Abstract
We present Log_{A}G, a weighted algebraic nonmonotonic logic for reasoning with graded beliefs. Log_{A}G 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 Log_{A}G is nonmonotonic 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 nonmonotonicity. We show how Log_{A}G can be utilised for modelling resourcebounded reasoning; simulating inconclusive reasoning with circular, liarlike 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 Log_{A}G does. We also show how Log_{A}G captures a wide variety of nonmonotonic logical formalisms. As such, Log_{A}G is a unifying framework for nonmonotonicity which is flexible enough to admit a wide array of potential uses.
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Notes
 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/KIK15 workshop, the Commonsense2017 symposium, the DKB/KIK18 workshop, and the reviewers of the TARK2019 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.
We prove the result by induction on the length l of the chain. If l = 1, then the chain is of the form 〈q_{0}〉 where q_{0} grades p. By Definition 5, since \(q_{0} \in \mathcal {Q}\) and q_{0} grades p, then \(\{q_{0}, p\} \subseteq E(\mathcal {Q})\). Now, assume that the observation holds for every grading chain of length kLog_{A}Geq1. Consider a chain C = 〈q_{0},…,q_{k− 1},q_{k}〉 of length k + 1. Since \(q_{0} \in \mathcal {Q}\), then, by the induction hypothesis, \(q_{i} \in E(\mathcal {Q})\), for 0 ≤ i ≤ k. Finally, since q_{k} grades p, then, by Definition 5, \(p \in E(\mathcal {Q})\).

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 〈q_{0},q_{1},…,q_{k− 1}〉 of q_{k} with \(q_{0} \in \mathcal {Q}\). But, then, 〈q_{0},q_{1},…,q_{k− 1},q_{k}〉 is a grading chain of p with \(q_{0} \in \mathcal {Q}\).
A.2 Proof of Proposition 1

1.
Follows from the pigeonhole principle and from \(\mathfrak {g}\)’s being a function on \(\mathcal {P}\).

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 pigeonhole principle, C’s length can be at most d. Now, we prove the statement by strong induction on 0 ≤ l ≤ d, where d − l 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 wellfounded. Assume that, for some 0 ≤ k ≤ d, every chain of length d − j, where j ≤ k, is wellfounded. Consider a chain C of length d − (k + 1) = (d − k) − 1. If C has no extension, then it is wellfounded. Otherwise, an extension of C has length d − j for some 0 ≤ j ≤ k. By the induction hypothesis, all extensions of C are wellfounded and, hence, C is wellfounded.

3.
Let \(\mathcal {Q}\) be depthbounded 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 〈q_{0},…,q_{d+k}〉 of p. But, then, there must be some 0 ≤ i < j ≤ d + k such that q_{i} = q_{j}. Thus, the chain 〈q_{0},…,q_{i},q_{j+ 1},…,q_{d+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.
Since \(\mathcal {Q}\) is nonexplosive with finitelymany 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 fanoutbounded, then each of said m grading propositions grades at most f_{out} propositions, where f_{out} is as indicated in Definition 9. Hence, the number of grading propositions in \(E^{k+1}(F(\mathcal {Q}))\) is at most mf_{out} + m. Further, if \(E(F(\mathcal {Q}))\) is depthbounded 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 finitelymany grading propositions.

