also illustrate if the searching direction is right. This kind of assessing
operationality is dynamic. It depends on the current status and the current goal of
performance. What is more, this assessment can generate measurement and
efficiency. But this cost much for the system must be tested once operationality is
assessed.
The operationality is vital to EBL system. However, current methods to detect
operationality depend on whether the performance hypothesis can be simplified
(these hypothesis are easy to be broken). Though researchers from home and
abroad are seeking for effective methods for dealing with simplification, most of
the research only treats with theory application while in real applications it is
hard to get a satisfied answer. There is still a long way to go for investigation.
9.9 EBL with imperfect domain theory
9.9.1
Imperfect domain theory
One of the most important problems in EBL is domain theory. As the premise of
EBL, domain theory should be complete and correct. These demands are often
hard to meet in real applications and in reality domain theory is always
incomplete and incorrect. If the domain theory can not explain the training
example, the existing EBG will be invalid.

Explanation-Based Learning
357
The imperfect of domain theory may involve the following conditions:
(1) Incomplete
：
Lack of rules and knowledge in the domain theory thus no
explanation of training example can be given.
(2) Incorrect
：
Some rules in the domain theory is unreasonable thus incorrect
explanation might be created.
(3) Intractable
：
Domain theory is too complex; the existing resource can not
afford to create an explanation tree for training examples.
In order to solve the problem of imperfect domain theory, we make some
attempts in inverting resolution and deep knowledge based approach.
9.9.2
Inverting Resolution
Resolution theorem in first order logic is the foundation of machine theorem
proving and the main way to construct explanation in EBL(Muggleton etal.,
1988).
Resolution theorem
：
Let C
1
,C
2
be two clauses without any common variables.
L
1
, L
2
are two literals of C
1
and C
2
respectively, if there is a most general unifier
σ
, then clause
C= (C
1
- {L
1
})
σ
∪
(C
2
- {L
2
})
σ
(9.2)
is the resolution clause of C
1
and C
2
.
Inverting resolution deals with that given C and C
1
how to obtain C
2
. In
propositional logic,
σ
＝
Φ
, so the inverting clause (9.2) can be converted into:
C =(C
1
∪
C
2
) - {L
1
,L
2
}
(9.3)
From formula (9.3),
the following formulas can be concluded
：
(1) if C
1
∩
C
2
=
∅
, then C
2
= (C-C
1
)
∪
{L
2
}
(2) if C
1
∩
C
2
≠
∅
, C
2
needs to contain the arbitrary sub-set of C
1
-{L
1
},
therefore, generally speaking:
C
2i
= (C - C
1
)
∪
{L
2
}
∪
S
1i
(9.4)
Here, S
1i
∈
P (C
1
-{L
1
}, P(
x
) represents the power set of set x.
It is obvious if there are n literals in C
1
, the number of solution for C
2
is 2n-1.

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