a point g, with $c < g < d$, such that the interval [f, g] satisfies $\langle O \rangle p$ as $[f,g] R_O [c,d]$. Thus, $M,[e,c] \Vdash \langle O \rangle \langle O \rangle p$, $M,[a',c] \Vdash [ E ] \langle O \rangle \langle O \rangle p$, and $M,[a,b] \Vdash \langle O \rangle [ E ] \langle O \rangle \langle O \rangle p$.

In order to prove the converse direction, suppose that $M,[a,b] \Vdash \langle O \rangle [ E ] \langle O \rangle \langle O \rangle p$ for an interval [a, b] in a model M. That means that there exists an interval [c, d] such that $[a,b] R_O [c,d]$ and $M,[c,d] \Vdash [ E ] \langle O \rangle \langle O \rangle p$. As a particular instance, the interval [e, d], for some e such that $b < e < d$ (the existence of such a point e is guaranteed by the density of M), satisfies $\langle O \rangle \langle O \rangle p$, that implies the existence of an interval [f, g], with $f > e$ $(> b)$, satisfying p. It immediately follows that $M,[a,b] \Vdash \langle L \rangle p$.

Consider now $\langle L \rangle p \equiv \langle O \rangle (\langle O \rangle \top \wedge [ O ] \langle B \rangle \langle O \rangle \langle O \rangle p )$. Suppose that $M,[a,b] \Vdash \langle L \rangle p$ for an interval [a, b] in a model M, which implies, as before, that there exists an interval [c, d] in M such that $b < c$ and $M,[c,d] \Vdash p$. Consider an interval $[a',c]$, with $a < a' < b$ (the existence of such a point $a'$ is guaranteed by the density of M). This interval is such that $[a,b] R_O [a',c]$ and it satisfies:

• $\langle O \rangle \top$, as $[a',c] R_O [b,d]$, and

• $[ O ] \langle B \rangle \langle O \rangle \langle O \rangle p$, as every interval [e, f], with $[a',c] R_O [e,f]$, is such that $e < c < f$; thus, the interval [e, c] is such that $[e,f] R_B [e,c]$, and, by the density of M, there exists an interval [g, h] such that $[e,c] R_O [g,h]$ and $[g,h] R_O [c,d]$, and this implies $M,[e,c] \Vdash \langle O \rangle \langle O \rangle p$, which in turn implies $M,[a',c] \Vdash [ O ] \langle B \rangle \langle O \rangle \langle O \rangle p$.

Hence, $M,[a',c] \Vdash \langle O \rangle \top \wedge [ O ] \langle B \rangle \langle O \rangle \langle O \rangle p$, and thus $M,[a,b] \Vdash \langle O \rangle (\langle O \rangle \top \wedge [ O ] \langle B \rangle \langle O \rangle \langle O \rangle p)$.

Conversely, suppose that $M,[a,b] \Vdash \langle O \rangle (\langle O \rangle \top \wedge [ O ] \langle B \rangle \langle O \rangle \langle O \rangle p )$ for an interval [a, b] in a model M. That means that there exists an interval [c, d] such that:

• $[a,b] R_O [c,d]$,

• $M,[c,d] \Vdash \langle O \rangle \top$, and thus there exists a point $f > d$, and

• $M,[c,d] \Vdash [ O ] \langle B \rangle \langle O \rangle \langle O \rangle p$.

By the density of M, there exists a point e, with $b < e < d$. The interval [e, f] is such that $[c,d] R_O [e,f]$, and thus, by the third condition above, it satisfies $\langle B \rangle \langle O \rangle \langle O \rangle p$, which implies the existence of an interval [g, h], with $g > e (> b)$, satisfying p. It immediately follows that $M,[a,b] \Vdash \langle L \rangle p$.

Finally, consider $\langle L \rangle p \equiv \langle O \rangle (\langle O \rangle \top \wedge [ O ] [ \overline{L} ] \langle O \rangle \langle O \rangle p)$. Suppose that M, [a,  $b] \Vdash \langle L \rangle p$ for an interval [a, b] in a model M. Thus, there exists an interval [c, d] in M such that $b < c$ and $M,[c,d] \Vdash p$. Consider an interval $[a',c]$, with $a < a' < b$ (the existence of such a point $a'$ is guaranteed by the density of M). This interval is such that $[a,b] R_O [a',c]$ and it satisfies both $\langle O \rangle \top$, as $[a',c] R_O [b,d]$, and $[ O ] [ \overline{L} ] \langle O \rangle \langle O \rangle p$, thanks to the following argument. Every interval [e, f], with $[a',c] R_O [e,f]$, is such that $e < c$. Thus, every interval [g, h], with $[e,f] R_{\overline{L}} [g,h]$, satisfies $\langle O \rangle \langle O \rangle p$ (by the density of M, there exist $g < i < h$ and $c < j < d$ such that both $[g,h] R_O [i,j]$ and $[i,j] R_O [c,d]$ hold). Therefore, we have that $M,[a',c] \Vdash [ O ] [ \overline{L} ] \langle O \rangle \langle O \rangle p$, which implies $M,[a,b] \Vdash \langle O \rangle (\langle O \rangle \top \wedge [ O ] [ \overline{L} ] \langle O \rangle \langle O \rangle p)$.

As for the other direction, suppose that $M,[a,b] \Vdash \langle O \rangle (\langle O \rangle \top \wedge [ O ] [ \overline{L} ] \langle O \rangle \langle O \rangle p)$ for an interval [a, b] in a model M. That means that there exists an interval [c, d] such that $[a,b] R_O [c,d]$, $M, [c,d] \Vdash \langle O \rangle \top$ (and thus, there exists a point $f > d$), and that $M, [c,d] \Vdash [ O ] [ \overline{L} ] \langle O \rangle \langle O \rangle p$. As a specific instance, consider the interval [e, f], for some e such that $b < e < d$ (the existence of such a point e is guaranteed by the density of M). Since $[c,d] R_O [e,f]$, then we have $M, [e,f] \Vdash [ \overline{L} ] \langle O \rangle \langle O \rangle p$, which in turn, together with the density assumption, implies the existence of an interval [g, h], with $b < g < h < e$, that satisfies $\langle O \rangle \langle O \rangle p$. Thus, there exists an interval [i, j], with $i > g (> b)$, which satisfies p. It immediately follows that $M,[a,b] \Vdash \langle L \rangle p$. $\square$

Appendix 2: Complete proof of Lemma 7

Lemma 7 $\langle L \rangle$ is not definable in $\mathsf {B E\overline{A E D}}$ relative to the class $\mathsf {Den}$.

Proof

Consider two interval models M and $M'$, defined as $M = M' = \langle \mathbb {I(R)}, V \rangle$, where $V(p) = \{[a,b] \mid a = f(b) \}$ and where f is the function defined at the beginning of Sect. 5.2. In addition, let $Z = \{ ([a,b],[a',b']) \mid a \sim f(b), a' \sim f(b') \text { where } \sim \in \{ <,=,> \} \}$ (see Fig. 2). It is immediate to check that $[-1,0] Z [0,1]$ (as $f(0) = -1$ and $f(1) = 0$), that $M,[-1,0] \Vdash \langle L \rangle p$ (as $M, [0.5, 2] \Vdash p$ because $f(2) = 0.5$) and that $M',[0,1] \Vdash \lnot \langle L \rangle p$ (as no interval [c, d], with $c>1$, satisfies p because c is not in the image of f for each $c>1$). Now, in order to show that Z is a $\mathsf {B E\overline{A E D}}$-bisimulation, consider a pair $([a,b],[a',b'])$ of Z-related intervals. The following chain of double implications holds:

$$\begin{aligned} M,[a,b] \Vdash p \Leftrightarrow a = f(b)\Leftrightarrow a' = f(b')\Leftrightarrow M',[a',b'] \Vdash p. \end{aligned}$$

This implies that the local condition holds. As for the forward condition, consider three intervals [a, b], $[a',b']$, and [c, d] such that $[a,b]Z[a',b']$ and $[a,b]R_X[c,d]$ for some $X \in \{ B, E, \overline{A}, \overline{E}, \overline{D} \}$. We need to exhibit an interval $[c',d']$ such that $[a',b'] R_X [c',d']$ and $[c,d] Z [c',d']$. We distinguish three cases.

• If $a > f(b)$ and $a' > f(b')$, then we distinguish the following sub-cases.

  • If $X = B$, then [c, d] is such that $a = c < d < b$. By the monotonicity of f, we have that $f(d) < f(b) < a = c$. Moreover, by the monotonicity of f, for every interval $[c',d']$, with $[a',b']R_B[c',d']$, $f(d') < c'$ holds, and thus $[c,d] Z [c',d']$.

  • If $X = E$, then [c, d] is such that $a < c < b = d$. Thus, $f(d) = f(b) < a < c$. For every interval $[c',d']$, with $[a',b']R_E[c',d']$, $f(d') < c'$ holds, and thus $[c,d] Z [c',d']$.

  • If $X = \overline{A}$, then [c, d] is such that $c < d = a$. Now, if $c < f(d) = f(a)$, then, by the definition of f and Lemma 6, there exists a point $c'$ such that $c' < f(a') < a'$. Thus, the interval $[c',d']$, with $d' = a'$, is such that $[a',b'] R_{\overline{A}} [c',d']$ and $[c,d] Z [c',d']$. If $c = f(d) = f(a)$, then take $c' =$ $f(a')$ $< a'$. The interval $[c',d']$, with $d' = a'$, is such that $[a',b'] R_{\overline{A}} [c',d']$ and $[c,d] Z [c',d']$. If $c > f(d) = f(a)$, then, by the density of ${\mathbb {R}}$, the definition of f, and Lemma 6, there exists a point $c'$ such that $f(a') < c' < a'$. The interval $[c',d']$, with $d' = a'$, is such that $[a',b'] R_{\overline{A}} [c',d']$ and $[c,d] Z [c',d']$.

  • If $X = \overline{E}$, then [c, d] is such that $c < a < b = d$. There are three possibilities. If $c < f(d)$, then, by the definition of f, there exists a point $c'$ such that $c' < f(b') < a'$. Thus, the interval $[c',d']$, with $d' = b'$, is such that $[a',b'] R_{\overline{E}} [c',d']$ and $[c,d] Z [c',d']$. If $c = f(d)$, then the interval $[c',d']$, with $d' = b'$ and $c' = f(d')$, is such that $[a',b'] R_{\overline{E}} [c',d']$ and $[c,d] Z [c',d']$. If $c > f(d)$, then, by the density of ${\mathbb {R}}$, there exists a point $c'$ such that $f(b') < c' < a'$, and the interval $[c',d']$, with $d' = b'$, is such that $[a',b'] R_{\overline{E}} [c',d']$ and $[c,d] Z [c',d']$.

  • If $X = \overline{D}$, then [c, d] is such that $c < a < b < d$. If $c < f(d)$, then, take $c' = f(a')$ and any $d' > b'$. The interval $[c',d']$ is such that $[a',b'] R_{\overline{D}} [c',d']$ and $[c,d] Z [c',d']$. If $c = f(d)$ (resp., $c > f(d)$), then, by the density of ${\mathbb {R}}$ and the monotonicity and the surjectivity of f, there exist two points $c',d'$ such that $c' < a' < b' < d'$ and $c' = f(d')$ (resp., $c' > f(d')$). Thus, the interval $[c',d']$ is such that $[a',b'] R_{\overline{D}} [c',d']$ and $[c,d] Z [c',d']$.

• If $a < f(b)$ and $a' < f(b')$, then we distinguish the following sub-cases.

