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ω-Groundedness of argumentation and completeness of grounded dialectical proof procedures

Abstract

Dialectical proof procedures in assumption-based argumentation are in general sound but not complete with respect to both the credulous and skeptical semantics (due to non-terminating loops). This raises the question of whether we could describe exactly what such procedures compute.

In a previous paper, we introduce infinite arguments to represent possibly non-terminating computations and present dialectical proof procedures that are both sound and complete with respect to the credulous semantics of assumption-based argumentation with infinite arguments.

In this paper, we study whether and under what conditions dialectical proof procedures are both sound and complete with respect to the grounded semantics of assumption-based argumentation with infinite arguments. We introduce the class of ω-grounded and finitary-defensible argumentation frameworks and show that finitary assumption-based argumentation is ω-grounded and finitary-defensible. We then present dialectical procedures that are sound and complete wrt finitary assumption-based argumentation.

1.Introduction

Dialectical proof procedures for assumption-based argumentation ([12,13,15,1719,22,24,32,33]) could be viewed as an integration of the dialectical procedures of abstract argumentation ([8,20,21,26,34,35]) with the process of argument constructions where the latter could get into a non-terminating loop leading to the incompleteness wrt both credulous and skeptical semantics.

A natural question here is: Can we give a precise semantical characterization of what dialectical proof procedures compute?

Representing possibly non-terminating loops as infinite arguments, we present dialectical proof procedures in [31], that are sound and complete wrt credulous semantics. Continuing this line of research, in this paper we present dialectical procedures that are sound and complete with respect to the grounded semantics.

A detailed analysis of the need for infinite arguments in understanding the semantics of dialectical proof procedures is given in [31].

To illustrate how some practical systems of dialectical procedures (like SWI-Prolog ([36])) handle infinite arguments, we analyze the working of SWI-Prolog on two examples taken from [31] below.11

F1:r:anot_αr:αf(0)rn:f(n)f(n+1),n0t:βF2:r:anot_αr:αnot_β,f(0)rn:f(n)f(n+1),n0t:β

The behaviors of F1 and F2 could be explained by the proof trees (viewed arguments) in Fig. 1.

Fig. 1.

Arguments of F1 and F2.

Arguments of F1 and F2.

SWI-prolog does not deliver any answer to the queries a? and “α?” because of infinite recursion in executing α.22

How could we interpret the outputs of SWI-prolog declaratively?

SWI-prolog could not overcome the non-termination of the process to construct an argument supporting α due to the “infinite-loop” represented by infinite argument C1.

We could interpret this observation as indicating that infinite arguments (representing “infinite-loop”) do not support their conclusions the way finite arguments do.

We could also say that the failure of SWI-prolog to deliver any answer to the query “a?” is due to the non-acceptability of argument A. It implies that though infinite argument C1 can not support its conclusion, it could still attack argument A.

SWI-prolog delivers respectively the answers “True”, “False”, “True” to the queries a?, “α?” and “β?” wrt program F2.33 The answer to “α?” is false because the only argument supporting it is C2 that is based on assumption not_β that is attacked by argument B. We could hence say that infinite arguments are attacked the same way finite arguments are attacked.

These observations suggest that infinite arguments still attack (or are attacked by) other arguments as finite arguments do though they do not support their own conclusions.

To capture this peculiar character of infinite arguments, [31] views infinite arguments as self-attacking arguments.

In [31], two proof procedures for credulous semantics are presented. In one, proof trees are constructed explicitly. A filtering mechanism to memorizing the attacked assumptions of both proponent and opponent is employed to prevent each player from constructing the same proof trees twice. The other procedure is simply a flattening of the first. The filtering mechanism in these procedures can not be applied for grounded semantics. The following example illustrates this point.

Example 1.

Consider the argumentation framework F represented by the simple logic program:

r:pnot_qt:qnot_p

To support p, the proof procedures in [31] would correctly deliver argument P (See Fig. 2).

The proponent first constructs argument P to support p. The opponent responds by constructing argument Q to attack P at assumption not_q. The filtering mechanism memorizes assumption not_q after the opponent attacks it and does not allow the opponent to attack it again. The procedures hence stop and deliver P as an admissible argument supporting p.

As the grounded extension of F is empty, procedures for grounded extension need to drop this filtering mechanism.

Fig. 2.

Illustration of Example 1.

Illustration of Example 1.

A consequence of dropping the filtering mechanism in the procedures for credulous semantics in [31] is that an argument could be constructed repeatedly many times at different stages in the computation. To distinguish between these distinct “copies” of the same argument, we consider them together with their histories.

The grounded extension of an argumentation framework AF=(Ar,att) coincides with the least fixed point of the characteristic function FAF. It hence follows that the grounded extension of AF contains the set FAFω()={FAFi()|i is a natural number}.44

To check whether an argument A is groundedly accepted, a dialectical proof procedure would compute the defenders of A (i.e. arguments attacking those attacking A) and then check whether such defenders could also be defended and so on. In cases where there are infinitely many arguments attacking A, such process may not succeed.

Example 2.

Let AF=(Ar,att) with Ar={B,A0,A1,,Ai,Ai+1,} and att={(A2k+1,B)|k0}{(Ak,Ak+1)|k0}. (A graphical illustration of this argumentation framework is in Fig. 3).

It is easy to see that for each natural number 1: FAFi()={A0,A2,,A2(i1)}. Hence Fω()={FAFi()|i is a natural number}={A0,A2,,A2i,}.

The grounded extension of AF is: FAF(FAFω())=FAFω(){B}.

To check whether B is groundedly accepted, a dialectical procedure would need to compute the infinite set {A2k+1|k0} of arguments attacking B and find the infinite set {A2k|k0} of defenders of B. In general, no dialectical proof procedures could carry out such task successfully.

Fig. 3.

An argumentation framework that is not ω-grounded.

An argumentation framework that is not ω-grounded.

To address this issue, we introduce the class of ω-grounded argumentation frameworks where the grounded extension is “computed” after at most ω steps, i.e. the grounded extension of AF coincides with FAFω()={FAFi()|i is a natural number}. We then present several key results:

  • We show that an argumentation framework is ω-grounded iff it is finitary defensible (i.e. all minimal defences of non-selfattacking arguments are finite);

  • We show that finitary assumption-based argumentation frameworks (see Definition 1) with infinite arguments are finitary defensible;

  • We present two dialectical proof procedures that are both sound and complete wrt finitary assumption-based frameworks with infinite arguments where in the first procedure, proof trees together with their histories are fully and explicitly represented to shed light on the construction process of arguments and counter arguments during a derivation and hence providing key insights into the soundness and completeness of the procedure. We then present another procedure that is simply the result of flattening the first where most of the data structures representing the proof tree structures of the arguments are striped away. Consequently, the second procedure is both sound and complete but with much simpler data structure and hence much more amenable to possible implementation.

The rest of this paper is organized as follows. In Section 2, we recall the basic notions of abstract and assumption-based argumentation as well as the machinery of infinite arguments. Section 3 proposes the concept of ω-groundedness of argumentation framework and show that finitary assumption-based argument systems are ω-grounded. We then present a dialectical proof procedure wrt grounded semantics in Section 4. The soundness and completeness of the grounded proof procedure are discussed in Sections 56. Further we present the flatten version of the proof procedure in Section 7. We conclude and discuss possible expansions of our works in Section 8.

2.Preliminaries: Argumentation with infinite arguments

In this section, we recall key basic concepts of abstract argumentation from [16] and [14] and infinite arguments together with some illustrating examples in assumption-based argumentation from [31] and [30].

2.1.Abstract argumentation

Following [14,16], an argumentation framework is a pair AF=(Ar,att) where Ar is a set of arguments, and attAr×Ar so that (X,Y)att specifies that X attacks Y. A set of arguments S attacks an argument X if some argument in S attacks X. S attacks another set of arguments S if S attacks some argument in S. Moreover we say that S defends X iff S attacks all arguments attacking X. We also say that an argument X is defensible if it is defended by some set of arguments.

A set SAr is said to be

  • conflict-free iff S does not attack itself; and

  • admissible iff S is conflict-free and S defends each argument in S; and

  • a preferred extension if S is maximally (wt set inclusion) admissible;55 and

  • a complete extension if S is admissible and contains each argument it defends.

The characteristic function of AF, denoted by FAF, is defined by

FAF:2Ar2Ar
where
FAF(S)={XAr|SdefendsX}

It is straightforward to see that FAF is monotonic (wrt set inclusion). Since 2Ar is a complete partial order (wrt set inclusion),66 FAF has a least fixed point.77

The grounded extension of AF denoted by GEAF, is defined as the least fixed point of FAF.88

As the set of complete extensions of AF is a complete semilattice ([16]),99 the grounded extension coincides with the least complete extension of AF.

2.2.Assumption-based argumentation

Given a logical language L, a standard assumption-based argumentation (ABA) framework ([6]) is a triple F=(R,A,) where R is a set of inference rules of the form l0l1,ln (n0 and l0,,lnL), and AL is a set of assumptions, and is a (total) one-one mapping from A into LA, where x is referred to as the contrary of x, and assumptions in A do not appear in the heads of rules.

Remark 1.

In non-standard ABA frameworks ([24]), the contrary α of an assumption α could be a set. Such non-standard frameworks could be translated into equivalent standard ones by introducing a new atom α for each assumption α and i) define α as the contrary of α; and ii) for each δα, add a new rule: αδ to R.

Remark 2.

Logic programming is a well-known instance of standard ABA where the contrary of a negation-as-failure assumption not_p is p.

Remark 3.

For each rule r of the form l0l1,ln, l0 and the set {l1,,ln} are referred respectively as the head and the body of r and denoted by hd(r), bd(r).

Further the set of assumptions (resp non-assumptions) appearing in the body of r is denoted by Ass(r) (resp. NAss(r)).

Definition 1

Definition 1(Finitary ABA).

An ABA framework F=(R,A,) is finitary if for each sentence δL, the set of rules with head δ is finite.1010

Convention 1.

From now on until the end of the paper,

  • we restrict our consideration to standard finitary ABA. Hence whenever we refer to an ABA framework, we mean a standard finitary one; and

  • if not otherwise mentioned, we assume an arbitrary but fixed finitary standard assumption-based argumentation framework F=(R,A,).

Definition 2

Definition 2(Partial Proof).

Given an ABA F, a partial proof supporting σ0 (wrt F) is a finite sequence of the form

(root,σ0).(r1,σ1).(rn,σn)
where riR, i1 such that σi1=hd(ri) and σibd(ri). If bd(ri)= then σi=true.

Example 3.

Let’s consider an argumentation framework F2 in the introduction.

F2:r:anot_αr:αnot_β,f(0)rn:f(n)f(n+1),n0t:β

Some partial proofs supporting a and α are described below and illustrated in Fig. 4.

ψ=(root,a)ψ=(root,a).(r,not_α)p0=(root,α)p1=(root,α).(r,not_β)p1=(root,α).(r,f(0))p2=(root,α).(r,f(0)).(r0,f(1))pn+1=(root,α).(r,f(0)).(r0,f(1))(rn1,f(n))

Fig. 4.

A graphical representation of partial proofs of Example 3.

A graphical representation of partial proofs of Example 3.

We next define partial proof trees where we identify the nodes in such trees with the partial proofs representing the unique paths from the root to them.

Definition 3

Definition 3(Partial Proof Trees).

A partial proof tree (or just proof tree for simplification) T for a sentence σ0 wrt F is a non-empty set of partial proofs supporting σ0 wrt F such that for each partial proof

p(root,σ0).(r1,σ1).(rn,σn),n>0
from T, the following properties hold:
  • The partial proof p(root,σ0).(r1,σ1).(rn1,σn1) also belongs to T and is referred to as the unique parent of p whereas p is referred to as a child of p;

  • Every partial proof of the form p.(rn,σ) with σbd(rn), also belongs to T and is a child of p;

  • p has no other children.

σ0 is often referred to as the conclusion of T, denoted by Cl(T) while the partial proof (root,σ0) is referred to as the root of T.

An example of partial proof trees is given in Fig. 5.

Remark 4.

For convenience, we often refer to a partial proof tree without mentioning its conclusion if there is no possibility for misunderstanding.

Notation 1

Notation 1(Nodes in Partial Proof Trees).

Abusing the notation for convenience, we often refer to a partial proof (root,σ0).(r1,σ1).(rn,σn) in a partial proof tree T as a node labeled by σn in T.

Fig. 5.

Some proof trees of F2.

Some proof trees of F2.
Notation 2.

