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A THEOREM-PROVER
Philippe
FOR A DECIDABLE
BESNARD
-
Rene
SUBSET
QUINIOU
-
OF DEFAULT
Patrice
LOGIC
QUINTON
IRISA - INRIA Rennes
Campus de Beaulieu
35042 RENNES Cedex
FRANCE
Abstract
II DEFAULT
Non-monotonic
such
logic
notions
evolution.
has
as
However
been
so
first-order
on the
of
proved
our
that
it
logic
deals
which
a
set
algorithm
and
801
we
study
the
and
the
can
of
A.I.,
complete
world
is an attempt
is
but
it
The
formulae.
in the field
decidability
problem
problem,
predefinite
subset
[Rei
unvalid
801
have
allows
can
resolution
First
and
of all,
we recall
which
III is devoted
to the
named
by
the
the
on
an
allow
more
of
default
main
goal
facilities
to
represent
example,
we know
default
principles
attractive
plausible
that
ostriches..)
as the
of
of our
default
theorem-prover
saturation
default
prover
is complete
We
which
generally
always
provide
some
knowledge.
can
a
then
derived
knowledge
by use
A default
of
particular
several
the
each
logic
can
Part
for
of
consistent
which
1: T=(
a proof
of the
a
cannot
be
rules.
(1I.T) where
the
theory
theories
A is a set
theory.
T
called
The use
by
building
extensions,
T.
A , -6
propose
a
heart
IV. This
terminates.
27
are
a conclusion
first-order
first-order
that
defaults
denotes
inference
A extends
the
default
is a couple
contains
So
as “until
a
fly(x)
is interpreted
v “C
)
A={
A :MB
B
, A :MC
C
This
default
theory
, C :MD
D
]
resolution
the
be
to suppose
fly”.
is.
and
default
x can
That
theory
default
if it is consistent
that
and T a consistent
of which
the
This
of classical
defaults
Example
of
schemata
inference.
of defaults
class,
and
infer
assume...”
of
inferences.
fly,
of inference
in default
in part
For
fly” (except
:Mfly(x)
fly(x)
CONS(G)).
patterns
problem
also
to
6:
prerequisite
contrary,
though
default
default
defaults
is to
“a bird
This
default
(noted
plausible
The consistency
constitutes
presented
and
the
Default
of default
the
as: “if x is a bird
apprehend
a
is
x can
the
class,
of a special
saturation.
generalisation
use
logic
common-sense
penguins,
consequent
801
avoid
logic.
A
ordinary
logic
of
represented
bird(x)
a
saturation.
presentation
the
[Al
of consistent
theory.
solved
by
determines
of default
is a crucial
named
focus
solvable
the inference
be
free-defaults,
technique
a
To
set
free-defaults,
some
the present
logic
which
context
which
technique
then
selected
in a first-order
nicely
logic
However,
of heuristics.
clauses,
of a set of formulae,
logic,
Non-monotonic
by use
of
supported
of non-monotonic
in the
formulae
be
to this fact.
variable
decidable
logic
we
feature
always
knowledge.
to remedy
realisations
this
a fundamental
cannot
to
by
he has
the
inferences.
applied
bird(x)
Decision-making
subset,
plausible
be
as proposed
free-defaults.
A Definition
I INTRODUCTION
fields
logic
proof-procedure
a default
additional
rule
default
have
it always
of first-order
[Rei
allow
obtain
to test the
that
in
rewriting
technique
We
we introduce
Next,
which
on the
formulae.
subset
up.
a
theorem-prover
designed
is complete
for the selected
such
of a resolution
first-order
Reiter
issue
propose
This
First
theory
with a restriction
with.
was initially
of
and
theorem-prover
We
generalisation
saturation,
consistency
terminates
decidable
for default
to take into account
knowledge
unexplored.
formulae
is based
named
the
far
theorem-prover
is an attempt
incomplete
LOGIC
has
2 extensions:
E,=Th(
A , -B
v “C
, B 1
E2=Th(
A , “B
v “C
, C , D 1
Definition:
A formula
to a default
Ao.
Al.
theory
.. . . Ak
U
(1) T
(2)
Q has a default
(A,T)
iff there
of finite
subsets
CONS(As)
For
I-
proof
exists
with
a finite
of A,
We have
respect
sequence
such
proved
Theorem
that
(A.T).
