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A
conc
(
)=
α
,
A
premise
(
)=∅
,
A
psm
(
)=
{
α
}
,
A
sent
(
)=
{α}
.
•Or
α
∈A
sm
DF
. Therefore,
α
is deduced by the set of premises
A
which, by definition, are all
assumptions in
A
sm
DF
and, by construction, there is only one rule r with
∼
deleted
(
r
)
∈
A
such as
head
(
.
-
Either r is a fact of
r
)=
α
−
1
A is a trivial argument defined such
T
(
A
)=
deducing
α
and so,
that:
A
conc
(
)=
α
,
A
premise
(
)=
{}
,
A
psm
(
)=∅
,
A
sent
(
)=
{α}
.
−
1
¯
Aisatree
T
(
A
)=
-
Otherwise r is a rule of
deducing
α
with a non-empty body and so,
A
argument built upon the subarguments
sbarg
(
)
defined such that:
A
conc
(
)=
α
,
A
premise
(
)=
body
(
)
r
,
A
)
A
psm
(
)=
∪
A
∈
sbarg
(
psm
(
,
A
)
A
∈
sbarg
(
A
)
A
sent
(
)=
body
(
r
)
∪{
head
(
r
)
}∪
sent
(
.
A
)
Proof 2
(Mapping between semantics)
.
Let
DF
=
DL
,
P
sm
,
I
,
T
,
P
,
RV
be a decision framework and
AF
=
A
(
DF
)
,
defeats
be our argumentation framework for decision making. Let us consider
G
∈G
.
• Let us consider a s-admissible set of structured arguments S
1
concluding
G
1
⊆
G
. Due to the
lemma 1, we can built the set of arguments
S
1
such that for any structured argument A
1
with
G
⊇RV
S
1
∈
A
1
)
thereisanargument
a
1
∈
S
1
,where
a
1
:
(
Lisinsome
PABFS
DF
(
G
)
. We consider here
pabf
1
DF
(
G
1
)
∈
PABFS
DF
(
G
)
where all the arguments appear. Due to the construction of
S
1
,
S
1
is conflict-free and defend itself within
pabf
1
A
(
(
))
DF
(
)
the set of arguments
. Therefore,
S
1
is an admissible set. Let us consider a different
pabf
2
DF
(
G
2
)
∈
PABFS
DF
(
G
)
G
such that
pabf
2
DF
(
G
2
)
P
pabf
1
DF
(
G
1
)
. Due to the definition 21,
G
2
⊇
G
1
and
∀
g
2
∈
G
2
\
G
1
there is no
g
2
. If we suppose that
pabf
2
contains an admissible set of arguments
deducing
G
2
, then the corresponding set of structured arguments concluding
G
2
is admissible. It is
not the case.
• Let us consider
pabf
1
g
1
∈
G
1
such that g
1
P
DF
(
G
)
∈
PABFS
DF
(
G
)
which contains an admissible set of assumptions
A
1
deducing
G
1
with
RV ⊆
G
1
⊆
G
. If we suppose that there is no
pabf
2
∈
PABFS
DF
(
G
)
,with
P
pabf
1
, which contains an admissible set of assumptions deducing
G
2
⊆
G
with
G
2
P
G
1
,
pabf
2
−
1
then the corresponding s-admissible set of structured arguments
(
A
1
)
concludes
G
1
and there is
no other s-admissible set of structured arguments S
2
concluding
G
2
.
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