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Title: Explorations in Artificial Intelligence


1
Explorations in Artificial Intelligence
  • Prof. Carla P. Gomes
  • gomes_at_cs.cornell.edu
  • Module 3-1-2
  • Logic Based Reasoning
  • Proof Methods

2
Proofs Methods
3
Proof methods
  • Proof methods divide into (roughly) two kinds
  • Application of inference rules
  • Legitimate (sound) generation of new sentences
    from old
  • Proof a sequence of inference rule
    applications Can use inference rules as
    operators in a standard search algorithm
  • Different types of proofs
  • Model checking
  • truth table enumeration (always exponential in n)
  • improved backtracking, e.g., Davis--Putnam-Logeman
    n-Loveland (DPLL) (including some inference rules)
  • heuristic search in model space (sound but
    incomplete)
  • e.g., min-conflicts-like hill-climbing
    algorithms

4
Proof
  • The sequence of wffs (w1, w2, , wn) is called a
    proof (or deduction) of wn from
  • a set of wffs ? iff each wi in the sequence is
    either in ? or can be inferred from a
  • wff (or wffs) earlier in the sequence by using
    a valid rule of inference.
  • If there is a proof of wn from ?, we say that wn
    is a theorem of the set ?.
  • ? wn
  • (read wn can be proved or inferred from ?)
  • The concept of proof is relative to a particular
    set of inference rules used. If we
  • denote the set of inference rules used by R, we
    can write the fact that wn can be
  • derived from ? using the set of inference rules
    in R
  • ? R wn
  • (read wn can be proved from ? using the
    inference rules in R)

5
Propositional logic Rules of Inference or
Methods of Proof
  • How to produce additional wffs (sentences) from
    other ones? What steps can we
  • perform to show that a conclusion follows
    logically from a set of hypotheses?
  • Example
  • Modus Ponens
  • P
  • P ?Q
  • ______________
  • ? Q
  • The hypotheses are written in a column and the
    conclusions below the bar
  • The symbol ? denotes therefore. Given the
    hypotheses, the conclusion follows.
  • The basis for this rule of inference is the
    tautology (P ? (P ? Q)) ? Q)

6
Propositional logic Rules of Inference or
Methods of Proof
  • How to produce additional wffs (sentences) from
    other ones? What steps can we
  • perform to show that a conclusion follows
    logically from a set of hypotheses?
  • Example
  • Modus Ponens
  • P
  • P ? Q
  • ______________
  • ? Q
  • The hypotheses (premises) are written in a column
    and the conclusions below the bar
  • The symbol ? denotes therefore. Given the
    hypotheses, the conclusion follows.
  • The basis for this rule of inference is the
    tautology (P ? (P ? Q)) ? Q)
  • aside check tautology with truth table to make
    sure
  • In words when P and P ? Q are True, then Q must
    be True also. (meaning of
  • second implication)

7
Propositional logic Rules of Inference or
Methods of Proof
  • Example
  • Modus Ponens
  • If you study the CS 372 material ? You will
    pass
  • You study the CS372 material
  • ______________
  • ? you will pass
  • Nothing deep, but again remember the formal
    reason is that
  • ((P (P ? Q)) ? Q is a tautology.

8
Propositional logic Rules of Inference or
Method of Proof
Rule of Inference Tautology (Deduction Theorem) Name
P ? P ? Q P ? (P ? Q) Addition
P ? Q ? P (P ? Q) ? P Simplification
P Q ? P ? Q (P) ? (Q) ? (P ? Q) Conjunction
P P?Q ? Q (P) ? (P? Q) ? (P ? Q) Modus Ponens
? Q P ? Q ? ?P (?Q) ? (P? Q) ? ?P Modus Tollens
P ? Q Q ? R ? P? R (P?Q) ? (Q ? R) ? (P?R) Hypothetical Syllogism (chaining)
P ? Q ?P ? Q (P ? Q) ? (?P) ? Q Disjunctive syllogism
P ? Q ?P ? R ? Q ? R (P ? Q) ? (?P ? R) ? (Q ? R) Resolution
9
Valid Arguments
  • An argument is a sequence of propositions. The
    final proposition is called the conclusion of
    the argument while the other proposition are
    called the premises or hypotheses of the
    argument.
  • An argument is valid whenever the truth of all
    its premises implies the truth of its conclusion.
  • How to show that q logically follows from the
    hypotheses (p1 ? p2 ? ?pn)?