E c
Q
there
the
following
result
1: for
any
extension
exists
an extension
[ Bes
E of
E’ of
831:
a default
(f(A),T)
theory
such
that
E’
lbi<k
T U CONS(Ai)
Prerequisites(Ai-,I
I-
III SATURATION
(3) Ak=O
( u CONS(
(4) T u
(2) expresses
be
true
the
to
that
let
the
the
is satisfiable.
prerequisite
consequent
In this
of a default
of this
default
must
to
be
in
extension.
Reiter
has
shown
i.e.
that
respect
to
(A,T)
iff
extension
order
completeness
Q has
there
exists
Q. However
This
logic,
an
us to
which
for
(4) show
the
saturation
by
which
will
of
(A.T)
without
Defaults
may
not
is
even
a subset
decision
“fly(Max)
The
only
T itself.
such
The
true,
a kind
Next,
generalisation
in
the
of resolution
we
of
present
the
definition
of
the
saturation
our
default
is
[ Bos
decidable
is a predefinite
clause
are
occur
also
-
the
81 ]
for
clause
of
problem
saturation
saturation
only
already
cited,
This
behaviours
([ Rei
we present
below
of
the
Max
in the
default
doesn’t
fly
the
({S]
that
while
,T)
we expect
generally
is
For
to
transformed
birds
all
the
type
of
are
predicate,
theory
resolution
variable
clause
symbols
clauses.
A
if:
that
occur
in
the
occuring
in a positive
literal
one.
for
context
an
application
of databases,
since
formulae
are
function--free
-
“bird(Max),
contain
particular
symbols.
is acceptable
-
a
variable
in a negative
restriction
which
bird(x)
:Mfly(x)
6=
fly(x)
and
extension
because
free-defaults
is
predefinite
function
constant
a default:
)
It doesn’t
or
counter-intuitive
to those
of
T={
prerequisite
have
In addition
example
be
a
-- all the variables
6 Defaults
an
introduce
saturation.
useful
of the
technique,
resolved.
811).
sets.
be
A Definition
with
clearlv , that
consider
the
default
proof
extension
problem
leads
for
result
a default
(1) and
membership
semi-decidable.
first
the
a formula
contains
the
we first
named
proof-procedure.
proofs,
which
chapter
technique
without
variables
the
prodefinite
example,
the
in
Type,(x,)
clauses
all
a...&
function
is
symbol.
defined
in
by
a
clauses.
clause
the
C(x
*.., Xn)
1’
Typen(xn)
can
variable
predefinite
=> C(x,,
be
clause
. . .. x,1.
do.
Consider
:Mbird(x)=>fly(x)
bird(x)=>fly(x)
6’=
6’
leads
to
becomes
cannot
default
later
the
formula
true,
then
the
default
be
used
theory
permanent
knowledge
and
(6’.
contains
&
“fly(Max)
6’ instantiated
is not
by
Max
in the
new
expresses
a
true
expressed
in
by
then
we
he flies.
6’
opposite
6.
he flies.
to
the
situational
By 6’ we always
By 6 only
We now
when
generalise
know
6=
f(6)
P(x)
:MR(x)
R(x)
is called
I--->
a free-default.
f(6)-
that
x is a bird
the
previous
transformation:
:MP(x)=>R(x)
P(x)=>R(x)
or
Theorem
clauses
The
2:
a
satisfiability
set
of
a saturated
variable
of
saturation
one
produced
clause.
subsumed
by one
A-ordering
leftmost
is
parent
The
to
produce,
the
tautologies
to
are
and
restrict
previous
test
clauses.
the
The
first.
all
most
recently
and
ancestors
for the literals
permit
to
variable
clauses
equivalent
to the
means
is
clause.
variable
Due
a
which
of their
[ Kow 691
literal
of
variable
empty
logically
clauses.
gives
is
is
of C.
the
of a set of predefinite
saturation
clauses
of C either
predefinite
it contains
of a set of predefinite
resolvants
inferences.
by a clause
iff
by resolution,
the
variable
of 2 clauses
saturated
of predefinite
principle
predefinite
is subsumed
saturation
theorem.
the
C of
resolvant
is inconsistent
produces,
set
prerequisite
knowledge
that
E’
bird(Max)
a set
if each
a tautology
T U { bird(Max)})
if x is a bird
know
“bird(Max)
without
default
The
and
If
saturated
extension
unique
a
E’=Th(“fly(Max),bird(Max)=>fly(Max))
“bird(Max).
