AI3391 Artificial intelligence Unit IV Notes _ merged.pdf

AI3391
ARTIFICAL INTELLIGENE
B.TECH
II YEAR – III SEM (R18)
(2023-2024)
Prepared
By
Asst.Prof.M.Gokilavani
Department of Artificial Intelligence and Data Science

DEPARTMENT OF ARTIFICIAL INTLLIGENCE AND DATA SCIENCE

UNIT – IV LOGICAL REASONING
Knowledge based agent- Propositional Logic- Propositional theorem proving- Propositional model checking –
agent based on Propositional. First-Order Logic-Syntax and Semantics – Knowledge representation and
Engineering- Inference in First-Order Logic-Forward Chaining-Backward Chaining-Resolution.
LOGIC AND KNOWLEDGE REPRESENTATION:
1. INTRODUCTION:
• In the topic of Propositional logic, we have seen that how to represent statements using propositional
logic.
• But unfortunately, in propositional logic, we can only represent the facts, which are either true or false.
• PL is not sufficient to represent the complex sentences or natural language statements.
• The propositional logic has very limited expressive power.
• Consider the following sentence, which we cannot represent using PL logic.
"Some humans are intelligent", or
"Sachin likes cricket."
• To represent the above statements, PL logic is not sufficient, so we required some more powerful logic,
such as first-order logic.
FIRST-ORDER LOGIC:
 First-order logic is another way of knowledge representation in artificial intelligence. It is an extension
to propositional logic.
 FOL is sufficiently expressive to represent the natural language statements in a concise way.
 First-order logic is also known as Predicate logic or First-order predicate logic. First-order logic is a
powerful language that develops information about the objects in a more easy way and can also express
the relationship between those objects.
 First-order logic (like natural language) does not only assume that the world contains facts like
propositional logic but also assumes the following things in the world:
i. Objects: A, B, people, numbers, colors, wars, theories, squares, pits, Wumpus,
ii. Relations: It can be unary relation such as: red, round, is adjacent, or n-any relation such
as: the sister of, brother of, has color, comes between
iii. Function: Father of, best friend, third inning of, end of,
o As a natural language, first-order logic also has two main parts:
a. Syntax
b. Semantics

2. SYNTAX OF FIRST-ORDER LOGIC:
 The syntax of FOL determines which collection of symbols is a logical expression in first-order logic.
The basic syntactic elements of first-order logic are symbols. We write statements in short-hand notation
in FOL.
Example: John likes all kind of food
Conversion of Facts into FOL.
Basic Elements of First-order logic:
Following are the basic elements of FOL syntax:
i. Atomic sentences:
 Atomic sentences are the most basic sentences of first-order logic. These sentences are formed from a
predicate symbol followed by a parenthesis with a sequence of terms.
 We can represent atomic sentences as Predicate (term1, term2... term n).

ii. Complex Sentences:
o Complex sentences are made by combining atomic sentences using connectives.
First-order logic statements can be divided into two parts:
o Subject: Subject is the main part of the statement.
o Predicate: A predicate can be defined as a relation, which binds two atoms together in a statement.
Consider the statement: "x is an integer.” it consists of two parts, the first part x is the subject of the
statement and second part "is an integer” is known as a predicate.
Quantifiers in First-order logic:
 A quantifier is a language element which generates quantification, and quantification specifies the
quantity of specimen in the universe of discourse.
 These are the symbols that permit to determine or identify the range and scope of the variable in the
logical expression. There are two types of quantifier:
a. Universal Quantifier, (for all, everyone, everything)
b. Existential quantifier, (for some, at least one).
a. Universal Quantifier:
 Universal quantifier is a symbol of logical representation, which specifies that the statement within its
range is true for everything or every instance of a particular thing.
 The Universal quantifier is represented by a symbol ∀, which resembles an inverted A.
 If x is a variable, then ∀x is read as:
o For all x
o For each x
o For every x.

