LogAG: An algebraic Non-Monotonic logic for reasoning with graded propositions

Abstract

We present LogAG, a weighted algebraic non-monotonic logic for reasoning with graded beliefs. LogAG is algebraic in that it is a language of only terms, some of which denote propositions and may be associated with ordered grades. The grades could be taken to represent a wide variety of phenomena including preference degrees, priority levels, trust ranks, and uncertainty measures. Reasoning in LogAG is non-monotonic and may give rise to contradictions. Belief revision is, hence, an integral part of reasoning and is guided by the grades. This yields a quite expressive language providing an interesting alternative to the currently existing approaches to non-monotonicity. We show how LogAG can be utilised for modelling resource-bounded reasoning; simulating inconclusive reasoning with circular, liar-like sentences; and reasoning about information arriving over a chain of sources each with a different degree of trust. While there certainly are accounts in the literature for each of these issues, we are not aware of any single framework that accounts for them all like LogAG does. We also show how LogAG captures a wide variety of non-monotonic logical formalisms. As such, LogAG is a unifying framework for non-monotonicity which is flexible enough to admit a wide array of potential uses.

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Notes

  1. 1.

    Filters are defined in Definition 3.

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Acknowledgements

We would like to express our gratitude to all the reviewers and the attendees of the DKB/KIK-15 workshop, the Commonsense-2017 symposium, the DKB/KIK-18 workshop, and the reviewers of the TARK-2019 conference to whom we submitted and presented parts of this research for their constructive criticism and invaluable advice.

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Appendices

Appendix A: Proofs

A.1 Proof of Observation 1

  1. 1.

    We prove the result by induction on the length l of the chain. If l = 1, then the chain is of the form 〈q0〉 where q0 grades p. By Definition 5, since \(q_{0} \in \mathcal {Q}\) and q0 grades p, then \(\{q_{0}, p\} \subseteq E(\mathcal {Q})\). Now, assume that the observation holds for every grading chain of length kLogAGeq1. Consider a chain C = 〈q0,…,qk− 1,qk〉 of length k + 1. Since \(q_{0} \in \mathcal {Q}\), then, by the induction hypothesis, \(q_{i} \in E(\mathcal {Q})\), for 0 ≤ ik. Finally, since qk grades p, then, by Definition 5, \(p \in E(\mathcal {Q})\).

  2. 2.

    We prove the result by induction on n. If n = 0, then \(\delta _{\mathcal {Q}}(p) = 1\). By Definition 6, there is some \(q \in G(p, E(\mathcal {Q}))\) with \(\delta _{\mathcal {Q}}(q)=0\). Hence, 〈q〉 is a grading chain of p. Moreover, \(q\in \mathcal {Q}\), by Definition 6. Now, assume that the statement holds for some \(k \in \mathbb {N}\) and let \(\delta _{\mathcal {Q}}(p) = k+1\). Hence, there must be some \(q_{k} \in G(p, E(\mathcal {Q}))\) with \(\delta _{\mathcal {Q}}(q_{k})=k\). By the induction hypothesis, there is a grading chain 〈q0,q1,…,qk− 1〉 of qk with \(q_{0} \in \mathcal {Q}\). But, then, 〈q0,q1,…,qk− 1,qk〉 is a grading chain of p with \(q_{0} \in \mathcal {Q}\).

A.2 Proof of Proposition 1

  1. 1.

    Follows from the pigeon-hole principle and from \(\mathfrak {g}\)’s being a function on \(\mathcal {P}\).

  2. 2.

    Let C be a chain in \(\mathcal {Q}\), and let d be the bound given in Definition 9. Since C is acyclic then, by the pigeon-hole principle, C’s length can be at most d. Now, we prove the statement by strong induction on 0 ≤ ld, where dl is the length of a grading chain. For l = 0, we have a chain C of length d. Since every extension of C will have a length greater than d, then C has no extension. By Definition 9, C is well-founded. Assume that, for some 0 ≤ kd, every chain of length dj, where jk, is well-founded. Consider a chain C of length d − (k + 1) = (dk) − 1. If C has no extension, then it is well-founded. Otherwise, an extension of C has length dj for some 0 ≤ jk. By the induction hypothesis, all extensions of C are well-founded and, hence, C is well-founded.

  3. 3.

    Let \(\mathcal {Q}\) be depth-bounded and let d be the bound given in Definition 9. Suppose that \(\delta _{\mathcal {Q}}(p) = d+k\), for k > 0 and \(p \in \mathcal {Q}\). By Observation 1, there is a grading chain 〈q0,…,qd+k〉 of p. But, then, there must be some 0 ≤ i < jd + k such that qi = qj. Thus, the chain 〈q0,…,qi,qj+ 1,…,qd+k〉 is a grading chain of p, implying \(\delta _{\mathcal {Q}}(p) < d+k\). Hence, a contradiction. It follows that, for every \(p \in \mathcal {Q}\), \(\delta _{\mathcal {Q}}(p) \leq d\).

  4. 4.

