Probabilistic Argument Graphs
for Argumentation Lotteries
Proofs of technical results
Anthony HUNTER a and Matthias THIMM b
of Computer Science, University College London, UK
b Institute for Web Science and Technology, University of Koblenz-Landau, Germany
a Department
Proposition 1. For every argument graph G and each semantics X ∈ {co, pr, gr, st},
for every division hΦ, Ψi ∈ DivisionsX (G) it is Φ ∩ Ψ = 0.
/
for every α ∈ Nodes(G), h{α}, {}i ∈ DivisionsX (G) or h0,
/ {α}i ∈ DivisionsX (G).
the empty division h0,
/ 0i
/ is always in DivisionsX (G).
for every extension E ∈ ExtensionsX (G) there is a division hE, (Nodes(G) \ Ei ∈
DivisionsX (G).
5. if hΦ, Ψi ∈ DivisionsX (G), Φ0 ⊆ Φ, and Ψ0 ⊆ Ψ, then hΦ0 , Ψ0 i ∈ DivisionsX (G)
1.
2.
3.
4.
Proof. (1) Let hΦ, Ψi be a division. Therefore, there is an extension E ∈ ExtensionsX (G)
such that Φ ⊆ E and E ∩ Ψ = 0.
/ Therefore, Φ ∩ Ψ = 0.
/
(2) Let α ∈ Nodes(G). Therefore, for each extension E ∈ ExtensionsX (G), either α ∈ E
or α 6∈ E. If α ∈ E, then h{α}, {}i ∈ DivisionsX (G), whereas if α 6∈ E, then h{}, {α}i ∈
DivisionsX (G). Therefore, for every α ∈ Nodes(G), h{α}, {}i ∈ DivisionsX (G) or
h{}, {α}i ∈ DivisionsX (G).
(3) For all E ∈ ExtensionsX (G), 0/ ⊆ E and E ∩ 0/ = 0.
/ Therefore, h0,
/ 0i
/ ∈ DivisionsX (G).
(4) Consider E ∈ ExtensionsX (G). Therefore, E ⊆ E and E ∩ (Nodes(G) \ E = 0.
/ Therefore, hE, (Nodes(G) \ Ei ∈ DivisionsX (G).
(5) Consider hΦ, Ψi in DivisionsX (G). Therefore, there is an extension E ∈ ExtensionsX (G)
such that Φ ⊆ E and E ∩ Ψ = 0.
/ Let Φ0 ⊆ Φ and Ψ0 ⊆ Ψ. Therefore, Φ0 ⊆ E and
0
0
0
E ∩ Ψ = 0.
/ Therefore, hΦ , Ψ i ∈ DivisionsX (G).
Proposition 2. For every argument graph G and each semantics X ∈ {co, pr, gr, st},
1. If Φ ∩ Ψ 6= 0,
/ then DividersG
/
X (hΦ, Ψi) = 0
2. If Φ = Ψ = 0,
/ then DividersG
(hΦ,
Ψi)
=
{G0 | G0 v G}
X
G
0
0
3. If Φ ⊆ Φ and Ψ ⊆ Ψ , then DividersX (hΦ0 , Ψ0 i) ⊆ DividersG
X (hΦ, Ψi)
Proof. (1) Let Φ ∩ Ψ 6= 0.
/ Therefore, for all G0 v G, for all E ∈ ExtensionsX (G), either
Φ 6⊆ E or Ψ ∩ E 6= 0.
/ Therefore, DividersG
/
X (hΦ, Ψi) = 0.
(2) Let Φ = 0/ and Ψ = 0.
/ Therefore, for all G0 v G, for all E ∈ ExtensionsX (G), Φ ⊆ E
0
0
and Ψ ∩ E = 0.
/ Therefore, DividersG
X (hΦ, Ψi) = {G | G v G}.
G
0
0
0
0
(3) Let Φ ⊆ Φ and Ψ ⊆ Ψ . If G ∈ DividersX (hΦ , Ψ0 i), then there is an extension
E ∈ ExtensionsX (G) such that Φ0 ⊆ E and Ψ0 ∩ E = 0.
