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First-Order Logic (FOL)
aka. predicate calculus
Tuomas Sandholm
Carnegie Mellon University
Computer Science Department
First-Order Logic (FOL)
Syntax
• User defines these primitives:
– Constant symbols (i.e., the "individuals" in the world)
E.g., Mary, 3
– Function symbols (mapping individuals to individuals)
E.g., father-of(Mary) = John, color-of(Sky) = Blue
– Predicate symbols (mapping from individuals to truth values)
E.g., greater(5,3), green(Grass), color(Grass, Green)
First-Order Logic (FOL)
Syntax…
• FOL supplies these primitives:
– Variable symbols. E.g., x,y
– Connectives. Same as in PL: not (~), and (^),
or (v), implies (=>), if and only if (<=>)
– Quantifiers: Universal (A) and Existential (E)
Quantifiers
• Universal quantification corresponds to conjunction ("and") in that (Ax)P(x)
means that P holds for all values of x in the domain associated with that variable.
– E.g., (Ax) dolphin(x)
=> mammal(x)
• Existential quantification corresponds to disjunction ("or") in that (Ex)P(x)
means that P holds for some value of x in the domain associated with that variable.
– E.g., (Ex)
mammal(x) ^ lays-eggs(x)
• Universal quantifiers are usually used with "implies" to form "if-then rules."
– E.g., (Ax) cs15-381-student(x) => smart(x) means "All cs15-381 students are
smart."
– You rarely use universal quantification to make blanket statements about every
individual in the world: (Ax)cs15-381-student(x) ^ smart(x) meaning that everyone
in the world is a cs15-381 student and is smart.
Quantifiers …
• Existential quantifiers are usually used with "and" to specify a list of
properties or facts about an individual.
– E.g., (Ex) cs15-381-student(x) ^ smart(x) means "there is a cs15-381
student who is smart."
– A common mistake is to represent this English sentence as the FOL sentence:
(Ex) cs15-381-student(x) => smart(x)
• Switching the order of universal quantifiers does not change the meaning:
(Ax)(Ay)P(x,y) is logically equivalent to (Ay)(Ax)P(x,y). Similarly, you
can switch the order of existential quantifiers.
• Switching the order of universals and existentials does change meaning:
– Everyone likes someone: (Ax)(Ey)likes(x,y)
– Someone is liked by everyone: (Ey)(Ax)likes(x,y)
First-Order Logic (FOL)
Syntax…
• Sentences are built up of terms and atoms:
– A term (denoting a real-world object) is a constant symbol, a variable
symbol, or a function e.g. left-leg-of ( ). For example, x and f(x1, ...,
xn) are terms, where each xi is a term.
– An atom (which has value true or false) is either an n-place predicate
of n terms, or, if P and Q are atoms, then ~P, P V Q, P ^ Q, P => Q,
P <=> Q are atoms
– A sentence is an atom, or, if P is a sentence and x is a variable, then
(Ax)P and (Ex)P are sentences
– A well-formed formula (wff) is a sentence containing no "free"
variables. I.e., all variables are "bound" by universal or existential
quantifiers.
• E.g., (Ax)P(x,y) has x bound as a universally quantified variable, but y is
free.
Translating English to FOL
• Every gardener likes the sun.
(Ax) gardener(x) => likes(x,Sun)
• You can fool some of the people all of the time.
(Ex)(At) (person(x) ^ time(t)) => can-fool(x,t)
• You can fool all of the people some of the time.
(Ax)(Et) (person(x) ^ time(t) => can-fool(x,t)
• All purple mushrooms are poisonous.
(Ax) (mushroom(x) ^ purple(x)) => poisonous(x)
Translating English to FOL…
• No purple mushroom is poisonous.
~(Ex) purple(x) ^ mushroom(x) ^ poisonous(x)
or, equivalently,
(Ax) (mushroom(x) ^ purple(x)) => ~poisonous(x)
• There are exactly two purple mushrooms.
(Ex)(Ey) mushroom(x) ^ purple(x) ^ mushroom(y) ^ purple(y) ^ ~(x=y) ^
(Az) (mushroom(z) ^ purple(z)) => ((x=z) v (y=z))
• Deb is not tall.
~tall(Deb)
• X is above Y if X is on directly on top of Y or else there is a pile of one or
more other objects directly on top of one another starting with X and
ending with Y.
