2. LA FACULTAD DE TEOLOGÍA Y EL PERFIL DE SU TEOLOGO
2.2. Bosquejos del Perfil de un Teólogo Javeriano
7.3
Consistency Properties of an AABA agent
We prove in this section some consistency properties of an AABA agent x defined as in Sec- tion 7.2. We assume Rp
x is such that if done(give(Y, R, Z), T ), done(give(Y ′ , R, Z′ ), T ) ∈ Rp x then Y = Y′ and Z = Z′
(i.e. resources are not given away or received by two different agents at the same point in time and agents have no facts as such). We assume also that facts done(give(Y, R, X), T ) are correct (i.e. such a fact originates and is communicated only if Y has given R to X at time T ).
Structuring the first two inference rules of Definition 7.4 for determining when an agent has a resource as they are avoids the possibility of holding conflicting beliefs in this regard. Notably, an agent x will never believe a resource to be held by two agents simultaneously, as in the lemma below. In addition, the need to manually remove facts from Rp
x as to which agent has a
given resource for the sake of maintaining consistency of the agent’s belief set (personal facts) is eliminated.
Lemma 7.1 (consistency of has) If |=AFx has(X, R) for some agent X ∈ names(AS) and resource R ∈ R, then 2AFx has(Y, R) for all agents Y ∈ names(AS) such that Y 6= X.
Proof. Assume, for the sake of contradiction, that
• {done(give( , R, X), T ), asm(¬gaveAwayAf ter(X, R, T ))} |=AFx has(X, R); and
• {done(give( , R, Y ), T′
), asm(¬gaveAwayAf ter(Y, R, T′
))} |=AFx has(Y, R)
for some T, T′
∈ N and Y ∈ names(AS) such that Y 6= X. Note that T 6= T′
since, by assumption, a resource R cannot be received by two agents at the same point in time and facts about resources changing hands are correct. Then, either
• T′
> T and hence {done(give( , R, Y ), T′
)} |=AFx gaveAwayAf ter(X, R, T ). But then asm(¬gaveAwayAf ter(X, R, T )) cannot be an admissible assumption defending the claim has(X, R); or
• T > T′
and hence {done(give( , R, X), T )} |=AFx gaveAwayAf ter(Y, R, T′). But then
asm(¬gaveAwayAf ter(Y, R, T′
)) cannot be an admissible assumption defending the claim has(Y, R).
Contradiction. 2
Likewise for has(X, R), structuring the inference rules for determining when an agent does not have some resource according to Definition 7.4 avoids the need to manually remove facts as to which agents do not have a a given resource for the sake of maintaining consistency of the agent’s belief set (personal facts). Also, the possibility of deriving conflicting claims regarding an agent both having and not having a certain resource is avoided, as follows:
Lemma 7.2 (consistency of has and ¬has) It can never be the case that both |=AFx has(X, R) and |=AFx ¬has(X, R) for any agent X ∈ names(AS) and resource R ∈ R.
Proof. Assume, for the sake of contradiction, that |=AFx has(X, R) and |=AFx ¬has(X, R)
for some X ∈ names(AS) and R ∈ R. Then, for it to be the case that |=AFx ¬has(X, R),
either |=AFx has(Y, R) for some Y 6= X (by the third inference rule of Definition 7.4) or
|=AFx asm(¬has(X, R)) (by the fourth and fifth inference rules of Definition 7.4). We look
at each case in turn. Firstly, if |=AFx has(Y, R) for some Y 6= X, then, by Lemma 7.1, 2AFx
has(X, R). Contradiction. Secondly, if |=AFx asm(¬has(X, R)), then it cannot be the case that the contrary has(X, R) is admissible, i.e. 2AFx has(X, R). Contradiction. 2
Similarly as for Lemma 7.2 whereby an agent will never claim an agent to both have and not have a particular resource, an agent will never claim an agent to both need and not need a particular set of resources, as follows:
Lemma 7.3 (consistency of needs and ¬needs) It can never be the case that both |=AFx
needs(X, Rs) and |=AFx ¬needs(X, Rs).
Proof. Assume, for the sake of contradiction, that both D1 |=AFx needs(X, Rs) and D2 |=AFx ¬needs(X, Rs) where D1and D2 are admissible defence sets. Then, according to Definition 7.4,
7.3. Consistency Properties of an AABA agent 165 D1 consists of a single (admissible) assumption asm(needs(X, Rs)). However, it cannot then
be that the contrary ¬needs(X, Rs) is an admissible claim, i.e. 2AFx ¬needs(X, Rs). Contra-
diction. 2 We define ⊢x
asm and ⊢x (see Definition 5.10) now in terms of |=AFx for the sentences of the
meta-language ML (see Definition 5.11) occurring in Definitions 5.12–5.15, as follows:
• ⊢x [Y has R] iff |=
AFx has(Y, R); • ⊢x ¬[Y has R] iff |=
AFx ¬has(Y, R);
• ⊢x
asm [Y has R] iff |=AFx asm(has(Y, R));
• ⊢x
asm ¬[Y has R] iff |=AFx asm(¬has(Y, R));
• ⊢x [the r.r.p. of Y is solved with R] iff |=
AFx usef ul(R, Y );
• ⊢x ¬[the r.r.p. of Y is solved with R] iff |=
AFx ¬usef ul(R, Y );
• ⊢x
asm ¬[the r.r.p. of Y is solved with R] iff |=AFx asm(¬usef ul(R, Y ));
• ⊢x [the r.r.p. of Y is solved] iff |=
AFx {usef ul(R, Y ), has(Y, R)} for some resource R; • ⊢x [the r.r.p. of Y would still be solved or unsolved by simply giving away R] iff |=
AFx ¬needs(Y, {R});
• ⊢x ¬[the r.r.p. of Y would still be solved or unsolved by simply giving away R] iff |= AFx needs(Y, {R});
• ⊢x
asm [the r.r.p. of Y would still be solved or unsolved by simply giving away R] iff |=AFx asm(¬needs(Y, {R}));
• ⊢x
asm¬[the r.r.p. of Y would still be solved or unsolved by simply giving away R] iff |=AFx asm(needs(Y, {R}));
• Given some set of resources Rs: ⊢x [after giving away each resource in Rs, the r.r.p. of
Further, we donote the following conditions occurring in Definitions 5.12–5.15 in terms of AFx:
• the r.r.p. of x is not solved and fulfils(R,G(x)) ∈ B(x) iff |=AFx {¬has(x, R),
needs(x, {R})};
• Y ∈ names(AS) and Y 6= x iff |=AFx isAgent(Y );
• It is the case that each of ⊢x [Y has R], ⊢x [the r.r.p. of Y is solved], ⊢x ¬[the r.r.p. of
Y would still be solved or unsolved by simply giving away R] and ⊢x [the r.r.p. of Y is
solved with R′
] holds iff |=AFx {has(Y, R), needs(Y, {R}), usef ul(R′, Y )};
• Given some set of resources Rs: (for all R ∈ Rs, fulfils(R,G(x)) /∈ B(x), or there is some R /∈ Rs such that R ∈ Res(x) and fulfils(R,G(x)) ∈ B(x)) iff (|=AFx asm(¬needs(x, Rs))).
The definition of AFx (Definitions 7.1–7.4) as well as the consistency properties demonstrated
in this section (Lemmas 7.1–7.3) make the mapping of ⊢x
asmand ⊢xto |=AFx as above legitimate
according to the requirements of Postulate 5.1. Notably, for some sentence [L] ∈ ML, it is not possible that ⊢x [L] and ⊢x¬[L], nor is it possible that ⊢x
asm [L] and ⊢x¬[L].