First-Order Logic in Artificial intelligence

First-Order Logic in Artificial intelligence

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:
    • Objects: A, B, people, numbers, colors, wars, theories, squares, pits, wumpus, ……
    • 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
    • Function: Father of, best friend, third inning of, end of, ……
  • As a natural language, first-order logic also has two main parts:
    1. Syntax
    2. Semantics

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.

Basic Elements of First-order logic:

Following are the basic elements of FOL syntax:How to find Nth Highest Salary in SQL

Constant1, 2, A, John, Mumbai, cat,….
Variablesx, y, z, a, b,….
PredicatesBrother, Father, >,….
Functionsqrt, LeftLegOf, ….
Connectives∧, ∨, ¬, ⇒, ⇔
Quantifier∀, ∃

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).

Example: Ravi and Ajay are brothers: => Brothers(Ravi, Ajay).
                Chinky is a cat: => cat (Chinky)

Complex Sentences:

  • Complex sentences are made by combining atomic sentences using connectives.

First-order logic statements can be divided into two parts:

  • Subject: Subject is the main part of the statement.
  • 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.

First-Order Logic in Artificial intelligence

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:
    1. Universal Quantifier, (for all, everyone, everything)
    2. Existential quantifier, (for some, at least one).

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.

Note: In universal quantifier we use implication “→”.

If x is a variable, then ∀x is read as:

  • For all x
  • For each x
  • For every x.


All man drink coffee.

Let a variable x which refers to a cat so all x can be represented in UOD as below:

First-Order Logic in Artificial intelligence

∀x man(x) → drink (x, coffee).

It will be read as: There are all x where x is a man who drink coffee.

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.

Note: In Existential quantifier we always use AND or Conjunction symbol (∧).

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.’


Some boys are intelligent.

First-Order Logic in Artificial intelligence

∃x: boys(x) ∧ intelligent(x)

It will be read as: There are some x where x is a boy who is intelligent.

Points to remember:

  • The main connective for universal quantifier  is implication .
  • The main connective for existential quantifier  is and .

Properties of Quantifiers:

  • In universal quantifier, ∀x∀y is similar to ∀y∀x.
  • In Existential quantifier, ∃x∃y is similar to ∃y∃x.
  • ∃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.

Next TopicKnowledg

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.


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“.

Note: First-order logic is capable of expressing facts about some or all objects in the universe.


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:

  • Universal Generalization
  • Universal Instantiation
  • Existential Instantiation
  • 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: Inference in First-Order Logic.
  • 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:Inference in First-Order Logic.


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:Inference in First-Order Logic


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: Inference in First-Order Logic
  • 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:

Inference in First-Order Logic


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.

  1. Here let say, p1′ is king(John)        p1 is king(x)  
  2. p2′ is Greedy(y)                       p2 is Greedy(x)  
  3. θ is {x/John, y/John}                  q is evil(x)  
  4. SUBST(θ,q).                                                        

Next TopicUnif

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

Knowledge Engineering in First-order logic

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. of Java

At the first level or highest level, we will examine the functionality of the circuit:

  • Does the circuit add properly?
  • 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:

  • Which gate is connected to the first input terminal?
  • 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:

  • Logic circuits are made up of wires and gates.
  • Signal flows through wires to the input terminal of the gate, and each gate produces the corresponding output which flows further.
  • In this logic circuit, there are four types of gates used: AND, OR, XOR, and NOT.
  • 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:

  • If two terminals are connected then they have the same input signal, it can be represented as:

  1. ∀  t1, t2 Terminal (t1) ∧ Terminal (t2) ∧ Connect (t1, t2) → Signal (t1) = Signal (2).   
  • Signal at every terminal will have either value 0 or 1, it will be represented as:

  1. ∀  t Terminal (t) →Signal (t) = 1 ∨Signal (t) = 0.  
  • Connect predicates are commutative:

  1. ∀  t1, t2 Connect(t1, t2)  →  Connect (t2, t1).       
  • Representation of types of gates:

  1. ∀  g Gate(g) ∧ r = Type(g) → r = OR ∨r = AND ∨r = XOR ∨r = NOT.   
  • Output of AND gate will be zero if and only if any of its input is zero.

