Logic of Justified Beliefs Based on Argumentation

Abstract

This manuscript presents a topological argumentation framework for modelling notions of evidence-based (i.e., justified) belief. Our framework relies on so-called topological evidence models to represent the pieces of evidence that an agent has at her disposal, and it uses abstract argumentation theory to select the pieces of evidence that the agent will use to define her beliefs. The tools from abstract argumentation theory allow us to model agents who make decisions in the presence of contradictory information. Thanks to this, it is possible to define two new notions of beliefs, grounded beliefs and fully grounded beliefs. These notions are discussed in this paper, analysed and compared with the existing notion of topological justified belief. This comparison revolves around three main issues: closure under conjunction introduction, the level of consistency and their logical strength.

Appendices

A Proofs

1.1 A.1 Proof of Proposition 4

For the first sufficient condition, assume that $\leftarrowtail$ is symmetric on the set of arguments. Observe that, then, ${{\,\mathrm{\mathsf {LFP}}\,}}_\tau = \{T \in \tau \mid \forall x \in \tau {\setminus } \{\varnothing \}: x \cap T \ne \varnothing \}$, which is closed under conjunction.

The second sufficient condition requires more details, and the following lemma will be useful.

Lemma 2

Let $M= (W, E_0, \tau _{E_0}, \leftarrowtail , V)$ be a TA model. Then, for all $F_1, F_2 \in {{\,\mathrm{\mathsf {LFP}}\,}}_\tau$,

$$\begin{aligned} \begin{array}{l} F_1, F_2 \in {{\,\mathrm{\mathsf {LFP}}\,}}_\tau \text { implies } F_1 \cap F_2 \in {{\,\mathrm{\mathsf {LFP}}\,}}_\tau \\ \text { if and only if } \\ \text {for all } T \in \tau , \text { if } F_1 \cap F_2 \leftarrowtail T \text { then there is } F \in {{\,\mathrm{\mathsf {LFP}}\,}}_\tau \text { such that } T \cap F = \varnothing . \end{array} \end{aligned}$$

Proof

Take arbitrary $F_1,F_2\in {{\,\mathrm{\mathsf {LFP}}\,}}_\tau$. From left to right, consider the contrapositive, and suppose there is an open $T \in \tau$ such that T attacks $F_1\cap F_2$ but is not in conflict with anybody in ${{\,\mathrm{\mathsf {LFP}}\,}}_\tau$. From the latter it follows that nobody in ${{\,\mathrm{\mathsf {LFP}}\,}}_\tau$ attacks T, and thus the attacked $F_1 \cap F_2$ is not defended by ${{\,\mathrm{\mathsf {LFP}}\,}}_\tau$; therefore, $F_1 \cap F_2$ is not in ${{\,\mathrm{\mathsf {LFP}}\,}}_\tau$.

From right to left, take an argument $T \in \tau$ such that $F_1 \cap F_2\leftarrowtail T$. Then, there is $F' \in {{\,\mathrm{\mathsf {LFP}}\,}}_\tau$ such that $T \cap F' = \varnothing$, and thus either $T \leftarrowtail F'$ or else $F' \leftarrowtail T$. The first case gives us an argument in ${{\,\mathrm{\mathsf {LFP}}\,}}_\tau$ attacking T, namely $F'$; the second case does that too, as $F'$ is in ${{\,\mathrm{\mathsf {LFP}}\,}}_\tau$, and thus there should be $F'' \in {{\,\mathrm{\mathsf {LFP}}\,}}_\tau$ such that $T \leftarrowtail F''$. Hence, for all $T \in \tau$ such that $F_1\cap F_2\leftarrowtail T$ there is $F \in {{\,\mathrm{\mathsf {LFP}}\,}}_\tau$ such that $T \leftarrowtail F$: the set $F_1\cap F_2$ is defended by ${{\,\mathrm{\mathsf {LFP}}\,}}_\tau$. Thus, $F_1\cap F_2\in {{\,\mathrm{\mathsf {LFP}}\,}}_\tau$. $\square$

Now, take any $F_1, F_2 \in {{\,\mathrm{\mathsf {LFP}}\,}}_\tau$ and, for the contrapositive, suppose $F_1 \cap F_2$ is not in ${{\,\mathrm{\mathsf {LFP}}\,}}_\tau$. By Lemma 2, there is an open $T \in \tau$ which attacks $F_1 \cap F_2$ (i.e., $F_1 \cap F_2 \leftarrowtail T$) and which is not in conflict with any elements of ${{\,\mathrm{\mathsf {LFP}}\,}}_\tau$ (i.e., $F \in {{\,\mathrm{\mathsf {LFP}}\,}}_\tau$ implies $T \cap F \ne \varnothing$). The goal is to show that $\leftarrowtail$ is not unambiguous, and in order to achieve that define the following $T_1, T_2$ and $T_3$:

$$\begin{aligned} T_1 := F_1 \cap \ T, \quad T_2 := F_2 \cap \ T, \quad T_3 := F_1 \cap F_2. \end{aligned}$$

Note how none of the sets are empty. Note also how, due to the fact that T attacks $F_1 \cap F_2$, T must be in conflict with $F_1 \cap F_2$ (that is, $(F_1\cap F_2)\cap T = \varnothing$); hence, $T_1 \cap T_2 = T_2 \cap T_3 = T_3 \cap T_1 = \varnothing$: the sets are in conflict with one another, and thus there should be pairwise attacks in at least one direction. Consider two cases, $T_1 \leftarrowtail T_2$ or $T_2\leftarrowtail T_1$.

• Suppose $T_1 \leftarrowtail T_2$. If $T_2\leftarrowtail T_3$ is also the case, then the fact that either $T_1\leftarrowtail T_3$ or else $T_3\leftarrowtail T_1$ should hold make $\leftarrowtail$ not unambiguous. Otherwise, $T_3\leftarrowtail T_2$ should hold and then, similarly, the fact that either $T_1\leftarrowtail T_3$ or else $T_3\leftarrowtail T_1$ hold make $\leftarrowtail$ not unambiguous.

• Suppose $T_2\leftarrowtail T_1$. By an analogous argument, $\leftarrowtail$ is not unambiguous.

Thus, the attack relation is not unambiguous.

B Proof of Theorem 1

The text has already argued for the soundness of the system of Table 1 within topological argumentation models (the validity of some of the interesting axioms is proved in Propositions 8, 9 and 10).

For completeness, It will be shown that any ${\mathsf {L}}_{\mathop {\Box }, \mathop {{\text {U}}}, \mathop {{\mathfrak {T}}}}$-consistent set of ${\mathcal {L}}_{\mathop {\Box }, \mathop {{\text {U}}}, \mathop {{\mathfrak {T}}}}$-formulas is satisfiable. Satisfiability will be proved in an Alexandroff qTA model (see below), which is ${\mathcal {L}}_{\mathop {\Box }, \mathop {{\text {U}}}, \mathop {{\mathfrak {T}}}}$-equivalent to its corresponding TA model.²⁸ In fact, we will prove a stronger completeness result than that stated in Theorem 1.

Theorem 2

The axiom system of Table 1 is strongly complete for the language ${\mathcal {L}}_{\mathop {\Box }, \mathop {{\text {U}}}, \mathop {{\mathfrak {T}}}}$ w.r.t. topological argumentation models which satisfies the following extra condition: for every $T, T_1, T'_1 \in \tau$: if $T_1 \leftarrowtail T$ and $T'_1 \subseteq T_1$, then $T'_1 \leftarrowtail T$.

Next we prove the theorem.

Definition 11

(qTA model) A quasi-topological argumentation model (qTA) is a tuple in which $(W, {\mathcal {E}}_0, \tau , \leftarrowtail , V)$ is a TA model (with $\tau$ generated by ${\mathcal {E}}_0$, as before) and ${\leqslant } \subseteq (W \times W)$ a preorder such that, for every $E \in {\mathcal {E}}_0$, if $u \in E$ and $u \leqslant v$, then $v \in E$.

