Transcript Ground Horn Logic

Automated Reasoning
Systems
For first order Predicate Logic
AR: general context.
 Given is a knowledge base in predicate logic: T
 is a set of formulae in first order logic
 formally also called: a “Theory’’
 Given is also an additional first order formula: F
 Is F a logical consequence of T ?
 Notation: T |= F
(T implies F)
 Find reasoning techniques that allow to decide on
this for EACH F and T.
 Requirements:

correctness -- completeness
-- efficiency
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AR: decidability.
 Theorem Church ‘36:
 There CANNOT EXIST AN ALGORITHM that decides
whether T |= F, for any theory T and any formula F.
 BUT: semi-decidable !
 Completeness Theorem of Goedel ‘31:
There exists a reasoning technique, such that for any
theory T and formula F, such that T |= F, the
reasoning technique proves T |= F.
 SO: if F follows from T, then we find a proof, else it is
possible that the procedure doesn’t terminate.
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Wait a second ...
 The theorems of Church and Goedel are contradicting:
We can try to prove F and ~F in parallel, and according
to Goedel’s theorem one of these must succeed after a
finite time.
 Wrong !
 Let: T = {smart(Kelly)} en F = strong(Kelly)
 Although strong(Kelly)  ~strong(Kelly) is always true,
we have:
 neither: {smart(Kelly)} |= strong(Kelly)
 nor: {smart(Kelly)} |= ~strong(Kelly)
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AR: general outline (1).
 First we sketch the most generally used approach
for automated reasoning in first-order logic:
 backward resolution
 The different technical components will only be
explained in full detail in a second pass (outline (2)).
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AR: general outline (2) .
 We study different subsets of predicate logic:




ground Horn clause logic
Horn clause logic
Clausal logic
full predicate logic
 In each case we study semi-deciding procedures.
 Each extension requires the introduction of new
techniques.
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Backward Reasoning
Resolution
… in a nutshell
Backward resolution:
0) The task: an example.
1) Proof by inconsistency.
2) Conversion to clausal form.
3) Unification.
4) The resolution step.
5) (Backward) resolution proofs.
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0) The TASK (example):
T
z ~ q(z)
y p(f(y))
x p(x)  q(x)  r(x)
p = parent
f = father
r = rich
q = old
 Are axioms: describe knowledge about some world.
 In this world, is
F
u r(f(u))
always true ?
 How to prove such theorems in general?
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1) Proof by inconsistency
 Don’t prove F directly:
F
u r(f(u))
 But add the negation of F to the axioms and prove
that this extension is inconsistent.
 NEW TASK:
is inconsistent.
~ u r(f(u))
z ~ q(z)
y p(f(y))
x p(x)  q(x)  r(x)
 the 4 axioms are never true in 1 same interpretation.
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2) Clausal form:
= Normalize the formulae to a (more simple) standard form.
 Each set of axioms can be transformed into a new set of
formulae, that contains only formulae of the form:
xy…z p(…)  q(…)  … r(…)  t(…)  s(…)  …  u(…)
  only left;  only right
 no ~ ; no 
which is inconsistent if and only if the original set was
inconsistent.
Notice: “x y … z“ can be dropped.
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Example:
 ~ u r(f(u))
u false  r(f(u))
 z ~ q(z)
z false  q(z)
 y p(f(y)) : is already in clausal form: ( P  P  true)
 x p(x)  q(x)  r(x)
x q(x)  r(x)  p(x)
Ps: usually requires much more work!
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3) Unification:
 Given 2 atomic formulae:
 Ex.:
p(f(A),y)
p(x, g(x))
find their most general common instance.
 Ex.:
x must become: f(A)
g(x) must become: g(f(A))
y must become: g(f(A))
p(f(A), g(f(A)))
 Most general unifier (mgu) :
x -> f(A)
y -> g(f(A))
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4) The resolution step
 Proposition logic:
P Q
P
Q
QP
P  true
Q  true
P Q
~Q
~P
QP
false  Q
false  P
P1  P2  …  Pn  Q1  ...  Qm
R1  …  Rk  P1  S1  …  Sl
P2 …  Pn  R1  …  Rk  Q1  ...  Qm  S1  …  Sl
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De resolutie step (2):
 Predicate logic:
 Example:
mgu(p(x,f(A)), p(B,z)) =
x -> B
z -> f(A)
p(x,f(A))  q(g(x))
r(z)  p(B,z)
r(f(A))  q(g(B))
= mgu applied to r(z)  q(g(x))
Clauses on which resolution is performed
must not have any variables in common.
