Knowledge Representation I

1
Horn Form
A sentence is in Horn form if and only if . It
is a conjunction of atomic sentences on the left
side, and a single atom on the right side i.e.
P ? Q ? R ? X A ? B ? P ? Q ? R ? X
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Overview
We plan to represent the world (or some subset
thereof, ie our domain) in FOL, and to TELL this
information into our Knowledge Base KB. We
will represent our problem (ie the statement we
want to prove) in FOL. We will sometimes call the
statement to be proved a goal. So proving a
statement is equivalent to seeing if the goal is
in the KB, or if not in it, can be deduced
(entailed) from it.
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The simplest example
Suppose our KB consists of P Q R
A simple goal is Q. Literally we are asking is
it the case that Q is true. In this case, the
answer is yes!
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Another simple example
Suppose our KB consists of P Q R
Now suppose the goal is P ? Q. We are
literally asking is it the case that P is true
and Q is true. In this case, the fact P ? Q is
not in our KB. But we can infer it by using the
And-Introduction rule.
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A more interesting example

• Suppose our KB consists of facts about
  containment
• General facts about containment
• ?x,y,z in(x,y) ? in(y,z) ? enclosedby(x,z)
• ?x,y in(x,y) ? enclosedby(x,y)
• Specific facts for our particular problem
• in(A,B) in(B,C) in(F,C)

Note, in means directly inside with nothing in
between, encloses means inside, possibly with
other items in between.
A
F
B
A goal might be ?x enclosedby(B,x) (ie, is
there any thing inside B?) or ?x,y,z
enclosedby(x,y) ? enclosedby(y,z)
C
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Models I
A model is simply a world in which a sentence
is true. Consider the sentence P ? Q , given
that P Means is tall. Q Means is rich. We
can now ask if the following interpretations of P
? Q are models Bill Gates is not a model of
P ? Q Eamonn Keogh is not a model of P ? Q
Michael Jordan is a model of P ? Q
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Models II
If we have the sentence P ? Q, and P Means is
astronaut. Q Means is female. Which of the
following interpretations of P ? Q are
models? Neil Armstrong, Eamonn Keogh, Sally
Ride, Dolly Parton, Bill Clinton, Valentina
Tereshkova,
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Models III
Models are important because they allow a simple
definition of entailment. A Knowledge Base KB
entails a sentence ? if and only if, all models
of KB are also models of ?. For example the KB
P ? Q can be said to entail the sentence P ?
Q, since all models of P ? Q are also models of P
? Q.
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In propositional logic, proof is easy I
Suppose we had P and P ? Q in our KB, and we have
a goal of Q We can always build a truth table
with an entry for each symbol and each item from
our KB
P ? Q P
Goal Q?
We need to look at all rows where the statements
in our KB are true
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In propositional logic, proof is easy I
Suppose we had P and P ? Q in our KB, and we have
a goal of Q We can always build a truth table
with an entry for each symbol and each item from
our KB
P ? Q P
Q?
We need to look at all rows where the statements
in our KB are true. Since in every case where our
KB statements are true, Q is also true, thus we
have proved Q! (I.e our KB entails Q) All
models of P ? Q , P are also models of Q.
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In FOL, proof is harder
Suppose we had P(A) and ?x P(x) ? Q(x) in our KB,
and we have a goal of Q(A) We can build a truth
table but we cannot fill it in.
?x P(x) ? Q(x) P(A)
Q(A)?
I.e. the general sentence never follows from any
of these instances. How are we to know if the KB
entails the sentence Q(A)?
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• In the next few slides we will look at some tools
  that will allow us to prove facts in First Order
  Logic
• This tools are
• Substitution
• Universal Elimination
• Existential Elimination
• Existential Introduction

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Substitution (Binding, Instantiation)
Substitution is a variable (or set of variables)
paired with a term. Example SUBSTx/Harpois a
substitution indicating that Harpo should be
substituted for x. Brother(x,Groucho)
becomes Brother(Harpo,Groucho) Example
SUBSTx/YoungestSonOF(Minnie)is a substitution
indicating that YoungestSonOF(Minnie) should be
substituted for x. Brother(x,Groucho)
becomes Brother(YoungestSonOF(Minnie),Grouch
o)
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Universal Elimination(Universal Instantiation)
?x ? SUBST(x/g, ?)
For any sentence ?, variable x, and term g.
For example, from ?x loves(x, Beatles) We can
use the substitution x/Nick to
infer loves(Nick, Beatles) Or can use the
substitution x/MotherOf(Joe) to
infer loves(MotherOf(Joe) , Beatles)

