Logic for knowledge representation

1
Logic for knowledge representation

• Points
• Propositional logic
• Predicate logic
• The pros and cons
• Syntax
• Semantics
• Inference
• Soundness and completeness
• The clausal form
• The conversion procedure
• Resolution
• Control of reasoning

2
Propositional logic

• We need to represent properties of objects in the
  world we model, and relations between objects.
• Propositional logic is not expressive enough. It
  allows no generality and no generalization just
  try to express the fact dogs have tails.
• This is the simplest form of logic. It has only
  one advantage there is an effective procedure
  for proving theorems in it. That is,
  propositional logic is decidable.

3
Predicate logic

• First-order predicate logic gives us much of what
  we need its various extensions offer a wide
  spectrum of knowledge representation
  possibilities.

• Predicate logic has a few obvious advantages
• it has been known and applied in practice for
  some 2500 years
• it has well-defined reasoning methods
• it is based on a solid theory that gives us for
  free the semantics of this knowledge
  representation method.

4
Predicate logic (2)

• There also are drawbacks of predicate logic
• it offers no support for hierarchical
  organization (e.g., no object orientation)
• it is absolute (frozen in time) -- there are
  problems with representing typical situations and
  objects, and expressing defaults
• it does not allow common-sense reasoning or
  approximate reasoning
• it has only undecidable or semidecidable proof
  procedures (but this is true of other
  representation methods).

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Predicate logic syntax -- a refresher
A very brief review the syntax (in a Prolog-style
notation). In an atom, the predicate denotes the
general nature of things, while the arguments
(terms) represent the details. Example onMarket(h
ouse(address(ottawa,main,10),
bedrooms(4)), price(225000)) Variables
allow us to generalize, for example onMarket(hous
e(A, B), price(P))
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... syntax -- a refresher (2)
The logical connectives ?, ?, ?, ?, ? produce
formulae out of formulae. Quantifiers allow us
to make general statements about classes of
individuals ?A ?N ( onMarket(house(A,
bedrooms(4)), price(N)) ? N lt
190000 ? makeOfferFor(A))
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First order?
First order means that we can only reason about
sets of individuals, that is, only individuals
can be quantified. Predicates are fixed. In
higher-order logics variables are allowed in
predicate positions. Two examples ?X ?Y
X(Y) each property has at least one instance ?X
(X(a) ? X(b)) a and b have the same properties
8
Predicate logic semantics
Quantifiers, logical connectives, atoms
(predicates, arguments), terms (functors,
constants, variables) are the syntactic elements
of first-order logic. They must be given a
semantic interpretation if we want them to
represent the world. Formally, the
termaddress(ottawa, main, 10)need not represent
a house. The symbols address, ottawa, main could
denote anything.
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Predicate logic semantics (2)
We interpret logical formulae by mapping their
elements into a domain. A term without variables
corresponds to an object in the domain. An atom
without variables corresponds to a true or false
statement about domain objects. For
example ottawa corresponds to the city of
Ottawa, main corresponds to Main Street in
Ottawa, address(ottawa, main, 10) corresponds to
a valid Ottawa address, onMarket(house(address(ott
awa, main, 10), bedrooms(4)), price(225000))
denotes a true or false statement about a
specific house.
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Predicate logic semantics (3)
Let S be a set of logical formulae that represent
an AI problem. Atoms in S correspond to true or
false statements about the problem. A model of S
is any subset of atoms -- all those that we
consider true. All atoms not in the model are by
definition false. A formula with logical
connectives has its truth value derived from the
values of its components. For example, S1 ? S2 is
true if both S1 and S2 are in the model. (This is
a hugely simplified approach to models in logic.
?)
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Predicate logic semantics (4)
A tautology is something that is true in every
model. A classic example is this formula true for
any H, P onMarket(H, P) ? ? onMarket(H, P) Two
formulae are model-equivalent if, in every model,
they are either both true or both false. Formula
F entails formula ? if ? is true in every model
in which F is true. This is written as F
? For example, (onMarket(H, P) ? ? sold(H)) ?
sold(H) ? onMarket(H, P)
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Inference

• Entailment allows us to represent certain aspects
  of rational behaviour
• a logically thinking agent draws conclusions from
  facts, and so in some way derives new knowledge.
• We need a procedure that will implement this kind
  of logical thinking.
• Inference is such a procedure, especially if it
  has two important properties
• it derives only good knowledge
• it derives all knowledge that can be derived.

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Inference rules

1. Tableaux-based methods. (Try to construct a model
   that makes a given formula F false. If we cannot
   construct any such model, we have proven F.)
2. The resolution rule (much more about it soon).
3. Natural deduction.