    Since \(\mathcal {Q}\) is non-explosive with finitely-many grading propositions, then so is \(F(\mathcal {Q})\). We establish the finiteness of the number of grading propositions in \(E^{n}(F(\mathcal {Q}))\) by induction on n. The basis, for n = 0, is obvious. Assuming the statement is true of \(E^{k}(F(\mathcal {Q}))\), we now show that it is true for \(E^{k+1}(F(\mathcal {Q}))\). Suppose that \(m \in \mathbb {N}\) is the number of grading propositions in \(E^{k}(F(\mathcal {Q}))\). Since \(\mathcal {P}\) is fan-out-bounded, then each of said m grading propositions grades at most fout propositions, where fout is as indicated in Definition 9. Hence, the number of grading propositions in \(E^{k+1}(F(\mathcal {Q}))\) is at most mfout + m. Further, if \(E(F(\mathcal {Q}))\) is depth-bounded then, by the third clause above, it follows that \(E(F(\mathcal {Q}))\) = \(E^{n}(F(\mathcal {Q}))\), for some \(n \in \mathcal {N}\). Hence, \(E(F(\mathcal {Q}))\) has finitely-many grading propositions.

  5. 5.

    Since \(\mathcal {Q}\) is depth-bounded, then there is some grading chain 〈q0,…,qk〉 of p, for some k < d (where d is as indicated in Definition 9), which is acyclic. Let C be the longest such chain. Note that (i) C is acyclic and, by designation, (ii) if C extends any chain, the resulting extension is not acyclic. Hence, C is a longest grading chain of p. If, further, \(\mathcal {Q}\) is fan-in-bounded then, by a simple combinatorial argument, the number of longest grading chains of p is at most (fin)d, where fin is as indicated in Definition 9.

A.3 Proof of Theorem 1

We need the following lemma in order to prove theorem 1.

Lemma 1

If \(\mathfrak {T}\) is a telescoping structure then, for every \(n \in \mathbb {N}\), if \(\mathcal {F}^{i}(\mathfrak {T})\) is defined and \(\mathcal {F}^{i}(\mathfrak {T}) \cap Range(\mathfrak {g}) = \tau ^{i}(\mathcal {T}) \cap Range(\mathfrak {g})\) for every in, then \(\mathcal {F}^{n}(\mathfrak {T}) = F(\mathcal {Q})\), for some \(\mathcal {Q} \subseteq E(\mathcal {T})\) and \(\mathcal {F}^{n}(\mathfrak {T}) \cap Range(\mathfrak {g}) \subseteq E^{n}(\mathcal {T})\).

We prove the lemma by induction on n. For n = 0, \(\mathcal {F}^{0}(\mathcal {T}) = F(\mathcal {T})\), where \(\mathcal {T} \subseteq E(\mathcal {T})\). Further, \(\mathcal {F}^{0}(\mathcal {T}) \cap Range(\mathfrak {g}) = \tau ^{0}(\mathcal {T}) \cap Range(\mathfrak (g)) = \mathcal {T} \cap Range(\mathfrak {g}) \subseteq E^{0}(\mathcal {T})\). Now assume that the statement holds for some \(k \in \mathbb {N}\). By the induction hypothesis, \(\mathcal {F}^{k}(\mathfrak {T}) \cap Range(\mathfrak {g}) \subseteq E^{k}(\mathcal {T})\). Therefore, all propositions graded in \(E^{1}(\mathcal {F}^{k}(\mathfrak {T}))\) are in \(E^{k+1}(\mathcal {T})\). Moreover, \(E^{1}(\mathcal {F}^{k}(\mathfrak {T})) \cap Range(\mathfrak {g}) \subseteq E^{k+1}(\mathcal {T})\). By Observation 3, \(\tau ^{k+1}_{\mathfrak {T}}(\mathcal {T}) = \tau _{\mathfrak {T}}(\mathcal {F}^{k}(\mathfrak {T})) = \varsigma (\kappa (E^{1}(\mathcal {F}^{k}(\mathfrak {T}))), \mathfrak {T}), \mathcal {T} = F(\mathcal {T}) \cup EG\), where EG is a set of propositions graded in \(\kappa (E^{1}(\mathcal {F}^{k}(\mathfrak {T})))\). By Definition 13, \(\kappa (E^{1}(\mathcal {F}^{k}(\mathfrak {T}))) \subseteq E^{1}(\mathcal {F}^{k}(\mathfrak {T}))\). Hence, \(EG \subseteq E^{k+1}(\mathcal {T})\). Thus, \(\mathcal {F}^{k+1}(\mathfrak {T}) = F([F(\mathcal {T}) \cup EG]) = F(\mathcal {T} \cup G)\), where \(\mathcal {T} \cup G \subseteq E(\mathcal {T})\). Moreover, \(\mathcal {F}^{k+1}(\mathfrak {T}) \cap Range(\mathfrak {g}) = \tau ^{k+1}(\mathcal {T}) \cap Range(\mathfrak {g}) \subseteq E^{k+1}(\mathcal {T})\).

We now proceed to proving the theorem.

By Definition 12 and Clause 4 of Proposition 1, \(E(\mathcal {T})\) has finitely-many grading propositions. In fact, since \(\mathcal {T}\) is finite, then \(E(\mathcal {T})\) is finite. Let \(b = |E(\mathcal {T})|\). Now, taking n = 2b + 1, suppose that, for every i ≤ 2b + 1, \(\mathcal {F}^{i}(\mathfrak {T})\) is defined and \(\mathcal {F}^{i}(\mathfrak {T}) \cap Range(\mathfrak {g}) = \tau ^{i}(\mathcal {T}) \cap Range(\mathfrak {g})\). By Lemma 4, \(\mathcal {F}^{i}(\mathfrak {T}) = F(\mathcal {Q}_{i})\), for some \(\mathcal {Q}_{i} \subseteq E(\mathcal {T})\). Since there are only 2b subsets of \(E(\mathcal {T})\), then \(\mathcal {F}^{n}(\mathfrak {T}) = \mathcal {F}^{j}(\mathfrak {T})\), for some j ≤ 2b. Further, by Proposition 2, for every \(j \in \mathbb {N}\), there is some kn such that \( \mathcal {F}^{n+j}(\mathfrak {T}) = \mathcal {F}^{k}(\mathfrak {T})\).