• Terms (A term is an expression that refers to
  an object).
• Either
• A symbol
• A function application
• Symbols can be constants John, Mary, 17, 56
• or variables F_name, L_name, X, Y
• Function applications are usually written as
  SomethingOf
• MotherOf(John)
• AgeOf(Mary)
• EyeColorOf(MotherOf(John))

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Existential Elimination(Existential
Instantiation)
For any sentence ?, variable x, and constant term
k. Where k does not already exist in our KB
?x ? SUBST(x/k, ?)
For example from ?x President(x, USA) We can
use substitution SUBSTx/someoldwhiteguy to
infer President(someoldwhiteguy, USA) Later on,
if we happen to learn the president is Bush we
can assign someoldwhiteguy Bush

• Terms (A term is an expression that refers to
  an object).
• Either
• A symbol
• A function application
• Symbols can be constants John, Mary, 17, 56
• or variables F_name, L_name, X, Y
• Function applications are usually written as
  SomethingOf
• MotherOf(John)
• AgeOf(Mary)
• EyeColorOf(MotherOf(John))

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Existential Introduction
For any sentence ?, term g that occurs in ? and x
that does not occur in ?.
? ?x SUBST(g/x, ?)
For example from President(Bush, USA) We
can infer ?x President(x, USA)
17

• We can now take these new tools
• Universal Elimination
• Existential Elimination
• Existential Introduction
• Combined with some tools we have already defined
• To prove facts in First Order Logic

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A sample proof in FOL
Bob is a buffalo Buffalo(Bob) Pat is a
pig Pig(Pat) Buffaloes outrun pigs ?x,y
Buffalo(x) ? Pig(y) ? Faster(x,y)
Can Bob outrun Pat? Faster(Bob,Pat)?
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A sample proof in FOL
Buffalo(Bob) Pig(Pat) ?x,y Buffalo(x)
? Pig(y) ? Faster(x,y) Buffalo(Bob) ?
Pig(Pat)
Can Bob outrun Pat? Faster(Bob,Pat)?
1. We can use And Introduction to add the
following to our database Buffalo(Bob) ?
Pig(Pat)
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A sample proof in FOL
Buffalo(Bob) Pig(Pat) ?x,y Buffalo(x)
? Pig(y) ? Faster(x,y) Buffalo(Bob) ?
Pig(Pat) Buffalo(Bob) ? Pig(Pat) ?
Faster(Bob, Pat)
Can Bob outrun Pat? Faster(Bob,Pat)?
1. We can use And Introduction to add the
following to our database Buffalo(Bob) ?
Pig(Pat) 2. We can do Universal Elimination to
get x/Bob, y/Pat Buffalo(Bob) ? Pig(Pat) ?
Faster(Bob, Pat)
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A sample proof in FOL
Buffalo(Bob) Pig(Pat) ?x,y Buffalo(x)
? Pig(y) ? Faster(x,y) Buffalo(Bob) ?
Pig(Pat) Buffalo(Bob) ? Pig(Pat) ?
Faster(Bob, Pat) Faster(Bob, Pat)
Can Bob outrun Pat? Faster(Bob,Pat)?
1. We can use And Introduction to add the
following to our database Buffalo(Bob) ?
Pig(Pat) 2. We can do Universal Elimination to
get x/Bob, y/Pat Buffalo(Bob) ? Pig(Pat) ?
Faster(Bob, Pat) 3. We then use Modus Ponens to
derive Faster(Bob, Pat)
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The need for a better proof mechanism
The proof mechanism we have just seen can be
imagined as search. The operators are inference
rules (Existential Introduction, Universal
Elimination, etc) The states are set of sentences
(The KB itself) The goal test is to see if a
state contains the query sentence.
This idea works, but It seriously slows down
when the KB gets larger, because the number of
possible substitutions grows exponentially. One
possible solution is to reduce the number of
operators This leads us to Generalized Modus
Ponens (GMP).
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Generalized Modus Ponens Combines
And-Introduction, Universal-Elimination, and
Modus Ponens
For atomic sentences p1, p2, and q, where there
is a substitution ? such that SUBST(?/ pi )
SUBST(?/ pi)
Example p1 Faster(Bob,Pat) p2
Faster(Pat,Steve) p1 ? p2 ? q
Faster(x,y) ? Faster(y,z) ? Faster(x,z)
SUBST(?/q) SUBST(x/Bob, y/Pat, z/Steve )
Then using GMP we can conclude
Faster(Bob,Steve) Note that GMP only works for a
KB of definite clauses (Exactly one positive
literal) either a single atom or (conjunction of
atomic sentences) ? atomic sentence Sentences in
this form are known as horn clauses. Note also
that all sentences are assumed to be universally
qualified
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Generalized Modus Ponens Combines
And-Introduction, Universal-Elimination, and
Modus Ponens
For atomic sentences p1, p2, and q, where there
is a substitution ? such that SUBST(?/ pi )
SUBST(?/ pi)
Example p1 Faster(Bob,Pat) p2
Faster(Pat,Steve) p1 ? p2 ? q
Faster(x,y) ? Faster(y,z) ? Faster(x,z)
SUBST(?/q) SUBST(x/Bob, y/Pat, z/Steve )
Then using GMP we can conclude
Faster(Bob,Steve) Faster(Bob,Pat) ?
Faster(Pat,Steve) ? Faster(Bob,Steve)
?
?
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Forward Chaining