  • If $X = B$, then [c, d] is such that $a = c < d < b$. Now, if $c < f(d)$ (resp., $c = f(d)$, $c > f(d)$), then, by the density of ${\mathbb {R}}$ and by the monotonicity and the surjectivity of f, there exists a point $d'$ such that $a' < d' < b'$ and $a' < f(d')$ (resp., $a' = f(d')$, $a' > f(d')$). Thus, the interval $[c',d']$, with $c' = a'$, is such that $[a',b'] R_{B} [c',d']$ and $[c,d] Z [c',d']$.

  • If $X = E$, then [c, d] is such that $a < c < b = d$. Now, if $c < f(d)$ (resp., $c = f(d)$, $c > f(d)$), then, by the density of ${\mathbb {R}}$, there exists a point $c'$ such that $a' < c' < b'$ and $c' < f(b')$ (resp., $c' = f(b')$, $c' > f(b')$). Thus, the interval $[c',d']$, with $d' = b'$, is such that $[a',b'] R_{E} [c',d']$ and $[c,d] Z [c',d']$.

  • If $X = \overline{A}$, then the same argument of the case when $a > f(b)$ and $a' > f(b')$ (and $X = \overline{A}$) applies.

  • If $X = \overline{E}$, then [c, d] is such that $c < a < b = d$. Thus, $c < a < f(b) = f(d)$. For every interval $[c',d']$, with $[a',b']R_{\overline{E}}[c',d']$, it holds $c' < f(d')$, and thus $[c,d] Z [c',d']$.

  • If $X = \overline{D}$, then [c, d] is such that $c < a < b < d$. Thus, by the monotonicity of f, $c < a < f(b) < f(d)$ holds. For every interval $[c',d']$, with $[a',b']R_{\overline{D}}[c',d']$, it holds, by the monotonicity of f, that $c' < f(d')$, and thus $[c,d] Z [c',d']$.

• If $a = f(b)$ and $a' = f(b')$, then we distinguish the following sub-cases.

  • If $X = B$, then [c, d] is such that $a = c < d < b$. Thus, $f(d) < f(b) = a = c$ holds by the monotonicity of f. For every interval $[c',d']$, with $[a',b']R_{B}[c',d']$, by the monotonicity of f, we have that $f(d') < c'$, and thus $[c,d] Z [c',d']$.

  • If $X = E$, then [c, d] is such that $a < c < b = d$. Thus, $c > a = f(b) = f(d)$ holds. For every interval $[c',d']$, with $[a',b']R_{E}[c',d']$, we have that $c' > f(d')$, and thus $[c,d] Z [c',d']$.

  • If $X = \overline{A}$, then the same argument of the case when $a > f(b)$ and $a' > f(b')$ (and $X = \overline{A}$) applies.

  • If $X = \overline{E}$, then [c, d] is such that $c < a < b = d$. Thus, $c < a = f(b) = f(d)$. For every interval $[c',d']$, with $[a',b']R_{\overline{E}}[c',d']$, $c' < f(d')$ holds, and thus $[c,d] Z [c',d']$.

  • If $X = \overline{D}$, then [c, d] is such that $c < a < b < d$. Thus, $c < a = f(b) < f(d)$ holds by the monotonicity of f. For every interval $[c',d']$, with $[a',b']R_{\overline{D}}[c',d']$, by the monotonicity of f, we have that $c' < f(d')$, and thus $[c,d] Z [c',d']$.

Since Z is symmetric, the backward condition immediately follows from Proposition 2. Therefore, Z is a $\mathsf {B E \overline{A E D}}$-bisimulation that violates $\langle L \rangle$, hence the thesis. $\square$

Appendix 3: Complete proof of Lemma 8

Lemma 8 $\langle L \rangle$ is not definable in $\mathsf {O \overline{B E O}}$ relative to the class $\mathsf {Den}$.

Proof

Consider the two interval models M and $M'$, defined as $M = M' = \langle \mathbb {I(R)}, V \rangle$, where $V(p) = \{[-a,a] \mid a \in {\mathbb {R}} \}$ (observe that no interval [c, d], with $c \ge 0$, satisfies p). Moreover, let $Z = \{ ([a,b],[a',b']) \mid -a \sim b \text { and } -a' \sim b' \text { for some } \sim \in \{ <, =, > \} \}$ (see Fig. 3). It is immediate to check that $[-4,-2] Z [-4,2]$, that $M,[-4,-2] \Vdash \langle L \rangle p$ (as $M, [-1, 1] \Vdash p$) and that $M',[-4,2] \Vdash \lnot \langle L \rangle p$ (as no interval [c, d], with $c>0$, satisfies p). In order to complete the proof for this fragment, we now proceed to show that Z is an $\mathsf {O\overline{B E O}}$-bisimulation. To this end, consider a pair $([a,b],[a',b'])$ of Z-related intervals. The following chain of equivalences hold:

$$\begin{aligned} M,[a,b] \Vdash p \Leftrightarrow -a = b\Leftrightarrow -a' = b' \Leftrightarrow M,[a',b'] \Vdash p. \end{aligned}$$

This implies that the local condition is satisfied. As for the forward condition, consider three intervals [a, b], $[a',b']$, and [c, d] such that $[a,b]Z[a',b']$ and $[a,b]R_X[c,d]$ for some $X \in \{ O, \overline{B}, \overline{E}, \overline{O} \}$. We need to exhibit an interval $[c',d']$ such that $[a',b'] R_X [c',d']$ and $[c,d] Z [c',d']$. We distinguish three cases.

• If $-a > b$ and $-a' > b'$, then, as a preliminary step, we show that the following facts hold: $(i)\, a < 0$ and $a' < 0$; $(ii)\,|a| > |b|$ and $|a'| > |b'|$. We only show the proofs for $a < 0$ and $|a| > |b|$ and we omit the ones for $a' < 0$ and $|a'| > |b|$, which are analogous. As for the former claim above, it is enough to observe that, if $a \ge 0$, then $a \ge 0 \ge -a > b$, which implies $b < a$, leading to a contradiction with the fact that [a, b] is an interval (thus $a < b$). Notice that, as an immediate consequence, we have that $|a| = -a$ holds. As for the latter claim above, firstly we suppose, by contradiction, that $|a| = |b|$ holds. Then, $-a = |a| = |b|$ holds and this implies either $b = -a$, contradicting the hypothesis that $-a > b$, or $b = a$, contradicting the fact that [a, b] is an interval. Secondly, we suppose, again by contradiction, that $|a| < |b|$ holds. Then, by the former claim, we have that $0 < -a = |a| < |b|$ holds, which implies $b \ne 0$. Now, we show that both $b < 0$ and $b > 0$ lead to a contradiction. If $b < 0$, then $|b| = -b$, and thus it holds $-a < -b$, which amounts to $a > b$, contradicting the fact that [a, b] is an interval. If $b > 0$, then $|b| = b$, and thus $-a < b$ holds, which contradicts the hypothesis that $-a > b$. This proves the two claims above. Now, we distinguish the following sub-cases.

  • If $X = O$, then [c, d] is such that $a < c < b < d$. We distinguish the following cases.

    • If $-c > d$, then take some $c'$ such that $a' < c' < -|b'| < 0$ (notice also that $c' < -|b'| \le b'$ trivially holds), and $d'$ such that $b' < d' < |c'| = -c'$ (the existence of such points $c',d'$ is guaranteed by the density of ${\mathbb {R}}$). The interval $[c',d']$ is such that $[a',b'] R_O [c',d']$ and $[c,d] Z [c',d']$.

    • If $-c = d$, then take some $c'$ such that $a' < c' < -|b'| < 0$, and $d' = -c'$ (the existence of such a point $c'$ is guaranteed by the density of ${\mathbb {R}}$). The interval $[c',d']$ is such that $[a',b'] R_O [c',d']$ and $[c,d] Z [c',d']$.

    • If $-c < d$, then take $c'$ such that $a' < c' < -|b'| < 0$, and any $d' > -c'$ (the existence of such a point $c'$ is guaranteed by the density of ${\mathbb {R}}$). The interval $[c',d']$ is such that $[a',b'] R_O [c',d']$ and $[c,d] Z [c',d']$.

  • If $X = \overline{B}$, then [c, d] is such that $a = c < b < d$. We distinguish the cases below.

    • If $-c > d$, then take $c' = a'$ and $d'$ such that $b' < d' < -a' = -c'$ (the existence of such a point $d'$ is guaranteed by the density of ${\mathbb {R}}$). The interval $[c',d']$ is such that $[a',b'] R_{\overline{B}} [c',d']$ and $[c,d] Z [c',d']$.

    • If $-c = d$, then take $c' = a'$ and $d' = -c' (= -a' > b')$. The interval $[c',d']$ is such that $[a',b'] R_{\overline{B}} [c',d']$ and $[c,d] Z [c',d']$.

    • If $-c < d$, then take $c' = a'$ and any $d' > -c' (= -a' > b')$. The interval $[c',d']$ is such that $[a',b'] R_{\overline{B}} [c',d']$ and $[c,d] Z [c',d']$.

  • If $X = \overline{E}$, then [c, d] is such that $c < a < b = d$. Notice that $|c| = -c > -a = |a|$ holds, because $c < a < 0$. Thus $-c > -a > b = d$ also holds. Then, take $d' = b'$ and any $c' < a'$. We have that $-c' > -a' > b' = d'$. The interval $[c',d']$ is therefore such that $[a',b'] R_{\overline{E}} [c',d']$ and $[c,d] Z [c',d']$.

  • If $X = \overline{O}$, then [c, d] is such that $c < a < d < b$. Notice that $|c| = -c > -a = |a|$ holds, because $c < a < 0$. Thus $-c > -a > b > d$ also holds. Then, take some $d'$ such that $a' < d' < b'$ and any $c' < a'$ (the existence of such a point $d'$ is guaranteed by the density of ${\mathbb {R}}$). Thus, it holds $-c' > -a' > b' > d'$. The interval $[c',d']$ is therefore such that $[a',b'] R_{\overline{O}} [c',d']$ and $[c,d] Z [c',d']$.

• If $-a = b$ and $-a' = b'$, then we have that $a < 0$ (resp., $a' < 0$) and $b > 0$ (resp., $b' > 0$). Indeed, if $a \ge 0$ held, then $b = -a \le 0 \le a$ would also hold, contradicting the fact that [a, b] is an interval (and thus $b > a$). From $a < 0$ and $-a = b$, it immediately follows that $b > 0$. The facts that $a' < 0$ and $b' > 0$ can be shown analogously. Notice also that, from $-a = b$ and $-a' = b'$, it follows that $|a| = |b|$ and $|a'| = |b'|$. Now, we distinguish the following sub-cases.

  • If $X = O$, then [c, d] is such that $a < c < b < d$. Notice that $-c \le |c| < |a| = |b| = b < d$ holds. Then, take $c' = 0$ and any $d' > b' (> 0)$. We have that $-c' < d'$. The interval $[c',d']$ is such that $[a',b'] R_{O} [c',d']$ and $[c,d] Z [c',d']$.

  • If $X = \overline{B}$, then [c, d] is such that $a = c < b < d$. Notice that $-c = - a = b < d$ holds. Then, take $c' = a'$ and any $d' > b'$. We have that $-c' = -a' = b' < d'$. The interval $[c',d']$ is such that $[a',b'] R_{\overline{B}} [c',d']$ and $[c,d] Z [c',d']$.

