CS621 : Artificial Intelligence Pushpak Bhattacharyya CSE Dept., IIT Bombay Lecture 12- Completeness Proof; Self References and Paradoxes 16 th August, 2010
Feb 23, 2016
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