CS621 : Artificial Intelligence

Pushpak BhattacharyyaCSE Dept., IIT Bombay

Lecture 12- Completeness Proof; Self References and Paradoxes

16th August, 2010

Soundness, Completeness &

Consistency

Syntactic World

----------Theorems,

Proofs

SemanticWorld

----------Valuation,Tautology

Soundness

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

Completeness Proof Statement

If V(A) = T for all V, then |--A i.e. A is a theorem.

In the example A is p(p V q)

We need prove p(p V q) is a theorem given the tautology.

Lemma:If P, Q├ A and P, ~Q├ Athen we show, P ├ A

Proof: To prove this lemma we need a theorem as follows.

Theorem:

( ) (( ) )A B A B B

Proof of Theorem We need to show, ( ) (( ) )A B A B B i.e. ├

i.e. ├ i.e. ├ i.e. ├

i.e. ├

( )A B (( ) )A B B

( ), ( )A B A B B

( ), ( ),A B A B B F

( ), ( ), ,A B A B B A F

( ), ( ), , ,A B A B B A B FBy Modus Ponens on and A B A

and derives F. Hence, the theorem is proved.B B

By Modus tollens on and B A B

Proof of Lemma, i.e. ( ) ---(i)

, i.e. ( )

i.e. ( ) ---(ii) B

P Q A P Q A

P Q A P Q A

P Q A A

├ ├├ ├

├y using the theorem

To prove Our hypothesis is P, so P is true. Hence by (i), we can write ( ) is true.Hence by (ii), we can write ( ) is true.From the above two sentence we can write A i

P A

Q AQ A A

├

s true.

Hence, P A├

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 proofFor every row of the truth table set up a proof:1. p, ~q |- p(p V q) ---(1)2. p, q |- p(p V q) ---(2)3. ~p, q |- p(p V q) ---(3)4. ~p, ~q |- p(p V q) ---(4)5. p |- p(p V q) ---(5) using (1) and (2)6. ~ p |- p(p V q) ---(6) using (3) and (4)7. |- p(p V q) ---(7) using (5) and (6)

Hence p(p V q) is a theorem

Completeness Proof

P1 P2 P3 . . . Pn AF F F . . . F TF F F . . . T T

. . .

T T T . . . T T

We have a truth table with 2n rows

We should show

P1’, P2’, …, Pn’ |- A’

For every row where

Pi’ = Pi if V(Pi) = T = ~Pi if V(Pi) = F

And A’ = A if V(A) = T = ~A if V(A) = F

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’, whereA’ = A if V(A) = true = ~A otherwiseSimilarly for each P’i

Proof for Lemma

Proof by induction on the number of ‘→’ symbols in ABasis: 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 ‘→’

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 Lemma (contd.)

Proof of Theorem A is a tautology. There are 2n models corresponding to P1, P2, …, Pn propositions. Consider,

P1, P2, …, Pn |-- Aand P1, P2, …, ~Pn |-- A

P1, P2, …, Pn-1 |-- Pn → Aand P1, P2, …, Pn-1 |-- ~Pn → A

RHS can be written as:|-- ((Pn → A) → ((~Pn → A) → A))|-- (~Pn → A) → A|-- A

Thus dropping the propositions progressively we show |-- A

Self Reference and Paradoxes

Paradox -1

“This statement is false” The truth of this cannot be decided

Paradox -2 (Russell Paradox or Barber Paradox)

“In a city, a barber B shaves all and only those who do not shave themselves”

Question: Does the barber shave himself?

Cannot be answered

Paradox -3 (Richardian Paradox)

Order the statements about properties of number in same order. E.g.,

1. “A prime no. is one that is divisible by itself of 1.”

2. “A square no. is one that is product of 2 identical numbers.”

.

.

Paradox -3 (Richardian Paradox)

Definition: A number is called Richardian if it does not have the property that it indexes.

For example, in the above arrangement 2 is Richardian because it is not a square no.

Paradox -3 (Richardian Paradox)

Now, suppose in this arrangement M is the number for the definition of Richardian

M : “A no. is called Richardian … “ Question : is M Richardian? Cannot be

answered.

Self Reference: source of paradoxes

All these paradoxes came because of

1. Self reference

2. Confusion between what is inside a system and what is outside