  • If $X = \overline{E}$, then [c, d] is such that $c < a < b = d$. Notice that $|c| = -c > -a = |a|$ holds, because $c < a < 0$. Thus $-c > -a = b = d$ also holds. Then, take $d' = b'$ and any $c' < a'$. We have that $-c' > -a' = b' = d'$. The interval $[c',d']$ is such that $[a',b'] R_{\overline{E}} [c',d']$ and $[c,d] Z [c',d']$.

  • If $X = \overline{O}$, then [c, d] is such that $c < a < d < b$. Notice that $|c| = -c > -a = |a|$ holds, because $c < a < 0$. Thus $-c > -a = b > d$ also holds. Then, take $d' = 0$ and any $c' < a' (< 0)$. We have that $-c' > d'$. The interval $[c',d']$ is such that $[a',b'] R_{\overline{O}} [c',d']$ and $[c,d] Z [c',d']$.

• If $-a < b$ and $-a' < b'$, then the following facts hold: $(i)\, b>0$ (otherwise, $-a < b \le 0$ would hold, which implies $a > 0 \ge b$, contradicting the fact that [a, b] is an interval), $(ii)\,|b| = b$ (this follows directly from $b>0$), and $(iii)\,|a| < |b|$ (otherwise, $|a| \ge |b| = b$ would hold, which implies either $a \ge b$, contradicting the fact that [a, b] is an interval, or $-a \ge b$, contradicting the hypothesis that $-a < b$). Now, we distinguish the following sub-cases.

  • If $X = \overline{O}$, then [c, d] is such that $c < a < d < b$. We distinguish the cases below.

    • If $-c < d$, then take some $d'$ and $c'$ such that $|a'| < d' < |b'| = b'$ and $-d' < c' < |a'| = -c$ (the existence of points $c', d'$ is guaranteed by the density of ${\mathbb {R}}$). The interval $[c',d']$ is such that $[a',b'] R_{\overline{O}} [c',d']$ and $[c,d] Z [c',d']$.

    • If $-c = d$, then take some $d'$ such that $|a'| < d' < |b'| = b'$ and $c' = -d'$ (the existence of such a point $d'$ is guaranteed by the density of ${\mathbb {R}}$). The interval $[c',d']$ is such that $[a',b'] R_{\overline{O}} [c',d']$ and $[c,d] Z [c',d']$.

    • If $-c > d$, then take some $d'$ and $c'$ such that $|a'| < d' < |b'| = b'$ and $c' < -d'$ (the existence of points $c', d'$ is guaranteed by the left-unboundedness and the density of ${\mathbb {R}}$, respectively). The interval $[c',d']$ is such that $[a',b'] R_{\overline{O}} [c',d']$ and $[c,d] Z [c',d']$.

  • If $X = \overline{E}$, then [c, d] is such that $c < a < b = d$. We distinguish the following cases.

    • If $-c < d$, then take $d' = b'$ and some $c'$ such that $-d' < c' < a'$ (the existence of such a point $c'$ is guaranteed by the density of ${\mathbb {R}}$). The interval $[c',d']$ is such that $[a',b'] R_{\overline{E}} [c',d']$ and $[c,d] Z [c',d']$.

    • If $-c = d$, then take $d' = b'$ and $c' = -d' (= -b' < a')$. The interval $[c',d']$ is such that $[a',b'] R_{\overline{E}} [c',d']$ and $[c,d] Z [c',d']$.

    • If $-c > d$, then take $d' = b'$ and any $c' < -d' (= -b' < a')$. The interval $[c',d']$ is such that $[a',b'] R_{\overline{E}} [c',d']$ and $[c,d] Z [c',d']$.

  • If $X = \overline{B}$, then [c, d] is such that $a = c < b < d$. Notice that $-d < -b < a = c$. Then, take $c' = a'$ and any $d' > b'$. It holds that $c' = a' > -b' > -d'$. The interval $[c',d']$ is such that $[a',b'] R_{\overline{B}} [c',d']$ and $[c,d] Z [c',d']$.

  • If $X = O$, then [c, d] is such that $a < c < b < d$. Notice that $-d < -b < a < c$. Then, take some $c'$ such that $a' < c' < b'$ (the existence of such a point $c'$ is guaranteed by the density of ${\mathbb {R}}$) and any $d' > b'$. It holds that $c' > a' > -b' > -d'$. The interval $[c',d']$ is such that $[a',b'] R_{O} [c',d']$ and $[c,d] Z [c',d']$.

Since the relation Z is symmetric, by Proposition 2 we have that the backward condition is verified, too. Therefore, Z is an $\mathsf {O \overline{B E O}}$-bisimulation that violates $\langle L \rangle$, and the thesis follows. $\square$

Appendix 4: Complete proof of Lemma 9

Lemma 9 $\langle E \rangle$ is not definable in $\mathsf {A B D O\overline{A B E}}$ relative to the classes $\mathsf {Lin}$ and $\mathsf {Den}$.

Proof

Let $M_1 = \langle {\mathbb {I}}({\mathbb {R}}), V_1 \rangle$ and $M_2 = \langle {\mathbb {I}}({\mathbb {R}}), V_2 \rangle$, where

• p is the only proposition letter of the language,

• the valuation function $V_1: {{\mathcal {AP}}} \rightarrow 2^{{\mathbb {I}}({\mathbb {R}})}$ is defined as: $[x,y] \in V_1(p) \Leftrightarrow x \in {\mathbb {Q}}$ if and only if $y \in {\mathbb {Q}}$, and

• the valuation function $V_2: {{\mathcal {AP}}} \rightarrow 2^{{\mathbb {I}}({\mathbb {R}})}$ is given by: $[w,z] \in V_2(p) \Leftrightarrow w \in {\mathbb {Q}}$ if and only if $z \in {\mathbb {Q}}$, and $([0,3],[w,z]) \notin R_E$.

Moreover, let Z be a relation between (intervals of) $M_1$ and $M_2$ defined as follows: $[x,y]Z[w,z] {\Leftrightarrow } [x,y] \in V_1(p)$ if and only if $[w,z] \in V_2(p)$. It is easy to verify that [0, 3]Z[0, 3], $M_1,[0,3] \Vdash \langle E \rangle p$, but $M_2,[0,3] \Vdash \lnot \langle E \rangle p$. We show now that Z is an $\mathsf {A B D O \overline{A B E}}$-bisimulation between $M_1$ and $M_2$. The local condition immediately follows from the definition. As for the forward condition, it can be checked as follows. Let [x, y] and [w, z] be two Z-related intervals, and let us assume that $[x,y] R_X [x',y']$ holds for some $X \in \{ A, B, D, O, \overline{A}, \overline{B}, \overline{E} \}$. We have to exhibit an interval $[w',z']$ such that $[x',y']$ and $[w',z']$ are Z-related, and [w, z] and $[w',z']$ are $R_X$-related. We proceed by considering each case in turn.

• If $X=A$, then $y = x'$. We can always find a point $z'$ such that $z' > \max \{ 3, z \}$ and $z' \in {\mathbb {Q}}$ if and only if $y' \in {\mathbb {Q}}$ (since both ${\mathbb {Q}}$ and $\overline{\mathbb {Q}}$ are right-unbounded). This implies that $[x',y']$ and $[z,z']$ are Z-related. Since [w, z] and $[z,z']$ are obviously $R_A$-related, we have the thesis.

• If $X=B$, the argument is similar to the previous one, but, in this case, the density of ${\mathbb {Q}}$ and $\overline{\mathbb {Q}}$ plays a major role. We choose a point $z'$ such that $w < z' < z$, $z' \ne 3$, and $z' \in {\mathbb {Q}}$ if and only if $y' \in {\mathbb {Q}}$. The interval $[w,z']$ is such that $[x',y']$ and $[w,z']$ are Z-related, and [w, z] and $[w,z']$ are $R_B$-related.

• If $X = D$, it suffices to choose two points $w'$ and $z'$ such that $w < w' < z' < z$, $z' \ne 3$, $w'$ belongs to ${\mathbb {Q}}$ if and only if $x'$ does, and $z'$ belongs to ${\mathbb {Q}}$ if and only if $y'$ does. The existence of such points is guaranteed by the density of ${\mathbb {Q}}$ and $\overline{\mathbb {Q}}$. The interval $[w',z']$ is such that $[w,z] R_D [w',z']$ and $[x',y'] Z [w',z']$.

• If $X=O$, then $w'$ and $z'$ are required to be such that $w < w' < z < z'$, and both density and right-unboundedness of ${\mathbb {Q}}$ and $\overline{\mathbb {Q}}$ must be exploited in order to choose a point $w'$ such that $w < w' < z$ and $w' \in {\mathbb {Q}}$ if and only if $x'$ does, and a point $z'$ such that $z' > \max \{ 3, z \}$ and $z'$ belongs to ${\mathbb {Q}}$ if and only if $y'$ does. The interval $[w',z']$ is such that $[w,z] R_O [w',z']$ and $[x',y'] Z [w',z']$.

• If $X=\overline{A}$, then there exists a point $w''$ such that $w'' < \min \{ 0, w \}$ and $w'' \in {\mathbb {Q}}$ if and only if w does (and thus $M', [w'',w] \Vdash p$) and there exists a point $w'''$ such that $w''' < w$ and $w''' \in {\mathbb {Q}}$ if and only if $w \in \overline{{\mathbb {Q}}}$ (and thus $M', [w''',w] \Vdash \lnot p$). We choose $w' = w''$ if $M,[x',y'] \models p$, otherwise we choose $w' = w'''$. The interval $[w',w]$ is such that $[w,z] R_{\overline{A}} [w',w]$ and $[x',y'] Z [w',w]$.

• If $X=\overline{B}$, then there exists a point $z''$ such that $z'' > \max \{ 3, z \}$ and $z'' \in {\mathbb {Q}}$ if and only if w does (and thus $M', [w,z''] \Vdash p$) and there exists a point $z'''$ such that $z''' > z$ and $z''' \in {\mathbb {Q}}$ if and only if $w \in \overline{{\mathbb {Q}}}$ (and thus $M', [w,z'''] \Vdash \lnot p$). We choose $z' = z''$ if $M,[x',y'] \models p$, otherwise we choose $z' = z'''$. The interval $[w,z']$ is such that $[w,z] R_{\overline{B}} [w,z']$ and $[x',y'] Z [w,z']$.

• If $X=\overline{E}$, then there exists a point $w''$ such that $w'' < \min \{ 0, w \}$ and $w'' \in {\mathbb {Q}}$ if and only if z does (and thus $M', [w'',z] \Vdash p$) and there exists a point $w'''$ such that $w''' < w$ and $w''' \in {\mathbb {Q}}$ if and only if $z \in \overline{{\mathbb {Q}}}$ (and thus $M', [w''',z] \Vdash \lnot p$). We choose $w' = w''$ if $M,[x',y'] \models p$, otherwise we choose $w' = w'''$. The interval $[w',z]$ is such that $[w,z] R_{\overline{E}} [w',z]$ and $[x',y'] Z [w',z]$.

The backward condition follows from Proposition 2. Therefore, Z is an $\mathsf {A B D O\overline{A B E}}$-bisimulation that violates $\langle E \rangle$, hence the thesis. $\square$

Appendix 5: Complete proof of Lemma 11

Lemma 11 $\langle A \rangle$ is not definable in $\mathsf {B E\overline{A B E}}$ relative to the classes $\mathsf {Lin}$ and $\mathsf {Den}$.