Let T be a partial proof tree and S be a set of partial proof trees,

  • A node N in T is said to be a leaf of T if N has no children.

    A leaf N of T is said to be final if N is labeled by true or by an assumption. N is non-final if it is not final.

  • The support of T, denoted by Sp(T), is the set of all sentences labeling the leaves of T and different from true.

  • The set of all assumptions appearing in T is denoted by Ass(T).

    The set of all assumptions appearing in partial proof trees in S is denoted by Ass(S).

  • The set of conclusions of partial proof trees in S is denoted by Cl(S).

Considering Fig. 5, the partial proof (root,α).(r,f(0)) is a leaf node of Tr0 while, (root,α).(r,f(0))(rn1,f(n)) is a leaf node of Trn.

Further, the support of Tr0, i.e. Sp(Tr0), is {not_β,f(0)} and Ass(Tr0) is {not_β}.

Definition 4

Definition 4(Arguments).

  • A partial proof tree T is a full proof tree if all leaves of T are final.

  • An argument for α is a full proof tree for α.

  • The set of all arguments wrt the ABA framework F is denoted by ArF.

Convention 2.

For short, we often simply say, proof trees instead of partial proof trees if there is no possibility of confusion.

When a player in a dialectical computation is attempting to construct an argument, the attempt may not terminate. Declaratively, we model a non-terminating attempt to construct an argument as an attempt to construct an infinite argument.

As infinite arguments do not provide support for their conclusions, they can not be accepted as an admissible argument. We model this intuition as self-attacks of infinite arguments.

Definition 5

Definition 5(Attacks).

  • An argument A attacks an argument B iff one of the following conditions holds:

    • (1) The conclusion of A is the contrary of some assumption in the support of B.

    • (2) A and B are identical and infinite.

  • The attack relation between arguments in ArF is denoted by attF. Define

    AFF=(ArF,attF)

Due to the fact that the infinite arguments always attack themselves, the following lemma holds obviously.

Lemma 1.

Let SArF be admissible wrt AFF=(ArF,attF). Then S contains only finite arguments.

Notation 3.

Let T, T be proof trees and N be a non-final leaf node in T labeled by a non-assumption σ.1111

  • T is an immediate expansion of T at N if there is a rule r of the form σb1,,bm such that T=T{N.(r,b1),,N.(r,bm)}.

    Note that if m=0, T=T{N.(r,true)}.1212

  • We write T=exp(T,N,r).

  • We say T is an immediate expansion of T if T is an immediate expansion of T at some leaf node N of T.

  • We define

    CE(T,N)={exp(T,N,r)|r is a rule s.t. hd(r)=σ}.

Considering Fig. 5, Tr1=exp(Tr0,N,r0) where N is (root,α).(r,f(0)).

Further the set of all immediate expansion of Tr0 at node N, i.e. CE(Tr0,N), is {Tr1}.

Notation 4.

Let T0, T1 be proof trees for σ0.

  • We say T, T are compatible iff TT is also a proof tree.

  • We say T0 is a prefix of T1 iff T0T1.1313 1414

  • We say T0 is a proper prefix of T1 if T0 is a prefix of T1 and T0T1.

Considering Fig. 5, Tr0 is a prefix of Tr1.

Remark 5.

It is worthwhile to note that two proof trees could be compatible without being in a prefix-relationship as illustrated below (see Fig. 6).

Consider an assumption-based framework with three rules:

r1:ab,c;r2:b;r3:c.

Consider two proof trees:

T0={p0,p1,p2,p3},T1={p0,p1,p2,p4}wherep0=(root,a),p1=(root,a).(r1,b),p2=(root,a).(r1,c)andp3=(root,a).(r1,b).(r2,true),p4=(root,a).(r1,c).(r3,true).T0,T1are compatible but neither is a prefix of the other.

Fig. 6.

Two compatible proof trees.

Two compatible proof trees.
Lemma 2.

Let T0, T1 be proof trees. The following statements hold:

  • (1) If T1 is an immediate expansion of T0 then T0 is a proper prefix of T1.

  • (2) Suppose T0 is a prefix of T1. It holds that

    • (a) the roots of T0, T1 coincide; and

    • (b) if N is a node in T0 then the parent and children (if exist) of N in T0 are respectively also the parent and children of N in T1.

Proof.

Obvious. □

Lemma 3.

Let T be an argument, T0 be a proof tree such that T0 is a proper prefix of T. Furthermore, let N be a leaf node in T0 such that CE(T0,N). Then there is an unique T1CE(T0,N) such that T1 is a prefix of T.

T1 is often simply denoted by exp(T0,N,T).

We also define exp(T0,T)={exp(T0,N,T)|N is a non-final leaf node of T0}.1515

Proof.

Lemma 3 in [31] proves the existence of T1. The uniqueness of T1 should be obvious. □

Definition 6.

An increasing sequence of proof trees T0T1Ti is said to be fair if for each Ti, for each non-final leaf node NTi there is a node MTj, j>i, such that N is a proper prefix of M.

Lemma 4.

Let sqT0T1Ti be an increasing sequence of proof trees. The following statements hold:

  • (1) T0T1Ti is a proof tree.

  • (2) If the sequence sq is fair then T0T1Ti is an argument.

Proof.

This is Lemma 4 in [31]. □

Notation 5.

Let T be a proof tree and N(root,σ0).(r1,σ1).(ri,σi) be a node in T.

The height of N in T, denoted by h(N,T), is defined by h(N,T)=i.1616

The minimum of the heights of the non-final leaf nodes in T are denoted by hi(T), i.e. hi(T)=min{h(N,T)|N is a non-final leaf node in T}

3.ω-Groundedness of argumentation frameworks

We first introduce the notion of ω-grounded argumentation frameworks. We then show that finitary ABA frameworks are ω-grounded.

3.1.ω-Grounded argumentation frameworks

Let AF=(Ar,att) be an argumentation framework.

Definition 7

Definition 7(ω-Groundedness).

An argumentation framework AF is ω-grounded if its grounded extension GEAF can be “computed” after at most ω steps, i.e.

GEAF=FAFω()=i=0FAFi()

It should be obvious that the argumentation framework in Fig. 1 is not ω-grounded.

We give now a sufficient condition for ω-groundedness.

Let D be a set of sets of arguments (i.e D2Ar). D is said to be directed iff for all S,SD, SSD.1717

Note that the least upper bound of D, lub(D), is lub(D)=D.

Definition 8

Definition 8(Admissibility Continuity).

A function Φ:2Ar2Ar is said to be admissibility-continuous (or just ad-continuous for short) if for each directed set of admissible sets D

lub{Φ(S)|SD}=Φ(lub(D))

Lemma 5.

If the characteristic function FAF is ad-continuous then AF is ω-grounded.

Proof.

Suppose FAF is ad-continuous. Let D={FAFi()|i is a natural number}. D is obviously directed and lub(D)=FAFω()=i=0FAFi().

Since FAF is ad-continuous, FAFω()=i=0FAFi()=lub{FAF(S)|SD}=FAF(lub(D))=FAF(FAFω()). Therefore FAFω() coincides with the grounded extension of AF. □

Definition 9.

A set S of arguments is said to be a minimal defense of an argument A if S defends A and no proper subset of S defends A.

Definition 10

Definition 10(Finitary Defensible).

  • A defensible argument A is finitary-defensible iff all minimal defenses of A are finite.1818

  • An argumentation framework AF is said to be finitary-defensible if all defensible arguments in AF that are not self-attacking1919 are finitary defensible.

Theorem 1

Theorem 1(Finitary Defensibility implies ad-Continuity).

Let AF be a finitary-defensible argumentation framework. Then FAF is ad-continuous and hence AF is ω-grounded.

Proof.

Let AF be a finitary-defensible argumentation framework, D be a directed set of admissible sets of arguments. We show that lub{FAF(S)|SD}=FAF(lub(D)).

It is obvious that for each SD: Slub(D). Since FAF is monotonic, it holds for each SD: FAF(S)FAF(lub(D)). Therefore lub{FAF(S)|SD}FAF(lub(D)).

It remains to show that lub{FAF(S)|SD}FAF(lub(D)).

Let AFAF(lub(D)). Hence lub(D) is a defense of A. Since D is a set of admissible sets of arguments, lub(D)=D is also admissible. Therefore FAF(lub(D)) is admissible. Thus A is not self-attacking. Since AF is a finitary-defensible, there is a finite minimal defense Mlub(D) of A. Hence for each XM there is SXD such that XSX. Since M is finite, S={SX|XM}D. Thus AFAF(S)lub{FAF(S)|SD}.

From Lemma 5, it follows immediately that AF is ω-grounded. □

3.2.Finitary ABA frameworks are finitary-defensible and ω-grounded

We present a novel insight into the semantic structure of finitary ABA systems by showing that finitary ABA frameworks are finitary-defensible and hence ω-grounded. We begin with a lemma stating that finite arguments of finitary ABA framework are finitary-defensible.

Lemma 6

Lemma 6(Finite Arguments are Finitary Defensible).

Let F be a finitary ABA framework. Then defensible finite arguments in AFF are finitary-defensible.

Proof.

Let B be a defensible finite argument in ArF.

Let S be a minimal defense of B wrt AFF=(ArF,attF). We show that S is finite.

Suppose the contrary that S is infinite. Hence S={D1,,Di,}. Let Si={D1,,Di}.

Because S is a minimal defense of B, it follows that the set AArF of attacks against B is infinite (if A is finite, pick for each XA, an argument DXS s.t. DX attacks X. Hence the set {DX|XA}S is a finite defense of B (wrt AFF). Contradiction to the fact that S is a infinite minimal defense of B).

Let PAn be the set of balanced proof trees of height n2020 that are prefixes of arguments in A and PANn are those elements in PAn that are not attacked by S.

Further let FAn be the set of the full arguments of height n in A.2121

Since F is finitary, both PAn and FAn are finite2222 and so are PANn.

  • We first show that for each n, PANn. Suppose the contrary that there is n such that PANn=, i.e. each partial argument in PAn is attacked by S. Since PAn is finite, there is some i such that Si attacks each partial argument in PAn. Therefore each full argument of height >n in A is attacked by Si.

    Since the set FA0FAn is also finite, there is some j such that Sj attacks each argument in this set. Therefore SiSj attacks each (full) argument in A. Hence SiSj is a finite defense of B, contrary to the assumption that S is a infinite minimal defense of B.

  • We show now that there is an infinite sequence T0T1 such that for each i0: TiPANi.

    It should be clear that each proof tree in PANi+1 is obtained by expanding a proof tree in PANi at all non-final leaf nodes of the later.

    Let T=(V,E) be a tree defined as follows:

    • The set of vertices of T is defined by: V={root}i=0PANi where root is the root of T.2323

    • The set of edges of T is defined by: (T,T)E iff

      • i s.t. TPANi and TPANi+1 and TT; or

      • T=root and TPAN0.

    Since PANi for all i, T is infinite and because F is finitary, each child of T has finitely many children. Thus there is an infinite path in T. In other words, there is an infinite sequence T0T1 such that for each i0: TiPANi. Since each Ti is a balanced tree of height i, the sequence T0T1 is obviously fair.

  • Let T=i=0Ti. T is therefore a full proof tree and hence an argument. T is hence an attack against B and not attacked by S. Contradiction.

 □

Since the set of finite arguments in AFF contains the set of all non-self-attacking arguments, it follows immediately from Theorem 1 and the above Lemma 6:

Theorem 2

Theorem 2(ω-Groundedness of finitary ABA frameworks).

Let F be a finitary ABA framework. Then the following statements hold:

  • AFF is finitary-defensible;

  • The characteristic function FAFF is ad-continuous;

  • AFF is ω-grounded.

4.Grounded dispute derivations

In this section we will present a proof procedure for grounded semantics where proof trees together with their histories are fully and explicitly represented to shed light on the construction process of arguments and counter arguments during a derivation and hence providing key insights into the soundness and completeness of the procedure. Later, in Section 7, we present another procedure that is simply the result of flattening the one presented in this section. Consequently, the second procedure is both sound and complete but with much simpler data structure and hence much more amenable to possible implementation.

The purpose of the dispute derivations is to construct arguments. So it is kind of natural to refer to proof trees representing partly constructed arguments in a dispute derivation as partial arguments. It also makes it more intuitive to talk about attack between partial arguments.

Remark 6

Remark 6(Partial Arguments).

  • Abusing the notation slightly for ease of reference, from now on until the end of the paper, we often refer to proof trees as partial arguments.

    To avoid any possibility of misunderstanding, we often refer to arguments as full arguments.