Example:
 All man drink coffee.
Solution:
Let a variable x which refers to a cat so all x can be represented in UOD as below:
b. Existential Quantifier:
 Existential quantifiers are the type of quantifiers, which express that the statement within its scope is
true for at least one instance of something.
 It is denoted by the logical operator ∃, which resembles as inverted E. When it is used with a predicate
variable then it is called as an existential quantifier.
 If x is a variable, then existential quantifier will be ∃x or ∃(x). And it will be read as:
There exists a 'x.'
For some 'x.'
For at least one 'x.'

3. USING FIRST ORDER LOGIC:
Points to remember:
o The main connective for universal quantifier ∀ is implication →.
o The main connective for existential quantifier ∃ is and ∧.
Properties of Quantifiers:
o In universal quantifier, ∀x∀y is similar to ∀y∀x.
o In Existential quantifier, ∃x∃y is similar to ∃y∃x.
o ∃x∀y is not similar to ∀y∃x.
Some Examples of FOL using quantifier:
1. All birds fly.
In this question the predicate is "fly(bird)."

And since there are all birds who fly so it will be represented as follows.
∀x bird(x) →fly(x).
2. Every man respects his parent.
In this question, the predicate is "respect(x, y)," where x=man, and y= parent.
Since there is every man so will use ∀, and it will be represented as follows:
∀x man(x) → respects (x, parent).
3. Some boys play cricket.
In this question, the predicate is "play(x, y)," where x= boys, and y= game. Since there are some boys so
we will use ∃, and it will be represented as:
∃x boys(x) → play(x, cricket).
4. Not all students like both Mathematics and Science.
In this question, the predicate is "like(x, y)," where x= student, and y= subject.
Since there are not all students, so we will use ∀ with negation, so following representation for this:
¬∀ (x) [ student(x) → like(x, Mathematics) ∧ like(x, Science)].
5. Only one student failed in Mathematics.
In this question, the predicate is "failed(x, y)," where x= student, and y= subject.
Since there is only one student who failed in Mathematics, so we will use following representation
for this:
∃(x) [student(x) → failed (x, Mathematics) ∧∀ (y) [¬(x==y) ∧ student(y) → ¬failed (x,
Mathematics)].
Free and Bound Variables:
The quantifiers interact with variables which appear in a suitable way. There are two types of variables
in First-order logic which are given below:
Free Variable: A variable is said to be a free variable in a formula if it occurs outside the scope of the
quantifier.
Example: ∀x ∃(y)[P (x, y, z)], where z is a free variable.
Bound Variable: A variable is said to be a bound variable in a formula if it occurs within the scope of the
quantifier.
Example: ∀x [A (x) B(y)], here x and y are the bound variables.

Example with solution:
Example with solution:

4. KNOWLEDGE ENGINEERING IN FIRST ORDER LOGIC:
What is knowledge-engineering?
 The process of constructing a knowledge-base in first-order logic is called as knowledge- engineering.
In knowledge-engineering, someone who investigates a particular domain, learns important concept of
that domain, and generates a formal representation of the objects, is known as knowledge engineer.
 In this topic, we will understand the Knowledge engineering process in an electronic circuit domain,
which is already familiar. This approach is mainly suitable for creating special-purpose knowledge
base.
The knowledge-engineering process:
 Following are some main steps of the knowledge-engineering process. Using these steps, we will
develop a knowledge base which will allow us to reason about digital circuit (One-bit full adder) which
is given below
1. Identify the task:
The first step of the process is to identify the task, and for the digital circuit, there are various reasoning
tasks.
At the first level or highest level, we will examine the functionality of the circuit:
o Does the circuit add properly?
o What will be the output of gate A2, if all the inputs are high?
At the second level, we will examine the circuit structure details such as:

