1. CS621: Artificial Intelligence
Pushpak Bhattacharyya CSE Dept., IIT Bombay
Lecture 5 – Logic

2. Logic and inferencing
Vision
NLP
Search
Reasoning
Learning
Knowledge
Robotics
Expert Systems
Planning
Obtaining implication of given facts and rules -- Hallmark of intelligence

3. Inferencing through Deduction Induction
Deduction (General to specific)
Induction (Specific to General)
Abduction (Conclusion to hypothesis in absence of any other evidence to contrary)
Deduction
Given: All men are mortal (rule)
Shakespeare is a man (fact)
To prove: Shakespeare is mortal (inference)
Induction
Given: Shakespeare is mortal
Newton is mortal (Observation)
Dijkstra is mortal
To prove: All men are mortal (Generalization)

4. If there is rain, then there will be no picnic
Fact1: There was rain
Conclude: There was no picnic
Deduction
Fact2: There was no picnic
Conclude: There was no rain (?)
Induction and abduction are fallible forms of reasoning. Their conclusions are susceptible to retraction
Two systems of logic
1) Propositional calculus
2) Predicate calculus

5. Propositions
Stand for facts/assertions
Declarative statements
As opposed to interrogative statements (questions) or imperative statements (request, order)
Operators
=> and ¬ form a minimal set (can express other operations)
- Prove it.
Tautologies are formulae whose truth value is always T, whatever the assignment is

6. Model
In propositional calculus any formula with n propositions has 2n models (assignments)
- Tautologies evaluate to T in all models.
Examples: 1) 2)
e Morgan with AND

7. Semantic Tree/Tableau method of proving tautology
Start with the negation of the formula
- α - formula
α-formula
β-formula
- β - formula
α-formula
- α - formula

8. Example 2: Contradictions in all paths X (α - formula) (α - formulae)B C
Contradictions in all paths B C

9. Exercise:
Prove the backward implication in the previous example

10. Inferencing in PC
Backward chaining
Resolution
Forward chaining

11. Knowledge
Procedural
Declarative
Declarative knowledge deals with factoid questions (what is the capital of India? Who won the Wimbledon in 2005? Etc.)
Procedural knowledge deals with “How”
Procedural knowledge can be embedded in declarative knowledge

12. Example: Employee knowledge base
Employee record
Emp id : 1124
Age : 27
Salary : 10L / annum
Tax : Procedure to calculate tax from basic salary, Loans, medical factors, and # of children

13. Text Knowledge Representation

14. A Semantic Graph bought student June computer new past tense agent
in: modifier
a: indefinite
the: definite
student
past tense
agent
bought
object
time
computer
new
June
modifier
The student bought a new computer in June.

15. Representation of Knowledge
UNL representation
Representation of Knowledge
Ram is reading the newspaper

16. UNL: a United Nations project
Dave, Parikh and Bhattacharyya, Journal of Machine Translation, 2002
Started in 1996
10 year program
15 research groups across continents
First goal: generators
Next goal: analysers (needs solving various ambiguity problems)
Current active language groups
UNL_French (GETA-CLIPS, IMAG)
UNL_Hindi (IIT Bombay with additional work on UNL_English)
UNL_Italian (Univ. of Pisa)
UNL_Portugese (Univ of Sao Paolo, Brazil)
UNL_Russian (Institute of Linguistics, Moscow)
UNL_Spanish (UPM, Madrid)

17. Knowledge Representation
UNL Graph - relations
read
agt
obj
Ram
newspaper

18. Knowledge Representation
UNL Graph - UWs
read(icl>interpret)
obj
agt
Ram(iof>person)
newspaper(icl>print_media)

19. Knowledge Representation
UNL graph - attributes
@entry
@present
@progress
read(icl>interpret)
obj
agt
@def
newspaper(icl>print_media)
Ram(iof>person)
Ram is reading the newspaper

20. The boy who works here went to school
Another Example
The boy who works here went to school
plt
agt
@ past
school(icl>institution)
go(icl>move)
boy(icl>person)
work(icl>do)
here
@ entry
plc
:01

21. Predicate Calculus

22. Predicate Calculus: well known examples
Man is mortal : rule
∀x[man(x) → mortal(x)]
shakespeare is a man
man(shakespeare)
To infer shakespeare is mortal
mortal(shakespeare)

23. Forward Chaining/ Inferencing
man(x) → mortal(x)
Dropping the quantifier, implicitly Universal quantification assumed
man(shakespeare)
Goal mortal(shakespeare)
Found in one step
x = shakespeare, unification

24. Backward Chaining/ Inferencing
man(x) → mortal(x)
Goal mortal(shakespeare)
x = shakespeare
Travel back over and hit the fact asserted
man(shakespeare)

25. Resolution - Refutation
man(x) → mortal(x)
Convert to clausal form
~man(shakespeare) mortal(x)
Clauses in the knowledge base
man(shakespeare)
mortal(shakespeare)

26. Resolution – Refutation contd
Negate the goal
~man(shakespeare)
Get a pair of resolvents

27. Resolution Tree

28. Search in resolution Heuristics for Resolution Search
Goal Supported Strategy
Always start with the negated goal
Set of support strategy
Always one of the resolvents is the most recently produced resolute

29. Inferencing in Predicate Calculus
Forward chaining
Given P, , to infer Q
P, match L.H.S of
Assert Q from R.H.S
Backward chaining
Q, Match R.H.S of
assert P
Check if P exists
Resolution – Refutation
Negate goal
Convert all pieces of knowledge into clausal form (disjunction of literals)
See if contradiction indicated by null clause can be derived

30. P
converted to
Draw the resolution tree (actually an inverted tree). Every node is a clausal form and branches are intermediate inference steps.

31. Terminology
Pair of clauses being resolved is called the Resolvents. The resulting clause is called the Resolute.
Choosing the correct pair of resolvents is a matter of search.

32. Predicate Calculus Rules
Introduction through an example (Zohar Manna, 1974):
Problem: A, B and C belong to the Himalayan club. Every member in the club is either a mountain climber or a skier or both. A likes whatever B dislikes and dislikes whatever B likes. A likes rain and snow. No mountain climber likes rain. Every skier likes snow. Is there a member who is a mountain climber and not a skier?
Given knowledge has:
Facts
Rules

33. Predicate Calculus: Example contd.
Let mc denote mountain climber and sk denotes skier. Knowledge representation in the given problem is as follows:
member(A)
member(B)
member(C)
∀x[member(x) → (mc(x) ∨ sk(x))]
∀x[mc(x) → ~like(x,rain)]
∀x[sk(x) → like(x, snow)]
∀x[like(B, x) → ~like(A, x)]
∀x[~like(B, x) → like(A, x)]
like(A, rain)
like(A, snow)
Question: ∃x[member(x) ∧ mc(x) ∧ ~sk(x)]
We have to infer the 11th expression from the given 10.
Done through Resolution Refutation.

34. Club example: Inferencing
member(A)
member(B)
member(C)
Can be written as

35. Negate–

36. Now standardize the variables apart which results in the following
member(A)
member(B)
member(C)

37. 10 7 12 5 4 13 14 2 11 15 16 13 2 17

38. Assignment
Prove the inferencing in the Himalayan club example with different starting points, producing different resolution trees.
Think of a Prolog implementation of the problem
Prolog Reference (Prolog by Chockshin & Melish)