  • We often say that a partial argument T attacks a partial argument T if there is an assumption αAss(T) such that Cl(T)=α.

  • We also often refer to a proof tree consisting only of the root and supporting a sentence δ by [δ], i.e. [δ]={(root,δ)}.

In a dispute derivation, an argument could be constructed repeatedly many times at different stages in the computation. To distinguish between these distinct “copies” of the same argument, we consider them together with their histories.

Definition 11

Definition 11(Histories of Partial Arguments).

A possible history of a partial argument T is represented by a sequence (T0,t0),,(Tk,tk) where T0,,Tk are partial arguments such that

  • T0=[Cl(T)] and T=Tk; and

  • Ti, 0<ik, is an immediate expansion of Ti1; and

  • t0<t1<<tk are natural numbers representing the stages in the dispute when such expansions happen.

Definition 12

Definition 12(Profiled Partial Arguments).

A profiled partial argument (ppa) is a pair π=(T,h) where T is a partial argument and h=(T0,t0),,(Tk,tk) is a history of T.

  • (1) We also often refer to T and h by Tπ and hπ respectively;

  • (2) t0 is often referred to as the starting time of π denoted by st(π) while k is referred to as the length of the history of π denoted by lh(π);

  • (3) An assumption α is often referred to as the target of π denoted by target(π), if Cl(T)=α;

Example 4.

Considering again the argumentation framework F2 in the introduction

F2:r:anot_αr:αnot_β,f(0)rn:f(n)f(n+1),n0.t:β

π0=(T0,h0) is an example of a profiled partial argument where T0=[a] and h0=(T0,0) where h0 says that T0 is obtained at stage 0.

Suppose T0 is expanded at stage 1 resulting in a new profiled partial argument π1=(T1,h1) (see Fig. 7) where h1=(T0,0),(T1,1). The history h1 says that T1 is obtained by expanding T0 at stage 1 and T0 is obtained a stage 0.

Fig. 7.

Proof trees of Example 4.

Proof trees of Example 4.

A dispute derivation is viewed as a game between a proponent and an opponent where the players take turn to develop their arguments. The goal of the proponent is to construct an argument to support some desired conclusion and other arguments to defend against the attacks from the opponent while the opponent’s goal is to construct arguments to attack proponent’s arguments. At each step, either player could choose to either expand their partly constructed arguments or start a new argument to attack the other’s argument.

Definition 13

Definition 13(Grounded Dispute Derivation).

A grounded dispute derivation for a sentence σ is a (possibly infinite) sequence of the form

PPT0,PTA0,OPT0,,PPTn,PTAn,OPTn
where 2424
    • (1) for each in, PPTi, OPTi are sets of profiled partial arguments; and

    • (2) PTAi is a set of assumptions appearing in the proponent partial arguments (up to stage i) that are to be attacked by the opponent; and

    • (3) PPT0 contains exactly one profiled partial argument consisting of only the root labeled by σ, i.e. PPT0={(T0,h0)} where T0=[σ] and h0=([σ],0), and

    • (4) PTA0= if σ is not an assumption, otherwise PTA0={σ}, and

    • (5) OPT0=.

  • at stage i>0, one of the dispute parties makes a move transforming the dispute from state i1 to state i as follows:

    • (1) Suppose the proponent makes a move at stage i. The proponent can choose one of the following two options:

      • (a) The proponent expands some profiled partial argument π=(T,h)PPTi1 by:

        • selecting a non-final leaf node N in T labeled by a non-assumption sentence δ and a rule r with hd(r)=δ; and

        • expanding (T,h) at N by r.

        The result will be:

        • PPTi=(PPTi1{π}){π} where π=(T,h) and T=exp(T,N,r) and h=h.(T,i);2525

        • PTAi=PTAi1Ass(r)

        • OPTi=OPTi1.

      • (b) The proponent attacks an opponent’s profiled partial argument π=(T,h)OPTi1 at an assumption αAss(T) resulting in:

        PPTi=PPTi1{(T,h)} where T=[α] and h=([α],i); and

        PTAi=PTAi1

        OPTi=OPTi1{π}

    • (2) Suppose the opponent makes a move at stage i. The opponent can choose one of the following two options:

      • (a) The opponent expands an opponent profiled partial argument π=(T,h)OPTi1 at a non-final leaf node NT labeled by a non-assumption sentence δ resulting in:

        PPTi=PPTi1

        PTAi=PTAi1

        OPTi=(OPTi1{π}){π|π=(T,h),TCE(T,N),h=h.(T,i)}2626

      • (b) The opponent attacks an assumption αPTAi resulting in:

        PPTi=PPTi1

        PTAi=PTAi1{α}

        OPTi=OPTi1{π} where π=(T,h) and T=[α] and h=([α],i)

Remark 7.

  • When opponent carries out step (2.a), it is possible that CE(T,N)=. We often refer to this case as a failed expansion of opponent profiled partial argument π.

  • Note that if the assumption-based framework is not finitary, the set OPTi in step (2a) could be infinite, and hence not implementable.

A dispute derivation is successful (for the proponent) if i) the proponent manages to construct in full her arguments to support her stated conclusion and to defend against the attacks from the opponent and ii) the opponent runs out of attacks against the proponent.

Definition 14

Definition 14(Successful Grounded Dispute Derivation).

A grounded dispute derivation

PPT0,PTA0,OPT0,,PPTn,PTAn,OPTn
is successful (for the proponent) if PTAn=OPTn= and for each (T,h)PPTn, T is a full argument.

Table 1

A successful grounded dispute derivation of F2

StageMovePPTPTAOPT
0π0=(T0,h0) h0=(T0,0)
11aπ1=(T1,h1) h1=(T0,0),(T1,1)not_α
22bπ1=(T1,h1), h1=(T0,0),(T1,1)π0=(T0,h0) h0=(T0,2)
32aπ1=(T1,h1), h1=(T0,0),(T1,1)π1=(T1,h1) h1=(T0,2),(T1,3)
41bπ1=(T1,h1), h1=(T0,0),(T1,1) π0=(T0,h0), h0=(T0,4)
51aπ1=(T1,h1), h1=(T0,0),(T1,1) π1=(T1,h1), h1=(T0,4),(T1,5)
Example 5.

A successful grounded dispute derivation

PPT0,PTA0,OPT0,,PPT5,PTA5,OPT5
for sentence a (wrt the argumentation framework F2 in the introduction) is illustrated in Table 1 and explained in details below. For convenience we recall F2 below.
F2:r:anot_αr:αnot_β,f(0)rn:f(n)f(n+1),n0t:β

At stage 0, we have PPT0={π0}, PTA0=OPT0= where π0=(T0,h0) and T0=[a] and h0=(T0,0).

  • At stage 1, the proponent makes a move to expand the ppa π0 by applying step (1a) using rule r.

    Hence PPT1=(PPT0{π0}){π1}={π1} where π1=(T1,h1), T1=exp(T0,root,r) and h1=(T0,0),(T1,1) (T1 is illustrated in Fig. 8).

    Since rule r has the assumption not_α (i.e. Ass(r)={not_α}), we have PTA1=PTA0Ass(r)=PTA0{not_α}={not_α}.

    OPT1=OPT0=.

  • Next, at stage 2, the opponent attacks the assumption not_αPTA1 by applying step (2b). Hence OPT2=OPT1{π0}={π0} where π0=(T0,h0) with T0=[α] and h0=(T0,2) stating that T0 is created at stage 2 (See Fig. 8).

    PPT2=PPT1={π1}.

    PTA2=PTA1{not_α}=.

  • At stage 3, the opponent applies step (2a) to expand T0 to T1. Since rule r is the only one with head α, it follows CE(T0,root)={T1} with T1=exp(T0,root,r) (See Fig. 8). Therefore

    OPT3=(OPT2{π0}){π1}={π1} where π1=(T1,h1) and h1=(T0,2),(T1,3).

    PPT3=PPT2={π1}

    PTA3=PTA2=

  • At stage 4, the proponent attacks opponent’s ppa π1 at assumption not_βT1 by applying step (1b).

    Therefore PPT4=PPT3{π0}={π1,π0} where π0=(T0,h0) and T0=[β] and h0=(T0,4) (See Fig. 8).

    PTA4=PTA3=

    OPT4=OPT3{π1}=.

  • Finally at stage 5, the proponent applies step (1a), using rule t, to expand T0 to argument T1=exp(T0,root,t) (See Fig. 8). Therefore

    PPT5=(PPT4{π0}){π1}={π1,π1} where π1=(T1,h1) with h1=(T0,4),(T1,5).

    PTA5=PTA4Ass(t)=

    OPT5=OPT4=

As there is no any other assumption in PTA5 and OPT5 is empty and all ppa in PPT5 are full, the derivation is successful.

Fig. 8.

Proof trees for Example 5.

Proof trees for Example 5.

Example 6.

Consider the following argumentation framework

F3:r:pnot_a

Table 2

A successful grounded dispute derivation of F3

StageMovePPTPTAOPT
0π0=(T0,h0) h0=(T0,0)
11aπ1=(T1,h1) h1=(T0,0),(T1,1)not_a
22bπ1=(T1,h1) h1=(T0,0),(T1,1)π0=(T0,h0) h0=(T0,2)
32aπ1=(T1,h1) h1=(T0,0),(T1,1)

Fig. 9.

Proof trees of the dispute derivation for sentence “p”.

Proof trees of the dispute derivation for sentence “p”.

A successful dispute derivation for sentence “p

PPT0,PTA0,OPT0,,PPT3,PTA3,OPT3
is illustrated in Table 2 and is explained in details below.

At stage 0, we have PPT0={π0}, PTA0=OPT0= where π0=(T0,h0) and T0=[p] and h0=(T0,0) (see Fig. 9).

  • At stage 1, the proponent makes a move to expand the ppa π0 by applying step (1a) using rule r.

    Hence PPT1=(PPT0{π0}){π1}={π1} where π1=(T1,h1), T1=exp(T0,root,r) (see Fig. 9) and h1=(T0,0),(T1,1).

    Since rule r has the assumption not_a (i.e. Ass(r)={not_a}), we have PTA1=PTA0Ass(r)=PTA0{not_a}={not_a}.

    OPT1=OPT0=.

  • Next, at stage 2, the opponent attacks the assumption not_aPTA1 by applying step (2b). Hence OPT2=OPT1{π0}={π0} where π0=(T0,h0) with T0=[a] and h0=(T0,2) stating that T0 is created at stage 2 (see Fig. 9).

    PPT2=PPT1={π1}.

    PTA2=PTA1{not_a}=.

  • At stage 3, the opponent applies step (2a) to expand T0. Since there is no rule with head a, it follows CE(T0,root)=. Therefore

    OPT3=(OPT2{π0})=.

    PPT3=PPT2={π1}

    PTA3=PTA2=

As there is no any other assumption in PTA3 and OPT3 is empty and all ppa in PPT3 are full, the derivation is successful.

5.Soundness of grounded dispute derivation

We first introduce some new notations.

Notation 6.

  • We say a ppa π is a continuation of another ppa π iff hπ is a prefix of hπ (and hence TπTπ and st(π)=st(π)).

    π is an immediate expansion of π if hπ=(T0,t0),,(Tk,tk) and hπ=hπ.(Tk+1,tk+1).2727

  • A ppa π is said to be full iff Tπ is an argument.

  • We say two ppas π, π are compatible if st(π)=st(π) and Tπ, Tπ are compatible.

  • We often refer to a ppa appearing in some PPTi (resp. OPTi) in a grounded dispute derivation as a proponent ppa (resp opponent ppa).

The following lemma expresses the intuition that when the proponent has partially constructed an argument, she is expected to finish it. And she should not construct two distinct arguments for the same purpose at the same time, i.e. at any stage there should be no distinct continuations of the same proponent ppa at some previous stage.

Lemma 7.

Let dd=PPT0,PTA0,OPT0,,PPTn,PTAn,OPTn be a grounded dispute derivation.

  • (1) Let π be some proponent ppa appearing in dd. The following statements hold:

    • (a) For each i:st(π)in, there is an unique continuation of π in PPTi.

    • (b) Let πi, πi+1 (st(π)i<n) be continuations of π in PPTi, PPTi+1 respectively. Then πi+1 is either an immediate expansion of πi or identical to πi.

  • (2) Let π, π be two proponent ppas in dd such that st(π)=st(π). Then one is a continuation of the other.

    Consequently, if π, π are full then they are identical.

Proof.