A.4 Proof of Corollary 1

We prove the case of k = 1 and the result follows by the same argument for all \(k \in \mathbb {N}\).

$$ \begin{array}{ll} \mathcal{F}^{n+1}(\mathfrak{T})\\ = F(\varsigma(\kappa(E^{1}(\mathcal{F}^{n}(\mathfrak{T})),\mathfrak{T}),\mathcal{T})) & (\text{Definitions~ 15~ and ~16}) \\ = F(\varsigma(\kappa(E^{1}(F(E(\mathcal{T}))),\mathfrak{T}),\mathcal{T})) \\ = F(\varsigma(\kappa(F(E(\mathcal{T}))),\mathfrak{T}),\mathcal{T}) & (\text{Definition ~of~ } E \text{~and~the~assumption~of~Theorem~1}) \\ = F(\varsigma(F(E(\mathcal{T})),\mathcal{T})) & \!\!\!\!\!\!(\text{Since~} F(E(\mathcal{T})) \text{~is~proper~given~that~}\mathcal{F}^{n+1}(\mathfrak{T}) \text{~is~defined})\\ = F(F(E(\mathcal{T}))) & \!\!\!\!\!\!\!(\text{Since~every~} p \in F(E(\mathcal{T})) \text{~is~supported~given~that~} \\ &F(E(\mathcal{T})) = \mathcal{F}^{n}(\mathfrak{T}) \text{~and~Definitions~15~and~16}) \\ = F(E(\mathcal{T})) \end{array} $$

A.5 Proof of Theorem 2

For n = 0, the statement is trivial, since \(\mathcal {F}^{0}(\mathfrak {T}) = F(\mathcal {T})\). Otherwise, the statement follows directly from Observation 2 since, by the definition of graded filters, \(\mathcal {F}^{k+1}(\mathfrak {T}) = F(K)\), for some \(K \subseteq \kappa (E^{1}(\mathcal {F}^{k}(\mathfrak {T})), \mathfrak {T})\).

A.6 Proof of Theorem 3

The following lemma will be useful in the proof of the theorem.

Lemma 2

For any proposition p, let \(\mathfrak {T}_{1}=\langle \mathcal {T}, \mathfrak {O}, \otimes , \oplus \rangle \), \(\mathfrak {T}_{2}=\langle \mathcal {T} \cup \{p\}, \mathfrak {O}, \otimes , \oplus \rangle \), and \(i \in \mathbb {N}\). If \(\mathcal {F}^{i}(\mathfrak {T}_{1})\) is a fixed point where \(p \in \mathcal {F}^{i}(\mathfrak {T}_{1}) \), then there is some \(j \in \mathbb {N}\) such that \(\mathcal {F}^{j}(\mathfrak {T}_{2})\) is a fixed point and \(\mathcal {F}^{i}(\mathfrak {T}_{1}) = \mathcal {F}^{j}(\mathfrak {T}_{2})\).

We now proceed to proving the theorem.

  1. 1.

    Reflexivity follows directly from Clause 1 in Definition 14 and the definition of graded filters.

  2. 2.

    If \(q \in \mathcal {F}^{m}(\mathfrak {T}_{2})\), then according to Lemma 5 it must be the case that \(q \in \mathcal {F}^{n}(\mathfrak {T}_{1})\). Accordingly, the graded consequence of LogAG relation observes cut.

  3. 3.

    Similarly, if \(p \in \mathcal {F}^{n}(\mathfrak {T}_{1})\) and \(q \in \mathcal {F}^{n}(\mathfrak {T}_{1})\), then it follow from Lemma 5 that \(q \in \mathcal {F}^{m}(\mathfrak {T}_{2})\). Accordingly, the graded consequence relation of LogAG observes cautious monotony.