o Which gate is connected to the first input terminal?
o Does the circuit have feedback loops?
2. Assemble the relevant knowledge:
In the second step, we will assemble the relevant knowledge which is required for digital circuits. So for
digital circuits, we have the following required knowledge:
o Logic circuits are made up of wires and gates.
o Signal flows through wires to the input terminal of the gate, and each gate produces the corresponding
output which flows further.
o In this logic circuit, there are four types of gates used: AND, OR, XOR, and NOT.
o All these gates have one output terminal and two input terminals (except NOT gate, it has one input
terminal).
3. Decide on vocabulary:
 The next step of the process is to select functions, predicate, and constants to represent the circuits,
terminals, signals, and gates.
 Firstly we will distinguish the gates from each other and from other objects. Each gate is represented as
an object which is named by a constant, such as, Gate(X1).
 The functionality of each gate is determined by its type, which is taken as constants such as AND, OR,
XOR, or NOT. Circuits will be identified by a predicate: Circuit (C1).
 For the terminal, we will use predicate: Terminal(x).
 For gate input, we will use the function In (1, X1) for denoting the first input terminal of the gate, and
for output terminal we will use Out (1, X1).
 The function Arity (c, i, j) is used to denote that circuit c has i input, j output.
 The connectivity between gates can be represented by predicate Connect (Out (1, X1), In (1, X1)).
We use a unary predicate On (t), which is true if the signal at a terminal is on.
4. Encode general knowledge about the domain:
To encode the general knowledge about the logic circuit, we need some following rules:
o If two terminals are connected then they have the same input signal, it can be represented as:
∀ t1, t2 Terminal (t1) ∧ Terminal (t2) ∧ Connect (t1, t2) → Signal (t1) = Signal (2).
o Signal at every terminal will have either value 0 or 1, it will be represented as:

∀ t Terminal (t) →Signal (t) = 1 ∨Signal (t) = 0.
o Connect predicates are commutative:
∀ t1, t2 Connect(t1, t2) → Connect (t2, t1).
o Representation of types of gates:
∀ g Gate(g) ∧ r = Type(g) → r = OR ∨r = AND ∨r = XOR ∨r = NOT.
o Output of AND gate will be zero if and only if any of its input is zero.
∀ g Gate(g) ∧ Type(g) = AND →Signal (Out(1, g))= 0 ⇔ ∃n Signal (In(n, g))= 0.
o Output of OR gate is 1 if and only if any of its input is 1:
∀ g Gate(g) ∧ Type(g) = OR → Signal (Out(1, g))= 1 ⇔ ∃n Signal (In(n, g))= 1
o Output of XOR gate is 1 if and only if its inputs are different:
∀ g Gate(g) ∧ Type(g) = XOR → Signal (Out(1, g)) = 1 ⇔ Signal (In(1, g)) ≠ Signal (In(2, g)).
o Output of NOT gate is invert of its input:
∀ g Gate(g) ∧ Type(g) = NOT → Signal (In(1, g)) ≠ Signal (Out(1, g)).
o All the gates in the above circuit have two inputs and one output (except NOT gate).
∀ g Gate(g) ∧ Type(g) = NOT → Arity(g, 1, 1)
∀ g Gate(g) ∧ r =Type(g) ∧ (r= AND ∨r= OR ∨r= XOR) → Arity (g, 2, 1).
o All gates are logic circuits:
∀ g Gate(g) → Circuit (g).
5. Encode a description of the problem instance:
 Now we encode problem of circuit C1, firstly we categorize the circuit and its gate components. This
step is easy if ontology about the problem is already thought.
 This step involves the writing simple atomics sentences of instances of concepts, which is known as
ontology.
 For the given circuit C1, we can encode the problem instance in atomic sentences as below:
 Since in the circuit there are two XOR, two AND, and one OR gate so atomic sentences for these gates
will be:
For XOR gate: Type(x1)= XOR, Type(X2) = XOR

For AND gate: Type(A1) = AND, Type(A2)= AND
For OR gate: Type (O1) = OR.
And then represent the connections between all the gates.
6. Pose queries to the inference procedure and get answers:
In this step, we will find all the possible set of values of the entire terminal for the adder circuit. The first
query will be:
What should be the combination of input which would generate the first output of circuit C1, as 0 and a second
output to be 1?
∃ i1, i2, i3 Signal (In(1, C1))=i1 ∧ Signal (In(2, C1))=i2 ∧ Signal (In(3, C1))= i3
∧ Signal (Out(1, C1)) =0 ∧ Signal (Out(2, C1))=1
7. Debug the knowledge base:
Now we will debug the knowledge base, and this is the last step of the complete process. In this step, we will try
to debug the issues of knowledge base. In the knowledge base, we may have omitted assertions like 1 ≠ 0.
5. PROPOSITIONAL VS. FIRST-ORDER INFERENCE:
i. Propositional logic
• Propositional logic (PL) is a statement made by propositions.
• It is a simple form of logic.
• Propositions are declarative statements which are either true or false.
• PL is a technique often used in logical and mathematical form.
• When we study PL, we usually start with formal natural language arguments, but they can also be
expressed mathematically.
• Propositional logic is based on formal logic, deductive reason and Boolean logic.
• We use symbolic variables to represent the logic. Often true and false can be symbolized by 1 and 0.
• Propositional logic also consists of an object, relations or function, and logical connectives.