  1. ∀  g Gate(g) ∧ Type(g) = AND →Signal (Out(1, g))= 0 ⇔  ∃n Signal (In(n, g))= 0.   
  • Output of OR gate is 1 if and only if any of its input is 1:

  1. ∀  g Gate(g) ∧ Type(g) = OR → Signal (Out(1, g))= 1 ⇔  ∃n Signal (In(n, g))= 1   
  • Output of XOR gate is 1 if and only if its inputs are different:

  1. ∀  g Gate(g) ∧ Type(g) = XOR → Signal (Out(1, g)) = 1 ⇔  Signal (In(1, g)) ≠ Signal (In(2, g)).  
  • Output of NOT gate is invert of its input:

  1. ∀  g Gate(g) ∧ Type(g) = NOT →   Signal (In(1, g)) ≠ Signal (Out(1, g)).  
  • All the gates in the above circuit have two inputs and one output (except NOT gate).

  1. ∀  g Gate(g) ∧ Type(g) = NOT →   Arity(g, 1, 1)   
  2. ∀  g Gate(g) ∧ r =Type(g)  ∧ (r= AND ∨r= OR ∨r= XOR) →  Arity (g, 2, 1).   
  • All gates are logic circuits:

  1. ∀  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:

  1. For XOR gate: Type(x1)= XOR, Type(X2) = XOR  
  2. For AND gate: Type(A1) = AND, Type(A2)= AND  
  3. For OR gate: Type (O1) = OR.    

And then represent the connections between all the gates.

Note: Ontology defines a particular theory of the nature of existence.

6. Pose queries to the inference procedure and get answers:

In this step, we will find all the possible set of values of all the 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?

  1. ∃ i1, i2, i3 Signal (In(1, C1))=i1  ∧  Signal (In(2, C1))=i2  ∧ Signal (In(3, C1))= i3  
  2.  ∧ 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.

Next TopicInference in 

Resolution in FOL


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. Ministers of India | List of Prime Minister of India (1947-2020)

Note: To better understand this topic, firstly learns the FOL in AI.

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.

Resolution in FOL

Where li and mj are complementary literals.

This rule is also called the binary resolution rule because it only resolves exactly two literals.


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.


  1. John likes all kind of food.
  2. Apple and vegetable are food
  3. Anything anyone eats and not killed is food.
  4. Anil eats peanuts and still alive
  5. Harry eats everything that Anil eats.
    Prove by resolution that:
  6. 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.

Resolution in FOL

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.

  • Eliminate all implication (→) and rewrite
    1. ∀x ¬ food(x) V likes(John, x)
    2. food(Apple) Λ food(vegetables)
    3. ∀x ∀y ¬ [eats(x, y) Λ ¬ killed(x)] V food(y)
    4. eats (Anil, Peanuts) Λ alive(Anil)
    5. ∀x ¬ eats(Anil, x) V eats(Harry, x)
    6. ∀x¬ [¬ killed(x) ] V alive(x)
    7. ∀x ¬ alive(x) V ¬ killed(x)
    8. likes(John, Peanuts).
  • Move negation (¬)inwards and rewrite
    1. ∀x ¬ food(x) V likes(John, x)
    2. food(Apple) Λ food(vegetables)
    3. ∀x ∀y ¬ eats(x, y) V killed(x) V food(y)
    4. eats (Anil, Peanuts) Λ alive(Anil)
    5. ∀x ¬ eats(Anil, x) V eats(Harry, x)
    6. ∀x ¬killed(x) ] V alive(x)
    7. ∀x ¬ alive(x) V ¬ killed(x)
    8. likes(John, Peanuts).
  • Rename variables or standardize variables
    1. ∀x ¬ food(x) V likes(John, x)
    2. food(Apple) Λ food(vegetables)
    3. ∀y ∀z ¬ eats(y, z) V killed(y) V food(z)
    4. eats (Anil, Peanuts) Λ alive(Anil)
    5. ∀w¬ eats(Anil, w) V eats(Harry, w)
    6. ∀g ¬killed(g) ] V alive(g)
    7. ∀k ¬ alive(k) V ¬ killed(k)
    8. likes(John, Peanuts).
  • 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.
  • 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.
    1. ¬ food(x) V likes(John, x)
    2. food(Apple)
    3. food(vegetables)
    4. ¬ eats(y, z) V killed(y) V food(z)
    5. eats (Anil, Peanuts)
    6. alive(Anil)
    7. ¬ eats(Anil, w) V eats(Harry, w)
    8. killed(g) V alive(g)
    9. ¬ alive(k) V ¬ killed(k)
    10. likes(John, Peanuts).