A.7 Proof of Observation 6

We prove this by induction on n. Base case (n = 0): By Definition 16, \(\mathcal {F}^{0}(\mathfrak {T}) = F(\tau ^{0}_{\mathfrak {T}}(\mathcal {Q}))= F(\mathcal {Q}) = F(E^{0}(\mathcal {Q}))\) since \(E^{0}(\mathcal {Q})=\mathcal {Q}\). Induction Hypothesis: For nLogAGe0, suppose that \(\mathcal {F}^{n}(\mathfrak {T})=F(E^{n}(\mathcal {Q}))\). Induction Step:\(\mathcal {F}^{n+1}(\mathfrak {T}) = F(\tau ^{n+1}_{\mathfrak {T}}(\mathcal {Q})) = F(\tau _{\mathfrak {T}}(\tau ^{n}_{\mathfrak {T}}(\mathcal {Q}))) = F \varsigma \kappa (E^{1}(F(\tau ^{n}_{\mathfrak {T}}(\mathcal {Q})))) = F\varsigma (\kappa (E^{1}(\mathcal {F}^{n}(\mathfrak {T}))))\). By the induction hypothesis, \(\mathcal {F}^{n+1}(\mathfrak {T}) = F(\varsigma (\kappa (E^{1}(F(E^{n}(\mathcal {Q}))))))\). Since \(\mathcal {Q}\) is consistent given it is the set of interpretations of LogAG which is consistent according to how it was constructed, then \(\mathcal {F}^{n}(\mathfrak {T})\) must be consistent according to Definition 16. Consequently, \(F(E^{n}(\mathcal {Q}))\) must be consistent too. Since \(E^{n}(\mathcal {Q})\) contains only the propositions in \(\mathcal {Q}\) (which are only grading propositions according to how LogAG was constructed) in addition to the propositions with embedding degrees less than or equal to n, then \(F(E^{n}(\mathcal {Q}))\) contains nothing other than the propositions in \(E^{n}(\mathcal {Q})\) in addition to their trivial consequences. Hence, \(E^{1}(F(E^{n}(\mathcal {Q}))) = F(E^{1}(E^{n}(\mathcal {Q}))) = F(E^{n+1}(\mathcal {Q}))\), and \(\mathcal {F}^{n+1}(\mathfrak {T}) = F(\varsigma (\kappa (F(E^{n+1}(\mathcal {Q})))))\). Let (ψ,a) ∈ΣPL where IncPL) = a. According to Observation 5, \(\delta (\psi ,\mathcal {Q})= {\Delta }(Inc({\Sigma }_{PL}),{\Sigma }_{PL})+1 > n+1\) (since n + 1 = Δ(IncPL),ΣPL)). It follows then that \(E^{n+1}(\mathcal {Q})\) is consistent. By Observation 4, \(\varsigma (\kappa (F(E^{n+1}(\mathcal {Q})))) = F(E^{n+1}(\mathcal {Q}))\). Therefore, \(\mathcal {F}^{n+1}(\mathfrak {T}) = F(F(E^{n+1}(\mathcal {Q}))) = F(E^{n+1}(\mathcal {Q}))\).

A.8 Proof of Theorem 4

In what follows, let \(\mathcal {Q}=\{{[\![ \gamma ]\!]}^{\mathcal {V}}~|~ {\Gamma } \in Log_A \mathbf {G}\}\) be the valuation of LogAG where \(\mathcal {V}\) is a natural valuation.

Suppose that ϕBSPL). Then, by definition of BSPL), it must be that ϕ is logically implied by a set \(\mathcal {S}=\{\psi ~|~ (\psi ,a) \in {\Sigma }_{PL} ~and~ a>Inc({\Sigma }_{PL})\}\). According to how LogAG was constructed by and Observation 5, for every formula \(\psi \in \mathcal {S}\) in \(\mathcal {Q}\), \(\delta _{\mathcal {Q}}({[\![ \psi ]\!]}^{\mathcal {V}})= {\Delta }(a,{\Sigma }_{PL})+1\) where a is the weight of ψ in ΣPL. Since a > IncPL), it follows that \(\delta _{\mathcal {Q}}({[\![ \psi ]\!]}^{\mathcal {V}}) \le n\) and \({[\![ \psi ]\!]}^{\mathcal {V}} \in E^{n}(\mathcal {Q})\). Hence, by Observation 6, \({[\![ \phi ]\!]}^{\mathcal {V}} \in \mathcal {F}^{n}(\mathfrak {T})\) for all telescoping structures \(\mathfrak {T}\) of \(\mathcal {Q}\). By Definition 17, it must be that \(Log_A \mathbf {G} Log_A \mathbf {G}con^{\mathcal {C}} \phi \).

Now suppose that \(Log_A \mathbf {G} Log_A \mathbf {G}con^{\mathcal {C}} \phi \), then according to Definition 17, it must be that \({[\![ \phi ]\!]}^{\mathcal {V}} \in \mathcal {F}^{n}(\mathfrak {T})\) for all telescoping structures \(\mathfrak {T}\) of \(\mathcal {Q}\). According to Observation 6, \({[\![ \phi ]\!]}^{\mathcal {V}} \in F(E^{n}(\mathcal {Q}))\). Let \(\mathcal {S}=\{p ~|~ p \in E^{n}(\mathcal {Q}) \cap \mathcal {P}_{\mathbf {B}}\} \). It follows that ϕ is logically implied by a set of propositional terms whose interpretations are in \(\mathcal {S}\), and each \(p \in \mathcal {S}\) has an embedding degree m less than or equal to n. Let ψ be the propositional term denoting p. It must be that (ψ,a) ∈ΣPL since the only embedded non-grading propositions in LogAG are the formulas appearing in ΣPL according to Definition 23. Further, by Observation 5, m = Δ(aPL) + 1. Therefore, m < n and a > IncPL). By the definition of BSPL), it must be that ϕBSPL).

A.9 Proof of Observation 8

Taking the ⊗ as the sum operator and the ⊕ as the max operator, since all grades in all grading chains in \(\mathbb {NM}^{I}_{R}\) are 1s, then the interpretation of the embedded rule \({[\![ \pi (r) ]\!]}^{\mathcal {V}}\) at level n has a grade of n. This is because, even if π(r) appears graded in shorter chains, the ⊕ operator will set the grade of \({[\![ \pi (r) ]\!]}^{\mathcal {V}}\) to the deeper depth n.