ii. First order Logic:
 First-order logic(FOL), Predicate logic or First-order predicate logic is an extension to propositional
logic.
 First-order logic expresses information about objects and expresses relationships between those objects
in a more functional way in comparison to propositional logic.
 First-order logic is more concise than propositional logic.
 In First-order logic, the statements are divided into two parts: the subject and the predicate.
 The predicate is not a proposition.
 It is neither true nor false.
 Predicates use variables and objects like people, colors, numbers, letters or ideas.
o They can also represent relationships and functions.
• Both syntax and semantics are important to first-order logic.
• Symbols are the basic syntax of FOL and can be written in shorthand.
• Syntax is the structure of the logical statements. Semantics gives meaning to the statements.
Examples:
• Every man has a heart = ∀x man(x) → have (x, heart).
• Some men are mean = ∃x: man(x) ∧ mean(x).
6. UNIFICATION AND LIFTING:
 Unification is a process of making two different logical atomic expressions identical by finding a
substitution.
 Unification depends on the substitution process.
 It takes two literals as input and makes them identical using substitution.
 Let Ψ1 and Ψ2 be two atomic sentences and 𝜎 be a unifier such that, Ψ1𝜎 = Ψ2𝜎, then it can be
expressed as UNIFY (Ψ1, Ψ2).
Example: Find the MGU for Unify {King(x), King (John)}
Let Ψ1 = King(x), Ψ2 = King (John),
Substitution: θ = {John/x} is a unifier for these atoms and applying this substitution, and both
expressions will be identical.
• The UNIFY algorithm is used for unification, which takes two atomic sentences and returns a unifier for
those sentences (If any exist).
• Unification is a key component of all first-order inference algorithms.
• It returns fail if the expressions do not match with each other.
• The substitution variables are called Most General Unifier or MGU.

Example: Let's say there are two different expressions, P(x, y), and P(a, f(z)).
In this example, we need to make both above statements identical to each other.
For this, we will perform the substitution.
P(x, y)......... (i)
P (a, f(z))......... (ii)
• Substitute x with a, and y with f(z) in the first expression, and it will be represented as a/x and f(z)/y.
• With both the substitutions, the first expression will be identical to the second expression and the
substitution set will be: [a/x, f (z)/y].
Conditions for Unification:
Following are some basic conditions for unification:
• Predicate symbol must be same, atoms or expression with different predicate symbol can never be
unified.
• Number of Arguments in both expressions must be identical.
• Unification will fail if there are two similar variables present in the same expression.
Algorithm:

Implementation of the Algorithm:
EXAMPLE 1:
Find the MGU of {p(f(a), g(Y)) and p(X, X)}.
Sol: S0 => Here,
• Ψ1 = p(f(a), g(Y)), and
• Ψ2 = p(X, X)
SUBST θ = {f(a) / X}
• S1 => Ψ1 = p(f(a), g(Y)), and Ψ2 = p(f(a), f(a))
SUBST θ = {f(a) / g(y)},
Unification failed.
Unification is not possible for these expressions.
EXAMPLE 2:
Find the MGU of {p(b, X, f(g(Z))) and p(Z, f(Y), f(Y))}
Sol: Here, Ψ1 = p(b, X, f(g(Z)))
Ψ2 = p(Z, f(Y), f(Y))
S0 => { p(b, X, f(g(Z))); p(Z, f(Y), f(Y))}
SUBST θ={b/Z}
S1 => { p(b, X, f(g(b))); p(b, f(Y), f(Y))}
SUBST θ={f(Y) /X}
S2 => { p(b, f(Y), f(g(b))); p(b, f(Y), f(Y))}
SUBST θ= {g(b) /Y}
S2 => { p(b, f(g(b)), f(g(b)); p(b, f(g(b)), f(g(b))}