Show that
(p1 ? p2 ? ?pn) ? q is a tautology
One can use the rules of inference to show the
validity of an argument.
10
Proof Tree
  • Proofs can also be based on partial orders we
    can represent them using a tree structure
  • Each node in the proof tree is labeled by a wff,
    corresponding to a wff in the original set of
    hypotheses or be inferable from its parents in
    the tree using one of the rules of inference
  • The labeled tree is a proof of the label of the
    root node.

Example Given the set of wffs P, R,
P?Q Give a proof of Q ? R
11
Tree Proof
P, P? Q, Q, R, Q ? R
MP
Conj.
What rules of inference did we use?
12
Length of Proofs
  • Why bother with inference rules? We could always
    use a truth table
  • to check the validity of a conclusion from a set
    of premises.

But, resulting proof can be much shorter than
truth table method.
Consider premises p_1, p_1 ? p_2, p_2 ? p_3
p_(n-1) ? p_n To prove conclusion p_n
Inference rules Truth
table
n-1 MP steps
2n
Key open question Is there always a short proof
for any valid conclusion? Probably not. The NP
vs. co-NP question.
13
Beyond Propositional LogicPredicates and
Quantifiers
14
Predicates
  • Propositional logic assumes the world contains
    facts that are true or false.
  • But lets consider a statement containing a
    variable
  • x gt 3 since we dont know the value of x we
    cannot say whether the expression is true or
    false
  • x gt 3 which corresponds to x is greater than 3

Predicate, i.e. a property of x
15
  • x is greater than 3 can be represented as P(x),
    where P denotes greater than 3
  • In general a statement involving n variables x1,
    x2, xn can be denoted by
  • P(x1, x2, xn )
  • P is called a predicate or the propositional
    function P at the n-tuple (x1, x2, xn ).

16
When all the variables in a predicate are
assigned values ? Proposition, with a certain
truth value.
Predicate On(x,y) Propositions ON(A,B) is
False (in figure) ON(B,A) is True Clear(B)
is True
17
Variables and Quantification
  • How would we say that every block in the world
    has a property say clear ? We would have to
    say
  • Clear(A) Clear(B) for all the blocks (it may
    be long or worse we may have an infinite number
    of blocks)
  • What we need is
  • Quantifiers
  • ? Universal quantifier
  • ?x P(x)
  • - P(x) is
    true for all the values x in the universe of
    discourse
  • ? Existential quantifier
  • ?x P(x)
  • - there
    exists an element x in the universe of discourse
  • such that
    P(x) is true.

18
Universal quantification
  • Everyone at Cornell is smart
  • ?x At(x,Cornell) ? Smart(x)
  • Implicity equivalent to the conjunction of
    instantiations of P
  • At(Mary,Cornell) ? Smart(Mary)
  • ? At(Richard,Cornell) ? Smart(Richard)
  • ? At(John,Cornell) ? Smart(John)
  • ?

19
A common mistake to avoid
  • Typically, ? is the main connective with ?
  • Common mistake using ? as the main connective
    with ?
  • ?x At(x,Cornell) ? Smart(x)
  • means Everyone is at Cornell and everyone is
    smart

20
Existential quantification
  • Someone at Cornell is smart
  • ?x (At(x,Cornell) ? Smart(x))
  • ?x P(x) There exists an element x in the
    universe of discourse such that P(x) is true
  • Equivalent to the disjunction of instantiations
    of P
  • (At(John,Cornell) ? Smart(John))
  • ? (At(Mary,Cornell) ? Smart(Mary))
  • ? (At(Richard,Cornell) ? Smart(Richard))
  • ? ...