Formulas in ${\mathcal {L}}_{\mathop {\Box }, \mathop {{\text {U}}}, \mathop {{\mathfrak {T}}}}$ are interpreted in qTA models just as in TA models. The only difference is $\mathop {\Box }$, which becomes a normal universal modality for $\leqslant$. More precisely, iff for all $v \in W$, if $w \leqslant v$ then . Now, two topological definitions, a refined qTA model, and the connection.

Definition 12

(Specification preorder) Let $(X,\tau )$ be a topological space. Its specification preorder ${\sqsubseteq _\tau } \subseteq (X \times X)$ is defined, for any $x, y \in X$, as $x \sqsubseteq _\tau y$ iff for all $T \in \tau$, $x \in T$ implies $y \in T$.

Definition 13

(Alexandroff space) A topological space $(X, \tau )$ is Alexandroff iff $\tau$ is closed under arbitrary intersections (i.e., $\bigcap T \in \tau$ for any $T \subseteq \tau$).

Definition 14

(Alexandroff qTA model) A qTA-model is called Alexandroff iff (i) $(W, \tau _{{\mathcal {E}}_0})$ is Alexandroff, and (ii) ${\leqslant } = {\sqsubseteq _\tau }$.

Proposition 16

Given an Alexandroff qTA model , take $M= (W, {\mathcal {E}}_0, \tau , \leftarrowtail , V)$. Then, for every $\varphi \in {{\mathcal {L}}_{\mathop {\Box }, \mathop {{\text {U}}}, \mathop {{\mathfrak {T}}}}}$.

Proof

Exactly as that of Özgün (2017, Prop. 5.6.14) for topological evidence models and ${\mathcal {L}}_{\mathop {\Box }, \mathop {{\text {U}}}}$, as $\mathop {{\mathfrak {T}}}$ has the same truth condition in qTA and TA models. $\square$

For notation, define $\varGamma ^\bigcirc = \{ \varphi \in {\mathcal {L}}_{\mathop {\Box }, \mathop {{\text {U}}}, \mathop {{\mathfrak {T}}}} \mid \bigcirc \varphi \in \varGamma \}$ for $\varGamma \subseteq {\mathcal {L}}_{\mathop {\Box }, \mathop {{\text {U}}}, \mathop {{\mathfrak {T}}}}$ and $\bigcirc \in \{\mathop {\Box }, \mathop {{\text {U}}}, \mathop {{\mathfrak {T}}}\}$. For the proof, let $\varPhi _0$ be a ${\mathsf {L}}_{\mathop {\Box }, \mathop {{\text {U}}}, \mathop {{\mathfrak {T}}}}$-consistent set of ${\mathcal {L}}_{\mathop {\Box }, \mathop {{\text {U}}}, \mathop {{\mathfrak {T}}}}$-formulas. A slightly modified version of Lindenbaum Lemma shows that it can be extended to a maximal consistent one. Let ${\text {MCS}}$ be the family of all maximally ${\mathsf {L}}_{\mathop {\Box }, \mathop {{\text {U}}}, \mathop {{\mathfrak {T}}}}$-consistent sets of ${\mathcal {L}}_{\mathop {\Box }, \mathop {{\text {U}}}, \mathop {{\mathfrak {T}}}}$-formulas; let $\varPhi$ be an element of ${\text {MCS}}$ extending $\varPhi _0$.

Definition 15

(Canonical qTA model) The canonical qTA model for $\varPhi$, , is defined as follows.

• $W^\varPhi := \{ \varGamma \in {\text {MCS}}\mid \varGamma ^{\mathop {{\text {U}}}} = \varPhi ^{\mathop {{\text {U}}}} \}$ and $V^\varPhi (p) := \{ \varGamma \in W^\varPhi \mid p \in \varGamma \}$.

• For $\varGamma , \varDelta \in W^\varPhi \;$, $\varGamma \leqslant ^\varPhi \varDelta \quad iff _ def \quad \text {for any } \varphi \in {\mathcal {L}}_{\mathop {\Box }, \mathop {{\text {U}}}, \mathop {{\mathfrak {T}}}} \;, \mathop {\Box }\varphi \in \varGamma \text { implies } \varphi \in \varDelta .$

• For any $\varGamma \in W^\varPhi$, define the set ${\leqslant ^{\varPhi }}[\varGamma ] := \{ \varOmega \in W^\varPhi \mid \varGamma \leqslant ^\varPhi \varOmega \}$. Then, let ${\mathcal {E}}_0^\varPhi := \{ \bigcup _{\varGamma \in U} {\leqslant ^{\varPhi }}[\varGamma ] \mid U \subseteq W^\varPhi \} {\setminus } \{\varnothing \}.$

While $\leqslant ^\varPhi$ and $V^{\varPhi }$ are standard (recall: $\mathop {\Box }$ is a normal universal modality for $\leqslant$), each $E \in {\mathcal {E}}_0^{\varPhi }$ is a non-empty union of the $\leqslant ^{\varPhi }$-upwards closure of the elements of some subset of $W^{\varPhi }$. The last component, the attack relation $\leftarrowtail ^{\varPhi }$, is the novel one in this model, and it requires more care. First, define $\{\!\vert \varphi \vert \!\}_M:= \{ \varGamma \in W^{\varPhi } \mid \varphi \in \varGamma \}.$ Then, by taking $\tau ^\varPhi$ to be the topology generated by ${\mathcal {E}}_0^\varPhi$ define, for any $T, T' \in \tau ^\varPhi$,

$$\begin{aligned} \bullet \;\; T \leftarrowtail ^\varPhi T' \quad iff _ def \quad \left\{ \begin{array}{l@{\quad }l@{}} T = \varnothing &{} \text {if } T' = \varnothing \\ T \cap T' = \varnothing \text { and there is no } \varphi \in {\mathcal {L}}_{\mathop {\Box }, \mathop {{\text {U}}}, \mathop {{\mathfrak {T}}}} \text { s.t. } &{} \text {otherwise } \\ \qquad \qquad \qquad \quad \text {both } \{\!\vert \mathop {{\mathfrak {T}}}\varphi \vert \!\} \subseteq T \text { and } \mathop {\widehat{{\text {U}}}}\mathop {{\mathfrak {T}}}\varphi \in \varPhi \\ \end{array} \right. \end{aligned}$$

When no confusion arises, the superscript $\varPhi$ will be omitted.

Note how is indeed a qTA model (Definition 11). First, it is clear that $\varnothing \notin {\mathcal {E}}_0$ and $W \in {\mathcal {E}}_0$. Moreover, $\leqslant$ is indeed a preorder (see its axioms) satisfying the extra condition. Finally, it can be proved that $\leftarrowtail$ satisfies the three conditions.

Lemma 3

Let be the model of Definition 15. Then,

• (i) for every $T_1, T_2 \in \tau$: $T_1 \cap T_2 = \varnothing$ if and only if $T_1 \leftarrowtail T_2$ or $T_2 \leftarrowtail T_1$;

• (ii) for every $T, T_1, T'_1 \in \tau$: if $T_1 \leftarrowtail T$ and $T'_1 \subseteq T_1$, then $T'_1 \leftarrowtail T$;

• (iii) for every $T \in \tau {\setminus } \{\varnothing \}$: $\varnothing \leftarrowtail T$ and $T \not \leftarrowtail \varnothing$.