A.10 Proof of Observation 9

Suppose that \(F\mathcal {V}(\mathbb {M_{R}} \cup ER_{n})\) is consistent. If \(chain(\pi (r),n) \in \mathbb {NM}^{I}_{R}\) and \(chain(\neg \pi (r),n) \not \in \mathbb {NM}^{I}_{R}\), then by Observation 8, the fused grade of \({[\![ \pi (r) ]\!]}^{\mathcal {V}}\) in \(E^{n}(\mathcal {T})\) is n. If \({[\![ \pi (\neg r) ]\!]}^{\mathcal {V}} \not \in E^{n}(F(\mathcal {T}))\), then \({[\![ \pi (r) ]\!]}^{\mathcal {V}}\) survives telescoping and is supported at level n and \({[\![ \pi (r) ]\!]}^{\mathcal {V}} \in \mathcal {F}^{n}(\mathfrak {T})\). Otherwise, if \({[\![ \pi (\neg r) ]\!]}^{\mathcal {V}} \in E^{n}(F(\mathcal {T}))\), we have three cases.

  1. 1.

    \({[\![ \pi (\neg r) ]\!]}^{\mathcal {V}} \in F(\mathcal {T})\). But this implies that \(F(\mathcal {V}(\mathbb {M}_{R})~ \cup ~ER_{n})\) is inconsistent. Hence, we get a contradiction; or

  2. 2.

    \({[\![ \pi (\neg r) ]\!]}^{\mathcal {V}}\) is embedded in a grading chain of length n. This can not be as \(chain(\neg \pi (r),n) \not \in \mathbb {NM}^{I}_{R}\); or

  3. 3.

    \({[\![ \pi (\neg r) ]\!]}^{\mathcal {V}}\) is supported by some graded propositions embedded at a degree of at most n. If \({[\![ \pi (\neg r) ]\!]}^{\mathcal {V}}\) is supported by at least a graded proposition with an embedding degree of n, we get a contradiction as \(F(\mathcal {V}(\mathbb {M}_{R})~\cup ~ER_{n})\) must be inconsistent. Then, it must be that \({[\![ \pi (\neg r) ]\!]}^{\mathcal {V}}\) is supported by graded propositions of embedding degrees less than n. In this case, however, all such graded propositions will not survive telescoping as \({[\![ \pi (\neg r) ]\!]}^{\mathcal {V}}\) has a higher grade depriving \({[\![ \pi (\neg r) ]\!]}^{\mathcal {V}}\) of its support. It follows then that \({[\![ \pi (r) ]\!]}^{\mathcal {V}}\) survives telescoping and is supported at level n. Hence, \({[\![ \pi (r) ]\!]}^{\mathcal {V}} \in \mathcal {F}^{n}(\mathfrak {T})\).

A.11 Proof of Proposition 3

We prove this by showing that \(F(\mathcal {V}(\mathbb {M}_{R})~\cup ~ ER_{n}~ \cup ~ G_{\mathcal {T}}) \subseteq \mathcal {F}^{n}(\mathfrak {T})\) and \(\mathcal {F}^{n}(\mathfrak {T})\subseteq F(\mathcal {V}(\mathbb {M}_{R})~\cup ~ ER_{n}~\cup ~\mathbf {G}_{\mathcal {T}})\). Suppose that \(F(\mathcal {V}(\mathbb {M}_{R})~ \cup ~ER_{n})\) is consistent. Let \({[\![ \pi (\phi ) ]\!]}^{\mathcal {V}} \in \mathcal {V}(\mathbb {M}_{R})~\cup ~ ER_{n} ~ \cup ~ G_{\mathcal {T}}\). Therefore, it must be one of the following cases.

  1. 1.

    \({[\![ \pi (\phi ) ]\!]}^{\mathcal {V}} \in \mathcal {V}(\mathbb {M}_{R})\). In this case, \({[\![ \pi (\phi ) ]\!]}^{\mathcal {V}}\) survives telescoping and is supported at level n since all members of the top theory \(\mathcal {T}\) survive telescoping and are supported at all levels.

  2. 2.

    \({[\![ \pi (\phi ) ]\!]}^{\mathcal {V}} \in ER_{n}\). By Observation 9, \({[\![ \pi (\phi ) ]\!]}^{\mathcal {V}}\) survives telescoping and is supported at level n.

  3. 3.

    \({[\![ \pi (\phi ) ]\!]}^{\mathcal {V}} \in G_{\mathcal {T}}\). The only possible grading propositions come from \(\mathbb {NM}^{I}_{R}\) or embedded grading propositions in \(\mathbb {NM}^{I}_{R}\). The interpretations of such grading propositions must survive telescoping and are supported at level n since such grading propositions are never members of ⊥-kernels given the construction of \(\mathbb {NM}^{I}_{R}\).

Hence, \(\mathcal {V}(\mathbb {M}_{R}) \cup ER_{n} \cup G_{\mathcal {T}}\) survive telescoping and is supported at level n. By the definition of graded filters then, \(F(\mathcal {V}(\mathbb {M}_{R})~\cup ~ER_{n}~\cup ~G_{\mathcal {T}}) \subseteq \mathcal {F}^{n}(\mathfrak {T})\).