Rule Given 1 Given 2 Inferred
modus ponens ? ? ? ? ?
modus tollens ? ? ? ? ? ? ?
negation (1) ? ? ? ?
negation (2) ? ? ? ?
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Inference rules (2)
Rule Given 1 Given 2 Inferred
conjunction introduced ? ? ? ? ?
conjunction eliminated ? ? ? ?
disjunction introduced ? ? ? ?
disjunction eliminated ? ? ? proof of ? from ? proof of ? from ? ?
Rule Given Inferred
conditional proof proof of ? from ? assuming ? ? ? ? assuming ?
proof by contradiction proof of ? ? ?? from ?? assuming ? ? assuming ?
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Good knowledge, all knowledge

• The notation F - ? means we can infer ? from
  F.
• An inference rule is sound if an inferred formula
  is also entailed
• If F - ? then F ?
• An inference rule is complete if every entailed
  formula can be inferred
• If F ? then F - ?
• We can say that, for every F and ?, a sound and
  complete inference rule ensures
• F - ? if and only if F ?
• or, even more neatly

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Resolution

• We begin with an advertisement...
• The resolution proof procedure is attractive one
  inference rule plus unification (to be explained
  soon).
• The resolution proof procedure is practical
  Prolog is based on resolution (meta-interpretation
  helps enrich the procedure with explanations).
• Downside the clausal form, required by
  resolution, is less natural than full first-order
  logic (but intuitive!), and the proof procedure
  is still costly unless we restrict the form of
  clauses.

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Resolution (2)
The clausal form

• A literal is either an atom, or a negated atom.
• A clause is a disjunction of literals
• A1 ? A2 ? ... ? An ?
• B1 ? B2 ? ... ? Bm
• In Prolog, only Horn clauses are allowed. A Horn
  clause has at most one positive literal.
• For example, the following Prolog clause
• a(X, Y) - b(X, Z), c(Z, Y).
• is really
• ?X ?Y ?Z (b(X, Z) ? c(Z, Y) ? a(X, Y))
• or, in the clausal form,
• a(X, Y) ? b(X, Z) ? c(Z, Y)

18
Resolution (3)
Unification

• Boolean function UNIFY(term T1, term T2)
• if ( T1 is a variable )
• T1 ? T2 / bind /
• elsif ( T2 is a variable )
• T2 ? T1 / bind /
• elsif ( T1 and T2 have different main functors
  )
• return false
• else
• Boolean unifiable true
• while ( unifiable, unmatched arguments
  remain )
• unifiable UNIFY(next argument of T1,
• next argument of T2)
• return unifiable

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Resolution (4)
Unification -- comments

• Variable bindings are accumulated on an initially
  empty list. If the terms are unifiable, the list
  is our result if they are not unifiable, no
  bindings will be performed.
• This unification procedure may create circular
  structures, as it does in Prolog
• ?- X f(X).
• X f(f(f(f(f(f(f(f(f(f(...))))))))))
• Yes
• We could check that, in Var ? Expr, Expr does
  not contain Var. The cost of this occurs-check
  is too high for practical systems such as Prolog.

20
Resolution (5)
Converting to clauses (i)

• Every first-order logic formula F can be
  converted into clauses in a model-equivalent way
  if F is mapped into a true statement about some
  world, then Fs clausal form is also true.
• We will show the conversion procedure on an
  example (see the textbook), probably with too
  many parentheses
• ?X ((a(X) ? b(X)) ?
• (c(X, k) ?
• ?Y (?Z c(Y, Z) ? d(X, Y)))
• ) ?
• ?X e(X)

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Resolution (6)
Converting to clauses (ii)

• Step 1 eliminate implication.
• ?????? ? ???????
• ?X (?(a(X) ? b(X)) ?
• (c(X, k) ?
• ?Y (?Z ?c(Y, Z) ? d(X, Y)))
• ) ?
• ?X e(X)

22
Resolution (7)
Converting to clauses (iii)

• Step 2 move negation inward.
• ?????? ??
• ?(??? ?)?????(???????) ?(??? ?)?????(???????)
• ?? X ??????? X ?? ?? X ??????? X ??
• ?X ((?a(X) ? ?b(X)) ?
• (c(X, k) ?
• ?Y (?Z ?c(Y, Z) ? d(X, Y)))
• ) ?
• ?X e(X)

23
Resolution (8)
Converting to clauses (iv)

• Step 3 rename variables bound by independent
  (not nested) quantifiers.
• ?X ((?a(X) ? ?b(X)) ?
• (c(X, k) ?
• ?Y (?Z ?c(Y, Z) ? d(X, Y)))
• ) ?
• ?W e(W)
• Step 4 move quantifiers to the front.
• ?X ?Y ?Z ?W
• ((?a(X) ? ?b(X)) ?
• (c(X, k) ?
• (?c(Y, Z) ? d(X, Y)))
• ) ? e(W)