(Ax)(Ay) above(x,y) <=> (on(x,y) v (Ez) (on(x,z) ^
above(z,y)))
Inference Rules for FOL
• Inference rules for PL apply to FOL as well. For example, Modus Ponens, AndIntroduction, And-Elimination, etc.
• New sound inference rules for use with quantifiers:
– Universal Elimination
If (Ax)P(x) is true, then P(c) is true, where c is a constant in the domain of x. For
example, from (Ax)eats(Ziggy, x) we can infer eats(Ziggy, IceCream).
• The variable symbol can be replaced by any ground term, i.e., any constant symbol or
function symbol applied to ground terms only.
– Existential Introduction
If P(c) is true, then (Ex)P(x) is inferred.
• For example, from eats(Ziggy, IceCream) we can infer (Ex)eats(Ziggy, x).
• All instances of the given constant symbol are replaced by the new variable symbol. Note
that the variable symbol cannot already exist anywhere in the expression.
– Existential Elimination
From (Ex)P(x) infer P(c).
• For example, from (Ex)eats(Ziggy, x) infer eats(Ziggy, Cheese).
• Note that the variable is replaced by a brand new constant that does not occur in this or
any other sentence in the Knowledge Base. In other words, we don't want to accidentally
draw other inferences about it by introducing the constant. All we know is there must be
some constant that makes this true, so we can introduce a brand new one to stand in for
that (unknown) constant.
Inference Rules for FOL
• Paramodulation
– Given two sentences (P1 v ... v PN) and
(t=s v Q1 v ... v QM) where each Pi and
Qi is a literal and Pj contains a term t, derive
new sentence (P1 v ... v Pj-1 v Pj[s] v
Pj+1 v ... v PN v Q1 v ... v QM) where
Pj[s] means a single occurrence of the term t
is replaced by the term s in Pj
– Example: From P(a) and a=b derive P(b)
Inference Rules for FOL
• Generalized Modus Ponens (GMP)
– Combines And-Introduction, Universal-Elimination, and Modus Ponens
– E.g.: from P(c), Q(c), and (Ax)(P(x) ^ Q(x)) => R(x), derive R(c)
– In general, given atomic sentences P1, P2, ..., PN, and implication
sentence (Q1 ^ Q2 ^ ... ^ QN) => R, where Q1, ..., QN and R are atomic
sentences, and subst(Theta, Pi) = subst(Theta, Qi) for i=1,...,N, derive
new sentence: subst(Theta, R)
– subst(Theta, alpha) denotes the result of applying a set of substitutions
defined by Theta to the sentence alpha
– A substitution list Theta = {v1/t1, v2/t2, ..., vn/tn} means to replace
all occurrences of variable symbol vi by term ti.
• Substitutions are made in left-to-right order in the list.
• E.g.: subst({x/IceCream, y/Ziggy}, eats(y,x)) = eats(Ziggy,
IceCream)
Automated Inference in FOL
• Automated inference in FOL is harder than in PL because variables can
take on potentially an infinite number of possible values from their
domain. Hence there are potentially an infinite number of ways to
apply the Universal-Elimination rule of inference
• Goedel's Completeness Theorem says that FOL entailment is
semidecidable. That is, if a sentence is true given a set of axioms, there
is a procedure that will determine this. However, if the sentence is
false, then there is no guarantee that a procedure will ever determine
this. In other words, the procedure may never halt in this case.
– Goedel's Incompleteness Theorem says that in a slightly extended
language (that enables mathematical induction), there are entailed
sentences that cannot be proved
• The truth table method of inference is not complete for FOL because
the truth table size may be infinite
• Natural deduction is complete for FOL but is not practical for
automated inference because the "branching factor" in a search is too
large, caused by the fact that we would have to potentially try every
inference rule in every possible way using the set of known sentences
Incompleteness of the generalized
modus ponens inference rule
Backward chaining
Generalized Modus Ponens
Forward chaining
x
x
x
x
P(x) => Q(x)
P(x) => Q(x)
Q(x) => S(x)
R(x) => S(x)
 Cannot be converted to Horn
Want to conclude S(A)
S(A) is true if Q(A) or R(A) is true, and one of those
must be true because either P(A) or P(A)
Incomplete: there are entailed sentences that the
procedure cannot infer.