Unified Successfully.
And Unifier = { b/Z, f(Y) /X , g(b) /Y}.
EXAMPLE 3:
Find the MGU of {p (X, X), and p (Z, f(Z))}
Sol: Here, Ψ1 = {p (X, X)
Ψ2 = p (Z, f(Z))
S0 => {p (X, X), p (Z, f(Z))}
SUBST θ= {X/Z}
S1 => {p (Z, Z), p (Z, f(Z))}
SUBST θ= {f(Z) / Z},
Unification Failed.
Therefore, unification is not possible for these expressions.
EXAMPLE 4:
UNIFY (knows (Richard, x), knows (Richard, John))
Sol: Here, Ψ1 = knows (Richard, x)
Ψ2 = knows (Richard, John)
S0 => { knows(Richard, x); knows(Richard, John)}
S SUBST θ= {John/x}
S1 => { knows(Richard, John); knows(Richard, John)},
Successfully Unified.
Unifier: {John/x}.
EXAMPLE 5:
Find the MGU of UNIFY (prime (11), prime(y)).
Sol: Here, Ψ1 = {prime (11) , and Ψ2 = prime(y)}
S0 => {prime(11) , prime(y)}
SUBST θ= {11/y}
S1 => {prime (11), prime (11)},
successfully unified.
Unifier: {11/y}.
5. Find the MGU of Q (a, g(x, a), f(y)), Q(a, g(f(b), a), x)}
Here, Ψ1 = Q(a, g(x, a), f(y)), and Ψ2 = Q(a, g(f(b), a), x)
S0 => {Q(a, g(x, a), f(y)); Q(a, g(f(b), a), x)}
SUBST θ= {f(b)/x}
S1 => {Q(a, g(f(b), a), f(y)); Q(a, g(f(b), a), f(b))}
SUBST θ= {b/y}
S1 => {Q(a, g(f(b), a), f(b)); Q(a, g(f(b), a), f(b))},

Unifier: [a/a, f(b)/x, b/y].
6. UNIFY (knows (Richard, x), knows(Richard, John))
Here, Ψ1 = knows (Richard, x), and Ψ2 = knows (Richard, John)
S0 => { knows (Richard, x); knows(Richard, John)}
SUBST θ= {John/x}
S1 => { knows(Richard, John); knows(Richard, John)},
Unifier: {John/x}.
7. INFERENCE IN FIRST-ORDER LOGIC:
 Inference in First-Order Logic is used to deduce new facts or sentences from existing sentences. Before
understanding the FOL inference rule, let's understand some basic terminologies used in FOL.
i. Substitution:
 Substitution is a fundamental operation performed on terms and formulas. It occurs in all inference
systems in first-order logic.
 The substitution is complex in the presence of quantifiers in FOL. If we write F [a/x], so it refers to
substitute a constant "a" in place of variable "x".
ii. Equality:
 First-Order logic does not only use predicate and terms for making atomic sentences but also uses
another way, which is equality in FOL. For this, we can use equality symbols which specify that the
two terms refer to the same object.
Example: Brother (John) = Smith.
As in the above example, the object referred by the Brother (John) is similar to the object referred
by Smith. The equality symbol can also be used with negation to represent that two terms are not the
same objects.
Example: ￢(x=y) which is equivalent to x ≠y.
FOL inference rules for quantifier:
As propositional logic we also have inference rules in first-order logic, so following are some basic
inference rules in FOL:
o Universal Generalization
o Universal Instantiation
o Existential Instantiation
o Existential introduction