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5) Resolution proofs:
 In order to prove a set of clauses inconsistent:
 select 2 of them, for which resolution is possible
 apply resolution and add the result to the set
 if you obtain the clause false  : STOP !
This means inconsistency of the last set
AND inconsistency of the original set
AND that F was implied by T
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Example:
q(x)  r(x)  p(x)
false  r(f(u))
x -> f(u)
q(f(u))  p(f(u))
false  q(z)
z -> f(u)
false  p(f(u))
p(f(y))
y -> u
So: inconsistent !
false 
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A deeper study:
Ground Horn Logic
Horn Logic
Clausal Logic
Full Predicate Logic
Modus ponens
Unification
Resolution
Normalization
Horn clause logic
 All formulae in T are of the form:
x1 … xk
A  B1  B2 …  Bn
where A, B1, B2,…,Bn are atoms.
 An atom is a formula of the form p(t1,…,tm), with p a
predicate symbol and t1,…,tm terms.
 Horn clause formulae are universally quantified over
all variables that occur in them.
 B1,…,Bn are called body-atoms of the Horn clause; A
is the head of the Horn clause.
 n may be 0: in this case we say that the Horn clause
is a fact.
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Wich kind of formulae
can we prove?
 In Horn clause logic, we limit ourselves to prove
formulae F of the form:
x1 … xk B1  B2  …  Bn
where B1, B2, …, Bn are again atoms.
 All variables are existentially quantified !
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A very simple example:
 Bosmans is a showmaster
(1)
 Showmasters are rich
(2)
 Rich people have big houses
(3)
 Big houses need a lot of maintenance
(4)
Goal: automatically deduce that Bosmans’ house
needs a lot of maintenance.
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Representation in Horn logica:
 Bosmans is a showmaster
showmaster(Bosmans)
 Showmasters are rich
p rich(p)  showmaster(p)
 Rich people have big houses
(1)
(2)
(3)
p big(house(p))  rich(p)
 Big houses need a lot of maintenance
(4)
p lot_maint(house(p))  big(house(p))
 To prove:
Lot_maint(house(Bosmans))
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AR for ground
Horn clause logic
Backward reasoning proof procedures
based on generalized Modus Ponens
Restricting to ground
Horn clauses:
 So, for now: Horn clauses without variables:
 Example:
showmaster(Bosmans)
rich(Bosmans)  showmaster(Bosmans)
big(house(Bosmans))  rich(Bosmans)
lot_maint(house(Bosmans))  big(house(Bosmans))
 Prove:
lot_maint(house(Bosman))
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Easy with modus ponens !
 3 applications of modus ponens:
showm(Bos)
rich(Bos)  showm(Bos)
rich(Bos)
big(house(Bos))  rich(Bos)
big(house(Bos))
lot_maint(house(Bos))  big(house(Bos))
lot_maint(house(Bos))
gives the desired conclusion.
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Modus ponens in AR:
 Modus ponens is correct:
B
AB
A
For any interpretation making both B
and A  B true (= any model of {B , A
 B} )
A is also true in this interpretation
(see truth tables)
 Problem: how to organize this into a procedure which
is also complete (for ground Horn clauses)?
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Inconsistency:
 A theory T is inconsistent if it has NO model.
 Theorem:
Let T be a theory and F a formula.
T implies F if and only if T  {~F} is inconsistent.
 Proof:
T implies F
iff
iff
iff
iff
Each model of T makes F true
Each model of T makes ~F false
T  {~F} has no model
T  {~F} is inconsistent
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The extended example:
 Prove that the theory:
showm(Bos)
belg(Bos)
european(Bos)  belg(Bos)
rich(Bos)  showm(Bos)  european(Bos)
big(house(Bos))  rich(Bos)
lot_maint(house(Bos))  big(house(Bos))
~ lot_maint(house(Bos))
is inconsistent.
 Problem: this is NOT a Horn clause theory !?