Definition:
( ~ :M(x) bird(x)=>fly(x))
(x1 bird(x)=>fly(x)
the
not
clauses
produced.
resolution
the
the
number
upon
of
Example
F=
(
-C
2: saturation
of
IV DEFAULT
F
= -C
v E
A The
v D
A is a free-default
1
=BvC
C
C
= -B
3
= rtc 2, c3)=
4
default
We
2
c
Q
C v D
now
prover
only
the
consider
= rtc
cd)=
D v E
1'
5
=AvC
C
default
set.
formula
Let
to
Prerequisites(A’)=0
C
BY SATURATION
v D , A v C , -A I
v E , B v C , -6
c
PROOF
theories
A’ be
be
(A.
T)
a subset
proved.
where
of
A and
Remark
that
thus
(2’)
T U CONS(A”)
(3’)
A” = 0
I-
Prerequisites(6’)
6
c
= -A
7
C
It is worth-noting
= rtc
8
c7)=
= rtc
has
not
cs)=
E
1'
9
set
C
6'
C
The
default
(cl,
The empty
.. .. cg } is saturated.
been
produced
then
the
(§ II) are
and
(4) of the
default
and
T
U CONS(A’)
by sets,
clause
these
I =I
B Saturation
It
is
that
later
in the
clause
possible
goal
Let
the
we
recursively
the
subsumed
in
sets
the
of
example
notion
of
predefinite
clause
2.
useless
production.
of F on E, produces
defined
a
It
is
clauses
To
S(E,F)
by
E U (c}
.c
, F’ -
(c}
1
the
.r E: F’ if
*or
and
r
r is
r
literal
process
Example
is
not
the
resolvant
A v B
{
}
a clause
upon
a clause
saturation
S(E,F)=S(
by
the
of
E
leftmost
E.
and
A v B
F={
}
“A
,
(
S(0.
(
A v B , -A
)
A v B , -A
, B
Theorem
3: S(E,F)
is finite
iff
Theorem
4: S(E,F)
is saturated
Corollary
5: S(0.C)
B
E and
, (
)
)
, 0
1
1
6i,
F are
finite.
equivalent
to C.
a
if E is saturated.
saturated
logically
T
It then
Then
for
proof
During
U ( -Qj
This
procedure
over
A’=( 61 ),
the set of the
successively
S(0.
T).
T U CONS(62)),
62))...
remark
remains
true,
so:
I= I
SG(0.T).
CONS(G1))
S(0,
T U CONS(G2))
I= I
S(S(0.T).
CONS(B2))
T U CONS(G1.
S(S(0,
S(0.
then,
the
every
T U CONS(A’))
also
CONEi
are
only
contains
or
A’ subset
contains
the
empty
it. Due to this
minimally
the
of A containing
clause
optimisation,
inconsistent
sets
U
T
computed.
By operating
a global
to compute
consistent
CONS(G2))
T U CONS(
for
it is useless
62))
T U CONS(GIH,
example
clause
we
never
evaluates
A’=0,
for
computes
be
example.
(A’ ranges
S(0.
can
it is useless
CONS(A’).
U
The
62}...
saturation
proof
of T U CONS(A’).
T U CONStA’)
of
the
T U CONS(6 1))
S(0.
that
of
first.
T U CONS(
default
be true).
( “Q])
computed
and
no
addition,
U
S(0.
If for
and
to S(0.
resolution,
T U CONS(G1.
empty
)
In
of T U CONEi
.. . . 6n}.
A).
I= I
only
is
linear
preceding
1
B
twice.
unsatisfiable
one
S(0.
}
is
(4) cannot
A’={61,
}
(
A v B , -A,
that
belongs
(because
T U CONS(G1)).
The
-A
=S(
clause
stopped
with
of
C-Q),
are
A={61,
S(0.
saturation
T U CONS(
consistency
by sets.
=S(
=(
of
the
of S(0,
discovered
subsets
subsumed
of c with
is called
3: E=(
E F
of
because
computed
of a refutation
A’={s2},
*either
(1)
T U CONEi
U ( “Q))
if Q is a consequence
Let
E F
is
the
T U CONSCA')
empty
search
use
(3)
of S(0.
given
have
By
be joined
of
To solve
elaboration
is
to test
the
clause
and
we saturate
T U CONS(A’)),
T U CONS(A’)
happens
S(E.F)=S(
This
So if the
can
(2)
verified.
T U CONS(A’)
no
clauses
found,
S(E.B)=E
where
the
process
clauses.
the set noted
way
steps
U (-Q}.