Proof

Let $M_1 = \langle {\mathbb {I}}({\mathbb {R}}), V_1 \rangle$ and $M_2 = \langle {\mathbb {I}}({\mathbb {R}}), V_2 \rangle$ be two models built on the only proposition letter p. In order to define the valuation functions $V_1$ and $V_2$, we make use of two partitions of the set ${\mathbb {R}}$, one for $M_1$ and the other for $M_2$, each of them consisting of four sets that are dense in ${\mathbb {R}}$. Formally, for $j = 1, 2$ and $i = 1, \ldots , 4$, let ${\mathbb {R}}_j^i$ be dense in ${\mathbb {R}}$. Moreover, for $j = 1, 2$, let ${\mathbb {R}} = \bigcup _{i=1}^4 {\mathbb {R}}_j^i$ and ${\mathbb {R}}_j^i \cap {\mathbb {R}}_j^{i'} = \emptyset$ for each $i,i' \in \{ 1,2,3,4 \}$, with $i \ne i'$. For the sake of simplicity, we impose the two partitions to be equal and thus we can safely omit the subscript, that is, ${\mathbb {R}}_1^{i} = {\mathbb {R}}_2^{i} = {\mathbb {R}}^{i}$ for each $i \in \{ 1,2,3,4 \}$. Thanks to this condition, the bisimulation relation Z, that we define below, is symmetric. We force points in ${\mathbb {R}}^1$ (resp., ${\mathbb {R}}^2$, ${\mathbb {R}}^3$, ${\mathbb {R}}^4$) to behave in the same way with respect to the truth of $p / \lnot p$ over the intervals they initiate and terminate by imposing the following constraints. For $j = 1,2$:

$$\begin{aligned}&\forall x, y \text { } (\text {if } x \in {\mathbb {R}}^1, \text { then } M_j,[x,y] \Vdash \lnot p); \\&\forall x, y \text { } (\text {if } x \in {\mathbb {R}}^2, \text { then } M_j,[x,y] \Vdash \lnot p); \\&\forall x, y \text { } (\text {if } x \in {\mathbb {R}}^3, \text { then } (M_j,[x,y] \Vdash p \text { iff } y \in {\mathbb {R}}^1 \cup {\mathbb {R}}^3)); \\&\forall x, y \text { } (\text {if } x \in {\mathbb {R}}^4, \text { then } (M_j,[x,y] \Vdash p \text { iff } y \in {\mathbb {R}}^2 \cup {\mathbb {R}}^4)). \end{aligned}$$

It can be easily shown that, from the given constraints, it immediately follows that:

$$\begin{aligned}&\forall x, y \text { } (\text {if } y \in {\mathbb {R}}^1, \text { then } (M_j,[x,y] \Vdash p \text { iff } x \in {\mathbb {R}}^3)); \\&\forall x, y \text { } (\text {if } y \in {\mathbb {R}}^2, \text { then } (M_j,[x,y] \Vdash p \text { iff } x \in {\mathbb {R}}^4)); \\&\forall x, y \text { } (\text {if } y \in {\mathbb {R}}^3, \text { then } (M_j,[x,y] \Vdash p \text { iff } x \in {\mathbb {R}}^3)); \\&\forall x, y \text { } (\text {if } y \in {\mathbb {R}}^4, \text { then } (M_j,[x,y] \Vdash p \text { iff } x \in {\mathbb {R}}^4)). \end{aligned}$$

The above constraints together induce the following definition of the valuation functions $V_j(p) : {{\mathcal {AP}}} \rightarrow 2^{{\mathbb {I}}({\mathbb {R}})}$:

$$\begin{aligned}{}[x,y] \in V_j(p) \Leftrightarrow (x \in {\mathbb {R}}^3 \wedge y \in {\mathbb {R}}^1 \cup {\mathbb {R}}^3) \vee (x \in {\mathbb {R}}^4 \wedge y \in {\mathbb {R}}^2 \cup {\mathbb {R}}^4). \end{aligned}$$

Now, let Z be the relation between (intervals of) $M_1$ and $M_2$ defined as follows. Two intervals [x, y] and [w, z] are Z-related if and only if at least one of the following conditions holds:

• 1. $x \in {\mathbb {R}}^1 \cup {\mathbb {R}}^2 \text { and } w \in {\mathbb {R}}^1 \cup {\mathbb {R}}^2;$

• 2. $x \in {\mathbb {R}}^3 \text {, } w \in {\mathbb {R}}^3, \text { and } (y \in {\mathbb {R}}^1 \cup {\mathbb {R}}^3 \text { iff } z \in {\mathbb {R}}^1 \cup {\mathbb {R}}^3);$

• 3. $x \in {\mathbb {R}}^3 \text {, } w \in {\mathbb {R}}^4, \text { and } (y \in {\mathbb {R}}^1 \cup {\mathbb {R}}^3 \text { iff } z \in {\mathbb {R}}^2 \cup {\mathbb {R}}^4);$

• 4. $x \in {\mathbb {R}}^4 \text {, } w \in {\mathbb {R}}^3, \text { and } (y \in {\mathbb {R}}^2 \cup {\mathbb {R}}^4 \text { iff } z \in {\mathbb {R}}^1 \cup {\mathbb {R}}^3);$

• 5. $x \in {\mathbb {R}}^4 \text {, } w \in {\mathbb {R}}^4, \text { and } (y \in {\mathbb {R}}^2 \cup {\mathbb {R}}^4 \text { iff } z \in {\mathbb {R}}^2 \cup {\mathbb {R}}^4).$

It is worth pointing out that two intervals [x, y] and [w, z] that are Z-related are such that if, for instance, both x and w belong to ${\mathbb {R}}^3$ (second clause), then either y and z both occur in odd-numbered partitions or they both occur in even-numbered partitions. Moreover, since the two partitions are equal, Z is symmetric.

Let us consider now two intervals [x, y] and [w, z] such that $x \in {\mathbb {R}}^1$, $w \in {\mathbb {R}}^1$, $y \in {\mathbb {R}}^3$, and $z \in {\mathbb {R}}^1$. By definition of Z, [x, y] and [w, z] are Z-related, and by definition of $V_1$ and $V_2$, there exists $y'> y$ such that $M_1,[y,y'] \Vdash p$, but there is no $z' > z$ such that $M_2,[z,z'] \Vdash p$. Thus, $M_1,[x,y] \Vdash \langle A \rangle p$ and $M_2,[w,z] \Vdash \lnot \langle A \rangle p$ hold.

To complete the proof, it suffices to show that the relation Z is a $\mathsf {B E \overline{A B E}}$-bisimulation. It can be easily checked that every pair ([x, y], [w, z]) of Z-related intervals is such that either $[x,y] \in V_1(p)$ and $[w,z] \in V_2(p)$, or $[x,y] \not \in V_1(p)$ and $[w,z] \not \in V_2(p)$.

In order to verify the forward condition, let [x, y] and [w, z] be two Z-related intervals. For each modality $\langle X \rangle$ of the language and each interval $[x',y']$ such that $[x,y] R_X [x',y']$, we have to exhibit an interval $[w',z']$ such that $[x',y']Z[w',z']$ and $[w,z]R_X[w',z']$. We proceed by considering each case in turn.

• Let $X=B$. If $x \in {\mathbb {R}}^1 \cup {\mathbb {R}}^2$ and $w \in {\mathbb {R}}^1 \cup {\mathbb {R}}^2$, then for any $z'$ such that $w < z' < z$, both $[x,y']Z[w,z']$ and $[w,z] R_B [w,z']$ hold. If $x \in {\mathbb {R}}^i$ and $w \in {\mathbb {R}}^i$, for some $i \in \{3,4\}$, and $y' \in {\mathbb {R}}^k$, for some $k \in \{1,2,3,4\}$, then for any $z'$ such that $w < z' < z$ and $z' \in {\mathbb {R}}^k$, it holds that $[x,y']Z[w,z']$ and $[w,z] R_B [w,z']$ (the existence of $z'$ is guaranteed by density of ${\mathbb {R}}^k$ in ${\mathbb {R}}$). Finally, if $x \in {\mathbb {R}}^i$ and $w \in {\mathbb {R}}^{i'}$ for $i,i' \in \{3,4\}$, with $i \ne i'$, and, in addition, $y' \in {\mathbb {R}}^1 \cup {\mathbb {R}}^3$ (resp., $y' \in {\mathbb {R}}^2 \cup {\mathbb {R}}^4$), then for any $z'$ such that $w < z' < z$ and $z' \in {\mathbb {R}}^2 \cup {\mathbb {R}}^4$ (resp., $z' \in {\mathbb {R}}^1 \cup {\mathbb {R}}^3$), it holds that $[x,y']Z[w,z']$ and $[w,z] R_B [w,z']$ (density of ${\mathbb {R}}^2$ and ${\mathbb {R}}^4$, resp., ${\mathbb {R}}^1$ and ${\mathbb {R}}^3$, in ${\mathbb {R}}$ is used).

• Let $X=E$. As $[x,y] R_E [x',y']$, we have that $y = y'$. We distinguish the following cases, where we tacitly use the density of the relevant sets in ${\mathbb {R}}$: (i) if $x' \in {\mathbb {R}}^1 \cup {\mathbb {R}}^2$, then we choose $w'$ such that $w < w' < z$ and $w' \in {\mathbb {R}}^1$; (ii) if either $x' \in {\mathbb {R}}^3$ and $y \in {\mathbb {R}}^1 \cup {\mathbb {R}}^3$, or $x' \in {\mathbb {R}}^4$ and $y \in {\mathbb {R}}^2 \cup {\mathbb {R}}^4$, then we choose $w'$ such that $w < w' < z$ and either $w' \in {\mathbb {R}}^3$ (if $z \in {\mathbb {R}}^1 \cup {\mathbb {R}}^3$), or $w' \in {\mathbb {R}}^4$ (if $z \in {\mathbb {R}}^2 \cup {\mathbb {R}}^4$); (iii) if either $x' \in {\mathbb {R}}^3$ and $y \in {\mathbb {R}}^2 \cup {\mathbb {R}}^4$, or $x' \in {\mathbb {R}}^4$ and $y \in {\mathbb {R}}^1 \cup {\mathbb {R}}^3$, then we choose $w'$ such that $w < w' < z$ and either $w' \in {\mathbb {R}}^3$ (if $z \in {\mathbb {R}}^2 \cup {\mathbb {R}}^4$), or $w' \in {\mathbb {R}}^4$ (if $z \in {\mathbb {R}}^1 \cup {\mathbb {R}}^3$). In all cases, we have that $[x',y]Z[w',z]$ and $[w,z] R_E [w',z]$.

• Let $X=\overline{A}$. As $[x,y] R_{\overline{A}} [x',y']$, we have that $x = y'$. We distinguish the following cases: (i) if $x' \in {\mathbb {R}}^1 \cup {\mathbb {R}}^2$, then we choose $w'$ such that $w' < w$ and $w' \in {\mathbb {R}}^1$; (ii) if either $x' \in {\mathbb {R}}^3$ and $x \in {\mathbb {R}}^1 \cup {\mathbb {R}}^3$, or $x' \in {\mathbb {R}}^4$ and $x \in {\mathbb {R}}^2 \cup {\mathbb {R}}^4$, then we choose $w'$ such that $w' < w$ and either $w' \in {\mathbb {R}}^3$ (if $w \in {\mathbb {R}}^1 \cup {\mathbb {R}}^3$), or $w' \in {\mathbb {R}}^4$ (if $w \in {\mathbb {R}}^2 \cup {\mathbb {R}}^4$); (iii) if either $x' \in {\mathbb {R}}^3$ and $x \in {\mathbb {R}}^2 \cup {\mathbb {R}}^4$, or $x' \in {\mathbb {R}}^4$ and $x \in {\mathbb {R}}^1 \cup {\mathbb {R}}^3$, then we choose $w'$ such that $w' < w$ and either $w' \in {\mathbb {R}}^3$ (if $w \in {\mathbb {R}}^2 \cup {\mathbb {R}}^4$), or $w' \in {\mathbb {R}}^4$ (if $w \in {\mathbb {R}}^1 \cup {\mathbb {R}}^3$). In all cases, we have that $[x',x]Z[w',w]$ and $[w,z] R_{\overline{A}} [w',w]$.