21
Another common mistake to avoid
  • Typically, ? is the main connective with
  • Common mistake using ? as the main connective
    with ?
  • ?x At(x,Cornell) ? Smart(x)
  • when is this true?

is true if there is anyone who is not at Cornell!
22
Quantified formulas
  • If ? is a wff and x is a variable symbol, then
    both ?x ? and ?x ? are
  • wffs.
  • x is the variable quantified over
  • ? is said to be within the scope of the
    quantifier
  • if all the variables in ? are quantified over in
    ?, we say that we have a closed wff or closed
    sentence.
  • Examples
  • ?x P(x) ? R(x)
  • ?x P(x)?(?y R(x,y) ? S(x))

23
Properties of quantifiers
  • ?x ?y is the same as ?y ?x
  • ?x ?y is the same as ?y ?x
  • ?x ?y is not the same as ?y ?x
  • ?x ?y Loves(x,y)
  • Everyone in the world loves at least one person
  • ?y ?x Loves(x,y)
  • Quantifier duality each can be expressed using
    the other
  • ?x Likes(x,IceCream) ??x ?Likes(x,IceCream)
  • ?x Likes(x,Broccoli) ??x ?Likes(x,Broccoli)
  • There is a person who is loved by everyone in
    the world ?

24
Statement When True When False
?x ?y P(x,y) ?y ?x P(x,y) P(x,y) is true for every pair There is a pair for which P(x.y) is false
?x ?y P(x,y) For every x there is a y for which P(x,y) is true There is an x such that P(x,y) is false for every y.
?x ?y P(x,y) There is an x such that P(x,y) is true for every y. For every x there is a y for which P(x,y) is false
?x ? y P(x,y) ?y ? x P(x,y) There is a pair x, y for which P(x,y) is true P(x,y) is false for every pair x,y.
25
Negation
Negation Equivalent Statement When is the negation True When is False
??x P(x) ?x ?P(x) For every x, P(x) is false There is an x for which P(x) is true.
? ?x P(x) ?x ?P(x) There is an x for which P(x) is false. For every x, P(x) is true.
26
Love Affairs Loves(x,y) x loves y
  • Everybody loves Jerry
  • ?x Loves (x, Jerry)
  • Everybody loves somebody
  • ?x ?y Loves (x, y)
  • There is somebody whom somebody loves
  • ?y ?x Loves (x, y)
  • Nobody loves everybody
  • ? ?x ?y Loves (x, y) ?x ?y ?Loves (x,
    y)
  • There is somebody whom Lydia doesnt love
  • ?y ?Loves (Lydia, y)

Note flipping quantifiers when moves in.
27
Love Affairscontinued
  • There is somebody whom no one loves
  • ?y ?x ?Loves (x, y)
  • There is exactly one person whom everybody loves
    (uniqueness)
  • ?y(?x Loves(x,y) ? ?z((?w Loves (w ,z)? zy))
  • There are exactly two people whom Lynn Loves
  • ?x ?y ((x?y) ? Loves(Lynn,x) Loves(Lynn,y) ?
  • ?z( Loves (Lynn ,z)? (zx ? zy)))
  • Everybody loves himself or herself
  • ?x Loves(x,x)
  • There is someone who loves no one besides herself
    or himself
  • ?x ?y Loves(x,y) ?(xy)
    (note biconditional )

28
  • Let Q(x,y) denote x?y 0 consider the domain
    of discourse the real
  • numbers
  • What is the truth value of
  • a) ?y ?x Q(x,y)?
  • b) ?x ?y Q(x,y)?