SG(0,
the
then
reach
saturation.
variable
this
during
SO
is useless.
of qhese
resolvant
extended
by
process
by c
number
of
F be
The saturation
clause
saturation
order
have
E and
a
subsumed
c
to reduie
by relaxing
this
In
obvious
produced
is the
by sets
the
immediatly
proof.
steps
S(0,
F is consistent,
set
that
proof
like
saturation,
some
this,
S(0.
subsets
a default
proof
T U CONS(A)
of
these
is nothing
U (-Q}).
clauses
but
knowing
are
really
computed.
Note
also
S(0.
I=I
S(0,
for
that
S(S(0.
and
saturation
remains
by sets
extensions
of
U {“Q},
T U CONS(A)),
T U CONS(A))
all
V CONCLUSION
T U CONS(A)
can
then
the
same
provides
a default
{“Q})
We
be computed
for
a kind
every
once
query.
of compilation
have
and
of default
The
contains
of the
axioms
theory.
Is
prover
with
is complete.
main
Theorem
7: the
B Example
prover
always
of a default
proof
terminates.
further
by saturation
the
T={ C v D, E]
6 =:M-C
v “E
1
“C v “E
The
This
default
theory
6 -:M-D
v “E
2- “D v “E
({ 6 1,
62,
U CONS(G1,
6s))
E2=Th(T
U CONS(G2,
6s))
S1=S(O.
theory
T)={
S2=SW0.
T),
=S1
U {
S3=S(S(0,
T),
4
=S2
U {
S5=S(s(0.
T),
S6=SWS(0,
=S2
CONS(G1)),
, 0
U {
[Al
801
prover:
[Bes
}
U {
“D
2
S1
U {
v “E
}
[ Bos
The
B v “E
)
computed
can
since
answer
a query
increasing
of the
For
S
set:
in
Ss=SIS4,CONS(6s))
contains
the
of clauses
to
take
a default
scsl.
{-D}>={
S(S2.
( -D})=S(S1.
are
empty
clause.
to answer
any
S(Si,
advantage
proof
of
C v D , E , “D
u
not
“C
the
efficiently
compilation
prover
is
not
a direction
the subset
deal
about
for
of formulae
with.
non-monotonic
vol.
13,
logic
1980
for
Note
198,
a non-monotonic
University
G. & Siegel,
au
of
de
R. & Hayes,
trees
in
automatic
Intelligence
4,
I,
1983
P.
secours
la non-monotonie
of Aix-Marseille
Kowalski,
logic
Rennes
pp.
II,
1981
P. J.
theorem-proving
87-101,
1969
(“Q))
[Rei
To
with
of the
D is given
v “E
, “E
Reiter,
R.
for
default
i
Intelligence
by:
}
30
Reiter,
interacting
Proc.
inclusion
, 0
811
On
}
, D v “E
803
A logic
reasoning
vol.
13,
pp.
81-132.
1980
Now we
query.
(“D})
t
691
Artificial
Si’s.
example,
is to deal
the
by sets
are
CONS(6s))
Q we evaluate
order
on
University
Machine
of
to
We
clause
P.
Bossu,
Semantic
}
clauses
use these
Besnard.
811
[Kow
U{Bv-E}
remaining
issue
La saturation
1)
[ Rei
=s3
is able
Intelligence.
831
CONS(6s)I
CONS(G2)),
saturation
provides
as extending
Special
Technical
CONS<6
CONS(G1)),
T).
This
Artificial
}
B v “E
S7=S(S(S(0,
updating.
Thesis,
CONS(6s))=
T),
to axiom
and
variable
once
However,
only
REFERENCES
}
= S1
prover
updating,
subset
=:MB v “E
B v “E
3
to the
, D v “E
CONS(B2))
“E
default
upon
realised.
A proof-procedure
v “E
T),
of the
a
subset
consequents
terminates.
been
theorem-prover
CONS(
“C
S =S(S(S(B,
is submitted
C v D , E
PROLOG.
6 3 } , T) has 2 extensions:
El=Th(T
default
6
the
based
It in
investigations
for
of this
in predefinite
always
and
has
adapted
all
and
advantage
querying
phase
theories
and
Our theorem-prover
complete
The
6: the
a theorem-prover
default
be transformed
implementing
Theorem
The
free-defaults
can
form.
proposed
logic.
IJCAI-81
R. & Criscuolo.
G.
defaults
pp.
270-276,
Vancouver,
1981