Generalized Modus Ponens in
Horn FOL
• Generalized Modus Ponens (GMP) is complete for
KBs containing only Horn clauses
– A Horn clause is a sentence of the form:
(Ax) (P1(x) ^ P2(x) ^ ... ^ Pn(x)) => Q(x)
where there are 0 or more Pi's, and the Pi's and Q are positive (i.e.,
un-negated) literals
– Horn clauses represent a subset of the set of sentences representable
in FOL. For example, P(a) v Q(a) is a sentence in FOL but is not a
Horn clause.
– Natural deduction using GMP is complete for KBs containing only
Horn clauses. Proofs start with the given axioms/premises in KB,
deriving new sentences using GMP until the goal/query sentence is
derived. This defines a forward chaining inference procedure
because it moves "forward" from the KB to the goal.
Example of forward chaining
•
Example: KB = All cats like fish, cats eat everything they like, and
Ziggy is a cat. In FOL, KB =
1.
2.
3.
•
(Ax) cat(x) => likes(x, Fish)
(Ax)(Ay) (cat(x) ^ likes(x,y)) => eats(x,y)
cat(Ziggy)
Goal query: Does Ziggy eat fish?
Proof: Data-driven
1.
Use GMP with (1) and (3) to derive: 4. likes(Ziggy,
2.
Use GMP with (3), (4) and (2) to derive eats(Ziggy,
3.
So, Yes, Ziggy eats fish.
Fish)
Fish)
Backward chaining
•
•
•
Backward-chaining deduction using GMP is complete for KBs containing only
Horn clauses. Proofs start with the goal query, find implications that would allow
you to prove it, and then prove each of the antecedents in the implication, continuing
to work "backwards" until we get to the axioms, which we know are true.
Example: Does Ziggy eat fish?
To prove eats(Ziggy, Fish), first see if this is known from one of the axioms
directly. Here it is not known, so see if there is a Horn clause that has the consequent
(i.e., right-hand side) of the implication matching the goal.
Proof: Goal Driven
1. Goal matches RHS of Horn clause (2), so try and prove new sub-goals cat(Ziggy)
and likes(Ziggy, Fish) that correspond to the LHS of (2)
2. cat(Ziggy) matches axiom (3), so we've "solved" that sub-goal
3. likes(Ziggy, Fish) matches the RHS of (1), so try and prove cat(Ziggy)
4. cat(Ziggy) matches (as it did earlier) axiom (3), so we've solved this sub-goal
5. There are no unsolved sub-goals, so we're done. Yes, Ziggy eats fish.
Resolution Procedure
(aka Resolution Refutation Procedure)
• Resolution procedure is a sound and complete inference procedure for FOL
(BFS is complete for Horn, DFS may go into infinite loops)
• Resolution procedure uses a single rule of inference: the Resolution Rule
(RR), which is a generalization of the same rule used in PL
• Resolution Rule for PL:
From sentence P1 v P2 v ... v Pn and sentence ~P1 v Q2 v ...
v Qm derive resolvent sentence: P2 v ... v Pn v Q2 v ... v Qm
– Examples
• From P and ~P v Q, derive Q (Modus Ponens)
• From (~P v Q) and (~Q v R), derive ~P v R
• From P and ~P, derive False
• From (P v Q) and (~P v ~Q), derive True
Resolution Procedure…
• Resolution Rule for FOL:
– Given sentence P1 v ... v Pn
– and sentence Q1 v ... v Qm
• where each Pi and Qi is a literal, i.e., a positive or negated predicate symbol
with its terms,
– if Pj and ~Qk unify with substitution list Theta,
– then derive the resolvent sentence:
subst(Theta, P1 v ... v Pj-1 v Pj+1 v ... v Pn v Q1 v
... Qk-1 v Qk+1 v ... v Qm)
• Example
– From clause P(x, f(a)) v P(x, f(y)) v Q(y)
– and clause ~P(z, f(a)) v ~Q(z),
– derive resolvent clause P(z, f(y)) v Q(y) v ~Q(z) using Theta
= {x/z}
Unification
• Unification is a "pattern matching" procedure that
takes two atomic sentences, called literals, as input,
and returns "failure" if they do not match and a
substitution list, Theta, if they do match.
– That is, unify(p,q) = Theta means subst(Theta, p)
subst(Theta, q) for two atomic sentences p and q.