• Let $X=\overline{B}$. Since $[x,y] R_{\overline{B}} [x',y']$, we have that $x = x'$. If $x \in {\mathbb {R}}^1 \cup {\mathbb {R}}^2$ and $w \in {\mathbb {R}}^1 \cup {\mathbb {R}}^2$, then for any $z' > z$, both $[x,y']Z[w,z']$ and $[w,z] R_{\overline{B}} [w,z']$ hold. If $x \in {\mathbb {R}}^i$ and $w \in {\mathbb {R}}^i$, for some $i \in \{3,4\}$, and $y' \in {\mathbb {R}}^k$, for some $k \in \{1,2,3,4\}$, then for any $z' > z$ such that $z' \in {\mathbb {R}}^k$, it holds that $[x,y']Z[w,z']$ and $[w,z] R_{\overline{B}} [w,z']$ (the existence of $z'$ is guaranteed by density of ${\mathbb {R}}^k$ in ${\mathbb {R}}$). Finally, if $x \in {\mathbb {R}}^i$ and $w \in {\mathbb {R}}^{i'}$ for $i,i' \in \{3,4\}$, with $i \ne i'$, and, in addition, $y' \in {\mathbb {R}}^1 \cup {\mathbb {R}}^3$ (resp., $y' \in {\mathbb {R}}^2 \cup {\mathbb {R}}^4$), then for any $z' > z$ such that $z' \in {\mathbb {R}}^2 \cup {\mathbb {R}}^4$ (resp., $z' \in {\mathbb {R}}^1 \cup {\mathbb {R}}^3$), it holds that $[x,y']Z[w,z']$ and $[w,z] R_{\overline{B}} [w,z']$ (density of ${\mathbb {R}}^2$ and ${\mathbb {R}}^4$, resp., ${\mathbb {R}}^1$ and ${\mathbb {R}}^3$, in ${\mathbb {R}}$ is used).

• Let $X=\overline{E}$. Since $[x,y] R_{\overline{E}} [x',y']$, we have that $y = y'$. We distinguish the following cases: (i) if $x' \in {\mathbb {R}}^1 \cup {\mathbb {R}}^2$, then we choose $w'$ such that $w' < w$ and $w' \in {\mathbb {R}}^1$; (ii) if either $x' \in {\mathbb {R}}^3$ and $y \in {\mathbb {R}}^1 \cup {\mathbb {R}}^3$, or $x' \in {\mathbb {R}}^4$ and $y \in {\mathbb {R}}^2 \cup {\mathbb {R}}^4$, then we choose $w'$ such that $w' < w$ and either $w' \in {\mathbb {R}}^3$ (if $z \in {\mathbb {R}}^1 \cup {\mathbb {R}}^3$), or $w' \in {\mathbb {R}}^4$ (if $z \in {\mathbb {R}}^2 \cup {\mathbb {R}}^4$); (iii) if either $x' \in {\mathbb {R}}^3$ and $y \in {\mathbb {R}}^2 \cup {\mathbb {R}}^4$, or $x' \in {\mathbb {R}}^4$ and $y \in {\mathbb {R}}^1 \cup {\mathbb {R}}^3$, then we choose $w'$ such that $w' < w$ and either $w' \in {\mathbb {R}}^3$ (if $z \in {\mathbb {R}}^2 \cup {\mathbb {R}}^4$) or $w' \in {\mathbb {R}}^4$ (if $z \in {\mathbb {R}}^1 \cup {\mathbb {R}}^3$). In all cases, we have that $[x',y]Z[w',z]$ and $[w,z] R_{\overline{E}} [w',z]$.

The backward condition follows from the forward one by Proposition 2. Therefore, Z is a $\mathsf {B E\overline{A B E}}$-bisimulation that violates $\langle A \rangle$, and the thesis immediately follows. $\square$

Appendix 6: Complete proof of Lemma 12

Lemma 12 $\langle D \rangle$ is not definable in $\mathsf {A B O\overline{A B E}}$ relative to the classes $\mathsf {Lin}$ and $\mathsf {Den}$.

Proof

As a first step, we define a pair of functions that will be used in the definition of the models involved in the bisimulation relation Z. Let ${\mathcal {P}}({\mathbb {Q}})= \{{\mathbb {Q}}_q \mid q \in {\mathbb {Q}}\}$ and ${\mathcal {P}}(\overline{\mathbb {Q}}) = \{\overline{\mathbb {Q}}_q \mid q \in {\mathbb {Q}}\}$ be countably infinite partitions of ${\mathbb {Q}}$ and $\overline{\mathbb {Q}}$, respectively, such that for every $q \in {\mathbb {Q}}$, both ${\mathbb {Q}}_q$ and $\overline{\mathbb {Q}}_q$ are dense in ${\mathbb {R}}$. For every $q \in {\mathbb {Q}}$, let ${\mathbb {R}}_q = {\mathbb {Q}}_q \cup \overline{\mathbb {Q}}_q$. We define a function $g: {\mathbb {R}} \rightarrow {\mathbb {Q}}$ that maps every real number x to the index q (a rational number) of the class ${\mathbb {R}}_{q}$ it belongs to. Formally, for every $x \in {\mathbb {R}}$, $g(x) = q$, where $q \in {\mathbb {Q}}$ is the unique rational number such that $x \in {\mathbb {R}}_{q}$. The two functions $f_1: {\mathbb {R}} \rightarrow {\mathbb {Q}}$ and $f_2: {\mathbb {R}} \rightarrow {\mathbb {Q}}$ are defined as follows:

$$\begin{aligned}&f_1(x) = \left\{ \begin{array}{l@{\quad }l} g(x) &{} \quad \text {if } x < g(x) \text {, } x \ne 1, \text { and } x \ne 0 \\ 2 &{} \quad \text {if } x = 1 \\ \lceil x+3 \rceil &{} \quad \text {otherwise} \end{array} \right. \\&f_2(x) = \left\{ \begin{array}{ll} g(x) &{} \quad \text {if } x < g(x) \text { and } x \not \in [0,3) \\ \lceil x+3 \rceil &{} \quad \text {otherwise} \end{array} \right. \end{aligned}$$

It is not difficult to check that the above-defined functions $f_i$ ($i \in \{ 1,2 \}$) satisfy the following properties:

• (i) for every $x \in {\mathbb {R}}$, $f_i(x) > x$,

• (ii) for every $x \in {\mathbb {Q}}$, both $f_i^{-1} (x) \cap {\mathbb {Q}}$ and $f_i^{-1} (x) \cap \overline{{\mathbb {Q}}}$ are left-unbounded (notice that surjectivity of $f_i$ immediately follows), and

• (iii) for every $x,y \in {\mathbb {R}}$, if $x < y$, then there exists $u_1 \in {\mathbb {Q}}$ (resp., $u_2 \in \overline{{\mathbb {Q}}}$) such that $x < u_1 < y$ (resp., $x < u_2 < y$) and $y < f_i(u_1)$ (resp., $y < f_i(u_2)$).

Now, we can define two models $M_1$ and $M_2$, built on the only proposition letter p, as follows: for each $i \in \{ 1,2 \}$, $M_i = \langle {\mathbb {I}}({\mathbb {R}}), V_i \rangle$, where $V_i : {{\mathcal {AP}}} \rightarrow 2^{{\mathbb {I}}({\mathbb {R}})}$ ($i \in \{ 1,2 \}$) is defined as follows: $[x,y] \in V_i(p)\Leftrightarrow y \ge f_i(x)$. Finally, we define the relation Z as:

$$\begin{aligned} ([x,y],[w,z]) \in Z \Leftrightarrow x \equiv w \text {, } y \equiv z \text {, and } [x,y] \equiv _l [w,z], \end{aligned}$$

where we define $u \equiv v \Leftrightarrow u \in {\mathbb {Q}}$ if and only if $v \in {\mathbb {Q}}$ and $[u,u'] \equiv _l [v,v'] \Leftrightarrow u' \sim f_1(u)$ and $v' \sim f_2(v)$, for $\sim \in \{<,=,>\}$.

Let us consider the interval [0, 3] in $M_{1}$ and the interval [0, 3] in $M_{2}$. It is immediate to see that these two intervals are Z-related. However, $M_{1},[0,3] \Vdash \langle D \rangle p$ (as $M_{1},[1,2] \Vdash p$), but $M_{2},[0,3] \Vdash \lnot \langle D \rangle p$.

To complete the proof, it suffices to show that Z is an $\mathsf {A B O\overline{A B E}}$-bisimulation between $M_{1}$ and $M_{2}$.

Let [x, y] and [w, z] be two Z-related intervals. By definition, $y \sim f_1(x)$ and $z \sim f_2(w)$ for some $\sim \in \{ <,=,$ $> \}$. If $\sim \in \{=,>\}$, then both [x, y] and [w, z] satisfy p; otherwise, both of them satisfy $\lnot p$. Thus, the local condition is satisfied.

As for the forward condition, let [x, y] and $[x',y']$ be two intervals in $M_{1}$ and [w, z] an interval in $M_{2}$. We have to prove that if [x, y] and [w, z] are Z-related, then, for each modality $\langle X \rangle$ of $\mathsf {A B O \overline{A B E}}$ such that $[x,y] R_X [x',y']$, there exists an interval $[w',z']$ such that $[x',y']$ and $[w',z']$ are Z-related and $[w,z] R_X [w',z']$. Once again, we proceed by examining each case in turn.

• Let $X = A$. By definition of $\langle A \rangle$, $x' = y$ and we are forced to choose $w' = z$. By $y \equiv z$, it immediately follows that $x' \equiv w'$. We must find a point $z'>z$ such that $y' \equiv z'$ and both $y' \sim f_1(y)$ and $z' \sim f_2(z)$ for some $\sim \in \{<,=,>\}$. Let us suppose that $y' < f_1(y)$. In such a case, we choose a point $z'$ such that $z < z' < f_2(z)$ and $y' \equiv z'$. The existence of such a point is guaranteed by property (i) of $f_2$ above and by the density of ${\mathbb {Q}}$ and $\overline{\mathbb {Q}}$ in ${\mathbb {R}}$. Otherwise, if $y' = f_1(y)$, we choose $z' = f_2(z)$. By definition of $f_1$ and $f_2$ (the codomain of $f_1$ and $f_2$ is ${\mathbb {Q}}$), both $y'$ and $z'$ belong to ${\mathbb {Q}}$ and thus $y' \equiv z'$. Finally, if $y' > f_1(y)$, we choose $z' > f_2(z)$ such that $y' \equiv z'$. The existence of such a point is guaranteed by right-unboundedness of ${\mathbb {Q}}$ and $\overline{\mathbb {Q}}$, and the interval $[z,z']$ is such that $[x',y']Z[z,z']$ and $[w,z] R_A [z,z']$.

• Let $X = B$. In this case, $x = x'$ and $y' < y$. We distinguish the following cases.

  • If $y' > f_1(x)$ and $y' \in {\mathbb {Q}}$ (resp., $y' \in \overline{{\mathbb {Q}}}$), then $y > f_1(x)$ holds as well (as $y' < y$), which implies $z > f_2(w)$. Thus, we can choose any point $z' \in {\mathbb {Q}}$ (resp., $z' \in \overline{{\mathbb {Q}}}$), with $f_2(w) < z' < z$ (the existence of such a point is guaranteed by density of ${\mathbb {Q}}$ and $\overline{{\mathbb {Q}}}$, respectively).