False
True (additive inverse)
29
  • The kinship domain
  • Brothers are siblings
  • ?x,y Brother(x,y) ? Sibling(x,y)
  • One's mother is one's female parent
  • ?m,c Mother(c) m ? (Female(m) ? Parent(m,c))
  • Sibling is symmetric
  • ?x,y Sibling(x,y) ? Sibling(y,x)

30
  • The set domain
  • Sets are empty sets or those made by adjoining
    something to a set
  • ?s Set(s) ? (s ) ? (?x,s2 Set(s2) ? s
    xs2)
  • The empty set has no element adjoined to it
  • ??x,s xs
  • Adjoining an element already in the set has no
    effect
  • ?x,s x ? s ? s xs
  • Only elements have been adjoined to it
  • ?x,s x ? s ? ?y,s2 (s ys2 ? (x y ? x ?
    s2))
  • Subset
  • ?s1,s2 s1 ? s2 ? (?x x ? s1 ? x ? s2)
  • Equality of sets
  • ?s1,s2 (s1 s2) ? (s1 ? s2 ? s2 ? s1)
  • Intersection
  • ?x,s1,s2 x ? (s1 ? s2) ?(x ? s1 ? x ? s2)
  • Uniion
  • ?x,s1,s2 x ? (s1 ? s2) ? (x ? s1 ? x ? s2)

31
Rules of Inference for Quantified Statements

(?x) P(x) ?P(c) Universal Instantiation
P(c) for an arbitrary c ?(?x) P(x) Universal Generalization
? ?(x) P(x) ? P(c) for some element c Existential Instantiation
P(c) for some element c ? ?(x) P(x) Existential Generalization
32
  • Example
  • Let CS372(x) denote x is taking CS372 class
  • Let CS(x) denote x has taken a course in CS
  • Consider the premises ?x (CS372(x) ? CS(x))
  • CS372(Ron)
  • We can conclude CS(Ron)

33
Arguments
  • Argument (formal)
  • Step Reason
  • 1 ?x (CS372(x) ? CS(x)) premise
  • 2 CS372(Ron) ? CS(Ron) Universal Instantiation
  • 3 CS372(Ron) Premise
  • 4 CS(Ron) Modus Ponens (2 and 3)

34
Example
  • Show that the premises
  • 1- A student in this class has not read the
    textbook
  • 2- Everyone in this class passed the first
    homework
  • Imply
  • Someone who has passed the first homework has not
    read the textbook

35
Example
  • Solution
  • Let C(x) x is in this class
  • T(x) x has read the textbook
  • P(x) x passed the first homework
  • Premises
  • ?x (Cx ? ?T(x))
  • ?x (C(x) ? P(x))
  • Conclusion we want to show ?x (P(x) ? ?T(x))

36
  • Step Reason
  • 1 ?x (Cx ??T(x))
    premise
  • 2 C(a) ? ?T(a) Existential Instantiation
    from 1
  • 3 C(a) Simplification 2
  • 4 ?x (C(x)?P(x))
    Premise
  • 5 C(a) ? P(a)
    Universal Instantiation from 4
  • 6 P(a) Modus ponens from 3 and 5
  • 7 ?T(a) Simplification from 2
  • 8 P(a) ?? T(a)
    Conjunction from 6 and 7
  • 9 ?x P(x) ??T(x) Existential generalization
    from 8

37
Resolution in Propositional Logic
38
Resolution (for CNF)
Very important inference rule several other
inference rules can be seen as special cases of
resolution.
Soundness of rule (validity of rule) (P ? Q) ?
(?P ? R) ? (Q ? R)

Resolution for CNF applied to a special type of
wffs conjunction of clauses. Literal either
an atom (e.g., P) or its negation (?P). Clause
disjunction of of literals (e.g., (P ? Q ?
?R)). Note Sometimes we use the notation of a
set for a clause e.g. P,Q,?R corresponds to
the clause (P?Q ??R) the empty clause
(sometimes written as Nil or ) is equivalent to
False
39
CNF
Conjunctive Normal Form (CNF) A wff is in CNF
format when it is a conjunction of disjunctions
of literals.
(?P ? Q ? R) ?(S ? P ? T ??R) ?(Q ? S)
Resolution for CNF applied to wffs in CNF
format.