/ Therefore, there is an extension
E ∈ ExtensionsX (G) such that Φ ⊆ E and Ψ ∩ E = 0.
/ Therefore, G0 ∈ DividersG
X (hΦ, Ψi).
0 , Ψ0 i) ⊆ DividersG (hΦ, Ψi).
Therefore, DividersG
(hΦ
X
X
Proposition 3. Tuples T = hΦ1 , Ψ1 i, T 0 = hΦ2 , Ψ2 i ∈ Tuples(G) with Φ1 ∩ Ψ1 = 0/ and
Φ2 ∩ Ψ2 = 0/ are disjoint iff either Ψ1 ∩ Φ2 6= 0/ or Ψ2 ∩ Φ1 6= 0.
/
Proof. Without loss of generality assume Ψ1 ∩ Φ2 6= 0.
/ Then there cannot be G0 ∈
0 ) such that in the grounded extension E of G0 we have both Φ ⊆ E and
DividersG
(T
2
X
Ψ1 ∩ E = 0.
/ Therefore, T and T 0 are disjoint. Assume now that neither Ψ1 ∩ Φ2 6= 0/ nor
Ψ2 ∩ Φ1 6= 0/ holds. Consider then the subgraph G0 that exactly contains the arguments in
Φ1 ∪ Φ2 and no attacks. Then G0 is a divider for both T and T 0 .
Proposition 4. Let G be an argument graph and define a set of divisions D via
D = {hΦ, Ψi | Φ ⊆ Nodes(G) and Ψ = Nodes(G) \ Φ}
Then D is exhaustive for G w.r.t. any semantics X and D is pairwise disjoint for G w.r.t.
grounded semantics.
Proof. (Exhaustiveness) Let D = {hΦ, Ψi | Φ ⊆ Nodes(G) and Ψ = Nodes(G) \ Φ}. For
each G0 v G, there is E ∈ ExtensionsX (G0 ), and there are Φ, Ψ ⊆ Nodes(G0 ) such that
Φ ⊆ E and Ψ ∩ E = 0.
/ Therefore, for each G0 v G, there is E ∈ ExtensionsX (G0 ), and
0
there are Φ, Ψ ⊆ Nodes(G0 ) such that G0 ∈ DividersG
X (hΦ, Ψi). Therefore, for each G v
G
0
G, there is T ∈ D such that G ∈ DividersX (T ). Therefore,
[
0
0
DividersG
X (T ) = {G | G v G}
T ∈D
(Disjointness) Let Φ, Ψ, Γ ⊆ Nodes(G) be pairwise disjoint. Therefore, Ti = hΦ ∪ Γ, Ψ
G
are T j = hΦ, Ψ ∪ Γ in D. Let Gi ∈ DividersG
gr (Ti ) and G j ∈ Dividersgr (T j ). Also let
Extensionsgr (Gi ) = {Gi } and Extensionsgr (G j ) = {G j }. Therefore, Φ ∪ Γ ⊆ Ei and
Ei ∩ Ψ = 0.
/ Similarly, Φ ⊆ E j and E j ∩ (Ψ ∪ Γ) = 0.
/ Therefore, Ei 6= E j . Therefore,
G
DividersG
(T
)
∩
Dividers
(T
)
=
0.
/
gr i
gr j
Proposition 5. Let D be an exhaustive set of divisions where the divisions in D are
pairwise disjoint. Assume T is a division not in D. Also let D0 ⊆ D be a subset of divisions.
If T subsumes D0 , then the set of divisions (D \ D0 ) ∪ {T } is exhaustive and pairwise
disjoint.
Proof. Assume D is exhaustive and pairwise disjoint, and D0 ⊆ D, and T 6∈ D, and T
subsumes D0 .