Proof

• (i) The right-to-left direction is immediate. From left to right, assume $T_1 \cap T_2 = \varnothing$; moreover, for a contradiction, suppose both $T_1 \not \leftarrowtail T_2$ and $T_2 \not \leftarrowtail T_1$. Then, there is $\varphi _1 \in {\mathcal {L}}_{\mathop {\Box }, \mathop {{\text {U}}}, \mathop {{\mathfrak {T}}}}$ such that $\{\!\vert \mathop {{\mathfrak {T}}}\varphi _1 \vert \!\} \subseteq T_1$ and $\mathop {\widehat{{\text {U}}}}\mathop {{\mathfrak {T}}}\varphi _1 \in \varPhi$, and there is $\varphi _2 \in {\mathcal {L}}_{\mathop {\Box }, \mathop {{\text {U}}}, \mathop {{\mathfrak {T}}}}$ such that $\{\!\vert \mathop {{\mathfrak {T}}}\varphi _2 \vert \!\} \subseteq T_2$ and $\mathop {\widehat{{\text {U}}}}\mathop {{\mathfrak {T}}}\varphi _2 \in \varPhi$. It follows that $\{\!\vert \mathop {{\mathfrak {T}}}\varphi _1 \vert \!\} \cap \{\!\vert \mathop {{\mathfrak {T}}}\varphi _2 \vert \!\} = \varnothing$; to finish the proof, it is enough to show

Lemma 4

For any $\varphi _1, \varphi _2 \in {\mathcal {L}}_{\mathop {\Box }, \mathop {{\text {U}}}, \mathop {{\mathfrak {T}}}}$, having $\{\!\vert \mathop {{\mathfrak {T}}}\varphi _1 \vert \!\} \cap \{\!\vert \mathop {{\mathfrak {T}}}\varphi _2 \vert \!\} = \varnothing$ and both $\mathop {\widehat{{\text {U}}}}\mathop {{\mathfrak {T}}}\varphi _1 \in \varPhi$ and $\mathop {\widehat{{\text {U}}}}\mathop {{\mathfrak {T}}}\varphi _2 \in \varPhi$ leads to a contradiction.

Proof

From $\{\!\vert \mathop {{\mathfrak {T}}}\varphi _1 \vert \!\} \cap \{\!\vert \mathop {{\mathfrak {T}}}\varphi _2 \vert \!\} = \varnothing$ and theorem $\mathop {{\mathfrak {T}}}\mathop {{\mathfrak {T}}}\varphi \rightarrow \mathop {{\mathfrak {T}}}\varphi$ it follows that $\{\!\vert \mathop {{\mathfrak {T}}}\mathop {{\mathfrak {T}}}\varphi _1 \vert \!\} \cap \{\!\vert \mathop {{\mathfrak {T}}}\mathop {{\mathfrak {T}}}\varphi _2 \vert \!\} = \varnothing$. Moreover: $\mathop {\widehat{{\text {U}}}}\mathop {{\mathfrak {T}}}\varphi _1 \in \varPhi$ and $\mathop {\widehat{{\text {U}}}}\mathop {{\mathfrak {T}}}\varphi _2 \in \varPhi$ imply, respectively, $\{\!\vert \mathop {{\mathfrak {T}}}\varphi _1 \vert \!\} \ne \varnothing$ and $\{\!\vert \mathop {{\mathfrak {T}}}\varphi _2 \vert \!\} \ne \varnothing$ (Proposition 17 below); this, together with axiom $\mathop {{\mathfrak {T}}}\varphi \rightarrow \mathop {{\mathfrak {T}}}\mathop {{\mathfrak {T}}}\varphi$, yields both $\{\!\vert \mathop {{\mathfrak {T}}}\mathop {{\mathfrak {T}}}\varphi _1 \vert \!\} \ne \varnothing$ and $\{\!\vert \mathop {{\mathfrak {T}}}\mathop {{\mathfrak {T}}}\varphi _2 \vert \!\} \ne \varnothing$, which imply (Proposition 17) $\mathop {\widehat{{\text {U}}}}\mathop {{\mathfrak {T}}}\mathop {{\mathfrak {T}}}\varphi _1 \in \varPhi$ and $\mathop {\widehat{{\text {U}}}}\mathop {{\mathfrak {T}}}\mathop {{\mathfrak {T}}}\varphi _2 \in \varPhi$.

Observe how $\{\!\vert \mathop {{\mathfrak {T}}}\varphi _1 \vert \!\} \cap \{\!\vert \mathop {{\mathfrak {T}}}\varphi _2 \vert \!\} = \varnothing$ also implies $\{\!\vert \mathop {{\mathfrak {T}}}\varphi _1 \vert \!\} \subseteq \{\!\vert \lnot \mathop {{\mathfrak {T}}}\varphi _2 \vert \!\}$, so $\mathop {{\text {U}}}(\mathop {{\mathfrak {T}}}\varphi _1 \rightarrow \lnot \mathop {{\mathfrak {T}}}\varphi _2) \in \varPhi$; then, from Proposition 7 it follows that $\mathop {{\text {U}}}(\mathop {{\mathfrak {T}}}\mathop {{\mathfrak {T}}}\varphi _1 \rightarrow \mathop {{\mathfrak {T}}}\lnot \mathop {{\mathfrak {T}}}\varphi _2) \in \varPhi$. From the latter and $\mathop {\widehat{{\text {U}}}}\mathop {{\mathfrak {T}}}\mathop {{\mathfrak {T}}}\varphi _1 \in \varPhi$ we get $\mathop {\widehat{{\text {U}}}}\mathop {{\mathfrak {T}}}\lnot \mathop {{\mathfrak {T}}}\varphi _2 \in \varPhi$; but then, axiom $\mathop {\widehat{{\text {U}}}}\mathop {{\mathfrak {T}}}\varphi \rightarrow \lnot \mathop {\widehat{{\text {U}}}}\mathop {{\mathfrak {T}}}\lnot \varphi$ and $\mathop {\widehat{{\text {U}}}}\mathop {{\mathfrak {T}}}\mathop {{\mathfrak {T}}}\varphi _2 \in \varPhi$ imply $\lnot \mathop {\widehat{{\text {U}}}}\mathop {{\mathfrak {T}}}\lnot \mathop {{\mathfrak {T}}}\varphi _2 \in \varPhi$. Thus, we have a contradiction, as $\varPhi$ is consistent.

• (ii) Take $T, T_1, T'_1 \in \tau$ with $T_1 \leftarrowtail T$ and $T'_1 \subseteq T_1$. If $T = \varnothing$, the case is trivial ($T_1$ should be $\varnothing$, and so $T'_1$), so suppose $T \ne \varnothing$. From $T_1 \leftarrowtail T$, there is no $\varphi \in {\mathcal {L}}_{\mathop {\Box }, \mathop {{\text {U}}}, \mathop {{\mathfrak {T}}}}$ such that $\{\!\vert \mathop {{\mathfrak {T}}}\varphi \vert \!\} \subseteq T_1$ and $\mathop {\widehat{{\text {U}}}}\mathop {{\mathfrak {T}}}\varphi \in \varPhi$. But $T'_1 \subseteq T_1$, so no such $\varphi$ exists for $T'_1$ either. Moreover, $T_1 \cap T = \varnothing$ so $T'_1 \cap T = \varnothing$; hence, $T'_1 \leftarrowtail T$.

• (iii) Immediate.$\square$

Thus, is a qTA model. The next proposition (standard proof) provides existence lemmas for the standard modality $\mathop {\Box }$ and the global modality $\mathop {\widehat{{\text {U}}}}$.

Proposition 17

For any $\varphi \in {\mathcal {L}}_{\mathop {\Box }, \mathop {{\text {U}}}, \mathop {{\mathfrak {T}}}}$ and any $\varGamma \in W$:

• $\mathop {\Diamond }\varphi \in \varGamma$ iff there is $\varDelta \in W$ s.t. $\varGamma \leqslant \varDelta$ and $\varphi \in \varDelta$.

• $\mathop {\widehat{{\text {U}}}}\varphi \in \varGamma$ iff there is $\varDelta \in W$ s.t. $\varphi \in \varDelta$;

Now, tools to prove a similar result for the operator $\mathop {{\mathfrak {T}}}$, whose truth clause relies on ${{\,\mathrm{\mathsf {LFP}}\,}}$, given by $\leftarrowtail$. First, some useful properties of the model.

Fact 2

(i) $\tau = {\mathcal {E}}_0\cup \{\varnothing \}$. (ii) If $\mathop {\widehat{{\text {U}}}}\mathop {\Box }\varphi \in \varPhi$, then $\{\!\vert \mathop {\Box }\varphi \vert \!\} \in \tau$. (iii) If $\mathop {\widehat{{\text {U}}}}\mathop {{\mathfrak {T}}}\varphi \in \varPhi$, then $\{\!\vert \mathop {{\mathfrak {T}}}\varphi \vert \!\} \in \tau$. (iv) For any $T \in \tau$ and any $\varphi \in {\mathcal {L}}_{\mathop {\Box }, \mathop {{\text {U}}}, \mathop {{\mathfrak {T}}}}$: if $T \subseteq \{\!\vert \varphi \vert \!\}$, then $T \subseteq \{\!\vert \mathop {\Box }\varphi \vert \!\}$.