Now, we proceed to proving that \(\mathcal {F}^{n}(\mathfrak {T})\subseteq F(\mathcal {V}(\mathbb {M}_{R})~\cup ~ ER_{n}~\cup ~G_{\mathcal {T}})\). According to Observation 3 and the definition of graded filters, \(\mathcal {F}^{n}(\mathfrak {T})=F(F(\mathcal {T})~\cup ~EG) = F(\mathcal {T}~\cup ~EG)= F(\mathcal {V}(\mathbb {M}_{R})~\cup ~\mathcal {V}(\mathbb {NM}^{I}_{R})~\cup ~ EG) \) for some set of embedded graded propositions EG in \(E^{n}(\mathcal {T})\) that survive telescoping and are supported at level n. Let \({[\![ \pi (\phi ) ]\!]}^{\mathcal {V}} \in \mathcal {V}(\mathbb {M}_{R})~\cup ~\mathcal {V}(\mathbb {NM}^{I}_{R})~ \cup ~ EG \). Therefore, one of the following cases is true.

  1. 1.

    \({[\![ \pi (\phi ) ]\!]}^{\mathcal {V}}\in \mathcal {V}(\mathbb {M}_{R})\). It follows trivially then that \({[\![ \pi (\phi ) ]\!]}^{\mathcal {V}} \in \mathcal {V}(\mathbb {M}_{R})~\cup ~ ER_{n}~\cup ~G_{\mathcal {T}}\).

  2. 2.

    \({[\![ \pi (\phi ) ]\!]}^{\mathcal {V}} \in \mathcal {V}(\mathbb {NM}^{I}_{R})\). Hence, \({[\![ \pi (\phi ) ]\!]}^{\mathcal {V}} \in G_{\mathcal {T}}\) by the definition of \(G_{\mathcal {T}}\) and the construction of \(\mathbb {NM}^{I}_{R}\).

  3. 3.

    \({[\![ \pi (\phi ) ]\!]}^{\mathcal {V}} \in EG\). If \({[\![ \pi (\phi ) ]\!]}^{\mathcal {V}}\) is a grading proposition, then \({[\![ \pi (\phi ) ]\!]}^{\mathcal {V}} \in G_{\mathcal {T}}\). Otherwise, \({[\![ \pi (\phi ) ]\!]}^{\mathcal {V}}\) is a graded proposition, then according to Definition 30 either \(chain(\pi (\phi ),n) \in \mathbb {NM}^{I}_{R}\) and \(chain(\neg \pi (\phi ),n) \not \in \mathbb {NM}^{I}_{R}\) or \(chain(\pi (\phi ),n) \in \mathbb {NM}^{I}_{R}\) and \(chain(\neg \pi (\phi ),n) \in \mathbb {NM}^{I}_{R}\). However, in the second case by Observation 8 both \({[\![ \pi (\phi ) ]\!]}^{\mathcal {V}}\) and \({[\![ \pi (\neg \phi ) ]\!]}^{\mathcal {V}}\) have the same grade of n in \(E^{n}(\mathcal {T})\). Accordingly, both do not survive telescoping at level n. It must then be that \(chain(\pi (\phi ),n) \in \mathbb {NM}^{I}_{R}\) and \(chain(\neg \pi (\phi ),n) \not \in \mathbb {NM}^{I}_{R}\) for \({[\![ \pi (\phi ) ]\!]}^{\mathcal {V}}\) to survive telescoping at level n according to Observation 9. Hence, \({[\![ \pi (\phi ) ]\!]}^{\mathcal {V}} \in ER_n\).

Thus, \(\mathcal {V}(\mathbb {M}_R)~\cup ~\mathcal {V}(\mathbb {NM}^{I}_{R})~\cup ~ EG \subseteq \mathcal {V}(\mathbb {M}_R)~\cup ~ ER_n~ \cup ~G_{\mathcal {T}}\). Since filters are monotonic, then \(F(\mathcal {V}(\mathbb {M}_R)~\cup ~\mathcal {V}(\mathbb {NM}^{I}_{R})~\cup ~ G) \subseteq F(\mathcal {V}(\mathbb {M}_R)~\cup ~ ER_n~\cup ~G_{\mathcal {T}})\). Hence, \(\mathcal {F}^n(\mathfrak {T}) = F(\mathcal {V}(\mathbb {M}_R)~\cup ~ ER_n~ \cup ~G_{\mathcal {T}})\).

A.12 Proof of Lemma 1

For any base fact AR (and hence in T by the definition of argument structures), π(A) is ungraded in the monotonic subtheory \(\mathbb {M}_R\) according to Definition 30. It follows from Observation 3 and the definition of graded filters that \({[\![ \pi (A) ]\!]}^{\mathcal {V}} \in \mathcal {F}^n({\mathfrak {T}})\) for every telescoping structure \(\mathfrak {T}=\langle \mathcal {T}, \mathfrak {O}, \otimes , \oplus \rangle \) since all members of the top theory \(\mathcal {T}\) survive telescoping and are supported at all levels.