? ? S1 ? ? ? S2 ? S1 ?
S2
Si- sets of literals i 1 ,2 ? atom
Resolution
Resolvent of the two clauses
atom resolved upon
40
ResolutionNotes
1 Rule of Inference Chaining
2 Rule of Inference Modus Ponens
41
ResolutionNotes
3 Unit Resolution
42
ResolutionNotes
4 No duplications in the resolvent set
only one instance of Q appears in the
resolvent, which is a set!
P ? Q ? R ? S ?P ? Q ? W ? Q ?
R ? S ? W
5 Resolving one pair at a time
DO NOT Resolve on Q and R
Resolving on Q
Resolving on R
True
43
ResolutionNotes
6 Same atom with opposite signs
False any set of wffs containing two
contradictory clauses is unsatisfiable. However,
a clause P, ?P is True.
44
Soundness of ResolutionValidity of the
Resolution Inference Rule
resolving on P
Validity (Tautology) (P ? Q) (?P ? R) ? (Q ?
R)
P Q R (P?Q) (?P?R) (P?Q)?(?P?R) (Q?R) (P ? Q) (?P ? R) ? (Q ? R)
0 0 0 0 1 0 0 1
0 0 1 0 1 0 1 1
0 1 0 1 1 1 1 1
1 0 0 1 0 0 0 1
1 1 0 1 0 0 1 1
1 0 1 1 1 1 1 1
0 1 1 1 1 1 1 1
0 0 0 0 1 0 0 1
45
Conversion to CNF
  • P ? (Q ? R)
  • 1.Eliminate ?, replacing a ? ß with (a ? ß)?(ß ?
    a).
  • (P ? (Q ? R)) ? ((Q? R) ? P)
  • 2. Eliminate ?, replacing a ? ß with ?a? ß.
  • (?P ? Q ? R) ? (?(Q? R) ? P)
  • 3. Move ? inwards using de Morgan's rules and
    double-negation
  • (?P? Q? R) ? ((?Q? ?R) ? P)
  • 4. Apply distributivity law (? over ?) and
    flatten
  • (?P ? Q ? R) ? (?Q? P) ? (?R ? P)

46
  • Converting DNF (Disjunctions of conjunctions)
    into CNF
  • 1 create a table each row corresponds to the
    literals in each conjunct
  • 2 - Select a literal in each row and make a
    disjunction of these literals

Example (P?Q ??R ) ?(S ?R ??P) ?(Q ?S ? P)
P Q ?R
S R ?P
Q S P
(P ? S ? Q) ? (P ? R ? Q)? (P ? ?P ? Q) (P ? S ?
S) (P ? R ? S) (P ? ?P ? S) (P ? ?P ? Q)
How many clauses?
47
ResolutionWumpus World
P?
  • P31 ? P2,2,
  • ?P2,2
  • ? P31

P?
48
Resolution Refutation
  • Resolution is sound but resolution is not
    complete e.g., (P? R) (P ? R) but
  • we cannot infer (P ? R) using resolution ?
  • we cannot use resolution directly to decide all
    logical entailments.
  • Resolution is Refutation Complete
  • We can show that a particular wff W is entailed
    from a given KB how?
  • Proof by contradiction
  • Write the negation of what we are trying to prove
    (?W) as a conjunction of clauses
  • Add those clauses (?W) to the KB (also a set of
    clauses), obtaining KB prove inconsistency for
    KB, i.e.,
  • Apply resolution to the KB until
  • No more resolvents can be added
  • Empty clause is obtained
  • To show that (P ? R) Res (P ? R) do (1) negate
    (P ? R), i.e. (?P)? (?R) (2) prove that
  • (P ? R) ? (?P)? (?R) is inconsistent

?!
?!
49
Propositional LogicProof by refutation or
contradiction
Satisfiability is connected to inference via the
following KB a if and only if (KB ??a) is
unsatisfiable One assumes ?a and shows that
this leads to a contradiction with the facts in KB
50
ResolutionRobot Domain
  • Example
  • BatIsOk
  • ?RobotMoves
  • BatIsOk ? BlockLiftable ?RobotMoves