Proof

• (i) Immediate, as $\{ \bigcup _{\varGamma \in U} {\leqslant ^{\varPhi }}[\varGamma ] \mid U \subseteq W^\varPhi \}$ is closed under intersections and unions.

• (ii) Analogous to the next one.

• (iii) Assume $\mathop {\widehat{{\text {U}}}}\mathop {{\mathfrak {T}}}\varphi \in \varPhi$; then, there is at least one $\varGamma \in W$ such that $\mathop {{\mathfrak {T}}}\varphi \in \varGamma$, that is, $\{\!\vert \mathop {{\mathfrak {T}}}\varphi \vert \!\} \ne \varnothing$. Now, note how $\{\!\vert \mathop {{\mathfrak {T}}}\varphi \vert \!\} = \bigcup _{\varGamma \in \{\!\vert \mathop {{\mathfrak {T}}}\varphi \vert \!\}} {\leqslant }[\varGamma ]$.²⁹ Moreover, $\{\!\vert \mathop {{\mathfrak {T}}}\varphi \vert \!\} \subseteq W$ so, by ${\mathcal {E}}_0$’s definition, $\bigcup _{\varGamma \in \{\!\vert \mathop {{\mathfrak {T}}}\varphi \vert \!\}} {\leqslant }[\varGamma ] = \{\!\vert \mathop {{\mathfrak {T}}}\varphi \vert \!\} \in {\mathcal {E}}_0$; then, by the first item, $\{\!\vert \mathop {{\mathfrak {T}}}\varphi \vert \!\} \in \tau$.

• (iv) First, note how $T = \bigcup _{\varGamma \in T} {\leqslant }[\varGamma ]$.³⁰ Then, from $T \subseteq \{\!\vert \varphi \vert \!\}$ it follows that $\bigcup _{\varGamma \in T} {\leqslant }[\varGamma ] \subseteq \{\!\vert \varphi \vert \!\}$, that is, ${\leqslant }[\varGamma ] \subseteq \{\!\vert \varphi \vert \!\}$ for every $\varGamma \in T$; therefore, $\mathop {\Box }\varphi \in \varGamma$ (Proposition 17) for every such $\varGamma$, and hence $T \subseteq \{\!\vert \mathop {\Box }\varphi \vert \!\}$.

$\square$

Here are the first steps towards locating ${{\,\mathrm{\mathsf {LFP}}\,}}$.

Definition 16

(Semi-acceptable and acceptable) Define ${\mathcal {C}}_1$ as

$$\begin{aligned} {\mathcal {C}}_1= \{ T \in \tau \mid \text {there exists } \varphi \in {\mathcal {L}}_{\mathop {\Box }, \mathop {{\text {U}}}, \mathop {{\mathfrak {T}}}} \text { such that } \{\!\vert \mathop {{\mathfrak {T}}}\varphi \vert \!\} \subseteq T \text { and } \mathop {\widehat{{\text {U}}}}\mathop {{\mathfrak {T}}}\varphi \in \varPhi \} \end{aligned}$$

• An open $T \in \tau$ is semi-acceptable if and only if, for any $\psi \in {\mathcal {L}}_{\mathop {\Box }, \mathop {{\text {U}}}, \mathop {{\mathfrak {T}}}}$ with $T \subseteq \{\!\vert \mathop {\Box }\psi \vert \!\}$, there is $\xi \in {\mathcal {L}}_{\mathop {\Box }, \mathop {{\text {U}}}, \mathop {{\mathfrak {T}}}}$ such that $\{\!\vert \mathop {{\mathfrak {T}}}\xi \vert \!\} \subseteq \{\!\vert \mathop {\Box }\psi \vert \!\}$ and $\mathop {\widehat{{\text {U}}}}\mathop {{\mathfrak {T}}}\xi \in \varPhi$.

• An open $T \in \tau$ is acceptable if and only if T is semi-acceptable and there is no $T' \in \tau$ such that $T \cap T' = \varnothing$ and $T' \cap T'' \ne \varnothing$ for all $T'' \in {\mathcal {C}}_1$.

Define ${\mathcal {C}}_2$ as ${\mathcal {C}}_2= \{T \in \tau {\setminus } {\mathcal {C}}_1\mid T \text { is acceptable} \}$.

Note that no element of ${\mathcal {C}}_1$ is attacked by elements of $\tau$. Moreover,

Fact 3

(i) For any $T \in \tau$, if $T \in {\mathcal {C}}_1$, then T is acceptable. (ii) If $T \in \tau$ is semi-acceptable, then $T \cap T' \ne \varnothing$ for all $T' \in {\mathcal {C}}_1$.

Proof

• (i) If $T \in {\mathcal {C}}_1$, then there is $\varphi \in {\mathcal {L}}_{\mathop {\Box }, \mathop {{\text {U}}}, \mathop {{\mathfrak {T}}}}$ with $\{\!\vert \mathop {{\mathfrak {T}}}\varphi \vert \!\} \subseteq T$ and $\mathop {\widehat{{\text {U}}}}\mathop {{\mathfrak {T}}}\varphi \in \varPhi$. For semi-acceptability, take any $\psi \in {\mathcal {L}}_{\mathop {\Box }, \mathop {{\text {U}}}, \mathop {{\mathfrak {T}}}}$ with $T \subseteq \{\!\vert \mathop {\Box }\psi \vert \!\}$; the initial $\varphi$ satisfies both $\{\!\vert \mathop {{\mathfrak {T}}}\xi \vert \!\} \subseteq \{\!\vert \mathop {\Box }\psi \vert \!\}$ and $\mathop {\widehat{{\text {U}}}}\mathop {{\mathfrak {T}}}\xi \in \varPhi$. For the second condition of acceptability, $T\in {\mathcal {C}}_1$, so for any $T'\in \tau$ such that $T'\cap T'' \ne \varnothing$ for all $T''\in {\mathcal {C}}_1$, it is the case that $T'\cap T \ne \varnothing$.

• (ii) It will be proved that, for any $\varphi \in {\mathcal {L}}_{\mathop {\Box }, \mathop {{\text {U}}}, \mathop {{\mathfrak {T}}}}$ such that $\mathop {\widehat{{\text {U}}}}\mathop {{\mathfrak {T}}}\varphi \in \varPhi$, any semi-acceptable open $T \in \tau$ satisfies that $T \cap \{\!\vert \mathop {{\mathfrak {T}}}\varphi \vert \!\} \ne \varnothing$. Fact 3.(ii) follows, because every open in ${\mathcal {C}}_1$ should be a superset of $\{\!\vert \mathop {{\mathfrak {T}}}\varphi \vert \!\}$ for some $\varphi \in {\mathcal {L}}_{\mathop {\Box }, \mathop {{\text {U}}}, \mathop {{\mathfrak {T}}}}$.