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Resolution (9)
Converting to clauses (v)
Step 5 skolemize (eliminate existential
quantifiers). ?A ... ?B ?C ?(A, ..., B, C) ? ?A
... ?B ?(A, ..., B, s(A, ..., B)) ?X ?W ((?a(X)
? ?b(X)) ? (c(X, k) ? (?c(f(X), g(X)) ?
d(X, f(X)))) ) ? e(W)
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Resolution (10)
Converting to clauses (vi)
Step 6 drop the prefix (all variables are
assumed to be universally quantified by
default). ((?a(X) ? ?b(X)) ? (c(X, k) ?
(?c(f(X), g(X)) ? d(X, f(X)))) ) ? e(W) Eliminate
parentheses, assuming the usual precedence. ?a(X)
? ?b(X) ? c(X, k) ? (?c(f(X), g(X)) ? d(X,
f(X))) ? e(W)
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Resolution (11)
Converting to clauses (vii)
Step 7 form a conjunction of disjunctions. ??????
????????????????)???(?????) (?a(X) ? ?b(X) ?
c(X, k) ? e(W)) ? (?a(X) ? ?b(X) ? ?c(f(X),
g(X)) ? d(X, f(X)) ? e(W))
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Resolution (12)
Converting to clauses (viii)
Step 8 split the conjunction into clauses (by
convention, they are treated as conjoined).
?a(X) ? ?b(X) ? c(X, k) ? e(W) ?a(X) ? ?b(X) ?
?c(f(X), g(X)) ? d(X, f(X)) ? e(W) Rewrite
into a neat clausal form with arrows. a(X), b(X)
? c(X, k), e(W) a(X), b(X), c(f(X), g(X)) ?
d(X, f(X)), e(W)
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Resolution (13)

• Lets see where this long story got us. We
  started here
• ?X ((a(X) ? b(X)) ?
• (c(X, k) ?
• ?Y (?Z c(Y, Z) ? d(X, Y)))
• ) ?
• ?X e(X)
• We ended up here
• (a(X) ? b(X) ? c(X, k) ? e(W))
• ?
• (a(X) ? b(X) ? c(f(X), g(X)) ? d(X, f(X)) ?
  e(W))

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Resolution (14)
The proof procedure (i)
Deduction is based on a single inference
rule. Let all Am, Bn be literals. Given are two
clauses A1 ?...? Ai-1 ? Ai ? Ai1 ?...? AM B1
?...? Bk-1 ? Bk ? Bk1 ?...? BN where Ai and Bk
are unifiable.
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Resolution (15)
The proof procedure (ii)
Unification is applied to both complete clauses,
the two unifiable literals are "cancelled out",
and a new clause is formed. A1 ?...? Ai-1 ? Ai1
?...? AM ? B1 ?...? Bk-1 ? Bk1 ?...? BN This is
the resolvent of the two given clauses. There is
one clause with a special meaning the empty
clause , containing no disjuncts. It is
interpreted as always false.
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Resolution (16)
The proof procedure (iii)
Here is how we prove a formula ?, given a set S
of axioms and theorems (we actually show that ?
is also a theorem) convert the given axioms and
theorems in S into the clausal form CS, convert
?? into the clausal form C?, try to deduce a
contradiction from CS ? C?, that is, to derive
the empty clause this is done by repeatedly
picking pairs of clauses and applying the
resolution inference rule.
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Resolution (17)
The proof procedure (iv)
Now, if the original theory S can produce ?, we
should be able to derive it somehow given also
??, we get exactly the empty clausethis proves ?
by contradiction. The order in which resolution
develops is dictated by the proof algorithm, or
by the operation of selecting two clauses that
give the next resolvent. Selection strategies can
lead to efficient proofs, but there is in general
no guarantee that a proof can even be found.
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Resolution (18)
An example (i)
Someone is happy if he passes a history exam and
wins a lottery. You pass exams if you study or if
you are lucky. You win a lottery if you are lucky
(studying does not help). John is lucky, but he
does not study. Is he happy?

• ?X pass(X, history) ?win(X, lottery) ? happy(X)
• ?X ?Y study(X) ? lucky(X)? pass(X, Y)
• ?X lucky(X) ? win(X, lottery)
• lucky(john)
• study(john)

34
Resolution (19)
An example (ii)
The same in the clausal form (the last clause
translates our question).

1. pass(X, history) ? win(X, lottery) ? happy(X)
2. study(Y) ? pass(Y, Z)
3. lucky(W) ? pass(W, V)
4. lucky(U) ? win(U, lottery)
5. lucky(john)
6. study(john)
7. happy(john)

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Resolution (20)
An example (iii)
pass(X, history) ? win(X, lottery) ?
happy(X) study(Y) ? pass(Y, Z) lucky(W) ?
pass(W, V) lucky(U) ? win(U, lottery) lucky(john
) study(john) happy(john)
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How to control resolution?

• The order in which we select clauses for the next
  resolution step decides how fast we reach the
  empty clause (if we ever do).
• There is a number of strategies... but we have to
  skip this fascinating topic ?.
• Please read section 13.2.4 of the textbook if you
  want to find out more.