• 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 KB All cats like fish, cats eat
  everything they like, and Ziggy is a cat. In FOL,
  KB
• ?x cat(x) ? likes(x, Fish)
• ?x ?y (cat(x) ? likes(x,y)) ? eats(x,y)
• cat(Ziggy)
• Goal query Does Ziggy eat fish?
• Proof
• Use GMP with (1) and (3) to derive 4.
  likes(Ziggy, Fish)
• Use GMP with (3), (4) and (2) to derive
  eats(Ziggy, Fish)
• So, Yes, Ziggy eats fish.

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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. Here,
• Proof
• 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)
• cat(Ziggy) matches axiom (3), so we've "solved"
  that sub-goal
• likes(Ziggy, Fish) matches the RHS of (1), so try
  and prove cat(Ziggy)
• cat(Ziggy) matches (as it did earlier) axiom (3),
  so we've solved this sub-goal
• There are no unsolved sub-goals, so we're done.
  Yes, Ziggy eats fish.

• ?x cat(x) ? likes(x, Fish)
• ?x ?y (cat(x) ? likes(x,y)) ? eats(x,y)
• cat(Ziggy)

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We have seen that we can use Generalized Modus
Ponens (GMP) combined with search to see if a
fact is entailed from a Knowledge
Base. Unfortunately, there are some facts that
we cannot prove using GMP Suppose we have in our
KB GoToSchool(Joe) ? GoToSchool(Sue)
?GoToSchool(Joe) ? GoToSchool(Sue) And we
want to know if GoToSchool(Sue) Although
humans can look at such a KB and immediately see
that GoToSchool(Sue) is true, GMP cannot entail
that fact!! So GMP is not complete.
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Is there any complete inference procedure for FOL?

• Yes! It is called resolution refutation (or just
  resolution).
• It is complete.
• It is sound.
• It is optimally efficient.
• It works for KB in any format unlike GMP which
  only works for KBs in Horn clause format
  (although we generally have to reorganize the
  format).