  For a contradiction, suppose there is $\varphi \in {\mathcal {L}}_{\mathop {\Box }, \mathop {{\text {U}}}, \mathop {{\mathfrak {T}}}}$ such that $\mathop {\widehat{{\text {U}}}}\mathop {{\mathfrak {T}}}\varphi \in \varPhi$ and $T \cap \{\!\vert \mathop {{\mathfrak {T}}}\varphi \vert \!\} = \varnothing$ for a semi-acceptable T. Now take any $\psi \in {\mathcal {L}}_{\mathop {\Box }, \mathop {{\text {U}}}, \mathop {{\mathfrak {T}}}}$ such that $T \subseteq \{\!\vert \mathop {\Box }\psi \vert \!\}$; then, $T\subseteq \{\!\vert \mathop {\Box }\psi \wedge \lnot \mathop {{\mathfrak {T}}}\varphi \vert \!\}$. By Item (iv) of Fact 2, $T \subseteq \{\!\vert \mathop {\Box }(\mathop {\Box }\psi \wedge \lnot \mathop {{\mathfrak {T}}}\varphi ) \vert \!\}$. But T is semi-acceptable, so there is $\xi \in {\mathcal {L}}_{\mathop {\Box }, \mathop {{\text {U}}}, \mathop {{\mathfrak {T}}}}$ such that $\{\!\vert \mathop {{\mathfrak {T}}}\xi \vert \!\} \subseteq \{\!\vert \mathop {\Box }(\mathop {\Box }\psi \wedge \lnot \mathop {{\mathfrak {T}}}\varphi ) \vert \!\}$ and $\mathop {\widehat{{\text {U}}}}\mathop {{\mathfrak {T}}}\xi \in \varPhi$. Now, $\vdash \mathop {\Box }(\mathop {\Box }\psi \wedge \lnot \mathop {{\mathfrak {T}}}\varphi ) \rightarrow \lnot \mathop {{\mathfrak {T}}}\varphi$, so $\{\!\vert \mathop {{\mathfrak {T}}}\xi \vert \!\} \subseteq \{\!\vert \lnot \mathop {{\mathfrak {T}}}\varphi \vert \!\}$, that is, $\{\!\vert \mathop {{\mathfrak {T}}}\xi \vert \!\} \cap \{\!\vert \mathop {{\mathfrak {T}}}\varphi \vert \!\} = \varnothing$. The contradiction then follows from Lemma 4.

$\square$

Lemma 5

Let ${\mathcal {C}}= {\mathcal {C}}_1\cup {\mathcal {C}}_2$. Then, ${{\,\mathrm{\mathsf {LFP}}\,}}= {\mathcal {C}}$.

Proof

$\varvec{(\supseteq })$ The proof of this direction consists of proving two cases. (i) If $T \in {\mathcal {C}}_1$, there is $\varphi \in {\mathcal {L}}_{\mathop {\Box }, \mathop {{\text {U}}}, \mathop {{\mathfrak {T}}}}$ such that $\{\!\vert \mathop {{\mathfrak {T}}}\varphi \vert \!\} \subseteq T$ and $\mathop {\widehat{{\text {U}}}}\mathop {{\mathfrak {T}}}\varphi \in \varPhi$. Hence, by $\leftarrowtail$’s definition, there is no $T' \in \tau$ with $T \cap T' = \varnothing$ such that $T \leftarrowtail T'$;³¹ thus, $T \in {{\,\mathrm{\mathsf {LFP}}\,}}$. Therefore, ${\mathcal {C}}_1\subseteq {{\,\mathrm{\mathsf {LFP}}\,}}$. (ii) Otherwise, $T \in {\mathcal {C}}_2$. Take any $T' \in \tau$ such that $T \leftarrowtail T'$. From $\leftarrowtail$’s definition, $T' \cap T = \varnothing$. From ${\mathcal {C}}_2$’s definition and $T \cap T' = \varnothing$, there is $T'' \in {\mathcal {C}}_1$ such that $T' \cap T'' = \varnothing$. From the previous case, $T''$ cannot be attacked, i.e., $T'' \not \leftarrowtail T'$; then, by the first (iii) on Lemma 3, $T' \leftarrowtail T''$. Summarising, for any $T' \in \tau$ attacking T, there is $T'' \in {\mathcal {C}}_1$ attacking $T'$; hence, $T \in d({\mathcal {C}}_1) \subseteq d({{\,\mathrm{\mathsf {LFP}}\,}}) = {{\,\mathrm{\mathsf {LFP}}\,}}$, that is, $T \in {{\,\mathrm{\mathsf {LFP}}\,}}$. Therefore, ${\mathcal {C}}_2\subseteq {{\,\mathrm{\mathsf {LFP}}\,}}$.

$\varvec{(\subseteq })$ Take now $T \in \tau$ such that $T \notin {\mathcal {C}}$; it will be shown that $T \notin {{\,\mathrm{\mathsf {LFP}}\,}}$. The case with $T = \varnothing$ is immediate, as $\varnothing \leftarrowtail \varnothing$. Thus, suppose $T \ne \varnothing$.

From $T \notin {\mathcal {C}}$ it follows that $T \notin {\mathcal {C}}_1$, so there is no $\phi \in {\mathcal {L}}_{\mathop {\Box }, \mathop {{\text {U}}}, \mathop {{\mathfrak {T}}}}$ such that $\{\!\vert \mathop {{\mathfrak {T}}}\phi \vert \!\} \subseteq T$ and $\mathop {\widehat{{\text {U}}}}\mathop {{\mathfrak {T}}}\phi \in \varPhi$; hence, from $\leftarrowtail$’s definition, every $T' \in \tau$ with $T \cap T' = \varnothing$ satisfies that $T \leftarrowtail T'$.

Note that there is at least one $T' \in \tau$ with $T \cap T' = \varnothing$, for suppose otherwise, i.e., suppose $T' \in \tau$ implies $T \cap T' \ne \varnothing$. Now, take any $\varGamma \in W$; since ${\leqslant }[\varGamma ] \in \tau$ (from ${\leqslant }[\varGamma ] \in {\mathcal {E}}_0$ and Item (i) of Fact 2), it follows that ${\leqslant }[\varGamma ] \cap T \ne \varnothing$, i.e., there is $\varDelta$ with $\varGamma \leqslant \varDelta$ and $\varDelta \in T$. Furthermore, take any $\varphi \in {\mathcal {L}}_{\mathop {\Box }, \mathop {{\text {U}}}, \mathop {{\mathfrak {T}}}}$ with $T \subseteq \{\!\vert \varphi \vert \!\}$; by Item (iv) of Fact 2, $T \subseteq \{\!\vert \mathop {\Box }\varphi \vert \!\}$ and then, from $\varGamma \leqslant \varDelta$ and $\varDelta \in \{\!\vert \mathop {\Box }\varphi \vert \!\}$, Proposition 17 gives us $\varGamma \in \{\!\vert \mathop {\Diamond }\mathop {\Box }\varphi \vert \!\}$, i.e., $\mathop {\Diamond }\mathop {\Box }\varphi \in \varGamma$. Thus, it has been shown that for any $\varGamma \in W$ and any $\varphi \in {\mathcal {L}}_{\mathop {\Box }, \mathop {{\text {U}}}, \mathop {{\mathfrak {T}}}}$ with $T \subseteq \{\!\vert \varphi \vert \!\}$, we have $\mathop {\Diamond }\mathop {\Box }\varphi \in \varGamma$. Then, from Proposition 17 it follows that any $\varphi \in {\mathcal {L}}_{\mathop {\Box }, \mathop {{\text {U}}}, \mathop {{\mathfrak {T}}}}$ with $T \subseteq \{\!\vert \varphi \vert \!\}$ satisfies that $\mathop {{\text {U}}}\mathop {\Diamond }\mathop {\Box }\varphi \in \varPhi$; thus, axiom $\mathop {{\text {U}}}\mathop {\Diamond }\mathop {\Box }\varphi \rightarrow \mathop {\widehat{{\text {U}}}}\mathop {{\mathfrak {T}}}\varphi$ implies $\mathop {\widehat{{\text {U}}}}\mathop {{\mathfrak {T}}}\varphi \in \varPhi$. Moreover, axiom $\mathop {{\mathfrak {T}}}\varphi \rightarrow \mathop {\Box }\varphi$ implies $\{\!\vert \mathop {{\mathfrak {T}}}\varphi \vert \!\} \subseteq \{\!\vert \mathop {\Box }\varphi \vert \!\}$. Thus, for any $\varphi \in {\mathcal {L}}_{\mathop {\Box }, \mathop {{\text {U}}}, \mathop {{\mathfrak {T}}}}$ with $T \subseteq \{\!\vert \mathop {\Box }\varphi \vert \!\}$ we have found a formula in ${\mathcal {L}}_{\mathop {\Box }, \mathop {{\text {U}}}, \mathop {{\mathfrak {T}}}}$, $\varphi$ itself, such that $\mathop {\widehat{{\text {U}}}}\mathop {{\mathfrak {T}}}\varphi \in \varPhi$ and $\{\!\vert \mathop {{\mathfrak {T}}}\varphi \vert \!\} \subseteq \{\!\vert \mathop {\Box }\varphi \vert \!\}$; hence, T is semi-acceptable. But $T \cap T' \ne \varnothing$ for all $T'\in \tau$, so there is no $T' \in \tau$ such that both $T \cap T' = \varnothing$ and, for any $T'' \in {\mathcal {C}}_1$, $T \cap T'' \ne \varnothing$: in other words, T is acceptable. Hence, $T \in {\mathcal {C}}_2$, i.e., $T \in {\mathcal {C}}$: a contradiction. Thus, indeed there must be $T' \in \tau$ such that $T \cap T' = \varnothing$.