A.13 Proof of Lemma 2

We prove this by induction on the height of the argument tree. Base case: The only argument trees containing single nodes in T are the base facts. Hence, the base case follows from Lemma 1. Induction hypothesis: If ϕ is the root of some argument tree ϕT of height at most h, then \({[\![ \pi (\phi ) ]\!]}^{\mathcal {V}} \in \mathcal {F}^n({\mathfrak {T}})\) for all n and for every telescoping structure \(\mathfrak {T}=\langle \mathcal {T}, \mathfrak {O}, \otimes , \oplus \rangle \). Induction step: Suppose that ϕ is the root of some argument tree ϕT of height h + 1. Then, there must be a monotonic rule \( r = A_1,...,A_m \rightarrow \phi \in R\) and, since T is monotonically closed, A1,...,Am are roots of argument trees of height at most h in T. Hence, by the induction hypothesis, \({[\![ \pi (A_i) ]\!]}^{\mathcal {V}} \in \mathcal {F}^n({\mathfrak {T}})\) for 1 ≤ im, all n, and every telescoping structure \(\mathfrak {T}=\langle \mathcal {T}, \mathfrak {O}, \otimes , \oplus \rangle \). According to Definition 30, \(\pi (r) \in \mathbb {M}_{R}\) and, hence, \({[\![ \pi (r) ]\!]}^{\mathcal {V}} \in \mathcal {F}^n({\mathfrak {T}})\) for all n by the definition of graded filters since all members of the top theory \(\mathcal {T}\) survive telescoping and are supported at all levels. It follows then that \({[\![ \pi (\phi ) ]\!]}^{\mathcal {V}} \in \mathcal {F}^n({\mathfrak {T}})\) as well.

A.14 Proof of Proposition 4

Let ϕR(T). If \(\mathcal {S}=\varnothing \), then \(R(T)=R^T_M\) is made up of only monotonic rules and \(\pi (\phi ) \in \mathbb {M}_R\) according to Definition 30. By the definition of graded filters, \({[\![ \pi (\phi ) ]\!]}^{\mathcal {V}}\in \mathcal {F}^n(\mathfrak {T})\) for all n. Otherwise, if \(\mathcal {S}\not = \varnothing \), then according to Definition 30, \(chain(r,I(\mathcal {S})) \in \mathbb {NM}^{I}_{R}\) and \(chain(\neg r,I(\mathcal {S})) \not \in \mathbb {NM}^{I}_{R}\). Since T is an argument structure, it must be that \(R(T)=R^T_M\cup \mathcal {S}\) is consistent. Accordingly, π(R(T)) must be consistent as well. Hence, it follows from Proposition 3 that \({[\![ \pi (\phi ) ]\!]}^{\mathcal {V}} \in \mathcal {F}^n(\mathfrak {T})\) with \(n=I(\mathcal {S})\).

A.15 Proof of Lemma 3

We prove this by induction on the height of the argument tree. Base case: The only argument trees containing single nodes in T are the base facts. Hence, the base case follows from Lemma 1. Induction hypothesis: If ϕ is the root of some argument tree ϕT of height at most h, then \({[\![ \pi (\phi ) ]\!]}^{\mathcal {V}} \in \mathcal {F}^n({\mathfrak {T}})\) for some degree \(n = I(\mathcal {S})\) and for every telescoping structure \(\mathfrak {T}=\langle \mathcal {T}, \mathfrak {O}, sum, max \rangle \). Induction step: Suppose that ϕ is the root of some argument tree ϕT of height h + 1 with direct children A1,...,Am. Since T is closed, A1,...,Am are roots of argument trees in T of height at most h. Hence, by the induction hypothesis, \({[\![ \pi (A_i) ]\!]}^{\mathcal {V}} \in \mathcal {F}^n({\mathfrak {T}})\) for 1 ≤ im, \(n = I(\mathcal {S})\), and every telescoping structure \(\mathfrak {T}=\langle \mathcal {T}, \mathfrak {O}, sum, max \rangle \). By Proposition 4, \(\pi (R(T)) \subset \mathcal {F}^n(\mathfrak {T})\) including the rules labelling the arcs from A1,...Am to ϕ. It follows then that \({[\![ \pi (\phi ) ]\!]}^{\mathcal {V}} \in \mathcal {F}^n(\mathfrak {T})\) for every telescoping structure \(\mathfrak {T}=\langle \mathcal {T}, \mathfrak {O}, sum, max \rangle \).

A.16 Proof of Theorem 5

Suppose that ϕWff(T), then by the definition of Wff(T) it must be one of the following three cases.

  1. 1.

    ϕ is a base fact. By lemma 1, \({[\![ \pi (\phi ) ]\!]}^{\mathcal {V}} \in \mathcal {F}^n({\mathfrak {T}})\) for all n and every telescoping structure \(\mathfrak {T}=\langle \mathcal {T}, \mathfrak {O}, \otimes , \oplus \rangle \). Hence, by Definition 16, \(\mathbb {T}^{I}_{R} Log_A \mathbf {G}con^{\mathcal {C}} \pi (\phi )\) for some grading canon \(\mathcal {C}=\langle sum, max,n\rangle \) where \(n=I(\mathcal {S})\).

  2. 2.

    ϕ is a root of some argument with all arcs labelled by monotonic rules. By lemma 2, \({[\![ \pi (\phi ) ]\!]}^{\mathcal {V}} \in \mathcal {F}^n({\mathfrak {T}})\) for all n and every telescoping structure \(\mathfrak {T}=\langle \mathcal {T}, \mathfrak {O}, \otimes , \oplus \rangle \). Hence, by Definition 16, \(\mathbb {T}^{I}_{R} Log_A \mathbf {G}con^{\mathcal {C}} \pi (\phi )\) for some grading canon \(\mathcal {C}=\langle sum, max,n\rangle \) where \(n=I(\mathcal {S})\).