5.
Since \(\mathcal {Q}\) is depthbounded, then there is some grading chain 〈q_{0},…,q_{k}〉 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 faninbounded then, by a simple combinatorial argument, the number of longest grading chains of p is at most (f_{in})^{d}, where f_{in} 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 i ≤ n, 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 finitelymany grading propositions. In fact, since \(\mathcal {T}\) is finite, then \(E(\mathcal {T})\) is finite. Let \(b = E(\mathcal {T})\). Now, taking n = 2^{b} + 1, suppose that, for every i ≤ 2^{b} + 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 2^{b} subsets of \(E(\mathcal {T})\), then \(\mathcal {F}^{n}(\mathfrak {T}) = \mathcal {F}^{j}(\mathfrak {T})\), for some j ≤ 2^{b}. Further, by Proposition 2, for every \(j \in \mathbb {N}\), there is some k ≤ n 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 \(k \in \mathbb {N}\).
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.
Reflexivity follows directly from Clause 1 in Definition 14 and the definition of graded filters.
 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 Log_{A}G relation observes cut.
 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 Log_{A}G 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 nLog_{A}Ge0, 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 Log_{A}G 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 Log_{A}G 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) ∈Σ_{PL} where Inc(Σ_{PL}) = a. According to Observation 5, \(\delta (\psi ,\mathcal {Q})= {\Delta }(Inc({\Sigma }_{PL}),{\Sigma }_{PL})+1 > n+1\) (since n + 1 = Δ(Inc(Σ_{PL}),Σ_{PL})). 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 Log_{A}G where \(\mathcal {V}\) is a natural valuation.
Suppose that ϕ ∈ BS(Σ_{PL}). Then, by definition of BS(Σ_{PL}), 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 Log_{A}G 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 Σ_{PL}. Since a > Inc(Σ_{PL}), 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) ∈Σ_{PL} since the only embedded nongrading propositions in Log_{A}G are the formulas appearing in Σ_{PL} according to Definition 23. Further, by Observation 5, m = Δ(a,Σ_{PL}) + 1. Therefore, m < n and a > Inc(Σ_{PL}). By the definition of BS(Σ_{PL}), it must be that ϕ ∈ BS(Σ_{PL}).
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.
\({[\![ \pi (\neg r) ]\!]}^{\mathcal {V}} \in F(\mathcal {T})\). But this implies that \(F(\mathcal {V}(\mathbb {M}_{R})~ \cup ~ER_{n})\) is inconsistent. Hence, we get a contradiction; or
 2.
\({[\![ \pi (\neg r) ]\!]}^{\mathcal {V}}\) 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.
\({[\![ \pi (\neg r) ]\!]}^{\mathcal {V}}\) 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.
\({[\![ \pi (\phi ) ]\!]}^{\mathcal {V}} \in \mathcal {V}(\mathbb {M}_{R})\). 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.
\({[\![ \pi (\phi ) ]\!]}^{\mathcal {V}} \in ER_{n}\). By Observation 9, \({[\![ \pi (\phi ) ]\!]}^{\mathcal {V}}\) survives telescoping and is supported at level n.
 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.
\({[\![ \pi (\phi ) ]\!]}^{\mathcal {V}}\in \mathcal {V}(\mathbb {M}_{R})\). It follows trivially then that \({[\![ \pi (\phi ) ]\!]}^{\mathcal {V}} \in \mathcal {V}(\mathbb {M}_{R})~\cup ~ ER_{n}~\cup ~G_{\mathcal {T}}\).
 2.
\({[\![ \pi (\phi ) ]\!]}^{\mathcal {V}} \in \mathcal {V}(\mathbb {NM}^{I}_{R})\). 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.
\({[\![ \pi (\phi ) ]\!]}^{\mathcal {V}} \in EG\). 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 A ∈ R (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 ϕ ∈ T 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 ϕ ∈ T 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, A_{1},...,A_{m} 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 ≤ i ≤ m, 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 ϕ ∈ R(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 ϕ ∈ T 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 ϕ ∈ T of height h + 1 with direct children A_{1},...,A_{m}. Since T is closed, A_{1},...,A_{m} 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 ≤ i ≤ m, \(n = I(\mathcal {S})\), 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 A_{1},...A_{m} 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 ϕ ∈ Wff(T), then by the definition of Wff(T) it must be one of the following three cases.
 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.
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.
ϕ is a root of some argument with all arcs labelled by nonmonotonic 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.
\({[\![ \pi (\phi ) ]\!]}^{\mathcal {V}} \in \mathcal {V}(\mathbb {M}_R)\). 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 ϕ ∈ R(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.
\({[\![ \pi (\phi ) ]\!]}^{\mathcal {V}} \in ER_n\). In this case, π(ϕ) must be a nonmonotonic rule embedded at level n whose negation is not embedded at level n. Hence, \(\phi \in \mathcal {S}\) and ϕ ∈ R(T) by the definition of R(T). Hence, π(ϕ) ∈ π(R(T)).
 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.
\({[\![ \phi ]\!]}^{\mathcal {V}} \in F(\mathcal {V}(\mathbb {M}_R)~\cup ~ ER_n~\cup ~G_{\mathcal {T}})\). It follows from the previous three cases and the monotonicity of filters that π(ϕ) is a logical consequence of \(\pi (R(T) \cup R_M^{\prime })\).
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Appendix B: Review of Boolean Algebra
A Boolean algebra is a sextuple \(\mathfrak {A}=<\mathcal {P}, +, \cdot , , \bot , \top >\) where \(\mathcal {P}\) is a nonempty 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:  a ⋅ b = b ⋅ a  
B2.1:  a + (b + c) = (a + b) + c  (Associativity) 
B2.2:  a ⋅ (b ⋅ c) = (a ⋅ b) ⋅ c  
B3.1:  a + (a ⋅ b) = a  (Absorption) 
B3.2:  a ⋅ (a + b) = a  
B4.1:  a ⋅ (b + c) = (a ⋅ b) + (a ⋅ c)  (Distribution) 
B4.2:  a + (b ⋅ c) = (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 nondegenerate.
The elements in B are partially ordered by the relation ≤, where a ≤ b if and only if a ⋅ b = a. Alternatively, a ≤ b 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 a ∈ B, exactly one of a and − a belong to U.
 U2.:

For every a,b ∈ B, a + b ∈ U if and only if a ∈ U or b ∈ U.
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Ehab, N., Ismail, H.O. Log_{A}G: An algebraic NonMonotonic logic for reasoning with graded propositions. Ann Math Artif Intell (2020). https://doi.org/10.1007/s10472020096970
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Keywords
 NonMonotonicity
 Weighted logics
 Uncertainty
 Graded propositions
 Unified framework for NonMonotonicity
Mathematics Subject Classification (2010)
 68T27