See proof of Lemma 7 in Appendix A.3. □

The following lemma states intuitively that at any stage in a derivation, it is not possible for the opponent to construct the same argument in different ways.

Lemma 8.

Let dd=PPT0,PTA0,OPT0,,PPTn,PTAn,OPTn be a grounded dispute derivation. It holds that for each 0in, for all π,πOPTi, if π, π are compatible then π, π are identical.

Proof.

See proof of Lemma 8 in Appendix A.3. □

Remark 8.

An opponent ppa π in a dispute derivation dd=PPT0,PTA0,OPT0,,PPTn,PTAn,OPTn is said to be discontinued at step i iff πOPTi1 and there is no continuation of π afterwards, i.e. ji there is no continuation of π in OPTj.

It is obvious that in a successful grounded dispute derivation, every opponent ppa is discontinued at some stage. The following lemma sheds light on the structure of the evolution of opponent ppas in a grounded dispute derivation.

Lemma 9.

Let dd be a dispute derivation dd=PPT0,PTA0,OPT0,,PPTn,PTAn,OPTn.

Further let A be an argument such that Cl(A)=α for some assumption α and π=([α],([α],i))OPTi for some 0in.

Then there is an unique sequence contdd(π)=π0,,πk (ii+kn) such that

  • (1) π0=π; and

  • (2) for each i<jk, πjOPTi+j and πj is a continuation of πj1 and TπjA; and

  • (3) if i+k<n then πk is discontinued at i+k+1 and there is no ppa at any OPTi+k+1,,OPTn that is compatible with πk.

Proof.

See proof of Lemma 9 in Appendix A.3. □

The following lemma states that each argument attacking a proponent argument in a successful dispute derivation is counter-attacked by some proponent argument.

Lemma 10.

Let dd be a successful grounded dispute derivation that terminates at PPTn,PTAn,OPTn and π=(T,h)PPTn and T be an argument attacking argument T. Then there is a opponent ppa π such that st(π)<st(π) and TπT and π is discontinued at some step between st(π) and n.

Proof.

See proof of Lemma 10 in Appendix A.3. □

The following theorem shows that in a successful dispute derivation, each proponent argument whose construction started at some stage i is defended by the proponent arguments constructed after stage i.

Theorem 3.

Let dd=PPT0,PTA0,OPT0,,PPTn,PTAn,OPTn be a successful grounded dispute derivation and π=(T,h)PPTn. Further let Tπ={Tπ|πPPTn and st(π)>st(π)}. It holds:

TFAFF(Tπ)

Proof.

Let TArF such that T attacks T. From Lemma 10 there is opponent ppa π (discontinued at stage i) such that st(π)<st(π) and TπT.

There are two cases:

  • (1) Case 1 The opponent makes a move at stage i. Because π is discontinued at step i, the opponent expands π at stage i. Because TπT, it follows from Lemma 3 that π has a continuation at stage i. Contradiction to the fact that π is discontinued at stage i. Hence this case is not possible.

  • (2) Case 2 The proponent makes a move at stage i. Since π is discontinued at stage i, it follows that the proponent attacks π at some assumption α at this stage. Since TπT, it follows that α appears in T.

    Hence a new proponent ppa π1=(T1,h1) is created in PPTi where T1=[α] and h1=(T1,i).

    Since dd is successful, there is π2PPTn that is a continuation of π1. Hence Tπ2 attacks T.

    Since st(π2)=st(π1)=i>st(π)>st(π), it follows that Tπ2Tπ. Thus Tπ defends T against T.

    Since T is arbitrary selected, it follows that Tπ defends T against all attacks against T.

    We have proved that TFAFF(Tπ).

 □

An immediate consequence of Theorem 3 is the soundness of the grounded dispute procedure as stated in the following theorem.

Theorem 4

Theorem 4(Soundness Theorem).

Let dd=PPT0,PTA0,OPT0,,PPTn,PTAn,OPTn be a successful grounded dispute derivation. It holds

{Tπ|πPPTn}GEAFF

Proof.

Let m=|PPTn|. Let πm,,π1 be an increasing (according to the starting times) listing of ppas in PPTn, i.e. for mi>1, st(πm)=0 and st(πi)<st(πi1).

Let Ti=Tπi. We first prove by induction on i, TiFAFFi().

Basic Step i=1.

From Theorem 3, it follows immediately that T1FAFF() (because Tπ1=).

Inductive Step Suppose for all j (i>j1), TjFAFFj().

We prove that TiFAFFi().

From FAFF()FAFF2()FAFFi1() and for each j (i>j1), TjFAFFj(), it follows that Tπi={Ti1,,T1}FAFF()FAFF2()FAFFi1()FAFFi1().

Thus from Theorem 3, TiFAF(Tπi)FAF(FAFFi1())=FAFFi().

Therefore {Tπ|πPPTn}={Tm,,T0}FAFFm()GEAFF. □

6.Completeness of grounded proof procedure

It should be clear that for each sentence σCl(GEAFF), there are always some arguments in GEAFF supporting σ. The completeness of the grounded proof procedure means that at least one of such arguments could be derived in a successful grounded dispute derivation.

In other words, completeness of grounded procedure is not about verifying whether a sentence is supported by an argument in the grounded extension. It is rather about showing that for each sentence in the grounded extension, an argument supporting it could be derived by a grounded derivation.

The proof is constructive by first constructing (according to the definition of strongly grounded dd) for each sentence σCl(GEAFF), a special kind of grounded dispute derivation, and then show that each of such special grounded derivation is always successful.

For each argument AGEAFF and sentence δCl(GEAFF), define

rank(A)=min{i|AFAFFi()} andrank(δ)=min{i|δCl(FAFFi())}

Remark 9.

  • (1) Since AFF is ω-grounded (because F is finitary (see Convention 1)), both rank(A) and rank(δ) are natural numbers.

  • (2) It is not difficult to see that for each AGEAFF, rank(Cl(A))rank(A).

  • (3) For each δCl(GEAFF), an argument AGEAFF with Cl(A)=δ and rank(δ)=rank(A), is often referred to as a ground support of δ.

    It is obvious that each sentence δCl(GEAFF) has a ground support.

  • (4) A mapping λ assigning to each sentence in δCl(GEAFF) a ground support λ(δ)GEAFF of δ is often referred to as a ground map of F.2828

Lemma 11.

Let AGEAFF and αAss(A). Further let B be an argument attacking α. The following statements hold:

  • (1) αCl(GEAFF) and rank(α)rank(A); and

  • (2) There is an assumption βAss(B) such that βCl(GEAFF) and rank(β)<rank(α).

Proof.

Since AGEAFF, it follows from Theorem 2 that AFAFFk() where k=rank(A) and k is a natural number. Hence, every attack against A is attacked by some argument in FAFFk1(). Thus every attack against α is attacked by some argument in FAFFk1() implying that αCl(FAFFk()). Hence rank(α)k=rank(A).

Let rank(α)=m. Hence αCl(FAFFm()). From Theorem 2, m is a natural number. Thus every attack against α is attacked by some argument in FAFFm1(). It follows that B is attacked by some argument CFAFFm1() at some assumption βAss(B). Thus Cl(C)=β and rank(β)rank(C)m1<m=rank(α)rank(A). □

We introduce the concept of strongly grounded dispute derivation below where the proponent arguments are grounded according to some ground map λ. This concept is inspired by the idea of the H-constrainted dispute derivations for admissibility semantics in [31].

The construction of strongly grounded dispute derivations mimics such derivations in abstract argumentation when arguments are assumed to be given and a key basic step is like: “pick some argument…”.

This step is elaborated in the strongly grounded dispute derivation by the proponents and opponents as follows:

  • Once the proponent has started to build up an argument, she will continue until getting the full argument constructed without being bothered to attack the other’s arguments. The opponent is not allowed to disrupt her even if he could attack her still partly constructed argument at some stage.

  • On the contrary, the opponent’s construction of his arguments will be disrupted by attacks from the proponent whenever she could launch such attacks. But the opponent is not allowed to interrupt the constructions of his own arguments to launch an attack against a proponent argument even if such attack is possible.

Definition 15

Definition 15(Strongly Grounded Dispute Derivation).

A strongly grounded dispute derivation for a sentence δ wrt a ground map λ is a grounded dispute derivation for δ (as defined in Definition 13) such that the following extra constraints are satisfied:

  • The proponent executes step (1.a) (to expand proponent ppa π=(T,h)) with an extra condition that exp(T,N,r)λ(Cl(T));

  • The proponent executes step (1.b) (to attack an opponent ppa π=(T,h)OPTi at an assumption αAss(T)) with two extra conditions:

    • αCl(GEAFF) and rank(α)<rank(target(π)); and

    • step (1.a) is not possible;

  • The opponent executes step (2.a) (to expand an opponent ppa π=(T,h)OPTi) with two extra conditions:

    • h(N,T)=hi(T);2929

    • It is not possible for the proponent to perform any of steps (1.a) or (1.b).

  • The opponent executes step (2.b) (to attack a proponent assumption αPTAi) with two extra conditions:

    • It is not possible for the proponent to perform any of steps (1.a) or (1.b);

    • It is not possible for the opponent to perform step (2.a).

Remark 10.

We often simply refer to a strongly grounded dispute derivation without explicitly mentioning the associated ground map if there is no possibilities for misunderstanding.

Definition 16.

A strongly grounded dispute derivation PPT0,PTA0,OPT0,,PPTn,PTAn,OPTn is said to be terminated if neither the proponent nor the opponent could make a move at stage n.

Theorem 5

Theorem 5(Completeness Theorem).

Let F be a finitary ABA framework and σCl(GEAFF). Then there is a successful strongly grounded dispute derivation for σ PPT0,PTA0,OPT0,,PPTn,PTAn,OPTn such that σCl(Tπ) for some πPPTn.

Proof.

See Section 6.1. □

The proof of the completeness theorem follows directly from two insights:

  • i) each terminated strongly grounded dispute derivation for σCl(GEAFF) is successful; and

  • ii) there is no infinite strongly grounded dispute derivation.

We first show below that each terminated strongly grounded dispute derivation for σCl(GEAFF) is successful.

Theorem 6.

Let sdd=PPT0,PTA0,OPT0,,PPTn,PTAn,OPTn be a terminated strongly grounded dispute derivation for δCl(GEAFF) wrt a ground map λ. Then sdd is successful.

Proof.

Let πPPTn and hπ=(T0,i0)(Tm,im) with T0=[δ]. From the design of the proof procedure in definitions 1315, it holds: T0T1Tmλ(δ). Since sdd is terminated, it is not possible to expand Tm further into λ(δ). Hence λ(δ)=Tm=Tπ. Hence Tπ is a full argument.

Since the opponent can not carry out step (2.b) (attack proponent assumptions in PTAn), even though all other steps are not possible, it follows that PTAn=.

Since the opponent can not carry out step (2.a) (expand opponent partial arguments) even though none of steps (1.a, 1.b, 2.b) can be executed, it follows that for each πOPTn: Tπ is a full argument and there is pPPTn s.t. Tπ attacks Tp. Since Tp=λ(Cl(Tp)), TpGEAFF.

From Lemma 11, it follows there is an assumption αAss(Tπ) such that αCl(GEAFF) and rank(α)<rank(target(π)). Thus it is possible for the proponent to execute step (1b) at stage n. Contradiction. Therefore π does not exist. Hence OPTn=.

We have proved that sdd is successful. □

To show the completeness theorem, it remains for us to show that there is no infinite strongly grounded dispute derivation.

Lemma 12.

Let sdd=PPT0,PTA0,OPT0,,PPTn,PTAn,OPTn be a strongly grounded dispute derivation wrt a ground map λ. The following statements hold:

  • (1) Each PPTi contains at most one not-full ppa;

  • (2) Let πPPTi, 0in. Then it holds that π is not full iff there is an unique immediate expansion of π in PPTi+1;

  • (3) Let π=(T,h)PPTi, 0in. Then h is of the form

    h=(T0,i0),(T1,i0+1),,(Tk,i0+k)
    such that

    • i0+ki; and

    • Tjλ(Cl(T))) for kj1; and

    • if π is not full then i0+k=i.

  • (4) Let πPPTi (0in) such that the proponent does not make a move to expand π at some stage k with ik>st(π). Then π is full.

Proof.

  • (1) By induction on i. The statement holds obviously for i=0.

    Suppose the statement holds for i. We show that it also holds for i+1. If the opponent makes a move at stage i+1 then PPTi+1=PPTi. The statement holds obviously.

    If the proponent executes step (1b) at stage i+1 meaning that step (1a) is not possible at this stage. Hence all ppas in PPTi are full. The statement holds obviously.