The rest of the proof is divided into two cases: either there is $T' \in \tau$ with $T \cap T' = \varnothing$ and $T' \in {\mathcal {C}}$ (at least one $T'$ contradicting T is in ${\mathcal {C}}$), or else for any $T'\in \tau$ with $T \cap T' = \varnothing$ we have $T'\notin {\mathcal {C}}$ (no $T'$ contradicting T is in ${\mathcal {C}}$). In both cases, it will be shown that $T \notin {{\,\mathrm{\mathsf {LFP}}\,}}$.

• (i) The first case is the simple one: take any $T' \in \tau$ such that $T \cap T' = \varnothing$ and $T' \in {\mathcal {C}}$. Then, as it has been argued, $T \leftarrowtail T'$; moreover, as it has been proved, ${\mathcal {C}}\subseteq {{\,\mathrm{\mathsf {LFP}}\,}}$. Thus, $T \notin {{\,\mathrm{\mathsf {LFP}}\,}}$, as ${{\,\mathrm{\mathsf {LFP}}\,}}$ has to be conflict-free.

• (ii) The second case requires more care. Since no element of ${\mathcal {C}}$ contradicts T, it follows that $C \in {\mathcal {C}}$ implies $T \cap C \ne \varnothing$. Now, consider the following two sub-cases: either T is semi-acceptable, or it is not. We will prove that, in both cases, $T\notin d({\mathcal {C}})$.

  If T is semi-acceptable, recall the initial assumption $T \notin C = {\mathcal {C}}_1\cup {\mathcal {C}}_2$. Then T cannot be acceptable, that is, there must be $T'' \in \tau$ such that $T \cap T'' = \varnothing$ and, for any $C_1 \in {\mathcal {C}}_1$, we have $T'' \cap C_1 \ne \varnothing$. Now, take any such $T''$. From $T \notin C$, $T \cap T'' = \varnothing$ and $T'' \ne \varnothing$ (as its intersection with $C_1$ is non-empty), it follows that $T \leftarrowtail T''$: this $T''$ attacks T. Since $T'' \cap C_1 \ne \varnothing$ for any $C_1 \in {\mathcal {C}}_1$, there is no $C_1 \in {\mathcal {C}}_1$ such that $T'' \leftarrowtail C_1$. Moreover, there is no $C_2 \in {\mathcal {C}}_2$ such that $T'' \leftarrowtail C_2$ – otherwise, $T'' \cap C_2 = \varnothing$ would imply $C_2 \notin {\mathcal {C}}_2$. So, in summary, if T is semi-acceptable, then there is $T'' \in \tau$ such that $T \leftarrowtail T''$; there is no $C \in {\mathcal {C}}$ such that $T'' \leftarrowtail C$. Therefore, $T \notin d({\mathcal {C}})$.

  If T is not semi-acceptable, there is $\varphi _T \in {\mathcal {L}}_{\mathop {\Box }, \mathop {{\text {U}}}, \mathop {{\mathfrak {T}}}}$ such that $T \subseteq \{\!\vert \mathop {\Box }\varphi _T \vert \!\}$ and there is no $\psi \in {\mathcal {L}}_{\mathop {\Box }, \mathop {{\text {U}}}, \mathop {{\mathfrak {T}}}}$ such that both $\{\!\vert \mathop {{\mathfrak {T}}}\psi \vert \!\} \subseteq \{\!\vert \mathop {\Box }\varphi _T \vert \!\}$ and $\mathop {\widehat{{\text {U}}}}\mathop {{\mathfrak {T}}}\psi \in \varPhi$. In particular, $\varphi _T$ itself cannot be such $\psi$, so either $\{\!\vert \mathop {{\mathfrak {T}}}\varphi _T \vert \!\} \not \subseteq \{\!\vert \mathop {\Box }\varphi _T \vert \!\}$ or else $\mathop {\widehat{{\text {U}}}}\mathop {{\mathfrak {T}}}\varphi _T \not \in \varPhi$. But axiom $\mathop {{\mathfrak {T}}}\varphi \rightarrow \mathop {\Box }\varphi$ implies $\{\!\vert \mathop {{\mathfrak {T}}}\varphi _T \vert \!\}\subseteq \{\!\vert \mathop {\Box }\varphi _T \vert \!\}$, so $\mathop {\widehat{{\text {U}}}}\mathop {{\mathfrak {T}}}\varphi _T \notin \varPhi$. Now, take any $C \in {\mathcal {C}}_1$; let $\varphi _C \in {\mathcal {L}}_{\mathop {\Box }, \mathop {{\text {U}}}, \mathop {{\mathfrak {T}}}}$ be one of the formulas satisfying both $\{\!\vert \mathop {{\mathfrak {T}}}\varphi _C \vert \!\} \subseteq C$ and $\mathop {\widehat{{\text {U}}}}\mathop {{\mathfrak {T}}}\varphi _C \in \varPhi$ (by ${\mathcal {C}}$’s definition, there is at least one). From theorem $\mathop {{\mathfrak {T}}}\varphi \rightarrow \mathop {\Box }\mathop {{\mathfrak {T}}}\varphi$, it follows that $(\mathop {\Box }\varphi _T \wedge \mathop {{\mathfrak {T}}}\varphi _C) \rightarrow (\mathop {\Box }\varphi _T \wedge \mathop {\Box }\mathop {{\mathfrak {T}}}\varphi _C)$ is a theorem too, and thus so are $(\mathop {\Box }\varphi _T \wedge \mathop {{\mathfrak {T}}}\varphi _C) \rightarrow (\mathop {\Box }\mathop {\Box }\varphi _T \wedge \mathop {\Box }\mathop {{\mathfrak {T}}}\varphi _C)$ (by axiom $\mathop {\Box }\varphi \rightarrow \mathop {\Box }\mathop {\Box }\varphi$) and $(\mathop {\Box }\varphi _T \wedge \mathop {{\mathfrak {T}}}\varphi _C) \rightarrow \mathop {\Box }(\mathop {\Box }\varphi _T \wedge \mathop {{\mathfrak {T}}}\varphi _C)$ (axiom K for $\mathop {\Box }$). Hence, by Proposition 17, $\mathop {{\text {U}}}\big ( (\mathop {\Box }\varphi _T \wedge \mathop {{\mathfrak {T}}}\varphi _C) \rightarrow \mathop {\Box }(\mathop {\Box }\varphi _T \wedge \mathop {{\mathfrak {T}}}\varphi _C) \big ) \in \varPhi$.