  3. 3.

    ϕ is a root of some argument with all arcs labelled by non-monotonic rules. By lemma 3, \({[\![ \pi (\phi ) ]\!]}^{\mathcal {V}} \in \mathcal {F}^n({\mathfrak {T}})\) for \(n=I(\mathcal {S})\) and every telescoping structure \(\mathfrak {T}=\langle \mathcal {T}, \mathfrak {O}, sum, max, \rangle \). Hence, by Definition 16, \(\mathbb {T}^{I}_{R} Log_A \mathbf {G}con^{\mathcal {C}} \pi (\phi )\) for some grading canon \(\mathcal {C}=\langle sum, max,n\rangle \).

A.17 Proof of Theorem 6

Proof

Suppose that \(\mathbb {T}^{I}_{R} Log_A \mathbf {G}con^{\mathcal {C}} \pi (\phi )\) with \(\mathcal {C}=\langle sum, max,n\rangle \). Then, \({[\![ \pi (\phi ) ]\!]}^{\mathcal {V}} \in \mathcal {F}^n(\mathfrak {T})\). According to Proposition 3, \(\mathcal {F}^n(\mathfrak {T}) = F(\mathcal {V}(\mathbb {M}_R)~\cup ~ ER_n~\cup ~G_{\mathcal {T}})\). We have four cases.

  1. 1.

    \({[\![ \pi (\phi ) ]\!]}^{\mathcal {V}} \in \mathcal {V}(\mathbb {M}_R)\). In this case, ϕ is either a base fact or a monotonic rule in R. If ϕ is a base fact, then it must be in Wff(T) by the definition of argument structures (and hence π(ϕ) a logical consequence of R(T)). If ϕ is a monotonic rule appearing as an arc label in T, then ϕR(T). Otherwise, if ϕ is a monotonic rule that does not appear as an arc label in T, then \(\phi \in R_M^{\prime }\). In the three cases, π(ϕ) is a logical consequence of \(\pi (R(T) \cup R_M^{\prime })\).

  2. 2.

    \({[\![ \pi (\phi ) ]\!]}^{\mathcal {V}} \in ER_n\). In this case, π(ϕ) must be a non-monotonic rule embedded at level n whose negation is not embedded at level n. Hence, \(\phi \in \mathcal {S}\) and ϕR(T) by the definition of R(T). Hence, π(ϕ) ∈ π(R(T)).

  3. 3.

    \({[\![ \pi (\phi ) ]\!]}^{\mathcal {V}} \in G_{\mathcal {T}}\). This can not be as ϕ must be a grading term and grading terms are never in R.

  4. 4.

    \({[\![ \phi ]\!]}^{\mathcal {V}} \in F(\mathcal {V}(\mathbb {M}_R)~\cup ~ ER_n~\cup ~G_{\mathcal {T}})\). It follows from the previous three cases and the monotonicity of filters that π(ϕ) is a logical consequence of \(\pi (R(T) \cup R_M^{\prime })\).

Appendix B: Review of Boolean Algebra

A Boolean algebra is a sextuple \(\mathfrak {A}=<\mathcal {P}, +, \cdot , -, \bot , \top >\) where \(\mathcal {P}\) is a non-empty set with \(\{\bot ,\top \} \subseteq \mathcal {P}\). \(\mathfrak {A}\) is closed under the two binary operators + and ⋅ and the unary operator −. The operators satisfy the following conditions.

B1.1:a + b = b + a(Commutativity)
B1.2:ab = ba 
B2.1:a + (b + c) = (a + b) + c(Associativity)
B2.2:a ⋅ (bc) = (ab) ⋅ c 
B3.1:a + (ab) = a(Absorption)
B3.2:a ⋅ (a + b) = a 
B4.1:a ⋅ (b + c) = (ab) + (ac) (Distribution)
B4.2:a + (bc) = (a + b) ⋅ (a + c) 
B5.1:a + −a = ⊤ (Complements)
B5.2:a ⋅−a = ⊥ 

Definition 31

A Boolean Algebra \(\mathfrak {A}=<B, +, \cdot , -, \bot , \top >\) is complete if, for every \(A \subseteq B\), \(\sum \limits _{a \in A}a \in B\) and \(\prod \limits _{a \in A}a \in B\).

Definition 32

A Boolean Algebra \(\mathfrak {A}=<B, +, \cdot , -, \bot , \top >\) is degenerate if ⊤ = ⊥. Otherwise, it is non-degenerate.

The elements in B are partially ordered by the relation ≤, where ab if and only if ab = a. Alternatively, ab if and only if a + b = b (this follows from B3.1 and B3.2).

Definition 33

An ultrafilter U is a filter of a Boolean algebra \(\mathfrak {A}\) which is maximal with respect to not including ⊥.Footnote 1

According to the definition of ultrafilters, the following observations hold.

U1.:

For every aB, exactly one of a and − a belong to U.

U2.:

For every a,bB, a + bU if and only if aU or bU.

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Ehab, N., Ismail, H.O. LogAG: An algebraic Non-Monotonic logic for reasoning with graded propositions. Ann Math Artif Intell (2020). https://doi.org/10.1007/s10472-020-09697-0

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Keywords

  • Non-Monotonicity
  • Weighted logics
  • Uncertainty
  • Graded propositions
  • Unified framework for Non-Monotonicity

Mathematics Subject Classification (2010)

  • 68T27