    Suppose the proponent executes step (1a) at stage i+1 (to expand a ppa πPPTi into a new ppa πPPTi+1). Hence π is the only non-full ppa in PPTi. Therefore, if π is full, there is no non-full ppa in PPTi+1. If π is not full, π is the only non-full ppa in PPTi+1. The statement holds.

  • (2) Suppose π is not full. It follows immediately from the definition of sdd that the proponent will execute step (1a) at stage i+1 to expand π into πPPTi+1. The uniqueness of π follows from Lemma 7.

    Suppose πPPTi+1 is the unique immediate expansion of π. Hence π is no-full obviously.

  • (3) By induction on i with ii0. The statement holds obviously for i=i0. Suppose the statement holds for ii0. We show that it also holds for i+1.

    Let πPPTi+1. There are two cases:

    Case 1 πPPTi. From statements (1,2) and Lemma 7, it follows that π is full. The statement follows directly from the induction hypothesis.

    Case 2 πPPTi. Hence the proponent executes step (1a) at stage i+1 to expand some pPPTi into π with hp=(T0,i0),(T1,i0+1),,(Tk,i0+k) and Tp=Tk. Hence p is not full. From the induction hypothesis, we have i0+k=i. Therefore hπ=hp.(Tπ,i+1) and TπCl(Tπ). Hence the statement holds.

  • (4) Suppose π is not full. From statement (3), hπ=(T0,i0),(T1,i0+1),,(Tk,i0+k) with i0+k=i where i0=st(π). It follows that the proponent makes the move (1a) at every stage j:ij>0. Contradiction. We have proved that π is full.

 □

Lemma 13.

Let sdd=PPT0,PTA0,OPT0,,PPTn,PTAn,OPTn be a strongly grounded dispute derivation for δCl(GEAFF) wrt a ground map λ.

It holds that for all i: PTAiCl(GEAFF).

Proof.

Let αPTAi, 0in. We show by induction on i that αCl(GEAFF).

Base Step: “i=0”. If δ is not an assumption then PTA0=. The lemma holds vacuously. Suppose δ is an assumption. The lemma holds since it is assumed that δCl(GEAFF).

Inductive Step: Suppose the lemma holds for i1. We show that it also holds for i.

The lemma follows directly from the inductive hypothesis if at step i, the opponent makes a move or the proponent attacks an opponent ppa.

Suppose now that the proponent expands some ppa π at step i. If αPTAi1, the lemma follows from the inductive hypothesis.

Suppose now αPTAiPTAi1. Therefore αAss(λ(Cl(Tπ)). Since λ(Cl(Tπ))GEAFF, it follows αCl(GEAFF). □

Lemma 14.

Let sdd=PPT0,PTA0,OPT0,,PPTn,PTAn,OPTn be a strongly grounded dispute derivation for δCl(GEAFF) wrt a ground map λ.

For each 0in, for all π,πOPTi, it holds that st(π)=st(π).

Proof.

Suppose there are π, π s.t. st(π)st(π); Without loss of generality, we assume that st(π)<st(π). We first show that π is full.

Suppose π is not full. That means that at stage st(π), the opponent attacks a proponent assumption (step 2.b) even though the opponent could have perform step (2a) (expanding a prefix of π). Hence sdd is not a strongly grounded dispute derivation. Contradiction.

We have proved that π is full.

Since step (2a) has higher priority than step (2b) in the definition of sdd, it follows that st(π)+lh(π)<st(π).

Since π is full, Tπ is an argument. Let α be an assumption such that Cl(Tπ)=α. From Lemma 13, it follows αCl(GEAFF). From Lemma 11, there is βAss(Tπ) such that rank(β)<rank(α) and βCl(GEAFF). It is thus possible for the proponent to execute step (1b) before st(π) to attack β. Hence πOPTi. Contradiction.

Therefore the assumption that st(π)st(π) is wrong. □

Definition 17.

Let π be an opponent ppa and π, π be proponent ppas in a strongly grounded dispute derivation sdd.

  • (1) We say π hits π at an assumption αAss(Tπ) if

    • π is full and α=Cl(Tπ); and

    • st(π)<st(π); and

    • there is no opponent ppa p in sdd such that α=Cl(Tp) and st(π)<st(p)<st(π).

    We say π hits π if π hits π at some assumption.

  • (2) We say π hits π (at an assumption αAss(Tπ)) if the proponent attacks π at assumption α at stage st(π).

  • (3) We say π supports π if there is an opponent ppa p such that p hits π and π hits p.

It is not difficult to see that the following lemma holds.

Lemma 15.

Let π be an opponent ppa and π be proponent ppa in a strongly grounded dispute derivation sdd.

  • (1) for each assumption αTπ, there is at most one proponent ppa hitting π at α.

  • (2) for each assumption αTπ, there is at most one opponent ppa hitting π at α.

  • (3) if π hits π then π is discontinued at stage st(π);3030

Lemma 16.

Let π, π be two proponent ppas in a strongly grounded dispute derivation sdd for δCl(GEAFF) such that π supports π. It holds rank(Cl(Tπ))>rank(Cl(Tπ)).

Proof.

Let p be an opponent ppa such that p hits π and π hits p.

From Lemma 11, it follows rank(Cl(Tπ))rank(target(p)). From the selection criteria for step (1b) in sdd, it follows rank(target(p))>rank(Cl(Tπ)).

Hence rank(Cl(Tπ))>rank(Cl(Tπ)). □

Theorem 7.

Let sdd=PPT0,PTA0,OPT0,,PPTn,PTAn,OPTn, be an infinite strongly grounded dispute derivation for δCl(GEAFF) wrt ground map λ.

Then for each k, the set OPTk={πi=0OPTi|st(π)=k} is finite.

Proof.

Suppose the contrary that there is k, such that OPTk is infinite. From Definition 13 (of grounded dispute derivation), it follows that the opponent attacks a proponent assumption α at stage k. From Lemma 14, it follows OPTk={π0} where π0=(T0,h0) with T0=[α] and h0=([α],k).

It follows that OPTk has the structure of a tree Tk as follows:

The root of Tk is π0. A ppa πOPTk is a child of πOPTk iff π is an immediate expansion of π.

Because F is finitary, each ppa has only finitely many immediate expansions. Hence each node in Tk has finitely many children. Since Tk is infinite (because OPTk is infinite), there is an infinite path labeled by π0,π1, , in Tk such that πi+1 is an immediate expansion of πi.

We show that T=i=0Tπi is a full argument. Suppose T is not a full argument. Then there is a non-final leaf node N in T. Therefore there is m such that for each jm, N is a non-final leaf node in Tπj.

For each jm, let Nj be the non-final leaf node in Tπj selected for expansion to Tπj+1. It follows from the extra condition for step (2a) in Definition 15 of sdd, that h(Nj,T)=h(Nj,Tπj)h(N,Tπj)=h(N,T). The set of Njs is therefore finite implying that the sequence π0,π1, is finite. Contradiction.

We have proved that T is a full argument.

It is clear that T attacks α. Since αCl(GEAFF) (Lemma 13), T is attacked by some XGEAFF such that rank(X)<rank(α). It follows there is βAss(T) such that β=Cl(X). Hence rank(β)<rank(X)<rank(α).

Let j=min{i|βAss(Tπi)}. From Definition 15 of sdd, it is possible for the proponent to execute step (1b) at stage j or some later stage >j.3131 As step (1b) has higher priority than step (2a), the infinite path π0,π1, , in Tk does not exist. Contradiction

Hence tree Tk is finite. Contradiction. Hence we have proved that OPTk is finite. □

Lemma 17.

Let sdd=PPT0,PTA0,OPT0,,PPTn,PTAn,OPTn, be an infinite strongly grounded dispute derivation for δCl(GEAFF) wrt ground map λ.

Let π{PPTi|i0} and π be a full ppa. Then the set of proponent ppas in sdd supporting π is finite.

Proof.

It is obvious that for each assumption αTπ there is at most one opponent ppa hitting π at α. Let P be the set of opponent ppas in sdd that hit π. Since Ass(Tπ) is finite, it follows directly from Lemma 15 that P is finite. Hence it also follows directly from Lemma 15 that there are only finite number of proponent ppas hitting the ppas in P. Therefore the lemma holds. □

Theorem 8.

Let sdd=PPT0,PTA0,OPT0,,PPTn,PTAn,OPTn, be an infinite strongly grounded dispute derivation wrt ground map λ.

The set of proponent profiled full arguments in sdd is finite.

Proof.

We first prove two relevant properties.

Property 1.

  • (1) Each opponent ppa in sdd hits some proponent ppa in sdd;

  • (2) Each proponent ppa in sdd that does not start at stage 0 hits some opponent ppa in sdd.

  • (3) Each full proponent ppa in sdd that does not start at stage 0 supports some other full proponent ppa in sdd.

Proof of Property 1.

Statement 2 holds obviously. Statement 3 follows directly from statements 1,2. We prove statement 1. Let π be an opponent ppa in sdd and i=st(π) and αPTAi1 such that α=Cl(Tπ).

Let π be a proponent full ppa in sdd satisfying the following condition:

  • αAss(Tπ);

  • st(π)=max{st(q)|αAss(Tq) and qPPTi1}

We show that π hits π. Suppose π does not hit π. Hence there is opponent ppa p such that st(π)<st(p)<st(π) s.t. α=Cl(Tp). Let j=st(p). It follows j<i and αPTAj. Since αPTAi1, it follows that there is proponent full ppa q such that st(π)<j<st(q)<i and αTq. Because st(q)<i, it follows qPPTi1. Contradiction to the construction of π. □

Let π0 be the unique proponent profiled full argument such that st(π0)=0 (the existence and uniqueness of π0 follows from Lemma 12).

A support path in sdd is a sequence of profiled full arguments π0,,πn (n0) in i=0PPTi such that πi+1 supports πi (0i<n).

Property 2.

Each full proponent ppa π in sdd appears as the last element of a support path.

The proof is by induction on st(π). The property holds obviously for st(π)=0. Suppose that the property holds for all full proponent ppas that start before π. Therefore π supports some full proponent ppa π (from above Property 1). Hence st(π)<st(π). From the induction hypothesis, it follows that π is the last element of some support path sq. Hence sq.π is a support path.

The set of all support paths in sdd forms a tree T where the root is π0 and the nodes in T are support paths and a support path π0,,πn,πn+1 is a child of π0,,πn.

Suppose the set of proponent profiled full argument in sdd is infinite. Therefore T is infinite.

From Lemma 17, each node in T has only finitely many children. Hence there is an infinite support path in T. Contrary to Lemma 16.

Hence the set of proponent profiled full arguments in sdd is finite. □

6.1.Proof of completeness theorem

We only need to show that there exists no infinite sdd for δ wrt λ.

Assume that there exists an infinite strongly grounded dispute derivation for δCl(GEAFF) wrt ground map λ

sdd=PPT0,PTA0,OPT0,,PPTi,PTAi,OPTi,

From Theorem 8, it follows that there is n such that PPTn contains all proponent profiled full arguments in sdd. Therefore the opponent makes a move at all stages m>n in sdd.

Let K0 be the set of starting times before n of opponent ppas in sdd while K1 be the set of starting times of all opponent ppas in sdd, i.e.

K0={j|πi=0OPTi s.t. j=st(π)<n},andK1={j|πi=0OPTi s.t. j=st(π)}.

From the Definition 15 (of strongly grounded dispute derivation), it follows that |K1||K0|+|PTAn|.

From i=0OPTi=kK1OPTk, it follows immediately from Theorem 7 that i=0OPTi is finite. Hence sdd is finite. Contradiction.

We have proved that there is no infinite sdd.

The completeness theorem follows directly from Theorem 6. □

7.Flatten dispute derivation

We have presented the grounded dispute derivations that are both sound and complete wrt finitary assumption-based frameworks where proof trees together with their histories are fully and explicitly represented to shed light on the construction process of arguments and counter arguments during a derivation and hence providing key structural insights into the soundness and completeness of the procedure.

In this section, we present another procedure that is simply the result of flattening the first where instead of the whole proof trees, only their supports are recorded. Consequently, the second procedure is both sound and complete but with much simpler date structure and hence much more amenable to possible implementation. The disadvantage of the flatten version (like Prolog for logic programming) is that it does not give the reasons how a returned answer is supported and defended.

In this section we propose the flatten grounded dispute derivations which focus on the supports of proof trees instead of the whole trees like other proposals including [1719,24].