Show that KB ?BlockLiftable
KB
  • BatIsOk ? ?BlockLiftable ? RobotMoves

BlockLiftable
  • BatIsOk
  • ?RobotMoves
  • BatIsOk ? ?BlockLiftable ? RobotMoves
  • BlockLiftable

?RobotMoves
KB
  • BatIsOk ? RobotMoves

BatIsOk
?BatIsOk
Nil
51
Resolution
  • Resolution is refutation complete (Completeness
    of resolution refutation)
  • If KB W, the resolution refutation procedure,
    i.e., applying resolution on KB, will produce
    the empty clause.
  • Decidability of propositional calculus by
    resolution refutation
  • If KB is a set of finite clauses and if KB W,
    then the resolution refutation procedure will
    terminate without producing the empty clause.
  • Ground Resolution Theorem
  • If a set of clauses is not satisfiable, then
    resolution closure of those clauses contains the
    empty clause.

In general, resolution for propositional logic
is exponential ?!
The resolution closure of a set of clauses W in
CNF, RC(W), is the set of all clauses derivable
by repeated application of the resolution rule to
clauses in W or their derivatives.
52
Resolution algorithm
  • Proof by contradiction, i.e., show KB??a
    unsatisfiable

Any complete search algorithm applying only the
resolution rule, can derive any conclusion
entailed by any knowledge base in propositional
logic resolution can always be used to either
confirm or refute a sentence refutation
completeness (Given A, its true we cannot use
resolution to derive A OR B but we can use
resolution to answer the question of whether A OR
B is true.)
53
Resolution exampleWumpus World
  • KB (B1,1 ? (P1,2? P2,1)) ?? B1,1 a
    ?P1,2

54
Resolution exampleWumpus World
  • KB (B1,1 ? (P1,2? P2,1)) ?? B1,1 a
    ?P1,2

KB (B11 ? (P1,2? P2,1)) ((P1,2? P2,1) ? B11)
?? B1,1 (?B11 ? P1,2? P2,1) (?(P1,2? P2,1) ?
B11) ?? B1,1 (?B11 ? P1,2? P2,1) ((? P1,2 ?
P2,1) ? B11)) ?? B1,1 (?B11 ? P1,2? P2,1) (?
P1,2 ? B11) (? P2,1 ? B11) ?? B1,1
55
Resolution Refutation Ordering Search
Strategies
  • Original clauses 0th level resolvents
  • Depth first strategy ?
  • Produce a 1st level resolvent
  • Resolve the 1st level resolvent with a 0th level
    resolvent to produce a 2nd level resolvent, etc.
  • With a depth bound, we can use a backtrack search
    strategy
  • Breadth first strategy ?
  • Generate all 1st level resolvents, then all 2nd
    level resolvents, etc.

Depth first strategy
56
Refinement Resolution Strategies
Set-of-support Resolution Strategy
  • Definitions
  • A clause ?2 is a descendant of a clause ?1 iif
  • Is a resolvent of ?1 with some other clause
  • Or is a resolvent of a descendant of ?1 with
    some other clause
  • If ?2 is a descendant of ?1, ?1 is an ancestor
    of ?2
  • Set-of-support set of clauses that are either
    clauses coming from the negation of the theorem
    to be proved or descendants of those clauses.
  • Set-of-support Strategy it allows only
    refutations in which one of the clauses being
    resolved is in the set of support.
  • Set-of-support Strategy is refutation complete.

Set-of-support Strategy
57
Refinement Strategies
  • Ancestry-filtered strategy allows only
    resolutions in which at least one member of the
    clauses being resolved either is a member of the
    original set of clauses or is an ancestor of the
    other clause being resolved
  • The ancestry-filtered strategy is refutation
    complete.