Note: Statements “food(Apple) Λ food(vegetables)” and “eats (Anil, Peanuts) Λ alive(Anil)” can be written in two separate statements.

  • 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:

Resolution in FOL

Hence the negation of the conclusion has been proved as a complete contradiction with the given set of statements.

Explanation of Resolution graph:

  • 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)
  • 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) .
  • 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) .
  • 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) .
  • In the last step of the resolution graph ¬ alive(Anil) and alive(Anil) get resolved.

Next TopicForward

What is Unification?

  • 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.

E.g. 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.History of Java

            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.

Unification Algorithm:

Algorithm: Unify(Ψ1, Ψ2)

Step. 1: If Ψ1 or Ψ2 is a variable or constant, then:
	a) If Ψ1 or Ψ2 are identical, then return NIL. 
	b) Else if Ψ1is a variable, 
		a. then if Ψ1 occurs in Ψ2, then return FAILURE
		b. Else return { (Ψ2/ Ψ1)}.
	c) Else if Ψ2 is a variable, 
		a. If Ψ2 occurs in Ψ1 then return FAILURE,
		b. Else return {( Ψ1/ Ψ2)}. 
	d) Else return FAILURE. 
Step.2: If the initial Predicate symbol in Ψ1 and Ψ2 are not same, then return FAILURE.
Step. 3: IF Ψ1 and Ψ2 have a different number of arguments, then return FAILURE.
Step. 4: Set Substitution set(SUBST) to NIL. 
Step. 5: For i=1 to the number of elements in Ψ1. 
	a) Call Unify function with the ith element of Ψ1 and ith element of Ψ2, and put the result into S.
	b) If S = failure then returns Failure
	c) If S ≠ NIL then do,
		a. Apply S to the remainder of both L1 and L2.
Step.6: Return SUBST. 

Implementation of the Algorithm

Step.1: Initialize the substitution set to be empty.

Step.2: Recursively unify atomic sentences:

  1. Check for Identical expression match.
  2. If one expression is a variable vi, and the other is a term ti which does not contain variable vi, then:
    1. Substitute ti / vi in the existing substitutions
    2. Add ti /vi to the substitution setlist.
    3. If both the expressions are functions, then function name must be similar, and the number of arguments must be the same in both the expression.

For each pair of the following atomic sentences find the most general unifier (If exist).

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.

2. Find the MGU of {p(b, X, f(g(Z))) and p(Z, f(Y), f(Y))}

Here, Ψ1 = p(b, X, f(g(Z))) , and 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}

3. Find the MGU of {p (X, X), and p (Z, f(Z))}

Here, Ψ1 = {p (X, X), and 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.

Hence, unification is not possible for these expressions.

4. Find the MGU of UNIFY(prime (11), prime(y))

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))}, Successfully Unified.

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)}, Successfully Unified.
Unifier: {John/x}.

Next TopicRes

Related Articles