  • If $y' = f_1(x)$, then $y' \in {\mathbb {Q}}$ (by definition of $f_1$) and $y > f_1(x)$ holds (as $y' < y$). The latter implies $z > f_2(w)$, and thus we choose $z' = f_2(w)$. Note that $f_2(w) \in {\mathbb {Q}}$ by the definition of $f_2$.

  • If $y' < f_1(x)$ and $y' \in {\mathbb {Q}}$ (resp., $y' \in \overline{{\mathbb {Q}}}$), then we choose a point $z' \in {\mathbb {Q}}$ (resp., $z' \in \overline{{\mathbb {Q}}}$) such that $w < z' < \min \{ z, f_2(w) \}$ (the existence of such a point is guaranteed by density of ${\mathbb {Q}}$ and $\overline{{\mathbb {Q}}}$, respectively).

  In all cases, the interval $[w,z']$ is such that $[x,y']Z[w,z']$ and $[w,z] R_B [w,z']$.

• Let $X = O$. Firstly, we choose a point $w'$ such that $w < w' < z$, $w' \in {\mathbb {Q}}$ iff $x' \in {\mathbb {Q}}$, and $f_2(w') > z$ (the existence of such a point is guaranteed by property (iii) of $f_2$ on page 19). Secondly, we choose a point $z'$ such that $z' \in {\mathbb {Q}}$ iff $y' \in {\mathbb {Q}}$, and

  • if $y' < f_1(x')$, then $z < z' < f_2(w')$ (density of ${\mathbb {Q}}$ and $\overline{{\mathbb {Q}}}$ is used here),

  • if $y' > f_1(x')$, then $z' > f_2(w')$ (right-unboundedness of ${\mathbb {Q}}$ and $\overline{{\mathbb {Q}}}$ is used here),

  • if $y' = f_1(x')$, then $z' = f_2(w')$.

  In all cases, the interval $[w',z']$ is such that $[x',y']Z[w',z']$ and $[w,z] R_O [w',z']$.

• Let $X = \overline{A}$. In this case, $y' = x$. We distinguish the following cases.

  • If $f_1(x') < y' (=x)$, then consider any point $\overline{w} \in {\mathbb {Q}}$, with $\overline{w} < w$. By property (ii) on page 19, there exist both a point $w'' \in {\mathbb {Q}}$ and a point $w''' \in \overline{{\mathbb {Q}}}$ such that $w'' < \overline{w}$, $w''' < \overline{w}$, and $f_2(w'') = f_2(w''') = \overline{w}$. We select $w' = w''$ if $x' \in {\mathbb {Q}}$, and $w' = w'''$ if $x' \in \overline{{\mathbb {Q}}}$.

  • If $f_1(x') > y' (=x)$, then consider any point $\overline{w} \in {\mathbb {Q}}$, with $w < \overline{w}$. By property (ii), there exist both a point $w'' \in {\mathbb {Q}}$ and a point $w''' \in \overline{{\mathbb {Q}}}$ such that $w'' < w$, $w''' < w$, and $f_2(w'') = f_2(w''') = \overline{w}$. We select $w' = w''$ if $x' \in {\mathbb {Q}}$, and $w' = w'''$ if $x' \in \overline{{\mathbb {Q}}}$.

  • If $f_1(x') = y' (=x)$, then $x,w \in {\mathbb {Q}}$. By property (ii), there exist both a point $w'' \in {\mathbb {Q}}$ and a point $w''' \in \overline{{\mathbb {Q}}}$ such that $w'' < w$, $w''' < w$, and $f_2(w'') = f_2(w''') = w$. We select $w' = w''$ if $x' \in {\mathbb {Q}}$, and $w' = w'''$ if $x' \in \overline{{\mathbb {Q}}}$.

  In all cases, the interval $[w',w]$ is such that $[x',x]Z[w',w]$ and $[w,z] R_{\overline{A}} [w',w]$.

• Let $X = \overline{B}$. In this case $x' = x$. We distinguish the following cases.

  • If $y' < f_1(x)$ and $y' \in {\mathbb {Q}}$ (resp., $y' \in \overline{{\mathbb {Q}}}$), then $y < f_1(x)$ holds as well (as $y < y'$), which implies $z < f_2(w)$. Thus, we can choose any point $z' \in {\mathbb {Q}}$ (resp., $z' \in \overline{{\mathbb {Q}}}$), with $z < z' < f_2(w)$ (the existence of such a point is guaranteed by density of ${\mathbb {Q}}$ and $\overline{{\mathbb {Q}}}$, respectively).

  • If $y' = f_1(x)$, then $y' \in {\mathbb {Q}}$ (by definition of $f_1$) and $y < f_1(x)$ holds (as $y < y'$). The latter implies $z < f_2(w)$, and thus we choose $z' = f_2(w)$. Note that $f_2(w) \in {\mathbb {Q}}$ by the definition of $f_2$.

  • If $y' > f_1(x)$ and $y' \in {\mathbb {Q}}$ (resp., $y' \in \overline{{\mathbb {Q}}}$), then we choose a point $z' \in {\mathbb {Q}}$ (resp., $z' \in \overline{{\mathbb {Q}}}$) such that $z' > \max \{ z, f_2(w) \}$ (the existence of such a point is guaranteed by right-unboundedness of ${\mathbb {Q}}$ and $\overline{{\mathbb {Q}}}$, respectively).

  In all cases, the interval $[w,z']$ is such that $[x',y']Z[w,z']$ and $[w,z] R_{\overline{B}} [w,z']$.

• Let $X = \overline{E}$. In this case $y = y'$ and $x' < x$. We distinguish the following cases.

  • If $f_1(x') < y' (=y)$, then consider any point $\overline{w} \in {\mathbb {Q}}$, with $\overline{w} < z$. By property (ii) on page 19, there exist both a point $w'' \in {\mathbb {Q}}$ and a point $w''' \in \overline{{\mathbb {Q}}}$ such that $w'' < w$, $w''' < w$, and $f_2(w'') = f_2(w''') = \overline{w}$. We select $w' = w''$ if $x' \in {\mathbb {Q}}$, and $w' = w'''$ if $x' \in \overline{{\mathbb {Q}}}$.

  • If $f_1(x') > y' (=y)$, then consider any point $\overline{w} \in {\mathbb {Q}}$, with $z < \overline{w}$. By property (ii) on page 19, there exist both $w'' \in {\mathbb {Q}}$ and $w''' \in \overline{{\mathbb {Q}}}$ such that $w'' < w$, $w''' < w$, and $f_2(w'') = f_2(w''') = \overline{w}$. We select $w' = w''$ if $x' \in {\mathbb {Q}}$, and $w' = w'''$ if $x' \in \overline{{\mathbb {Q}}}$.

  • If $f_1(x') = y' (=y)$, then $y,z \in {\mathbb {Q}}$. By property (ii) on page 19, there exist both $w'' \in {\mathbb {Q}}$ and $w''' \in \overline{{\mathbb {Q}}}$ such that $w'' < w$, $w''' < w$, and $f_2(w'') = f_2(w''') = z$. We select $w' = w''$ if $x' \in {\mathbb {Q}}$, and $w' = w'''$ if $x' \in \overline{{\mathbb {Q}}}$.

  In all cases, the interval $[w',z]$ is such that $[x',y]Z[w',z]$ and $[w,z] R_{\overline{E}} [w',z]$.

The backward condition can be verified in a very similar way and thus the details of the proof are omitted. Hence, Z is an $\mathsf {A B O\overline{A B E}}$-bisimulation that violates $\langle D \rangle$, hence the thesis. $\square$

Appendix 7: Complete proof of Lemma 14

Lemma 14 $\langle O \rangle$ is not definable in $\mathsf {A B E\overline{A E D}}$ relative to the classes $\mathsf {Lin}$ and $\mathsf {Den}$.

Proof

The bisimulation we use here is very similar to those constructed for the operators $\langle E \rangle$ and $\langle \overline{E} \rangle$ in the proofs of Lemma 9 and Lemma 10, respectively. Let $M_1 = \langle {\mathbb {I}}({\mathbb {R}}), V_1 \rangle$ and $M_2 = \langle {\mathbb {I}}({\mathbb {R}}), V_2 \rangle$ be two models over the set of proposition letters ${{\mathcal {AP}}} = \{p\}$, where the valuation functions $V_1: {{\mathcal {AP}}} \rightarrow 2^{{\mathbb {I}}({\mathbb {R}})}$ and $V_2: {{\mathcal {AP}}} \rightarrow 2^{{\mathbb {I}}({\mathbb {R}})}$ are, respectively, defined as follows:

• $[x,y] \in V_1(p) \Leftrightarrow x \in {\mathbb {Q}}$ iff $y \in {\mathbb {Q}}$ and

• $[w,z] \in V_2(p) \Leftrightarrow w \in {\mathbb {Q}}$ iff $z \in {\mathbb {Q}}$, and $[0,3]R_O[w,z]$ does not hold (that is, it is not the case that $0 < w < 3 < z$).

Then, we define the relation Z between intervals of $M_1$ and intervals of $M_2$ as: $[x,y]Z[w,z] \Leftrightarrow [x,y] \in V_1(p)$ iff $[w,z] \in V_2(p)$. It is immediate to see that [0, 3]Z[0, 3], $M_1,[0,3] \Vdash \langle O \rangle p$, but $M_2,[0,3] \Vdash \lnot \langle O \rangle p$.

We show that Z is an $\mathsf {A B E \overline{A E D}}$-bisimulation between $M_1$ and $M_2$. The local condition immediately follows from the definition. As for the forward condition, it can be checked as follows. Let [x, y] and [w, z] be two Z-related intervals, and let us assume that $[x,y] R_X [x',y']$ holds for some $X \in \{ A, B, E, \overline{A}, \overline{E}, \overline{D} \}$. We have to exhibit an interval $[w',z']$ such that $[x',y']$ and $[w',z']$ are Z-related, and [w, z] and $[w',z']$ are $R_X$-related. We proceed by a case analysis on $X \in \{ A, B, E, \overline{A}, \overline{E}, \overline{D} \}$.

• If $X=A$, then we distinguish the following cases: (a) if $0 < z < 3$, then we select a point $z'$ such that $z < z' < 3$ and $z' \in {\mathbb {Q}}$ iff $y' \in {\mathbb {Q}}$ (the existence of such a point is guaranteed by density of ${\mathbb {Q}}$ and $\overline{\mathbb {Q}}$); (b) otherwise, we select a point $z'$ such that $z' > z$ and $z' \in {\mathbb {Q}}$ iff $y' \in {\mathbb {Q}}$ (the existence of such a point is guaranteed by right-unboundedness of ${\mathbb {Q}}$ and $\overline{\mathbb {Q}}$). In both cases, the interval $[z,z']$ is such that $[x',y']Z[z,z']$ and $[w,z]R_A[z,z']$.

• If $X=B$, the argument is similar to the previous one. We distinguish the following cases: (a) if $0 < w < 3$, then we choose a point $z'$ such that $w < z' < \min \{ 3,z \}$ and $z' \in {\mathbb {Q}}$ iff $y' \in {\mathbb {Q}}$; (b) otherwise, we choose a point $z'$ such that $w < z' < z$ and $z' \in {\mathbb {Q}}$ iff $y' \in {\mathbb {Q}}$. In both cases, the interval $[w,z']$ is such that $[x',y']Z[w,z']$ and $[w,z]R_B[w,z']$.