The representation of flatten grounded dispute derivations is based on multisets. To keep the paper self-contained, we recall in Appendix A.1 a brief but formal introduction of multisets from [31].3232

Definition 18

Definition 18(Flatten Grounded Dispute Derivation).

A flatten grounded dispute derivation for a sentence δ is a sequence of the form

PS0,OS0,,PSn,OSn
where
  • for each i, PSi is a multiset of sentences while OSi is a multisets of multisets of sentences; and

  • PS0={δ}, and OS0=, and

  • at stage i (i>0), one of the dispute parties makes a move satisfying the following properties:

    • (1) Suppose the proponent makes a move at stage i. The proponent has two options:

      • (a) The proponent selects a non-assumption σPSi1, a rule r with hd(r)=σ and,

        PSi=(PSi1{σ})bd(r)

        OSi=OSi1.

      • (b) The proponent selects SOSi1 and an assumption αS and,

        PSi=PSi1{α}

        OSi=OSi1{S}

    • (2) Suppose the opponent makes a move at stage i. The opponent has two options:

      • (a) The opponent selects SOSi1 and a non-assumption σS and proceeds as follows:

        PSi=PSi1

        OSi=(OSi1{S}){(S{σ})bd(r)|hd(r)=σ}.

      • (b) The opponent selects an assumption αPSi1 and,

        PSi=PSi1{α}

        OSi=OSi1{{α}}.

Definition 19.

A flatten grounded dispute derivation PS0,OS0,,PSn,OSn is successful if PSn=OSn=.

Table 3

A successful flatten grounded dispute derivation

StageMovePSOS
0{a}
11a{not_α}
22b{{α}}
32a{{not_β,f(0)}}
41b{β}
51a
Example 7.

A successful flatten grounded dispute derivation PS0,OS0,,PS5,OS5 for sentence a (wrt the argumentation framework F2 in the introduction) is illustrated in Table 3 and explained in details below. For convenience we recall F2 below.

F2:r:anot_αr:αnot_β,f(0)rn:f(n)f(n+1),n0t:β

At stage 0, we have PS0={a} and OS0=.

  • At stage 1, the proponent makes a move to expand the non-assumption a by applying step (1a) using rule r.

    Hence PS1=(PS0{a})bd(r)={not_α} where bd(r)={not_α}.

    OS1=OS0=.

  • Next, at stage 2, the opponent attacks the assumption not_αPS1 by applying step (2b). Hence OS2=OS1{{α}}={{α}}.

    PS2=PS1{not_α}=.

  • At stage 3, the opponent applies step (2a), selecting S={α} from OS2 and expanding α. Since rule r is the only one with head α, it follows

    OS3=(OS2{S}){(S{α})bd(r)}={{not_β,f(0)}}.

    PS3=PS2=

  • At stage 4, the proponent applies step (1b) by selecting S={not_β,f(0)}OS3 and assumption not_βS for attack. Therefore

    PS4=PS3{β}={β}.

    OS4=OS3{S}=.

  • Finally at stage 5, the proponent applies step (1a), using rule t, to expand β. Therefore

    PS5=(PS4{β})bd(r)= (since bd(r)=).

    OS5=OS4=

As both PS and OS are empty, the derivation is successful.

Definition 20.

  • (1) Let dd be a grounded dispute derivation of length n for a sentence δ (as defined in Definition 13).

    Define a sequence of multisets of assumptions

    M(dd)=MPTA0,,MPTAn
    by:

    • MPTA0=PTA0.

    • Suppose MPTAi1 is defined. Then

      • MPTAi=MPTAi1 if the move at stage i is (1b) or (2a);

      • MPTAi=MPTAi1Ass(r) if the move at stage i is (1a);

      • MPTAi=MPTAi1{α} if the move at stage i is (2b);

  • (2) For any proof tree T, NSpm(T) and Spm(T) denote respectively the multiset of the non-assumptions and the multiset of sentences labeling the leaf nodes of T and different to true.

    NSpm(T) and Spm(T) are often referred to as the multiset non-assumption support or multiset support of T, respectively.

    For a set S of ppas, define

    NSpm(S)={NSpm(Tπ)|πS};

Lemma 18.

Let dd be a grounded dispute derivation of length n for a sentence δ and M(dd)=MPTA0,,MPTAn. It holds that for all 0in, MPTAi and PTAi are set-equivalent.

Proof.

We prove by induction that for each i, MPTAi and PTAi are set-equivalent.

It is obvious that PTA0=MPTA0.

Let i>0. Suppose for each j<i, PTAj and MPTAj are set-equivalent. We show PTAi and MPTAi are set-equivalent. There are four cases according to four possible moves of the players at stage i.

  • The move at stage i is (1a). Hence PTAi=PTAi1Ass(r) and MPTAi=MPTAi1Ass(r). From the induction hypothesis, PTAi1 and MPTAi1 are set-equivalent, it follows obviously that PTAi and MPTAi are set-equivalent.

  • The move at stage i is (1b) or (2a). From the induction hypothesis, PTAi1 and MPTAi1 are set-equivalent and MPTAi=MPTAi1 and PTAi=PTAi1, it follows obviously that PTAi and MPTAi are set-equivalent.

  • The move at stage i is (2b). Hence PTAi=PTAi1{α} and MPTAi=MPTAi1{α}. From the induction hypothesis, PTAi1 and MPTAi1 are set-equivalent, it follows obviously that PTAi and MPTAi are set-equivalent.

 □

The following lemmas show that each grounded dispute derivation could be translated into an equivalent flatten one and vice versa.

Lemma 19.

Let dd=PPT0,PTA0,OPT0,,PPTn,PTAn,OPTn be a grounded dispute derivation for δ.

Further let M(dd)=MPTA0,,MPTAn.

Then the sequence PS0,OS0,,PSn,OSn where for each 0in,

  • PSi=NSpm(PPTi)MPTAi; and

  • OSi={Spm(Tπ)|πOPTi};

is a flatten grounded dispute derivation for δ.

Proof.

See Appendix A.4 □

Lemma 20.

Let fdd=PS0,OS0,,PSn,OSn be a flatten grounded dispute derivation for δ.

There is a grounded dispute derivation dd=PPT0,PTA0,OPT0,,PPTn,PTAn,OPTn for δ such that for all 0in, the following properties hold:

  • PSi=NSpm(PPTi)MPTAi; and

  • OSi={Spm(Tp)|pOPTi};

where M(dd)=MPTA0,,MPTAn.

Proof.

See Appendix A.5. □

Theorem 9

Theorem 9(Soundness Theorem for Flatten Grounded Dispute Derivation).

Let dd=PS0,OS0,,PSn,OSn be a successful flatten grounded dispute derivation for δ. Then δCl(GEAFF).

Proof.

From Lemma 20, there exists a grounded dispute derivation dd=PPT0,PTA0,OPT0,,PPTn,PTAn,OPTn for δ such that PSi=NSpm(PPTi)MPTAi; and OSi={Spm(Tp)|pOPTi};

Since PSn=, the sets NSpm(PPTn) are empty. Therefore all ppas in PPTn are full.

As OSn={Spm(Tp)|pOPTn} and OSn=, it follows OPTn=.

Because PSn= and PSn=NSpm(PPTn)MPTAn, MPTAn is empty.

Since PTAn and MPTAn are set-equivalent (see Lemma 18) and MPTAn is empty, PTAn is empty.

Therefore dd is a successful grounded dispute derivation. The theorem follows from Theorem 4. □

Theorem 10

Theorem 10(Completeness Theorem for Flatten Grounded Dispute Derivation).

Let F be a finitary ABA framework and σCl(GEF). Then there is a successful flatten dispute derivation PS0,OS0,,PSn,OSn for σ.

Proof.

From the completeness Theorem 5, there is a successful grounded dispute derivation sdd=PPT0,PTA0,OPT0,,PPTn,PTAn,OPTn for σ.

It holds that OPTn= and all ppas in PPTn are full.

Let sdd=PS0,OS0,,PSn,OSn be the flatten dispute derivation for σ as defined in Lemma 19.

Since PTAn=, MPTAn= (see Lemma 18).

Since all ppas in PPTn are full, NSpm(PPTn)=. It hence follows PSn=NSpm(PPTn)MPTAn=.

Since OPTn=, OSn={Spm(Tp)|pOPTn}=, sdd is successful. □

8.Discussion

We study the soundness and completeness of dialectical proof procedures for assumption-based argumentation with respect to the skeptical semantics. We have presented two proof procedures. In one, proof trees together with their histories are fully and explicitly represented. The other procedure is simply the result of flattening the first. We show the soundness and completeness of both proof procedures wrt grounded semantics of assumption-based argumentation frameworks where possibly non-terminating computation is represented by infinite arguments.

Assumption-based argumentation is an instance of abstract argumentation. Another well-known instance of abstract argumentation is the ASPIC+ system [27,28]. It would be interesting to see how to apply our proof procedures to ASPIC+.

In many applications, a sentence is (universally) accepted if it is accepted in every preferred extensions (or stable extensions) [2]. It would be interesting to see how our proof systems in this paper as well as in [31] could be extended for the universal acceptance.

Attacks may have different strength [16]. [7] proposed a dialectical proof procedure for abstract argumentation frameworks with attacks having different strength. It would be interesting to see how to integrate the ideas of [7]’s into our proof system for assumption-based argumentation with attacks of different strength.

Dialectical proof procedures are not the only approach to proof theory of logical argumentation. [1] proposes dynamic proof systems for logical argumentation. It would be intriguing to see the relationship between these rather distinct approaches.

Grounded semantics of an assumption-based argumentation could be viewed as a form of nonmonotonic inductive definition of the ABA framework [10]. In that sense, our proof system can be viewed as an example of a dialectical proof procedure for a nonmonotonic inductive definition. It may be worth exploring further whether dialectical proof systems could be developed for other nonmonotonic inductive definition systems as defined in [10].

There are recently several interesting works on the semantics of infinite argumentation frameworks [3,4]. It is worth noting that the inclusion of infinite arguments into the argumentation frameworks of assumption-based argumentation does not transform it into infinite argumentation frameworks as illustrated in the introduction. This leads to an interesting question of how to extend the results in [3,4] to frameworks with “two kinds of infinity”: infinite number of arguments and arguments of infinite size.

Dialectical procedures could be viewed as a form of dialogue games where for both players the sets of rules as well as assumptions are given and fixed at the beginning. A more general form of dialogue games is proposed in [23] where rules and assumptions are introduced during the dialogue. While the soundness of such games has been studied in [23], their completeness is not. The completeness results in this paper, especially the completeness of the flatten procedure could shed light on the completeness of such dialogue games.

Notes

1 The written codes in SWI-prolog of both F1, F2 can be found in appendix A.2.

2 See Fig. 10 in appendix A.2.

3 See Fig. 11 in appendix A.2.

4 Readers who are familiar with theory of ordinals could recognize that the grounded extension could be “computed” as the least upper bound of the chain (FAFθ())θ where θ is an ordinal and

FAF0()=,FAFθ+1()=FAF(FAFθ()),FAFθ()={FAFγ()|γ<θ}if θ is a limit ordinal.
Note that ω is the least limit ordinal.

5 Assuming the axiom of choice (or equivalently the maximality principles) it follows immediately that for each admissible set S of arguments, there exists a preferred extensions E containing S (i.e SE). More on this topic, see [9,29].

6 A complete partial order is a partial order that has a bottom element and each directed subset has a least upper bound.

7 More on fixpoints and least fixpoints, see Knaster–Tarski fixpoint theorem and fixpoint theorems for complete partial oder in [9].

8 It is not difficult to show that grounded extension is also admissible (see [16] for more about semantics of argumentation frameworks).

9 A partial order is a complete semilattice if each nonempty subset has a glb and each increasing sequence has a lub.

10 As we will see later, the finitariness guarantees that the corresponding argumentation framework is ω-grounded (see Theorem 2) and hence all possible attacks against proponent arguments could be considered (see Remark 7).

11 See Notations 1 and 2.

12 I.e. T is obtained from T by adding m children N.(r,b1),,N.(r,bm) to N (for m=0, a child node N.(r,true) is added to N).

13 It may be worthwhile to note that if T0 is a prefix of T1 then obviously T0, T1 are compatible.

14 One may wonder what is the prefix of a partial proof (nodes). Well, a partial proof (or node) is in essence a sequence. And so a prefix of a partial proof (root,σ0).(r1,σ1).(rn,σn) is a partial proof of the form (root,σ0).(r1,σ1).(rk,σk) where kn.