  So far we have $\mathop {\widehat{{\text {U}}}}\mathop {{\mathfrak {T}}}\varphi _T \notin \varPhi$ and, for every $C \in {\mathcal {C}}_1$, not only $\mathop {\widehat{{\text {U}}}}\mathop {{\mathfrak {T}}}\varphi _C \in \varPhi$ but also $\mathop {{\text {U}}}\big ( (\mathop {\Box }\varphi _T \wedge \mathop {{\mathfrak {T}}}\varphi _C) \rightarrow \mathop {\Box }(\mathop {\Box }\varphi _T \wedge \mathop {{\mathfrak {T}}}\varphi _C) \big ) \in \varPhi$. The first and theorem $\mathop {{\mathfrak {T}}}\varphi \leftrightarrow \mathop {{\mathfrak {T}}}\mathop {\Box }\varphi$ imply $\mathop {\widehat{{\text {U}}}}\mathop {{\mathfrak {T}}}\mathop {\Box }\varphi _T \notin \varPhi$; the second and axiom $\mathop {{\mathfrak {T}}}\varphi \rightarrow \mathop {{\mathfrak {T}}}\mathop {{\mathfrak {T}}}\varphi$ imply $\mathop {\widehat{{\text {U}}}}\mathop {{\mathfrak {T}}}\mathop {{\mathfrak {T}}}\varphi _C \in \varPhi$. These two, the third, and axiom $\big (\mathop {\widehat{{\text {U}}}}\mathop {{\mathfrak {T}}}\varphi \wedge \lnot \mathop {\widehat{{\text {U}}}}\mathop {{\mathfrak {T}}}\psi \wedge \mathop {{\text {U}}}((\varphi \wedge \psi ) \rightarrow \mathop {\Box }(\varphi \wedge \psi ))\big ) \rightarrow \mathop {\widehat{{\text {U}}}}\mathop {\Box }(\varphi \wedge \lnot \psi )$ imply $\mathop {\widehat{{\text {U}}}}\mathop {\Box }(\mathop {{\mathfrak {T}}}\varphi _C \wedge \lnot \mathop {\Box }\varphi _T) \in \varPhi$. For the final part, take S to be the union of $\{\!\vert \mathop {\Box }(\mathop {{\mathfrak {T}}}\varphi _C \wedge \lnot \mathop {\Box }\varphi _T) \vert \!\}$ for all $C \in {\mathcal {C}}_1$, that is,

  $$\begin{aligned} S := \bigcup _{C \in {\mathcal {C}}_1} \{\!\vert \mathop {\Box }(\mathop {{\mathfrak {T}}}\varphi _C \wedge \lnot \mathop {\Box }\varphi _T) \vert \!\} \end{aligned}$$

The following two facts about S are crucial for proving $T \notin d({\mathcal {C}})$:

• (a) $S \cap T = \varnothing$. For its proof, from $T \subseteq \{\!\vert \mathop {\Box }\varphi _T \vert \!\}$ and $\{\!\vert \mathop {\Box }(\mathop {{\mathfrak {T}}}\varphi _C \wedge \lnot \mathop {\Box }\varphi _T) \vert \!\} \subseteq \{\!\vert \lnot \mathop {\Box }\varphi _T \vert \!\}$ for all $C \in {\mathcal {C}}_1$ (for the latter, use axiom $\mathop {\Box }\varphi \rightarrow \varphi$), it follows that $T \cap \{\!\vert \mathop {\Box }(\mathop {{\mathfrak {T}}}\varphi _C \wedge \lnot \mathop {\Box }\varphi _T) \vert \!\} = \varnothing$ for all such C. Therefore, $S \cap T = \varnothing$.

• (b) For any $C' \in {\mathcal {C}}$, we have $C' \cap S \ne \varnothing$. For its proof, take any $C' \in {\mathcal {C}}$. If $C' \in {\mathcal {C}}_1$ then, as $\{\!\vert \mathop {\Box }(\mathop {{\mathfrak {T}}}\varphi _{C'} \wedge \lnot \mathop {\Box }\varphi _T) \vert \!\} \subseteq \{\!\vert \mathop {{\mathfrak {T}}}\varphi _{C'} \vert \!\}$ (same as before) and $\{\!\vert \mathop {\Box }(\mathop {{\mathfrak {T}}}\varphi _{C'} \wedge \lnot \mathop {\Box }\varphi _T) \vert \!\}\ne \varnothing$ (by $\mathop {\widehat{{\text {U}}}}\mathop {\Box }(\mathop {{\mathfrak {T}}}\varphi _C \wedge \lnot \mathop {\Box }\varphi _T) \in \varPhi$), it follows that $\{\!\vert \mathop {\Box }(\mathop {{\mathfrak {T}}}\varphi _{C'} \wedge \lnot \mathop {\Box }\varphi _T) \vert \!\} \cap \{\!\vert \mathop {{\mathfrak {T}}}\varphi _{C'} \vert \!\} \ne \varnothing$. But $\{\!\vert \mathop {{\mathfrak {T}}}\varphi _{C'} \vert \!\} \subseteq C'$ for any $C' \in {\mathcal {C}}_1$; hence, $S \cap C' \ne \varnothing$. Otherwise, $C' \in {\mathcal {C}}_2$ and, for a contradiction, assume $S \cap C' = \varnothing$. Then, since $S\cap C''\ne \varnothing$ for all $C''\in {\mathcal {C}}_1$ as we have proved, there is an open in $\tau$, S, such that $S\cap C' = \varnothing$ and $S\cap C''\ne \varnothing$ for all $C''\in {\mathcal {C}}_1$. Thus, $C'$ violates the second condition of acceptance, and hence $C'\notin {\mathcal {C}}_2$: contradiction. So $S\cap C' \ne \varnothing$ for any $C'\in {\mathcal {C}}_2$. Therefore, in sum, $C' \in {\mathcal {C}}$ implies $C' \cap S \ne \varnothing$

Now, since $S \cap T = \varnothing$ and T is not semi-acceptable (so there is no $\varphi \in {\mathcal {L}}_{\mathop {\Box }, \mathop {{\text {U}}}, \mathop {{\mathfrak {T}}}}$ s.t. both $\{\!\vert \mathop {{\mathfrak {T}}}\varphi \vert \!\} \subseteq T$ and $\mathop {\widehat{{\text {U}}}}\mathop {{\mathfrak {T}}}\varphi \in \varPhi$), we have found an open S in $\tau$ with $T \leftarrowtail S$, according to the definition of $\leftarrowtail$. But $S \cap C \ne \varnothing$ for all $C \in {\mathcal {C}}$, so $S \not \leftarrowtail C$ for all $C \in {\mathcal {C}}$: no open in ${\mathcal {C}}$ attacks S. Hence, $T \notin d({\mathcal {C}})$.

Therefore, regardless of whether T is semi-acceptable or not, we have proved that $T\notin {\mathcal {C}}$ implies that $T \notin d({\mathcal {C}})$.

By the fact that ${\mathcal {C}}_1$ is not attacked and ${\mathcal {C}}_2\subseteq d({\mathcal {C}}_1)$, which we have proved in our proof of ${\mathcal {C}}\subseteq {{\,\mathrm{\mathsf {LFP}}\,}}$, it follows that ${\mathcal {C}}\subseteq d({\mathcal {C}})$. Together with $d({\mathcal {C}})\subseteq {\mathcal {C}}$, it follows that ${\mathcal {C}}= d({\mathcal {C}})$. By the fact that ${\mathcal {C}}\subseteq {{\,\mathrm{\mathsf {LFP}}\,}}$ and ${{\,\mathrm{\mathsf {LFP}}\,}}$ is the least fixed point, it follows that ${\mathcal {C}}= {{\,\mathrm{\mathsf {LFP}}\,}}$.

Thus, in both cases $T \notin {\mathcal {C}}$ implies $T \notin {{\,\mathrm{\mathsf {LFP}}\,}}$. This completes the proof. $\square$

Proposition 18

(Truth lemma) For any $\varphi \in {\mathcal {L}}_{\mathop {\Box }, \mathop {{\text {U}}}, \mathop {{\mathfrak {T}}}}$ and any $\varGamma \in W$,

Proof

The proof proceeds by induction, with the cases for atomic propositions and Boolean connectives being routine, and those for and $\mathop {\Box }$ and $\mathop {{\text {U}}}$ relying on Proposition 17. Here we focus on the case for $\mathop {{\mathfrak {T}}}$.