• If $X = E$, then we distinguish the following cases: (a) if $z > 3$, then we choose a point $w'$ such that $\max \{ 3,w \} < w' < z$ and $w' \in {\mathbb {Q}}$ iff $x' \in {\mathbb {Q}}$; (b) otherwise, we choose a point $w'$ such that $w < w' < z$ and $w' \in {\mathbb {Q}}$ iff $x' \in {\mathbb {Q}}$. In both cases, the interval $[w',z]$ is such that $[x',y']Z[w',z]$ and $[w,z]R_E[w',z]$.

• If $X = \overline{A}$, then we choose a point $w'$ such that $w' < \min \{ 0,w \}$ and $w' \in {\mathbb {Q}}$ iff $x' \in {\mathbb {Q}}$. The interval $[w',w]$ is such that $[x',y']Z[w',w]$ and $[w,z]R_{\overline{A}}[w',w]$.

• If $X = \overline{E}$, then we choose a point $w'$ such that $w' < \min \{ 0,w \}$ and $w' \in {\mathbb {Q}}$ iff $x' \in {\mathbb {Q}}$. The interval $[w',z]$ is such that $[x',y']Z[w',z]$ and $[w,z]R_{\overline{E}}[w',z]$.

• If $X = \overline{D}$, then we first choose a point $w'$ such that $w' < \min \{ 0,w \}$ and $w' \in {\mathbb {Q}}$ iff $x' \in {\mathbb {Q}}$. Next, we choose a point $z'$ such that $z' > z$ and $z' \in {\mathbb {Q}}$ iff $y' \in {\mathbb {Q}}$. The interval $[w',z']$ is such that $[x',y']Z[w',z']$ and $[w,z]R_{\overline{D}}[w',z']$.

The backward condition can be verified in a very similar way and thus we omit the details of the proof. Therefore, Z is an $\mathsf {A B E\overline{A E D}}$-bisimulation that violates $\langle O \rangle$. The thesis immediately follows. $\square$

Appendix 8: Complete proof of Lemma 15

Lemma 15 $\langle O \rangle$ is not definable in $\mathsf {A B D\overline{A B E}}$ relative to the classes $\mathsf {Lin}$ and $\mathsf {Den}$.

Proof

The $\mathsf {A B D\overline{A B E}}$-bisimulation that we present here has some similarities with the $\mathsf {A B O\overline{A B E}}$-bisimulation that violates $\langle D \rangle$, presented in the proof of Lemma 12. However, we need to ‘rearrange’ the partitions of ${\mathbb {Q}}$ and $\overline{{\mathbb {Q}}}$ that we exploited to prove Lemma 12. More precisely, we still need two infinite countable partitions ${\mathcal {P}}({\mathbb {Q}})$ of ${\mathbb {Q}}$ and ${\mathcal {P}}(\overline{\mathbb {Q}})$ of $\overline{\mathbb {Q}}$, whose elements are dense in ${\mathbb {R}}$, but it is useful to provide a more suitable enumeration for both of them, as follows: ${\mathcal {P}}({\mathbb {Q}}) = \{ {\mathbb {Q}}_q^c \mid c \in \{a,b\}, q \in {\mathbb {Q}} \}$ and ${\mathcal {P}}(\overline{\mathbb {Q}}) = \{ \overline{\mathbb {Q}}_q^c \mid c \in \{a,b\}, q \in {\mathbb {Q}} \}$. Analogously to Lemma 12, we require these partitions to be such that, for each $c \in \{a,b\}$ and $q \in {\mathbb {Q}}$, sets ${\mathbb {Q}}_q^c$ and $\overline{\mathbb {Q}}_q^c$ are dense in ${\mathbb {R}}$. Now, we define the partition ${\mathcal {P}}({\mathbb {R}})$ of ${\mathbb {R}}$ as: ${\mathcal {P}}({\mathbb {R}}) = \{ {\mathbb {R}}_q^c \mid c \in \{a,b\}, q \in {\mathbb {Q}}\}$, where ${\mathbb {R}}_q^c = {\mathbb {Q}}_q^c \cup \overline{\mathbb {Q}}_q^c$, for each $c \in \{a,b\}$ and $q \in {\mathbb {Q}}$. We use ${\mathbb {Q}}^c$ (resp., $\overline{\mathbb {Q}}^c$, $\mathbb {R}^c$) as an abbreviation for $\bigcup _{q \in {\mathbb {Q}}} {\mathbb {Q}}_q^c$ (resp., $\bigcup _{q \in {\mathbb {Q}}} \overline{\mathbb {Q}}_q^c$, $\bigcup _{q \in {\mathbb {Q}}} {\mathbb {R}}_q^c$), for each $c \in \{a,b\}$. In addition, we define ${\mathcal {S}}_1,{\mathcal {S}}_2 \subseteq {\mathbb {I}}({\mathbb {R}})$ as follows:

$$\begin{aligned} {\mathcal {S}}_1= & {} \{[x,y] \mid x,y \in {\mathbb {R}}^c, c \in \{a,b\} \} \quad \text {and} \\ {\mathcal {S}}_2= & {} \{[w,z] \mid w,z \in {\mathbb {R}}^c, c \in \{a,b\} \} {\setminus } \{[w,z] \mid 0 < w < 3 < z \}. \end{aligned}$$

Finally, for each $i \in \{1,2\}$, we use $\overline{\mathcal {S}}_i$ to denote the set ${\mathbb {I}}({\mathbb {R}}) \setminus {\mathcal {S}}_i$. It is easy to verify that, for every pair of points $x,y \in {\mathbb {I}}({\mathbb {R}})$, if $x < y$, then there exist $y_1,y_2,y_3,y_4 \in {\mathbb {R}}$ such that $x < y_i < y$, for each $i \in \{1,2,3,4\}$, and:

$$\begin{aligned}&y_1 \in {\mathbb {Q}} \text { and } [x,y_1] \in {\mathcal {S}}_1 (\text {resp., } {\mathcal {S}}_2 ), \nonumber \\&y_2 \in {\mathbb {Q}} \text { and } [x,y_2] \in \overline{\mathcal {S}}_1 (\text {resp., } \overline{\mathcal {S}}_2 ), \nonumber \\&y_3 \in \overline{\mathbb {Q}} \text { and } [x,y_3] \in {\mathcal {S}}_1 (\text {resp., } {\mathcal {S}}_2 ), \nonumber \\&y_4 \in \overline{\mathbb {Q}} \text { and } [x,y_4] \in \overline{\mathcal {S}}_1 (\text {resp., } \overline{\mathcal {S}}_2 ). \end{aligned}$$

(4)

We define now a pair of functions that will be used in the definition of the models involved in the bisimulation relation Z. Let $g: {\mathbb {R}} \rightarrow {\mathbb {Q}}$ be a function defined as follows (notice the strong similarity with the definition of g in Lemma 12): for each $x \in {\mathbb {R}}$, $g(x) = q$, where $q \in {\mathbb {Q}}$ is the unique rational number such that $x \in {\mathbb {R}}_{q}^a \cup R_{q}^b$. The functions $f_1: {\mathbb {R}} \rightarrow {\mathbb {Q}}$ and $f_2: {\mathbb {R}} \rightarrow {\mathbb {Q}}$ are defined as follows:

$$\begin{aligned}&f_1(x) = \left\{ \begin{array}{l@{\quad }l} g(x) &{} \text {if } x < g(x) \\ \lceil x+3 \rceil &{} \text {otherwise} \end{array} \right. \\&f_2(x) = \left\{ \begin{array}{l@{\quad }l} g(x) &{} \text {if } x < g(x) \text { and } ([0,3],[x,g(x)]) \not \in R_O \\ \lceil x+3 \rceil &{} \text {if } x \ge g(x) \text { and } x \not \in (0,3) \\ a_{n'} &{} \text {otherwise} \end{array} \right. \end{aligned}$$

where $a_{n'}$ is the least element of the series $a_n = 3-(\frac{1}{n})$ ($n\ge 1$) such that $x < a_{n'}$. It is not hard to verify that the functions $f_i$ ($i \in \{ 1,2 \}$) fulfill the following conditions:

(i):

$f_i(x) > x$ for every $x \in {\mathbb {R}}$;

(ii):

for each $x \in {\mathbb {Q}}$, $f_i^{-1} (x) \cap {\mathbb {Q}}^a$, $f_i^{-1} (x) \cap {\mathbb {Q}}^b$, $f_i^{-1} (x) \cap \overline{\mathbb {Q}}^a$, and $f_i^{-1} (x) \cap \overline{{\mathbb {Q}}}^b$ are left-unbounded (notice that surjectivity of $f_i$ immediately follows);

(iii):

for each $x,y \in {\mathbb {R}}$, if $x < y$, then there exist:

• $u_1 \in {\mathbb {Q}}^a$ such that $x < u_1 < y$ and $y > f_i(u_1)$,

• $u_2 \in {\mathbb {Q}}^b$ such that $x < u_2 < y$ and $y > f_i(u_2)$,

• $u_3 \in \overline{{\mathbb {Q}}}^a$ such that $x < u_3 < y$ and $y > f_i(u_3)$, and

• $u_4 \in \overline{{\mathbb {Q}}}^b$ such that $x < u_4 < y$ and $y > f_i(u_4)$.

In addition, function $f_2$ satisfies the following property:

(iv):

for each $w \in (0,3)$, $f_2(w) < 3$.

At this point, we are ready to define the models $M_1$ and $M_2$, and the bisimulation relation between their intervals. Let $i \in \{1,2\}$ and $M_{i} = \langle {\mathbb {I}}({\mathbb {R}}), V_{f_i} \rangle$, where the valuation functions $V_{i} : {{\mathcal {AP}}} \rightarrow 2^{{\mathbb {I}}({\mathbb {R}})}$ is defined as follows:

$$\begin{aligned}{}[x,y] \in V_i(p) \Leftrightarrow \text {either } y = f_i(x) \text { or both } y < f_i(x) \text { and } [x,y] \in {\mathcal {S}}_i. \end{aligned}$$

The relation Z is defined as follows:

$$\begin{aligned}{}[x,y]Z[w,z] \Leftrightarrow x \equiv w \text {, } y \equiv z \text {, and } [x,y] \equiv _l [w,z], \end{aligned}$$

where the relations $\equiv$ and $\equiv _l$ are defined, respectively, in the following way:

$$\begin{aligned}&x \equiv w \Leftrightarrow x \in {\mathbb {Q}} \text { iff } w \in {\mathbb {Q}} \\&[x,y] \equiv _l [w,z] \Leftrightarrow \left\{ \begin{array}{l@{\quad }l} \text {either} &{} y > f_1(x) \text { and } z > f_2(w) \\ \text {or} &{} y = f_1(x) \text { and } z = f_2(w) \\ \text {or} &{} y < f_1(x) \text {, } z < f_2(w) \text {, and } ([x,y] \in {\mathcal {S}}_1 \text { iff } [w,z] \in {\mathcal {S}}_2) \end{array} \right. \end{aligned}$$

Now, by the definition of Z, we have that [0, 3]Z[0, 3] (notice that this is also a consequence of the facts that $f_1(0) = f_2(0)$ and that $[0,3]R_O,[0,3]$ does not hold). Moreover, it is easy to see that $M_{1},[0,3] \Vdash \langle O \rangle p$, while $M_{2},[0,3] \Vdash \lnot \langle O \rangle p$ (this is a direct consequence of property (iv) of $f_2$ and of the fact that $f_1(x) > 3$ for some $x \in (0,3)$).