15 I.e. exp(T0,T) is the set of all immediate expansions of T0 that are prefixes of T.

16 Hence the height of the root is 0.

17 The set 2Ar is ordered wrt set inclusion.

18 Note that A is defensible if it is defended by some set of arguments.

19 An argument A is self-attacking if A attacks itself.

20 A balanced proof tree of height n is a proof tree such that the heights of all non-final leaf nodes are n.

21 The height of a full argument is the length of the longest path from the root to a leaf.

22 We could show by induction. Since PA0FA0 consists of the contraries of the assumptions in B and B is a finite argument, both PA0 and FA0 are obviously finite. From the finitariness of F and the finiteness of PAn, it should be clear that the set PAn+1FAn+1 is finite since it is the set of the expansions of PAn obtained by expanding each non-final leaf node in each proof tree in PAn one step.

23 Note that rooti=0PANi.

24 PPT, PTA, OPT stand respectively for “proponent profiled trees”, “proponent to-be-attacked assumptions “and “opponent profiled trees”.

25 See Notation 3 for the definition of exp(T,N,r).

26 See Notation 3 for the definition of CE(T,N).

27 I.e. π is a continuation of π and Tπ is an immediate expansion of Tπ.

28 We assume the axiom of choice or equivalently the maximality principles (see [9] for more details). Hence such mapping always exists.

29 See Notation 5 for the introduction of hi(T).

30 Since π is discontinued at stage i, there are two possibilities: π is attacked by the proponent or the expansion of π by the opponent fails. Since π starts at i, it is the proponent who makes a move at stage i. Hence π is attacked by the proponent.

31 From statements (1,2,3) in Lemma 12, it follows that at any point during a sdd, the proponent has at most one ppa to expand and such ppa is expanded in consecutively stages and hence finished after finitely many steps (1a). Afterwards, if step (1b) is executable, it will be executed.

32 Intuitively, a multiset is like a set where its member may occur multiple times. For example, A={2,2,2,5} is a multiset with 2 occurs 3 times and 5 occurs one time. Some readers may find it sensible to imagine A as a bag of three 2-dollars notes and one 5-dollar note (if such notes are ever in circulation). So if you have another bag B={2,5,5} of money containing one 2-dollar note and two 5-dollar notes then putting them together (their union) gives you a bag C=AB={2,2,2,2,5,5,5} of four 2-dollar notes and 3 5-dollars notes. If you removes one 2-dollar note from C (C{2}), you would get a bag of 3 2-dollar notes and 3 5-dollar notes (C{2}={2,2,2,5,5,5}. But if you remove all 2-dollar note from C (C{2}), what remains is a bag of 3-dollar notes (C{2}={5,5,5}.

33 Further details can be seen in [5,11,25].

34 {(2,3),(5,1)} represents the function assigning 3 to 2 and 1 to 5.

35 Note that (S{σ})bd(r)=Spm(exp(Tπ,Nσ,r)).

Appendices

Appendix

A.1.Appendix: Multisets

This section recalls a brief introduction of multisets from [31].3333

Intuitively, a multiset is like a set but allowing each element to have many instances.

Formally, a multiset is a pair A=(B,μ) where B is a set referred to as the base set of A, and μ is a function from B into the set of positive integers. The function μ is referred to as the multiplicity function of the multiset A and often denoted referred to by μA.

Two multisets M=(B,μ) and M=(B,μ) are set-equivalent if B, B are identical.

For simplicity we represent a multiset as a set where its member may occur multiple times. For example, the prime factorization of 40 can be represented as a multiset A={2,2,2,5}. Another representation of the prime factorization of 40 is A=({2,5},μA) where μA={(2,3),(5,1)}.3434

We introduce the major types of the multisets operations performing on two multisets as follows.

Definition 21.

The union and intersection of two multisets M=(B,μ), M=(B,μ) are defined by:

  • (1) MM=(BB,μ+μ) where for xBB, (μ+μ)(x)=μ(x)+μ(x).

  • (2) MM=(BB,μ) where μ(x)=min{μ(x),μ(x)}.

We also introduce two notions of difference and strong difference between multisets and sets in the definition below.

Definition 22.

Let M=(B,μ), M=(B,μ) be multisets and S be a set.

  • (1) The difference between M, M is defined by:

    MM=(B,μ) where the following conditions are satisfied:

    • (a) B=(BB){xBB|μ(x)>μ(x)}.

    • (b)

      μ(x)=μ(x)if xBBμ(x)μ(x)if xBB

  • (2) The strong difference between M and S is defined by

    MS=(BS,μ)
    where for each xBS, μ(x)=μ(x).

A.2.Appendix: Execution of programs on SWI-prolog

See Fig. 10 and Fig. 11 .

Fig. 10.

The execution of F1 on SWI-prolog.

The execution of F1 on SWI-prolog.
Fig. 11.

The execution of F2 on SWI-prolog.

The execution of F2 on SWI-prolog.

A.3.Appendix: Proofs of lemmas supporting soundness theorem

A.3.1.Proof of Lemma 7

Lemma.

Let dd=PPT0,PTA0,OPT0,,PPTn,PTAn,OPTn be a grounded dispute derivation.

  • (1) Let π be some proponent ppa appearing in dd. The following statements hold:

    • (a) For each i:st(π)in, there is an unique continuation of π in PPTi.

    • (b) Let πi, πi+1 (st(π)i<n) be continuations of π in PPTi, PPTi+1 respectively. Then πi+1 is either an immediate expansion of πi or identical to πi.

  • (2) Let π, π be two proponent ppas in dd such that st(π)=st(π). Then one is a continuation of the other.

    Consequently, if π, π are full then they are identical.

Proof.

Statement 2 follows directly from statements (1a,1b) and the fact that at any stage, at most one new proponent ppa is created (according to Definition 13 of dispute derivation).

Let k=st(π). We prove statements (1a,1b) by induction on i>st(π).

Base case: i=k. It follows directly from the fact that at any stage, at most one new proponent ppa is created (according to Definition 13 of dispute derivation).

Inductive step. Suppose the lemma holds for i:ki<n. We show that it also holds for i+1.

If it is the opponent who makes a move at stage i+1. Then PPTi=PPTi+1. The lemma follows directly from the induction hypothesis.

If the proponent makes a move according to step (1b) (in Definition 13 of dispute derivation) at stage i+1 or step (1a) where πi is not selected. Then πiPPTi+1. The lemma follows directly from the induction hypothesis.

Suppose the proponent makes a move according to step (1a) (in Definition 13 of dispute derivation) by expanding πi at stage i+1. The lemma follows directly from the definition of step (1a) of dispute derivation. □

A.3.2.Proof of Lemma 8

Lemma.

Let dd=PPT0,PTA0,OPT0,,PPTn,PTAn,OPTn be a grounded dispute derivation. It holds that for each 0in, for all π,πOPTi, if π, π are compatible then π, π are identical.

Proof.

By induction on i.

Basic step: i=0. Since OPT0=, the lemma holds obviously.

Inductive step: Suppose the lemma holds for i1. Let π, π be compatible ppas from OPTi. We show that π, π are identical.

There are two cases:

  • Both π, π belong to OPTi1. The lemma follows immediately from the induction hypothesis.

  • Suppose πOPTiOPTi1.

    It follows immediately that the opponent moves at stage i. Hence there are two cases:

    • The opponent makes a move according to step (2a). Hence there exists π0OPTi1 such that π is an immediate expansion of π0 and π0OPTi.

      There are two cases:

      • πOPTi1. Since π, π are compatible, it follows that π, π0 are also compatible. From the induction hypothesis, it is obvious that π=π0. Hence πOPTi. Contradiction. This case hence can not happen.

      • πOPTiOPTi1. Hence π is also an immediate expansion of π0. Since π, π are compatible, Tπ, Tπ are compatible. From Lemma 3, Tπ=Tπ. Hence π=π.

    • The opponent makes a move according to step (2b). Therefore the starting time of π is i. Since π, π are compatible, the starting time of π is also i. Hence π=π.

 □

A.3.3.Proof of Lemma 9

Lemma.

Let dd be a dispute derivation

dd=PPT0,PTA0,OPT0,,PPTn,PTAn,OPTn

Further let A be an argument such that Cl(A)=α for some assumption α and π=([α],([α],i))OPTi for some 0in.

Then there is an unique sequence contdd(π)=π0,,πk (ii+kn) such that

  • (1) π0=π; and

  • (2) for each i<jk, πjOPTi+j and πj is a continuation of πj1 and TπjA; and

  • (3) if i+k<n then πk is discontinued at i+k+1 and there is no ppa at any OPTi+k+1,,OPTn that is compatible with πk.

Proof.

Let m=ni. We prove by induction on m. Note that i is fixed.

Base Step: m=0. The lemma holds obviously (as the statements 2,3 do not apply).

Inductive Step: Suppose the lemma holds for m0. We show that it also holds for m+1 (i.e. for n+1).

Let dd=dd,PPTn+1,PTAn+1,OPTn+1, and πOPTi and contdd=π0,,πk where π0=π.

There are two cases:

  • i+k<n. Let contdd(π)=contdd(π).

    From the induction hypothesis, it follows that πk is discontinued at i+k+1 and there is no ppa at any OPTi+k+1,,OPTn that is compatible with πk.

    We show that there is no ppa in OPTn+1 that is compatible with πk. Suppose there is πOPTn+1 that is compatible with πk. Therefore πOPTn+1OPTn. Hence it is the opponent who makes a move at stage n+1. There are two cases:

    • The opponent makes a move according to step (2a) to expand some of her ppa pOPTn. Since π, πk are compatible, p, πk are hence compatible. Contradiction!

    • The opponent makes a move according to step (2b). Therefore the starting time of π is n+1. π is hence not compatible with πk. Contradiction.

    The uniqueness of contdd(π) comes from the uniqueness of contdd(π).

  • i+k=n. There are six cases.

    • (1) At stage n+1, the proponent makes a move to expand some proponent ppa. Hence OPTn+1=OPTn. The lemma holds obviously for contdd(π)=contdd(π).πk.

    • (2) At stage n+1, the proponent makes a move to attacks some opponent ppa that is different to πk. Hence πkOPTn+1. The lemma holds obviously for contdd(π)=contdd(π).πk.

    • (3) At stage n+1, the proponent makes a move to attacks πk. Hence πkOPTn+1. The lemma holds obviously for contdd(π)=contdd(π).

    • (4) At stage n+1, the opponent moves to attack some proponent ppa and create a new ppa π. It is obvious that st(π)>st(πk). Hence the lemma holds obviously for contdd(π)=contdd(π).πk.

    • (5) At stage n+1, the opponent moves to expand some opponent ppa, and the selected ppa is not πk. Hence πkOPTn+1. Thus the lemma holds obviously for contdd(π)=contdd(π).πk.

    • (6) At stage n+1, the opponent moves to expand πk at a non-final leaf node NTπk labeled by a non-assumption sentence δ and OPTn+1=(OPTn{πk}){(T,h)|TCE(Tπk,N),h=hπk.(T,n+1)}

      If the expansion fails at stage n+1 (i.e. CE(Tπk,N)=), the lemma holds obviously for contdd(π)=contdd(π).

      Suppose now that CE(Tπk,N). Hence from Lemma 3, there is an unique immediate expansion of πk+1OPTn+1 of πk and Tπk+1A.

      Let contdd(π)=contdd(π).πk+1. It remains to show that contdd(π) is unique.

      Suppose the contrary that contdd(π) is not unique. From the uniqueness of contdd(π), it follows that there is πOPTn+1 such that π is a continuation of πk and TπA and ππk+1. There are two cases:

      • πOPTn. From Lemma 8, it follows that π=πk. Hence πkOPTn+1 Contradiction. This case can not happen.

      • πOPTn+1OPTn. Hence π is an immediate expansion of πk. Since TπA, it follows from Lemma 3 that π=πk+1. Contradiction. Hence this case is also not possible.

      Thus contdd(πk+1) is unique.

 □

A.3.4.Proof of Lemma 10

The following lemma states that each argument attacking a proponent argument in a successful dispute derivation is counter-attacked by some proponent argument.

Lemma.

Let dd be a successful grounded dispute derivation that terminates at PPTn,PTAn,OPTn and π=(T,h)PPTn and T be an argument attacking argument T. Then there is a opponent ppa π such that st(π)<st(π) and TπT and π is discontinued at some step between st(π) and n.

Proof.

Let αAss(T) such that α=Cl(T). Therefore there is mst(π) s.t. αPTAm and there is im s.t. the opponent attack α at stage i. Hence the ppa π0=([α],([α],i))OPTi.