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Refutation proofs:
the “false” predicate
 We introduce a new predicate symbol:
false
 We agree that false has the truth value ‘false’
under every interpretation.
 Imagine that we defined false as :
false  p  ~p
for some predicate p
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“definite” goals:
 In the Horn logic setting F has the form:
x1 … xm B1  B2  …  Bn
 So what is the form of ~F?
~(x1 … xm B1  B2  …  Bn)
 x1 … xm ~(B1  B2  …  Bn)
 x1 … xm false  ~(B1  B2  …  Bn)
 x1 … xm false  B1  B2  …  Bn
A  ~B  A  B
 Observe: ~F is again a Horn clause !!
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Back to the example
 The extended theory (to be proven inconsistent)
now is:
showm(Bos)
belg(Bos)
european(Bos)  belg(Bos)
rich(Bos)  showm(Bos)  european(Bos)
big(house(Bos))  rich(Bos)
lot_maint(house(Bos))  big(house(Bos))
false  lot_maint(house(Bos))
a ground Horn clause theory !
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Modus ponens
generalized:
A  B1  B2  …  Bi  …  Bn
Bi  C1  C2  …  Cm
A  B1  B2  …  C1  C2  …  Cm  …  Bn
 Ordinary Modus ponens is the special case with:
 n = i = 1 and m =0
 Correctness: via truth tables
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Some backward reasoning
steps in the example:
false  lot_maint(house(Bos))
lot_maint(house(Bos))  big(house(Bos))
false  big(house(Bos))
false  big(house(Bos))
big(house(Bos))  rich(Bos)
false  rich(Bos)
and so on ...
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The backward procedure:
the idea
 Convert F into a definite goal:
false  B1  B2  …  Bi  …  Bn
 Apply generalized modus ponens to the body-atoms Bi
of the goal, using the Horn clauses of T
until:
false 
is deduced.
 Then: a false formula ia a consequence of T  {~F}
we have proven inconsistency of T  {~F}
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Backwards procedure
Goal := false  B1  B2  …  Bn ;
Repeat
Select some Bi atom from the body of Goal
Select some clause Bi  C1  C2  …  Cm from T
Replace Bi in the body of Goal by C1  C2  …  Cm
Until Goal = false  or no more Selections possible
 On top of this you need to apply backtracking over
the selected clauses and the selected body atoms.
 If the algorithm stops because it has tried all these
alternatives: F was not implied!
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Back to the example
Step 0: Goal := false  lot_maint(house(Bos))
select: lot_maint(house(Bos))  big(house(Bos))
Step 1: Goal := false  big(house(Bos))
select: big(house(Bos))  rich(Bos)
Step 2: Goal := false  rich(Bos)
select: rich(Bos)  showm(Bos)  european(Bos)
Step 3: Goal := false  showm(Bos)  european(Bos)
select: showm(Bos)
Step 4: Goal := false  european(Bos)
select: european(Bos)  belg(Bos)
Step 5: Goal := false  belg(Bos)
select: belg(Bos)
Step 6: Goal := false 
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Another example
(propositional)
pqr
qt
qs
rn
ro
s
o
n
 Prove: p
 Observe: non-determinism on both atom selection
and on clause selection !
we only illustrate the clause selection here
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Search tree traversed by
the backward procedure
false  p
false  q  r
false  t  r
false  s  r
false  r
pqr
qt
qs
rn
ro
s
o
n
false  n
false  o
false 
false 
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The backward
procedure is efficient
 The proof is goal directed towards the theorem.
 no exploration of irrelevant rules
 Different search methodes can be used to traverse
this search tree.
 Atom-selection may influence efficiency too:
 ex.: by detecting a failing branch sooner
but has no impact on whether or not we find a solution
(in case there are only finitely many ground Horn clauses)
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Completeness:
 Example:
false  p
pp
p
(1)
(2)
 Possible derivations:
(2)
false 
false  p
(1)
false  p
(1)
false  p
 Is only complete if the search tree is
traversed using a complete search method.
(1)
……
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Representation-power of
ground Horn clauses
 Is  a subset of propositional logic.
 Example:
showm(Bos)
showm_Bos
big(house(Bos))
big_house_Bos
 In general, more expressive logics are needed.
 Essence: with variables, one formula may be equivalent
to a very large number of propositional formulae.
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