– Theta is called the most general unifier (mgu)
=
• All variables in the given two literals are implicitly
universally quantified
• To make literals match, replace (universallyquantified) variables by terms
Unification Algorithm
procedure unify(p, q, theta)
Scan p and q left-to-right and find the first
corresponding terms where p and q "disagree" ; where p
and q not equal
If there is no disagreement, return theta ; success
Let r and s be the terms in p and q, respectively, where
disagreement first occurs
If variable(r) then theta = union(theta, {r/s})
unify(subst(theta, p), subst(theta, q), theta)
else if variable(s) then theta = union(theta, {s/r})
unify(subst(theta, p), subst(theta, q), theta)
else return "failure" end
Unification…
• Examples
Literal 1
Literal 2
Literal 3
parents(x, father(x),
mother(Bill))
parents(Bill,
father(Bill), y)
{x/Bill,
y/mother(Bill)}
parents(x, father(x),
mother(Bill))
parents(Bill,
father(y), z)
{x/Bill, y/Bill,
z/mother(Bill)}
parents(x, father(x),
mother(Jane))
parents(Bill,
father(y), mother(y))
Failure
Unification…
• Unify is a linear time algorithm that returns the most
general unifier (mgu), i.e., a shortest length
substitution list that makes the two literals match.
– (In general, there is not a unique minimum length
substitution list, but unify returns one of them.)
• A variable can never be replaced by a term containing
that variable. For example, x/f(x) is illegal. This
"occurs check" should be done in the above pseudocode before making the recursive calls.
Resolution Procedure…
•
•
•
•
•
•
Proof by contradiction method
Given a consistent set of axioms KB and goal sentence Q, we want to show that KB |=
Q. This means that every interpretation I that satisfies KB, satisfies Q. But we know
that any interpretation I satisfies either Q or ~Q, but not both. Therefore if in fact KB
|= Q, an interpretation that satisfies KB, satisfies Q and does not satisfy ~Q. Hence KB
union {~Q} is unsatisfiable, i.e., that it's false under all interpretations.
In other words, (KB |- Q) <=> (KB ^ ~Q |- False)
What's the gain? If KB union ~Q is unsatisfiable, then some finite subset is
unsatisfiable
Resolution procedure can be used to establish that a given sentence Q is entailed by
KB; however, it cannot, in general, be used to generate all logical consequences of a
set sentences. Also, the resolution procedure cannot be used to prove that Q is not
entailed by KB.
Resolution procedure won't always give an answer since entailment is only
semidecidable. And you can't just run two proofs in parallel, one trying to prove Q
and the other trying to prove ~Q, since KB might not entail either one.
Resolution Algorithm
procedure resolution-refutation(KB, Q)
;; KB is a set of consistent, true FOL sentences
;; Q is a goal sentence that we want to derive
;; return success if KB |- Q, and failure otherwise
KB = union(KB, ~Q)
while false not in KB do
pick 2 sentences, S1 and S2, in KB that contain
literals that unify (if none, return "failure“)
resolvent = resolution-rule(S1, S2)
KB = union(KB, resolvent)
return "success"
Resolution example (using PL sentences)
• From “Heads I win, tails you lose” prove that “I win”
• First, define the axioms in KB:
1. "Heads I win, tails you lose."
(Heads => IWin) or, equivalently, (~Heads v IWin)
(Tails => YouLose) or, equivalently, (~Tails v YouLose)
2. Add some general knowledge axioms about coins, winning, and losing:
(Heads v Tails)
or, equivalently, (~YouLose v IWin)
YouLose) or, equivalently, (~IWin v YouLose)
(YouLose => IWin)
(IWin =>
• Goal: IWin
Resolution example (using PL sentences)…
Sentence 1
Sentence 2
Resolvent
~IWin
~Heads v IWin
~Heads
~Heads
Heads v Tails
Tails
Tails
~Tails v YouLose
YouLose
YouLose
~YouLose v Iwin
IWin
IWin
~IWin
False
Problems yet to be addressed
• Resolution rule of inference is only applicable
with sentences that are in the form P1 v P2 v ...
v Pn, where each Pi is a negated or nonnegated
predicate and contains functions, constants, and
universally quantified variables, so can we convert
every FOL sentence into this form?
• Resolution strategy
– How to pick which pair of sentences to resolve?
– How to pick which pair of literals, one from each
sentence, to unify?