1. Universal Generalization:
 Universal generalization is a valid inference rule which states that if premise P(c) is true for any
arbitrary element c in the universe of discourse, then we can have a conclusion as ∀ x P(x).
It can be represented as:
 This rule can be used if we want to show that every element has a similar property.
 In this rule, x must not appear as a free variable.
Example: Let's represent, P(c): "A byte contains 8 bits", so for ∀ x P(x) "All bytes contain 8 bits.",
It will also be true.
2. Universal Instantiation:
 Universal instantiation is also called as universal elimination or UI is a valid inference rule. It can be
applied multiple times to add new sentences.
 The new KB is logically equivalent to the previous KB.
 As per UI, we can infer any sentence obtained by substituting a ground term for the variable.
 The UI rule state that we can infer any sentence P(c) by substituting a ground term c (a constant within
domain x) from ∀ x P(x) for any object in the universe of discourse.
 It can be represented as:
Example: 1
IF "Every person like ice-cream"=> ∀x P(x) so we can infer that
"John likes ice-cream" => P(c)
Example: 2
Let's take a famous example,
"All kings who are greedy are Evil." So let our knowledge base contains this detail as in the form
of FOL:
∀x king(x) ∧ greedy (x) → Evil (x),
So from this information, we can infer any of the following statements using Universal Instantiation:
 King(John) ∧ Greedy (John) → Evil (John),
 King(Richard) ∧ Greedy (Richard) → Evil (Richard),
 King(Father(John)) ∧ Greedy (Father(John)) → Evil (Father(John)),
3. Existential Instantiation:
 Existential instantiation is also called as Existential Elimination, which is a valid inference rule in
first-order logic.

 It can be applied only once to replace the existential sentence.
 The new KB is not logically equivalent to old KB, but it will be satisfiable if old KB was satisfiable.
 This rule states that one can infer P(c) from the formula given in the form of ∃x P(x) for a new
constant symbol c.
 The restriction with this rule is that c used in the rule must be a new term for which P(c) is true.
It can be represented as:
Example:
From the given sentence: ∃x Crown(x) ∧ OnHead (x, John),
So we can infer: Crown (K) ∧ OnHead (K, John), as long as K does not appear in the knowledge base.
 The above used K is a constant symbol, which is called Skolem constant.
 The Existential instantiation is a special case of Skolemization process.
4. Existential introduction:
 An existential introduction is also known as an existential generalization, which is a valid inference rule
in first-order logic.
 This rule states that if there is some element c in the universe of discourse which has a property P, then
we can infer that there exists something in the universe which has the property P.
It can be represented as
Example: Let's say that, "Priyanka got good marks in English."
"Therefore, someone got good marks in English.
Generalized Modus Ponens Rule:
 For the inference process in FOL, we have a single inference rule which is called Generalized Modus
Ponens. It is lifted version of Modus ponens.
 Generalized Modus Ponens can be summarized as, “P implies Q and P is asserted to be true, therefore Q
must be true."
 According to Modus Ponens, for atomic sentences pi, pi', q. Where there is a substitution θ such that
SUBST (θ, pi',) = SUBST(θ, pi), it can be represented as:

Example:
We will use this rule for Kings are evil, so we will find some x such that x is king, and x is greedy so we
can infer that x is evil.
Here let say, p1' is king(John) p1 is king(x)
p2' is Greedy(y) p2 is Greedy(x)
θ is {x/John, y/John} q is evil(x)
SUBST (θ,q).
8. FORWARD AND BACKWARD CHAINING: (Refer UNIT II notes)
9. RESOLUTION IN FOL:
Resolution
 Resolution is a theorem proving technique that proceeds by building refutation proofs, i.e., proofs by
contradictions. It was invented by a Mathematician John Alan Robinson in the year 1965.
 Resolution is used, if there are various statements are given, and we need to prove a conclusion of those
statements.
 Unification is a key concept in proofs by resolutions. Resolution is a single inference rule which can
efficiently operate on the conjunctive normal form or clausal form.
 Clause: Disjunction of literals (an atomic sentence) is called a clause. It is also known as a unit clause.
 Conjunctive Normal Form: A sentence represented as a conjunction of clauses is said to
be conjunctive normal form or CNF.
THE RESOLUTION INFERENCE RULE:
 The resolution rule for first-order logic is simply a lifted version of the propositional rule. Resolution
can resolve two clauses if they contain complementary literals, which are assumed to be standardized
apart so that they share no variables.
Where li and mj are complementary literals.
 This rule is also called the binary resolution rule because it only resolves exactly two literals.
Example:
We can resolve two clauses which are given below:
[Animal (g(x) V Loves (f(x), x)] and [￢ Loves(a, b) V ￢Kills(a, b)]
Where two complimentary literals are: Loves (f(x), x) and ￢ Loves (a, b)
These literals can be unified with unifier θ= [a/f(x), and b/x] , and it will generate a resolvent clause:
[Animal (g(x) V ￢ Kills(f(x), x)].