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How resolution refutation works I
Suppose our KB consists the following sentences
P ? Q ?P ? R ?Q ? R And the goal is to prove R
? S We negate the goal, so R ? S becomes ?(R ?
S) ?R ? ?S (by De Morgans law)
?R , ?S (And-Elimination) Now we add ?R , ?S to
our knowledge base and look for a
contradiction.
P ? Q ?P ? R ?Q ? R ?R ?S
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How resolution refutation works II
P ? Q ?P ? R ?Q ? R ?R ?S
P ? Q
Q ? P
?P ? R
Q ? R
?Q ? R
R
?R
R ??R Contradiction!
That's all there is to resolution for
propositional logic, for first order logic we
just have to put the KB in the right format,
otherwise it works the same.
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In order for resolution to work for FOL, the KB
must be in Conjunctive Normal Form (CNF)
Conjunctive Normal Form The KB is a conjunction
of disjunctions
?P(w) ? Q(w) P(x) ? R(x) ?Q(y) ? S(y) ?R(z)
? S(z)
?P(w) ? Q(w)
?Q(y) ? S(y)
SUBSTy/w
?P(w) ? S(w) S(w) ? ?P(w)
The Goal is S(A), so negate it and add it to the
KB
P(x) ? R(x)
SUBSTw/x
?P(w) ? Q(w) P(x) ? R(x) ?Q(y) ? S(y) ?R(z)
? S(z) ?S(A)
S(x) ? R(x)
?R(z) ? S(z)
SUBSTx/A, z/A
S(A)
?S(A)
Now lets look for a contradiction
We will address the problem of translating an
arbitrary KB into CNF in the next few slides.
S(A) ? ?S(A)
Contradiction!
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• We have seen that we can do resolution when the
  KB is in CNF.
• But most KB are not is that format.
• Theorem We can convert any KB to CNF
• Eliminate Implications
• Move ? inwards
• Standardize Variables
• Move Quantifiers left
• Skolemize Eliminate Existential Quantifiers,
  Drop Universal Quantifiers
• Distribute ? over ?
• Flatten nested conjunctions and disjunctions.
• Note that we must do the steps in order!

Gloss over!
33
Eliminate Implications Change all implication
sentences to disjunctions So A ? B becomes
?A ? B C ? ?D becomes ?C ? ?D etc
Gloss over!
34
Move ? inwards Negation is only allow on atom,
not whole sentences. So ?(A ? B) becomes ?A
? ?B ?(A ? B) becomes ?A ? ?B ??x P
becomes ?x ?P ??x P becomes ?x ?P
??P becomes P
Gloss over!
35
Standardize Variables Make sure that you are
not using the same variable name twice in a
single sentence (unless you really meant
to). So ( ?x P(x) ) ? ( ?x Q(x)
) becomes ( ?x P(x) ) ? ( ?z Q(z) )
Gloss over!
36
Move Quantifiers left Move all quantifiers left,
but keep them in order! So ?x P(x) ? ?y
Q(y) becomes ?x ?y P(x) ? Q(y)
Gloss over!
37
Skolemize Eliminate Existential Quantifiers,
Drop Universal Quantifiers
Existential quantifiers can be eliminated by the
introduction of a new constant that does not
appear elsewhere in the database. ?x P(x) ?
Q(x) becomes P(A) ? Q(A) SUBSTx/A One
possible complication occurs if we also have
Universal quantifiers Consider ?x person(x) ?
?y heart(y) ? has(x,y) Becomes by
SUBSTy/H ?x person(x) ? heart(H) ?
has(x,H) !!! Instead we have to create a new
(Skolem) function to map from a person to their
heart F(x) ?x person(x) ? heart(F(x)) ?
has(x,F(x))
Gloss over!
Drop Universal Quantifiers
38
Distribute ? over ? (A ? B) ? C becomes (A ?
C) ? (A ? B) Just like distribution in
arithmetic (5 4) 6 becomes (5 6 )
(4 6 )
Gloss over!
39
Flatten nested conjunctions and
disjunctions. (A ? B) ? C becomes (A ? B ?
C) (A ? B) ? C becomes (A ? B ? C)
Gloss over!
40
Done!!

• Eliminate Implications
• Move ? inwards
• Standardize Variables
• Move Quantifiers left
• Skolemize Eliminate Existential Quantifiers,
  Drop Universal Quantifiers
• Distribute ? over ?
• Flatten nested conjunctions and disjunctions.

Once we have done all the above steps, our KB is
in CNF. We can now do resolution
refutation. There is still an element of search,
we have to decide which pairs of sentences to
resolve, and which substitutions we should use.
The goal is simply to find a contradiction. There
are many heuristics for the search (Our textbook
lists Unit preference, Set of support, Input
resolution )
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Bad News

• Even with all the heuristics resolution is
  exponential in general.
• Resolution is complete, but only semidecidable.
• If our KB implies A (or it implies ?A),
  resolution will find the proof in finite time.
  But if the KB does not implies A (or ?A),
  resolution will simply run forever.