Converting FOL sentences to clause form
•
•
Every FOL sentence can be converted to a logically equivalent
sentence that is in a "normal form" called clause form
Steps to convert a sentence to clause form:
1.
2.
3.
Eliminate all <=> connectives by replacing each instance of the form (P
<=> Q) by expression ((P => Q) ^ (Q => P))
Eliminate all => connectives by replacing each instance of the form (P
=> Q) by (~P v Q)
Reduce the scope of each negation symbol to a single predicate by
applying equivalences such as converting
to P
–
~~P
–
~(P v Q)
–
–
–
to ~P ^
~(P ^ Q) to ~P v
~(Ax)P to (Ex)~P
~(Ex)P to (Ax)~P
~Q
~Q
Converting FOL sentences to clause form…
4.
Standardize variables: rename all variables so that each quantifier has its
own unique variable name. For example, convert (Ax)P(x) to (Ay)P(y)
if there is another place where variable x is already used.
Eliminate existential quantification by introducing Skolem functions.
For example, convert (Ex)P(x) to P(c) where c is a brand new constant
symbol that is not used in any other sentence. c is called a Skolem
constant. More generally, if the existential quantifier is within the scope
of a universal quantified variable, then introduce a Skolem function that
depends on the universally quantified variable.
5.
•
For example, (Ax)(Ey)P(x,y) is converted to (Ax)P(x, f(x)). f is called
a Skolem function, and must be a brand new function name that does not
occur in any other sentence in the entire KB.
Example: (Ax)(Ey)loves(x,y) is converted to (Ax)loves(x,f(x)) where in
this case f(x) specifies the person that x loves.
•
•
(If we knew that everyone loved their mother, then f could stand for the motherof function.)
Converting FOL sentences to clause form…
6.
Remove universal quantification symbols by first moving them all
to the left end and making the scope of each the entire sentence, and
then just dropping the "prefix" part. E.g., convert (Ax)P(x) to P(x)
7.
Distribute "and" over "or" to get a conjunction of disjunctions called
conjunctive normal form.
•
•
convert (P
convert (P
to (P
R to (P
^ Q) v R
v R) ^ (Q v R)
v Q) v
v Q v R)
8.
Split each conjunct into a separate clause, which is just a disjunction
("or") of negated and nonnegated predicates, called literals
9.
Standardize variables apart again so that each clause contains
variable names that do not occur in any other clause
Converting FOL sentences to clause form…
Examples:
•
Convert the sentence
•
(Ax)(P(x) => ((Ay)(P(y) => P(f(x,y))) ^ ~(Ay)(Q(x,y) => P(y))))
1.
Eliminate <=>
Nothing to do here.
2.
Eliminate =>
(Ax)(~P(x) v ((Ay)(~P(y) v P(f(x,y))) ^ ~(Ay)(~Q(x,y) v P(y))))
3.
Reduce scope of negation
(Ax)(~P(x) v ((Ay)(~P(y) v P(f(x,y))) ^ (Ey)(Q(x,y) ^ ~P(y))))
Converting FOL sentences to clause form…
4.
Standardize variables
(Ax)(~P(x) v ((Ay)(~P(y) v P(f(x,y))) ^ (Ez)(Q(x,z) ^ ~P(z))))
5.
Eliminate existential quantification
(Ax)(~P(x) v ((Ay)(~P(y) v P(f(x,y))) ^ (Q(x,g(x)) ^ ~P(g(x)))))
6.
Drop universal quantification symbols
(~P(x) v ((~P(y) v P(f(x,y))) ^ (Q(x,g(x)) ^ ~P(g(x)))))
7.
Convert to conjunction of disjunctions
(~P(x) v ~P(y) v P(f(x,y))) ^ (~P(x) v Q(x,g(x))) ^ (~P(x) v
~P(g(x)))
Converting FOL sentences to clause form…
8. Create separate clauses
–
–
–
~P(x) v ~P(y) v P(f(x,y))
~P(x) v Q(x,g(x))
~P(x) v ~P(g(x))
9. Standardize variables
–
–
–
~P(x) v ~P(y) v P(f(x,y))
~P(z) v Q(z,g(z))
~P(w) v ~P(g(w))
Example: Hoofers Club [example from Charles Dyer]
• Problem Statement: Tony, Shi-Kuo and Ellen belong to
the Hoofers Club. Every member of the Hoofers Club is
either a skier or a mountain climber or both. No mountain
climber likes rain, and all skiers like snow. Ellen dislikes
whatever Tony likes and likes whatever Tony dislikes.