STEPS FOR RESOLUTION:
1. Conversion of facts into first-order logic.
2. Convert FOL statements into CNF
3. Negate the statement which needs to prove (proof by contradiction)
4. Draw resolution graph (unification).
To better understand all the above steps, we will take an example in which we will apply resolution.
EXAMPLE: 1
a. John likes all kind of food.
b. Apple and vegetable are food
c. Anything anyone eats and not killed is food.
d. Anil eats peanuts and still alive
e. Harry eats everything that Anil eats.
Prove by resolution that:
John likes peanuts.
Step-1: Conversion of Facts into FOL
In the first step we will convert all the given statements into its first order logic.
Step-2: Conversion of FOL into CNF
In First order logic resolution, it is required to convert the FOL into CNF as CNF form makes easier for
resolution proofs.

o Eliminate existential instantiation quantifier by elimination.
In this step, we will eliminate existential quantifier ∃, and this process is known as Skolemization. But
in this example problem since there is no existential quantifier so all the statements will remain same in
this step.
o Drop Universal quantifiers.
In this step we will drop all universal quantifier since all the statements are not implicitly quantified so
we don't need it.
o Distribute conjunction ∧ over disjunction ¬.
This step will not make any change in this problem.
Step-3: Negate the statement to be proved
In this statement, we will apply negation to the conclusion statements, which will be written as ¬likes
(John, Peanuts)
Step-4: Draw Resolution graph:
Now in this step, we will solve the problem by resolution tree using substitution. For the above problem, it will
be given as follows:

Hence the negation of the conclusion has been proved as a complete contradiction with the given set of
statements.
EXPLANATION OF RESOLUTION GRAPH:
o In the first step of resolution graph, ¬likes(John, Peanuts) , and likes(John, x) get resolved(canceled)
by substitution of {Peanuts/x}, and we are left with ¬ food(Peanuts)
o In the second step of the resolution graph, ¬ food (Peanuts), and food (z) get resolved (canceled) by
substitution of {Peanuts/z}, and we are left with ¬ eats(y, Peanuts) V killed(y).
o In the third step of the resolution graph, ¬ eats(y, Peanuts) and eats (Anil, Peanuts) get resolved by
substitution {Anil/y}, and we are left with Killed (Anil).
o In the fourth step of the resolution graph, Killed (Anil) and ¬ killed (k) get resolve by
substitution {Anil/k}, and we are left with ¬ alive (Anil).
o In the last step of the resolution graph ¬ alive (Anil) and alive (Anil) get resolved.

First-Order Logic
Chapter 8
Examples
[04PBT 04PFT 04PLT 04PNT 04PRT]
Genap 2011-2012
[dks0638]

First-order logic
• Whereas propositional logic assumes the
world contains facts,
• first-order logic (like natural language)
assumes the world contains
– Objects: people, houses, numbers, colors,
baseball games, wars, …
– Relations: red, round, prime, brother of, bigger
than, part of, comes between, …
– Functions: father of, best friend, one more
than, plus, …

Syntax of FOL
• Constants KingJohn, 2, UofA,...
• Predicates Brother, >,...
• Functions Sqrt, LeftLegOf,...
• Variables x, y, a, b,...
• Connectives , , , , 
• Equality =
• Quantifiers , 

Example Knowledge Base
The law says that it is a crime for an
American to sell weapons to hostile
nations. The country Nono, an enemy of
America, has some missiles, and all of its
missiles were sold to it by Colonel West,
who is American.
Prove that Col. West is a criminal!

