Propositional Argumentation Systems and Symbolic Evidence Theory

2.3 Doubt and Plausibility 11
An argument a supporting some h ′ ∈ C(h) is an argument against h (or
in favour of h ❁ ). Let DB(h) ⊆ A be the set of arguments a ∈ A, which
allow to assert any h ❁ . Such arguments are called refuting arguments.
Again, DB(h) forms a downward–set with a unique maximal element.
This maximal element is given by the coarsest refuting argument db(h).
The function db(h), which assigns to each hypothesis h ∈ H the coarsest
refuting argument, is called doubt in h. It defines a mapping d : H −→ A
called allocation of doubt. It satisfies the following conditions (Kohlas,
1995):
(D1) db(△) = ▽ (with △ ∈ H and ▽ ∈ A),
(D2) db(h 1 ⊔ h 2 ) = db(h 1 ) ⊓ db(h 2 ),
(D3) if h 1 ⊑ h 2 , then db(h 2 ) ⊑ db(h 1 ),
(D4) db(h 1 ⊓ h 2 ) ⊒ db(h 1 ) ⊔ db(h 2 ).
For a body of arguments B = (H, A, qs) allocation of doubt db can be derived
from allocation of quasi–support qs and vice versa. Let h be a hypothesis
in H and h ❁ its complementary hypothesis, then we have:
db(h) = qs(h ❁ ), qs(h) = db(h ❁ ). (2.5)
Given two finite Boolean algebras H and A of hypotheses and arguments, it
can be interesting to look at arguments for which a hypothesis h is not excluded.
Such arguments are called plausible arguments for h. In general,
plausible arguments do not necessarily allow to infer the hypothesis h, but in
all cases h remains possible. A hypothesis h is possible with an argument
a, if a does not support h ❁ . Let P L(h) ⊆ A be the set of plausible arguments
which do not allow to infer h ❁ . P L(h) is the complement of DB(h).
Note that the complement of an downward–set is always an upward–set.
Therefore, P L(h) forms an upward–set, and the set of its minimal elements
(lower bound) is denoted by νP L(h) (there is no unique minimal element).
We call
pl(h) =⊔ {a : a ∈ νP L(h)} (2.6)
plausibility of h. This mapping from H to A is called allocation of
plausibility. It satisfies the following properties (Kohlas, 1995):
(P1) pl(△) = △ (the first △ is in H, the second in A),
(P2) pl(h 1 ⊔ h 2 ) = pl(h 1 ) ⊔ pl(h 2 ),