Tony likes rain and snow.
• Query: Is there a member of the Hoofers Club who is a
mountain climber but not a skier?
Example: Hoofers Club…
•
•
Translation into FOL Sentences
Let S(x) mean x is a skier, M(x) mean x is a mountain climber, and L(x,y)
mean x likes y, where the domain of the first variable is Hoofers Club
members, and the domain of the second variable is snow and rain. We can
now translate the above English sentences into the following FOL wffs:
1.
2.
3.
4.
5.
6.
(Ax) S(x) v M(x)
~(Ex) M(x) ^ L(x, Rain)
(Ax) S(x) => L(x, Snow)
(Ay) L(Ellen, y) <=> ~L(Tony, y)
L(Tony, Rain)
L(Tony, Snow)
7.
8.
Query: (Ex) M(x) ^ ~S(x)
Negation of the Query: ~(Ex) M(x) ^ ~S(x)
Example: Hoofers Club…
•
Conversion to Clause Form
1.
2.
3.
4.
5.
6.
7.
8.
S(x1) v M(x1)
~M(x2) v ~L(x2, Rain)
~S(x3) v L(x3, Snow)
~L(Tony, x4) v ~L(Ellen, x4)
L(Tony, x5) v L(Ellen, x5)
L(Tony, Rain)
L(Tony, Snow)
Negation of the Query: ~M(x7) v S(x7)
Write these on blackboard
Example: Hoofers Club…
• Resolution Refutation Proof
Clause 1
Clause 2
Resolvent
MGU (i.e.,
Theta)
8
1
9. S(x1)
{x7/x1}
9
3
10. L(x1, Snow)
{x3/x1}
10
4
11. ~L(Tony, Snow)
{x4/Snow, x1/Ellen}
11
7
12. False
{}
Example: Hoofers Club…
• Answer Extraction
Clause 1
Clause 2
Resolvent
MGU (i.e.,
Theta)
~M(x7) v S(x7) v
(M(x7) ^ ~S(x7))
1
9. S(x1) v (M(x1) ^ ~S(x1))
{x7/x1}
9
3
10. L(x1, Snow) v (M(x1) ^
~S(x1))
{x3/x1}
10
4
11. ~L(Tony, Snow) v (M(Ellen) ^
~S(Ellen))
{x4/Snow,
x1/Ellen}
11
7
12. M(Ellen) ^ ~S(Ellen)
{}
• Answer to the query:
Ellen!
Resolution procedure as search
• Resolution procedure can be thought of as the bottom-up
construction of a search tree, where the leaves are the
clauses produced by KB and the negation of the goal.
When a pair of clauses generates a new resolvent clause,
add a new node to the tree with arcs directed from the
resolvent to the two parent clauses. The resolution
procedure succeeds when a node containing the False
clause is produced, becoming the root node of the tree.
• A strategy is complete if its use guarantees that the empty
clause (i.e., false) can be derived whenever it is entailed
Some strategies for controlling
resolution's search
•
Breadth-First
– Level 0 clauses are those from the original axioms and the negation of the goal. Level
k clauses are the resolvents computed from two clauses, one of which must be from
level k-1 and the other from any earlier level.
– Compute all level 1 clauses possible, then all possible level 2 clauses, etc.
– Complete, but very inefficient.
•
Set-of-Support
– At least one parent clause must be from the negation of the goal or one of the
"descendents" of such a goal clause (i.e., derived from a goal clause)
– Complete (assuming all possible set-of-support clauses are derived)
•
Unit Resolution
– At least one parent clause must be a "unit clause," (clause containing a single literal)
– Not complete in general, but complete for Horn clause KBs
•
Input Resolution
– At least one parent from the original KB (axioms and the negation of the goal)
– Not complete in general, but complete for Horn clause KBs
•
Linear Resolution
– P and Q can be resolved together if either P is in the original KB or if P is an ancestor
of Q in the proof tree
– Complete
Subsumption
• Motivation: Helps keep the KB small =>
less search
• Never keep more specific (“subsumed”)
sentences in the KB than more general ones
• E.g., if the KB has P(x), then eliminate
– P(c)
– P(a) v P(b)