From left to right, suppose $\varGamma \in \{\!\vert \mathop {{\mathfrak {T}}}\varphi \vert \!\}$. Then, $\mathop {{\mathfrak {T}}}\varphi \in \varGamma$ so, by Proposition 17, $\mathop {\widehat{{\text {U}}}}\mathop {{\mathfrak {T}}}\varphi \in \varPhi$ which, by Item (iii) of Fact 2, implies $\{\!\vert \mathop {{\mathfrak {T}}}\varphi \vert \!\} \in \tau$. Now, let $T = \{\!\vert \mathop {{\mathfrak {T}}}\varphi \vert \!\}$. Then, (i) from $\{\!\vert \mathop {{\mathfrak {T}}}\varphi \vert \!\} \subseteq T$ and $\mathop {\widehat{{\text {U}}}}\mathop {{\mathfrak {T}}}\varphi \in \varPhi$, it follows that $T \in {\mathcal {C}}_1$ which, by Lemma 5, implies $T \in {{\,\mathrm{\mathsf {LFP}}\,}}$; (ii) $\varGamma \in T$, as $\varGamma \in \{\!\vert \mathop {{\mathfrak {T}}}\varphi \vert \!\}$; (iii) from axiom $\mathop {{\mathfrak {T}}}\varphi \rightarrow \varphi$ it follows that $T \subseteq \{\!\vert \varphi \vert \!\}$ which, by inductive hypothesis $\{\!\vert \varphi \vert \!\} = \llbracket \varphi \rrbracket$, implies $T \subseteq \llbracket \varphi \rrbracket$. Hence, by $\mathop {{\mathfrak {T}}}$’s truth condition, $\varGamma \in \llbracket \mathop {{\mathfrak {T}}}\varphi \rrbracket$.

From right to left, suppose $\varGamma \in \llbracket \mathop {{\mathfrak {T}}}\varphi \rrbracket$. Then, by $\mathop {{\mathfrak {T}}}$’s truth condition, there is $T \in {{\,\mathrm{\mathsf {LFP}}\,}}$ with $\varGamma \in T$ and $T \subseteq \llbracket \varphi \rrbracket$.

The inductive hypothesis implies $\llbracket \varphi \rrbracket = \{\!\vert \varphi \vert \!\}$. By Item (iv) of Fact 2 and $T \subseteq \{\!\vert \varphi \vert \!\}$, we have $T \subseteq \{\!\vert \mathop {\Box }\varphi \vert \!\}$. So we have proved that $\varGamma \in T$ and $T \subseteq \{\!\vert \mathop {\Box }\varphi \vert \!\}$.

By Lemma 5, ${{\,\mathrm{\mathsf {LFP}}\,}}= {\mathcal {C}}_1\cup {\mathcal {C}}_2$; thus, $T \in {\mathcal {C}}_1\cup {\mathcal {C}}_2$. Suppose $T \in {\mathcal {C}}_1$; then there is $\psi \in {\mathcal {L}}_{\mathop {\Box }, \mathop {{\text {U}}}, \mathop {{\mathfrak {T}}}}$ with $\{\!\vert \mathop {{\mathfrak {T}}}\psi \vert \!\} \subseteq T$ and $\mathop {\widehat{{\text {U}}}}\mathop {{\mathfrak {T}}}\psi \in \varPhi$. Thus, $\{\!\vert \mathop {{\mathfrak {T}}}\psi \vert \!\} \subseteq T \subseteq \{\!\vert \mathop {\Box }\varphi \vert \!\}$, so $\mathop {{\text {U}}}(\mathop {{\mathfrak {T}}}\psi \rightarrow \mathop {\Box }\varphi ) \in \varPhi$. Now, take any $\varDelta \in \{\!\vert \mathop {{\mathfrak {T}}}\psi \vert \!\}$. The fact that $\mathop {{\text {U}}}(\mathop {{\mathfrak {T}}}\psi \rightarrow \mathop {\Box }\varphi ) \in \varDelta$, together with theorem $\mathop {{\text {U}}}(\varphi \rightarrow \psi ) \rightarrow (\mathop {{\mathfrak {T}}}\varphi \rightarrow \mathop {{\mathfrak {T}}}\psi )$ (Proposition 7), implies $\mathop {{\mathfrak {T}}}\mathop {{\mathfrak {T}}}\psi \rightarrow \mathop {{\mathfrak {T}}}\mathop {\Box }\varphi \in \varDelta$. Moreover: $\varDelta \in \{\!\vert \mathop {{\mathfrak {T}}}\psi \vert \!\}$ implies $\varDelta \in \{\!\vert \mathop {{\mathfrak {T}}}\mathop {{\mathfrak {T}}}\psi \vert \!\}$, so $\varDelta \in \{\!\vert \mathop {{\mathfrak {T}}}\mathop {\Box }\varphi \vert \!\}$, that is, $\mathop {{\mathfrak {T}}}\mathop {\Box }\varphi \in \varDelta$. The latter, together with theorem $\mathop {{\mathfrak {T}}}\varphi \leftrightarrow \mathop {{\mathfrak {T}}}\mathop {\Box }\varphi$ and axiom $\mathop {{\mathfrak {T}}}\varphi \rightarrow \mathop {{\text {U}}}(\mathop {\Box }\varphi \rightarrow \mathop {{\mathfrak {T}}}\varphi )$, imply $\mathop {{\text {U}}}(\mathop {\Box }\varphi \rightarrow \mathop {{\mathfrak {T}}}\varphi ) \in \varDelta$, and thus $\mathop {{\text {U}}}(\mathop {\Box }\varphi \rightarrow \mathop {{\mathfrak {T}}}\varphi ) \in \varPhi$. Hence, $\{\!\vert \mathop {\Box }\varphi \vert \!\} \subseteq \{\!\vert \mathop {{\mathfrak {T}}}\varphi \vert \!\}$ and thus, since $\varGamma \in T$ and $T \subseteq \{\!\vert \mathop {\Box }\varphi \vert \!\}$, we have $\varGamma \in \{\!\vert \mathop {{\mathfrak {T}}}\varphi \vert \!\}$. Otherwise, $T \in {\mathcal {C}}_2$, and hence for any $\psi \in {\mathcal {L}}_{\mathop {\Box }, \mathop {{\text {U}}}, \mathop {{\mathfrak {T}}}}$ with $T \subseteq \{\!\vert \mathop {\Box }\psi \vert \!\}$ there is $\xi \in {\mathcal {L}}_{\mathop {\Box }, \mathop {{\text {U}}}, \mathop {{\mathfrak {T}}}}$ with $\{\!\vert \mathop {{\mathfrak {T}}}\xi \vert \!\} \subseteq \{\!\vert \mathop {\Box }\psi \vert \!\}$ and $\mathop {\widehat{{\text {U}}}}\mathop {{\mathfrak {T}}}\xi \in \varPhi$. Thus, since $\varphi$ satisfies that $T \subseteq \{\!\vert \mathop {\Box }\varphi \vert \!\}$, there is $\eta \in {\mathcal {L}}_{\mathop {\Box }, \mathop {{\text {U}}}, \mathop {{\mathfrak {T}}}}$ such that $\{\!\vert \mathop {{\mathfrak {T}}}\eta \vert \!\} \subseteq \{\!\vert \mathop {\Box }\varphi \vert \!\}$ and $\mathop {\widehat{{\text {U}}}}\mathop {{\mathfrak {T}}}\eta \in \varPhi$. From here we can repeat the argument used in the case of $T \in {\mathcal {C}}_1$ in order to get $\varGamma \in \{\!\vert \mathop {{\mathfrak {T}}}\varphi \vert \!\}$ again. Thus, in both cases, $\varGamma \in \{\!\vert \mathop {{\mathfrak {T}}}\varphi \vert \!\}$, which completes the proof. $\square$

Lemma 6

is Alexandroff.

Proof

Whether is Alexandroff has nothing to do with $\leftarrowtail$; thus, we can apply Prop. 5.6.15 in Özgün (2017), which states that if $\tau = \{\bigcup _{\varGamma \in U} {\leqslant }[\varGamma ] \mid U \subseteq W\}$ then is Alexandroff. But Item (i) of Fact 2 and the definition of ${\mathcal {E}}_0$ imply the required condition; then, is Alexandroff. $\square$

Since is Alexandroff, Proposition 16 tells us it has a modally equivalent topological argumentation model. Hence, the ${\mathsf {L}}_{\mathop {\Box }, \mathop {{\text {U}}}, \mathop {{\mathfrak {T}}}}$-consistent set of ${\mathcal {L}}_{\mathop {\Box }, \mathop {{\text {U}}}, \mathop {{\mathfrak {T}}}}$-formulas $\varPhi _0$ is satisfiable in a topological argumentation model which satisfies the following condition: for every $T, T_1, T'_1 \in \tau$: if $T_1 \leftarrowtail T$ and $T'_1 \subseteq T_1$, then $T'_1 \leftarrowtail T$ (see Lemma 3.2).