58
Refinement Strategies
  • Linear Input Resolution Strategy at least one
    of the clauses being resolved is a member of the
    original set of clauses (including the theorem
    being proved).
  • Linear Input Resolution Strategy is not
    refutation complete.

Example (P? Q) (?P? Q) (P ? ?Q) (?P ?
?Q) This set of clauses is inconsistent but
there is no linear-input refutation strategy
but there is a resolution refutation strategy
(P? Q)
(?P? Q)
(P ? ?Q) (?P ? ?Q)
Q
?Q
This is NOT Linear Input Resolution Strategy
Nil
59
Horn Clauses
60
Horn Clauses
  • Definition
  • A Horn clause is a clause that has at most one
    positive literal.
  • Examples
  • P P ? ?Q ? P ? ?Q ? P ? ?Q ? R

Types of Horn Clauses Fact single atom
e.g., P Rule implication, whose antecendent
is a conjunction of positive literals and
whose consequent consists of a single
positive literal e.g., P?Q ? R Head is R
Tail is (P?Q ) Set of negative literals - in
implication form, the antecedent is a
conjunction of positive literals and the
consequent is empty. e.g., P?Q ?
equivalent to ? P ? ?Q.
61
Forward chainingHORN (Expert Systems and Logic
Programming)
  • Horn Form (restricted)
  • KB conjunction of Horn clauses
  • Horn clause
  • proposition symbol or
  • (conjunction of symbols) ? symbol
  • E.g., C ? (B ? A) ? (C ? D ? B)
  • Modus Ponens (for Horn Form) complete for Horn
    KBs
  • a1, ,an, a1 ? ? an ? ß
  • ß
  • Can be used with forward chaining ?

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Forward ChainingDiagnosis systems
  • Example diagnostic system
  • IF the engine is getting gas and the engine turns
    over THEN the problem is spark plugs
  • IF the engine does not turn over and the lights
    do not come onTHEN the problem is battery or
    cables
  • IF the engine does not turn over and the lights
    come onTHEN the problem is starter motor
  • IF there is gas in the fuel tank and there is gas
    in the carburator
  • THEN the engine is getting gas

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Forward chaining(Data driven reasoning)
  • Idea fire any rule whose premises are satisfied
    in the KB,
  • add its conclusion to the KB, until query is found

AND-OR graph
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Forward chaining algorithm
  • Forward chaining is sound and complete for Horn KB

65
Count
Agenda
Inferred
A B
P gt Q 1 L and M gt P 2 B and L gt M 2 A and
P gt L 2 A and B gt L 2
P F L F M F B F A F
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Forward chaining example
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Forward chaining example
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Forward chaining example
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Forward chaining example
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Forward chaining example
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Forward chaining example
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Forward chaining example
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Forward chaining example
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Proof of completeness
  • FC derives every atomic sentence that is entailed
    by KB
  • FC reaches a fixed point where no new atomic
    sentences are derived
  • Consider the final state as a model m, assigning
    true/false to symbols
  • Every clause in the original KB is true in m
  • a1 ? ? ak ? b
  • Hence m is a model of KB
  • If KB q, q is true in every model of KB,
    including m

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Backward chaining
  • Idea work backwards from the query q
  • to prove q by BC,
  • check if q is known already, or
  • prove by BC all premises of some rule concluding
    q
  • Avoid loops check if new subgoal is already on
    the goal stack
  • Avoid repeated work check if new subgoal
  • has already been proved true, or
  • has already failed

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Backward chaining example
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Backward chaining example
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Backward chaining example
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Backward chaining example
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Backward chaining example
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Backward chaining example
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Backward chaining example
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Backward chaining example
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Backward chaining example
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Backward chaining example
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Forward vs. backward chaining
  • FC is data-driven, automatic, unconscious
    processing,
  • e.g., object recognition, routine decisions
  • May do lots of work that is irrelevant to the
    goal
  • BC is goal-driven, appropriate for
    problem-solving,
  • e.g., Where are my keys? How do I get into a PhD
    program?
  • Complexity of BC can be much less than linear in
    size of KB
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