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CS344 : Introduction to Artificial
Intelligence
Pushpak Bhattacharyya
CSE Dept.,
IIT Bombay
Lecture 9- Completeness proof;
introducing knowledge representation
Soundness, Completeness &
Consistency
Soundness
Semantic
World
----------
Syntactic
World
---------Theorems,
Proofs
*
Valuation,
Tautology
Completeness
*
An example to illustrate the
completeness proof
p
q
p(p V q)
T
F
T
T
T
T
F
T
T
F
F
T
Running the completeness
proof
For every row of the truth table set up a
proof:
1.
2.
3.
4.
p, ~q |- p(p V q)
p, q |- p(p V q)
~p, q |- p(p V q)
~p, ~q |- p(p V q)
Completeness Proof
Completeness of Propositional
Calculus


Statement
If V(A) = T for all V,
then |--A i.e. A is a theorem.
Lemma:
If A consists of propositions P1, P2, …,
Pn then P’1, P’2, …, P’n |-- A’, where
A’ = A
if V(A) = true
= ~A otherwise
Similarly for each P’i
Proof for Lemma

Proof by induction on the number of ‘→’
symbols in A
Basis: Number of ‘→’ symbols is zero.
A is ℱ or P. This is true as, |-- (A
→ A)
i.e. A → A is a theorem.
Hypothesis: Let the lemma be true for
number of ‘→’ symbols ≤ n.
Induction: Let A which is B → C contain
n+1 ‘→’
Proof of Lemma (contd.)

Induction:
By hypothesis,
P’1, P’2, …, P’n |-- B’
P’1, P’2, …, P’n |-- C’
If we show that B’, C’ |-- A’ (A is B → C), then the proof is
complete.
For this we have to show:
•
•
•
•

B, C
|-- B → C
True as B, C, B |-- C
B, ~C |-- ~(B → C)
True since B, ~C, B → C |-- ℱ
~B, C |-- B → C
True since ~B, C, B |-- C
~B, ~C |-- B → C
True since ~B, ~C, B, C → ℱ |-- ℱ
Hence the lemma is proved.
Proof of Theorem



A is a tautology.
There are 2n models corresponding to P1, P2, …, Pn propositions.
Consider,
P1, P2, …, Pn
and P1, P2, …, ~Pn
|-|--
P1, P2, …, Pn-1 |-and P1, P2, …, Pn-1 |--

A
A
Pn → A
~Pn → A
RHS can be written as:
|-((Pn → A) → ((~Pn → A) → A))
|-(~Pn → A) → A
|-A
Thus dropping the propositions progressively we show |-- A
Assignment-2

Create a system to prove syntactically
the theoremhood of any propositional
calculus expression. This will be an
automatic theorem prover for
propositional calculus. The input to the
system will be any propositional
calculus expression and the output a
yes/no answer depending on whether
the expression is a theorem or not.
Detour

Reasoning

two types:



Monotonic: Adding inferred knowledge monotonically to the system but not retracting
from the knowledge base.
Nonmonotonic: Retracts knowledge which becomes false in the face of new evidence
Types of Sentences in English: 3 kinds of sentences important from
Natural Language Processing point of view. Useful to remember in
knowledge extraction.


Simple Sentence: Single verb, e.g., Ram plays cricket
Compound Sentence: Two independent clauses joined by coordinator. Two verbs
are present.
e.g. Ram went to school and Shyam played cricket

Complex Sentence: Independent clauses joined by one or more dependent
clauses. More than one verb
e.g. Ram who sings well is performing in the festival today
Predicate Calculus

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
Predicate Calculus: Example
contd.

Let mc denote mountain climber and sk denotes skier.
Knowledge representation in the given problem is as follows:
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.


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.
Interpretation in Logic
Logical expressions or formulae are “FORMS”
(placeholders) for whom contents are created
through interpretation.
Example:

F F (a)  b xP( x)  F ( x)  g x, F h( x)


This is a Second Order Predicate Calculus
formula.
Quantification on ‘F’ which is a function.
Examples


Interpretation – 1
a and b ∈ N
In particular, a = 0 and b = 1
x∈N
P(x) stands for x > 0
g(m,n) stands for (m x n)
h(x) stands for (x – 1)
Above interpretation defines Factorial
Examples (contd.)

Interpretation – 2
a=b=λ
P(x) stands for “x is a non empty string”
g(m, n) stands for “append head of m
to n”
h(x) stands for tail(x)

Above interpretation defines “reversing a
string”
More Examples


∀x [ P(x) → Q(x)]
Following interpretations conform to
above expression:
man(x) → mortal(x)
dog(x) → mammal(x)
prime(x) → 2_or_odd(x)
CS(x) → bad_hand_writing(x)
Structure of Interpretation


All interpretations begin
with a domain ‘D’,
constants (0-order
functions) and functions
pick values from there.
With respect to ∀x [ P(x) → Q(x)]
D = {living beings}
P: D → {T, F}
Q: D → {T, F}

P can be looked upon
as a table shown here:
Elements of
D i.e. x
P(x)
Ram
T
Pushpak
T
Virus1201
F
Factorial: interpretation of the
structure





D = {0, 1, 2, …, ∞}
a = 0, b = 1
g(m, n) = m x n and g: D x D → D
h(x) = (x – 1) and h(0) = 0, h: D →D
P(x) is x > 0 and P: D → {T, F}
Steps in Interpretation
1.
2.
3.
4.



Fix Domain D
Assign values to constants
Define functions
Define predicates
An expression which is true for all interpretations
is called valid or tautology.
Interpretations and their validity in “Herbrand’s
Universe” is sufficient for proving validity in
Predicate Calculus.
Note:- Possible seminar topic – “Herbrand’s
Interpretation and Validity”