From Lemma 9 and the fact that OPTn=, it follows that there is an unique contdd(p)=π0,,πk, i+k<n, such that for each j (0jk), πjOPTi+j and πj is a continuation of πj1 and TπjA; and

πk is discontinued at i+k+1 and there is no ppa at any OPTi+k+1,,OPTn that is compatible with πk.

The lemma holds obviously for π=πk. □

A.4.Appendix: Proof of Lemma 19

Lemma.

Let dd=PPT0,PTA0,OPT0,,PPTn,PTAn,OPTn be a grounded dispute derivation for δ.

Further let M(dd)=MPTA0,,MPTAn.

Then the sequence PS0,OS0,,PSn,OSn where for each 0in,

  • PSi=NSpm(PPTi)MPTAi; and

  • OSi={Spm(Tπ)|πOPTi};

is a flatten grounded dispute derivation for δ.

Proof.

We prove by induction on n. The base case (n=0) holds obviously.

Suppose the lemma holds for n1. We show that it also holds for n.

Let dd=PS0,OS0,,PSn1,OSn1 be a flatten grounded dispute derivation such that 0in1,

PSi=NSpm(PPTi)MPTAi;

OSi={Spm(Tπ)|πOPTi};

There are four cases:

  • (1) The proponent executes step (1a) at stage n by expanding some profiled partial argument π=(T,h)PPTn1 by expanding T at a non-final leaf node NT labeled by a non-assumption sentence σ and a rule r with hd(r)=σ such that

    PPTn=(PPTn1{π}){π} where π=(T,h) and T=exp(T,N,r) and h=h.(T,n);

    MPTAn=MPTAn1Ass(r)

    OPTn=OPTn1.

    It is clear that NSpm(Tπ)=(NSpm(Tπ){σ})(bd(r)Ass(r)).

    From the induction hypothesis, we have

    PSn1=NSpm(PPTn1)MPTAn1;

    OSn1={Spm(Tp)|πOPTn1};

    It is clear that σPSn1.

    Let OSn=OSn1 and PSn=(PSn1{σ})bd(r).

    It is obvious that the sequence dd, PSn,OSn is a flatten grounded dispute derivation.

    We now proceed to show that it satisfies the lemma.

    Since OPTn=OPTn1 and OSn=OSn1, it holds obviously that OSn={Spm(Tπ)|πOPTn};

    Let PPTX=(PPTn1{π})

    It follows:

    PSn=(PSn1{σ})bd(r)=((NSpm(PPTn1)MPTAn1){σ})bd(r)(from induction hypothesis)=((NSpm(PPTX)NSpm(Tπ)MPTAn1){σ})bd(r)=((NSpm(PPTX)MPTAn1)NSpm(Tπ)){σ})bd(r)=((NSpm(PPTX)MPTAn1Ass(r))(NSpm(Tπ){σ})(bd(r)Ass(r))=NSpm(PPTX)MPTAnNSpm(Tπ)=NSpm(PPTn)MPTAn

  • (2) The proponent executes step (1b) at stage n by attacking an opponent’s profiled partial argument π=(T,h)OPTn1 at an assumption αAss(T) resulting in:

    PPTn=PPTn1{(T,h)} where T=[α] and h=([α],n); and

    PTAn=PTAn1

    OPTn=OPTn1{π}

    Let S=Spm(T). It is clear αS.

    From the induction hypothesis, we have

    PSn1=NSpm(PPTn1)MPTAn1;

    OSn1={Spm(Tp)|pOPTn1};

    Let OSn=OSn1{S}.

    PSn=PSn1{α}

    It is obvious that the sequence dd, PSn,OSn is a flatten grounded dispute derivation.

    We now proceed to show that it satisfies the lemma.

    It holds:

    OSn=OSn1{S}OSn={Spm(Tp)|pOPTn1}{Spm(T)}(induction hypothesis)OSn={Spm(Tp)|pOPTn};PSn=PSn1{α}PSn=NSpm(PPTn1)MPTAn1{α}(induction hypothesis)PSn=NSpm(PPTn1){α}MPTAnPSn=NSpm(PPTn)MPTAn

  • (3) The opponent expands at stage n an opponent ppa π=(T,h)OPTn1 at a non-final leaf node NT labeled by a non-assumption sentence σ resulting in:

    PPTn=PPTn1

    PTAn=PTAn1

    OPTn=OPT0OPT1 where OPT0=OPTn1{π} and

    OPT1={(T,h)|TCE(T,N),h=h.(T,n)}

    From the induction hypothesis, we have

    PSn1=NSpm(PPTn1)MPTAn1;

    OSn1={Spm(Tp)|pOPTn1};

    Let S=Spm(T) and

    PSn=PSn1 and

    OSn=(OSn1{S}){(S{σ})bd(r)|hd(r)=σ}.

    It is obvious that the sequence dd, PSn,OSn is a flatten grounded dispute derivation.

    We now proceed to show that it satisfies the lemma.

    For T=exp(T,N,r), it holds Spm(T)=(Spm(T){σ})bd(r)=(S{σ})bd(r).

    It holds obviously that

    OSn=(OSn1{S}){(S{δ})bd(r)|hd(r)=δ}OSn={Spm(Tp|pOPT0}{Spm(Tp|pOPT1}OSn={Spm(Tp)|pOPTn}PSn=PSn1PSn=NSpm(PPTn1)MPTAn1(induction hypothesis)PSn=NSpm(PPTn)MPTAn

  • (4) The opponent attacks an assumption αPTAn1 resulting in:

    PPTn=PPTn1

    PTAn=PTAn1{α}

    OPTn=OPTn1{π} where π=([α],([α],n))

    From the induction hypothesis, we have

    PSn1=NSpm(PPTn1)MPTAn1;

    OSn1={Spm(Tp)|pOPTn1};

    Let PSn=PSn1{α} and

    OSn=OSn1{{α}}.

    It is obvious that the sequence dd, PSn,OSn is a flatten grounded dispute derivation.

    We now proceed to show that it satisfies the lemma.

    It holds obviously that

    PSn=PSn1{α}OSn=(NSpm(PPTn1)MPTAn1){α}(induction hypothesis)OSn=NSpm(PPTn1)(MPTAn1{α})OSn=NSpm(PPTn)MPTAnOSn=OSn1{{α}}OSn={Spm(Tp)|pOPTn1}{{α}}(induction hypothesis)OSn={Spm(Tp)|pOPTn1}{Spm(Tπ)}OSn={Spm(Tp)|pOPTn}

 □

A.5.Appendix: Proof of Lemma 20

Lemma.

Let fdd=PS0,OS0,,PSn,OSn be a flatten grounded dispute derivation for δ.

There is a grounded dispute derivation dd=PPT0,PTA0,OPT0,,PPTn,PTAn,OPTn for δ such that for all 0in, the following properties hold:

  • PSi=NSpm(PPTi)MPTAi; and

  • OSi={Spm(Tp)|pOPTi};

where M(dd)=MPTA0,,MPTAn.

Proof.

By induction on n.

The lemma holds obviously for n=0.

Suppose that the lemma holds for n1. We show that it also holds for n.

Let dd=PPT0,PTA0,OPT0,,PPTn1,PTAn1,OPTn1 be a grounded dispute derivation for δ and M(dd)=MPTA0,,MPTAn1. such that for all 0in1, the following properties hold:

  • PSi=NSpm(PPTi)MPTAi; and

  • OSi={Spm(Tp)|pOPTi};

We construct a triple PPTn,PTAn,OPTn such that dd=dd,PPTn,PTAn,OPTn is a grounded dispute derivation and PSn=NSpm(PPTn)MPTAn and OSn={Spm(Tp)|pOPTn} where MPTAn is defined from MPTAn1 as in Definition 20.

There are four cases:

  • (1) The proponent selects a non-assumption σPSn1, a rule r with hd(r)=σ and,

    PSn=(PSn1{σ})bd(r)

    OSn=OSn1.

    Since PSn1=NSpm(PPTn1)MPTAn1 (from the induction hypothesis), it follows there exists π=(T,h)PPTn1 such that σNSpm(T).

    Let N be a leaf-node in T labeled by σ and T=exp(T,N,r).

    Let PPTn=(PPTn1{π}){(T,h)} where h=h.(T,n);

    PTAn=PTAn1Ass(r)

    OPTn=OPTn1.

    The sequence dd, PPTn,PTAn,OPTn is hence a grounded dispute derivation. We now proceed to show that it satisfies the lemma.

    It holds directly that OSn=OSn1={Spm(Tp)|pOPTn1}={Spm(Tp)|pOPTn};

    Let PPTX=PPTn1{π}.

    It is easy to see that NSpm(T)=(NSpm(T){σ})NAss(r). Therefore

    PSn=(PSn1{σ})bd(r)=((NSpm(PPTn1)MPTAn1){σ})bd(r)(from induction hypothesis)=((NSpm(PPTX)NSpm(T)MPTAn1){σ})(Ass(r)NAss(r))=NSpm(PPTX)((NSpm(T){σ})NAss(r))MPTAn1Ass(r)=NSpm(PPTX)NSpm(T)MPTAn1Ass(r)=NSpm(PPTn)MPTAn.

  • (2) The proponent selects SOSn1 and an assumption αS and,

    PSn=PSn1{α}

    OSn=OSn1{S}

    From OSn1={Spm(Tp)|pOPTn1} (induction hypothesis), there is π=(T,h)OPTn1 such that Spm(T)=S. Hence αAss(T).

    Let PPTn=PPTn1{(T,h)} where T=[α] and h=([α],n); and

    PTAn=PTAn1

    OPTn=OPTn1{π}

    The sequences dd, PPTn,PTAn,OPTn is hence a grounded dispute derivation. We now proceed to show that it satisfies the lemma.

    It holds:

    OSn=OSn1{S}OSn={Spm(Tp)|pOPTn1}{Spm(T}(induction hypothesis)OSn={Spm(Tp)|pOPTn}.PSn=PSn1{α}OSn=NSpm(PPTn1)MPTAn1{α}(induction hypothesis)OSn=NSpm(PPTn)MPTAn

  • (3) The opponent selects SOSn1 and a non-assumption σS and proceeds as follows:

    PSn=PSn1

    OSn=(OSn1{S}){(S{σ})bd(r)|hd(r)=σ}.

    From the induction hypothesis, we have

    PSn1=NSpm(PPTn1)MPTAn1; and

    OSn1={Spm(Tp)|pOPTn1}.

    Therefore there is πOPTn1 such that Spm(Tπ)=S. Further let Nσ be a non-final leaf node in Tπ labeled by σ.

    Let PPTn=PPTn1

    PTAn=PTAn1

    OPTn=(OPTn1{π}){(T,h)|TCE(Tπ,Nσ),h=h.(T,n)}

    The sequences dd, PPTn,PTAn,OPTn is hence a grounded dispute derivation. We now proceed to show that it satisfies the lemma.

    It holds

    OSn=(OSn1{S}){(S{σ})bd(r)|hd(r)=σ}

    OSn={Spm(Tp)|pOPTn1{π}}{(S{σ})bd(r)|hd(r)=σ}

    OSn={Spm(Tp)|pOPTn1{π}}{Spm(exp(Tπ,Nσ,r))|hd(r)=σ} 3535

    OSn={Spm(Tp)|pOPTn1{π}}{Spm(T)|TCE(Tπ,Nσ)}.

    OSn={Spm(Tp)|pOPTn}.

    Further, we have:

    PSn=PSn1=NSpm(PPTn1)MPTAn1(induction hypothesis)=NSpm(PPTn)MPTAn;

  • (4) The opponent selects an assumption αPSn1 and,

    PSn=PSn1{α}

    OSn=OSn1{{α}}

    Let PPTn=PPTn1

    PTAn=PTAn1{α}

    OPTn=OPTn1{(T,h)} where T=[α] and h=([α],n)

    The sequences dd, PPTn,PTAn,OPTn is hence a grounded dispute derivation. We now proceed to show that it satisfies the lemma.

    From the induction hypothesis, we have

    PSn1=NSpm(PPTn1)MPTAn1;

    OSn1={Spm(Tp)|pOPTn1};

    It holds

    OSn=OSn1{{α}}OSn={Spm(Tp)|pOPTn1}{{α}}(induction hypothesis)OSn={Spm(Tp)|pOPTn}PSn=PSn1{α}OSn=(NSpm(PPTn1)MPTAn1){α}OSn=NSpm(PPTn1)(MPTAn1{α})OSn=NSpm(PPTn)MPTAn

 □

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