(Exhaustiveness) Since T subsumes D0 ,
DividersG
X (T ) =
[
0
DividersG
X (T )
T 0 ∈D0
Therefore,
[
Ti ∈D
[
DividersG
X (Ti ) =
Tj
DividersG
X (T j )
∈((D\D0 )∪{T })
Therefore, (D \ D0 ) ∪ {T } is exhaustive.
(Pairwise disjoint) D is pairwise disjoint. Therefore, D \ D0 is pairwise disjoint and D0 is
pairwise disjoint. Therefore, for Ti ∈ D \ D0 and T j ∈ D0 , Ti and T j is pairwise disjoint.
Therefore,
(
[
DividersG
gr (Ti )) ∩ (
Ti ∈D\D0
[
Tj
DividersG
/
gr (T j )) = 0
∈D0
Since T subsumes D0 ,
(
[
G
DividersG
/
gr (Ti )) ∩ Dividersgr (T ) = 0
Ti ∈D\D0
Therefore, (D \ D0 ) ∪ {T } is pairwise disjoint.
Proposition 6. For every argument graph G and each semantics X ∈ {co, pr, gr, st},
1. If Φ ∩ Ψ 6= 0,
/ then PX (hΦ, Ψi) = 0
2. If Φ = Ψ = 0,
/ then PX (hΦ, Ψi) = 1
3. If Φ ⊆ Φ0 and Ψ ⊆ Ψ0 , then PX (hΦ0 , Ψ0 i) ≤ PX (hΦ, Ψi)
Proof. (1) Let Φ ∩ Ψ 6= 0.
/ Therefore, for all G0 v G, for all E ∈ ExtensionsX (G), either
Φ 6⊆ E or Ψ ∩ E 6= 0.
/ Therefore, DividersG
/ Therefore, PX (hΦ, Ψi) = 0.
X (hΦ, Ψi) = 0.
(2) Let Φ = 0/ and Ψ = 0.
/ Therefore, for all G0 v G, for all E ∈ ExtensionsX (G),
0
0
Φ ⊆ E and Ψ ∩ E = 0.
/ Therefore, DividersG
X (hΦ, Ψi) = {G | G v G}. Therefore,
PX (hΦ, Ψi) = 1.
0
0
(3) Let Φ ⊆ Φ0 and Ψ ⊆ Ψ0 . If G0 ∈ DividersG
X (hΦ , Ψ i), then there is an exten0
0
sion E ∈ ExtensionsX (G) such that Φ ⊆ E and Ψ ∩ E = 0.
/ Therefore, there is an
extension E ∈ ExtensionsX (G) such that Φ ⊆ E and Ψ ∩ E = 0.
/ Therefore, G0 ∈
G
G
G
0
0
DividersX (hΦ, Ψi). Therefore, DividersX (hΦ , Ψ i) ⊆ DividersX (hΦ, Ψi). Therefore,
PX (hΦ0 , Ψ0 i) ≤ PX (hΦ, Ψi).
Proposition 7. Let G be an argument graph, P a probability distribution on G, and
{T1 , . . . , Tk } = {hΦ, Ψi | Φ ⊆ Nodes(G), Ψ = Nodes(G)\Φ}. Then [P(T1 ), T1 ; . . . ; P(Tk ), Tk ]
is an argumentation lottery for G with respect to grounded semantics.
Proof. Assume D = {hΦ, Ψi | Φ ⊆ Nodes(G) and Ψ = Nodes(G) \ Φ}. So by Proposition 4, D is exhaustive and pairwise disjoint w.r.t. grounded semantics. Hence, we satisfy
condition 2 of Definition 9. Since D is exhaustive, we get the following by Definition 4.
[
0
0
DividersG
gr (T ) = {G | G v G}
T ∈D
Therefore, by Definition 8, ∑T ∈D P(T ) = ∑T ∈D ∑G0 ∈Dividers(T ) P(G0 ) = ∑G0 vG P(G0 ) = 1.
Hence, we satisfy condition 1 of Definition 9. Therefore, [P(T1 ), T1 ; . . . ; P(Tk ), Tk ] is an
argumentation lottery for G with respect to grounded semantics.