Properties of forward chaining
9
Sound and complete for first-order definite clauses (proof similar to
propositional proof)
Datalog = first-order definite clauses + no functions (e.g., crime KB)
FC terminates for Datalog in poly iterations: at most p ∙ nk literals
May not terminate in general if  is not entailed
This is unavoidable: entailment with definite clauses is semidecidable

Efficiency of forward chaining
Simple observation: no need to match a rule on iteration k
if a premise wasn't added on iteration k - 1
 match each rule whose premise contains a newly added
literal
Matching itself can be expensive
Database indexing allows O(1) retrieval of known facts
e.g., query Missile(x) retrieves Missile(M1)
Matching conjunctive premises against known facts is NP-hard
Forward chaining is widely used in deductive databases

Properties of backward chaining
18
Depth-first recursive proof search: space is linear in size of proof
Incomplete due to infinite loops
 fix by checking current goal against every goal on stack
Inefficient due to repeated subgoals (both success and failure)
 fix using caching of previous results (extra space!)
Widely used (without improvements!) for logic programming

Resolution refutation proofs involves
the following steps:
1. Put the premises or axioms into clause form.
2. Add the negation of what is to be proved, in
clause form, to the set of axioms.
3. Resolve these clauses together, producing new
clauses that logically follow from them.
4. Produce a contradiction by generating the empty
clause.
5. The substitutions used to produce the empty
clause are those under which the opposite of the
negated goal is true.

A Facts in Propositional Logic
23
Given Axioms Clause Form
P P (1)
(P  Q)  R  P   Q  R (2)
(S  T)  Q  S  Q (3)
 T  Q (4)
T T (5)
Prove R!

Resolution in Propositional Logic
24
 P   Q  R (2)  R
 P   Q P (1)
 T  Q (4)  Q
 T T (5)

Prove fido will die!
1. All dogs are animal.
2. Fido is a dog.
3. All animals will die.

Resolution proof for the “dead dog” problem.

Lucky Student
1. Anyone passing his history exams and
winning the lottery is happy.
2. Anyone who studies or is lucky can
pass all his exams.
3. John did not study but he is lucky.
4. Anyone who is lucky wins the lottery.
Prove that John is happy!

One resolution refutation for the “happy student” problem.
[1] [6]
[5]
[3]
[5]
[7]
[10]
[9]
[9]
[8]

Exciting life
1. All people that are not poor and are smart
are happy.
2. Those people that read are not stupid.
3. John can read and is wealthy.
4. Happy people have exiting lives.
Can anyone be found with an exciting life?

Resolution proof for the “exciting life” problem.
[6] [5]
[1]
[2]
[4]
[3]

Resolution Example
1. John likes all kinds of food.
x: food(x)  likes(john, x)
2. Apples are food.
food(apple)
3. Chicken is food.
food(chicken)
4. Anything anyone eats and isn't killed by is food.
x:(y: eats(y, x)   killedby(y, x))  food(x)
5. Bill eats peanuts and is still alive.
A. eats(Bill, peanuts) B. alive(Bill)
6. Sue eats everything Bill eats.
x:eats(Bill, x) eats(Sue, x)
7. x:y: alive(x)  killedby (x,y)
35

Clause form:
1.  food(x1)  likes(John, x1)
2. food(apples)
3. food(chicken)
4.  eats(y4,x4)  killedby(y4 , x4)  food(x4)
5. Eats (Bill, peanuts)
6. Alive (Bill)
7.  eats(Bill,x7)  eats(Sue,x7)
8.  alive(x8)   killedby(x8, y8)
36

37
Resolution proof that John likes peanuts
 likes (John, peanuts) (1)  food(x1)  likes (John, x1)
 food(peanuts) (4)  eats(y4,x4)  killedby(y4 . x4)  food(x4)
 eats(y4,peanuts)  killedby(y4. peanuts) (5) eats(Bill.peanuts)
killedby(Bill. peanuts) (8)  alive(x8)   killedby(x8 y8)
 alive(Bill) (6) alive(Bill)
peanuts/x1
peanuts/x4
Bill/x8, peanuts/y8
Bill/y4

AI3391 Artificial intelligence Unit IV Notes _ merged.pdf

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AI3391 Artificial intelligence Unit IV Notes _ merged.pdf