We show that Z is an $\mathsf {A B D\overline{A B E}}$- bisimulation. For the local condition, consider two intervals [x, y] and [w, z] such that [x, y]Z[w, z]. First, we assume that $[x,y] \in V_1(p)$ and we show that $[w,z] \in V_2(p)$ follows. Since $[x,y] \in V_1(p)$, either $y = f_1(x)$ holds or both $y < f_1(x)$ and $[x,y] \in {\mathcal {S}}_1$ hold. In the former case, by the definition of Z, it must be $z = f_2(w)$, which implies $[w,z] \in V_2(p)$. In the latter case, by the definition of Z, both $z < f_2(w)$ and $[w,z] \in {\mathcal {S}}_2$ hold, and thus $[w,z] \in V_2(p)$. Second, we assume that $[w,z] \in V_2(p)$ and we show that $[x,y] \in V_1(p)$ follows. Since $[w,z] \in V_2(p)$, either $z = f_2(w)$ holds or both $z < f_2(w)$ and $[w,z] \in {\mathcal {S}}_2$ hold. In the former case, by the definition of Z, it must be $y = f_1(x)$, which implies $[x,y] \in V_1(p)$. In the latter case, by the definition of Z, both $y < f_1(x)$ and $[x,y] \in {\mathcal {S}}_1$ hold, and thus $[x,y] \in V_1(p)$.

In order to prove that the forward condition is satisfied, we assume that [x, y]Z[w, z] and $[x,y] R_X [x',y']$, for some $X \in \{ A, B, D, \overline{A}, \overline{B}, \overline{E} \}$ and some $[x,y], [w,z], [x',y'] \in {\mathbb {I}}({\mathbb {R}})$, and we show the existence of an interval $[w',z']$ such that $[x',y']Z[w',z']$ and $[w,z] R_X [w',z']$. As usual, we proceed by considering each case in turn.

• If $X = A$, then we distinguish three cases.

  • If $y' > f_1(x')$ and $y' \in {\mathbb {Q}}$ (resp., $y' \in \overline{{\mathbb {Q}}}$), then we select $z'$ such that $z' > f_2(z)$ and $z' \in {\mathbb {Q}}$ (resp., $z' \in \overline{{\mathbb {Q}}}$).

  • If $y' = f_1(x')$, then $y' \in {\mathbb {Q}}$, and we select $z' = f_2(z)$.

  • If $y' < f_1(x')$ and $y' \in {\mathbb {Q}}$ (resp., $y' \in \overline{{\mathbb {Q}}}$), then, by property (3) on page 21, there exists a point $z' \in {\mathbb {Q}}$ (resp., $z' \in \overline{{\mathbb {Q}}}$) such that $z < z' < f_2(z)$ and $[x',y'] \in {\mathcal {S}}_1$ iff $[z,z'] \in {\mathcal {S}}_2$ (notice that property (i) of $f_2$ plays a role here).

  In all cases, the interval $[z,z']$ is such that $[x',y']Z[z,z']$ and $[w,z] R_A [z,z']$.

• If $X = B$, then we distinguish three cases.

  • If $y' > f_1(x')$ and $y' \in {\mathbb {Q}}$ (resp., $y' \in \overline{{\mathbb {Q}}}$), then it must be $y > f_1(x)$ (as $y > y'$ and $x = x'$), which implies $z > f_2(w)$, and we select $z'$ such that $f_2(w) < z' < z$ and $z' \in {\mathbb {Q}}$ (resp., $z' \in \overline{{\mathbb {Q}}}$).

  • If $y' = f_1(x')$, then $y' \in {\mathbb {Q}}$ and $y > f_1(x)$ (as $y > y'$ and $x = x'$), which implies $z > f_2(w)$, and we select $z' = f_2(w)$.

  • If $y' < f_1(x')$ and $y' \in {\mathbb {Q}}$ (resp., $y' \in \overline{{\mathbb {Q}}}$), then, by property (3) on page 21, there exists a point $z' \in {\mathbb {Q}}$ (resp., $z' \in \overline{{\mathbb {Q}}}$) such that $w < z' < f_2(w)$ and $[x',y'] \in {\mathcal {S}}_1$ iff $[w,z'] \in {\mathcal {S}}_2$ (notice that property (i) of $f_2$ plays a role here).

  In all cases, the interval $[w,z']$ is such that $[x',y']Z[w,z']$ and $[w,z] R_B [w,z']$.

• If $X = D$, then we first select a point $w'$ such that $w < w' < z$, $w' \in {\mathbb {Q}}$ iff $x' \in {\mathbb {Q}}$, and $f_2(w') < z$ (the existence of such a point is guaranteed by property (iii) of $f_2$). Then, we select a point $z'$ as follows.

  • If $y' > f_1(x')$ and $y' \in {\mathbb {Q}}$ (resp., $y' \in \overline{{\mathbb {Q}}}$), then we select $z'$ such that $f_2(w) < z' < z$ and $z' \in {\mathbb {Q}}$ (resp., $z' \in \overline{{\mathbb {Q}}}$).

  • If $y' = f_1(x')$, then $y' \in {\mathbb {Q}}$, and we select $z' = f_2(w)$. Notice that $z' \in {\mathbb {Q}}$ as well.

  • If $y' < f_1(x')$ and $y' \in {\mathbb {Q}}$ (resp., $y' \in \overline{{\mathbb {Q}}}$), then, by property (3) on page 21, there exists a point $z' \in {\mathbb {Q}}$ (resp., $z' \in \overline{{\mathbb {Q}}}$), such that $w' < z' < f_2(w')$ and $[x',y'] \in {\mathcal {S}}_1$ iff $[w',z'] \in {\mathcal {S}}_2$ (notice that property (i) of $f_2$ plays a role here).

  In all cases, the interval $[w',z']$ is such that $[x',y']Z[w',z']$ and $[w,z] R_D [w',z']$.

• If $X = \overline{A}$, then we distinguish three cases.

  • If $y' > f_1(x')$ and $x' \in {\mathbb {Q}}$ (resp., $x' \in \overline{{\mathbb {Q}}}$), then consider a point $\overline{z} \in {\mathbb {Q}}$ such that $\overline{z} < w$. We select a point $w' \in {\mathbb {Q}}$ (resp., $w' \in \overline{{\mathbb {Q}}}$) such that $w' < \overline{z} < w$ and $f_2(w') = \overline{z}$ (the existence of such a point is guaranteed by property (ii) of $f_2$).

  • If $y' = f_1(x')$ and $x' \in {\mathbb {Q}}$ (resp., $x' \in \overline{{\mathbb {Q}}}$), then $y' = x \in {\mathbb {Q}}$, which implies $w \in {\mathbb {Q}}$. We select a point $w' \in {\mathbb {Q}}$ (resp., $w' \in \overline{{\mathbb {Q}}}$) such that $w' < w$ and $f_2(w') = w$ (the existence of such a point is guaranteed by property (ii) of $f_2$).

  • If $y' < f_1(x')$ and $x' \in {\mathbb {Q}}$ (resp., $x' \in \overline{{\mathbb {Q}}}$), then consider a point $\overline{z} \in {\mathbb {Q}}$ such that $\overline{z} > w$. We select a point $w' \in {\mathbb {Q}}$ (resp., $w' \in \overline{{\mathbb {Q}}}$) such that $w' < \min \{ 0,w \}$, $f_2(w') = \overline{z}$, and $[w',w] \in {\mathcal {S}}_2$ iff $[x',y'] \in {\mathcal {S}}_1$ (the existence of such a point is guaranteed by property (ii) of $f_2$). Notice that, since $w' < 0$, it is not the case that $[0,3] R_O [w',w]$.

  In all cases, the interval $[w',w]$ is such that $[x',y']Z[w',w]$ and $[w,z] R_{\overline{A}} [w',w]$.

• If $X = \overline{B}$, then we distinguish three cases.

  • If $y' > f_1(x')$ and $y' \in {\mathbb {Q}}$ (resp., $y' \in \overline{{\mathbb {Q}}}$), then we select $z'$ such that $z' > z$ and $z' \in {\mathbb {Q}}$ (resp., $z' \in \overline{{\mathbb {Q}}}$),

  • If $y' = f_1(x')$, then $y' \in {\mathbb {Q}}$ and $y < f_1(x)$ (as $y < y'$ and $x = x'$), which implies $z < f_2(w)$, and we select $z' = f_2(w)$.

  • If $y' < f_1(x')$ and $y' \in {\mathbb {Q}}$ (resp., $y' \in \overline{{\mathbb {Q}}}$), then it must be the case that $y < f_1(x)$ (as $y < y'$ and $x = x'$). This yields $z < f_2(w)$, and we select $z' \in {\mathbb {Q}}$ (resp., $z' \in \overline{{\mathbb {Q}}}$) such that $z < z' < f_2(w)$ and $[x',y'] \in {\mathcal {S}}_1$ iff $[w,z'] \in {\mathcal {S}}_2$ Notice that the existence of such a point strongly depends on the fact that it is not the case that $[0,3] R_O [w,z']$. By way of contradiction, suppose that $[0,3] R_O [w,z']$ holds. Then, we have $0 < w < 3 < z'$. By property (iv) of $f_2$, $0 < w < 3$ implies $f_2(w) < 3$, and thus $z' < 3$ (as $z' < f_2(w)$), contradicting the fact that $0 < w < 3 < z'$ holds.

  In all cases, the interval $[w,z']$ is such that $[x',y']Z[w,z']$ and $[w,z] R_{\overline{B}} [w,z']$.

• If $X = \overline{E}$, then we distinguish three cases.

  • If $y' > f_1(x')$ and $x' \in {\mathbb {Q}}$ (resp., $x' \in \overline{{\mathbb {Q}}}$), then consider a point $\overline{z} \in {\mathbb {Q}}$ such that $\overline{z} < z$. We select a point $w' \in {\mathbb {Q}}$ (resp., $w' \in \overline{{\mathbb {Q}}}$) such that $w' < w$ and $f_2(w') = \overline{z}$ (the existence of such a point is guaranteed by property (ii) of $f_2$).

  • If $y' = f_1(x')$ and $x' \in {\mathbb {Q}}$ (resp., $x' \in \overline{{\mathbb {Q}}}$), then $y' = y \in {\mathbb {Q}}$, which implies $z \in {\mathbb {Q}}$. We select a point $w' \in {\mathbb {Q}}$ (resp., $w' \in \overline{{\mathbb {Q}}}$) such that $w' < w$ and $f_2(w') = z$ (the existence of such a point is guaranteed by property (ii) of $f_2$).

  • If $y' < f_1(x')$ and $x' \in {\mathbb {Q}}$ (resp., $x' \in \overline{{\mathbb {Q}}}$), then consider a point $\overline{z} \in {\mathbb {Q}}$ such that $\overline{z} > z$. We select a point $w' \in {\mathbb {Q}}$ (resp., $w' \in \overline{{\mathbb {Q}}}$) such that $w' < \min \{ 0,w \}$, $f_2(w') = \overline{z}$, and that $[w',w] \in {\mathcal {S}}_2$ if and only if $[x',y'] \in {\mathcal {S}}_1$ (the existence of such a point is guaranteed by property (ii) of $f_2$). Notice that, since $w' < 0$, it is not the case that $[0,3] R_O [w',z]$.

  In all cases, the interval $[w',z]$ is such that $[x',y']Z[w',z]$ and $[w,z] R_{\overline{E}} [w',z]$.

The backward condition can be verified in a very similar way and thus we omit the details of the proof. Therefore, Z is an $\mathsf {A B D\overline{A B E}}$-bisimulation that violates $\langle O \rangle$, hence the thesis. $\square$