CS344 : Introduction to Artificial Intelligence

1
CS344 Introduction to Artificial Intelligence

• Pushpak BhattacharyyaCSE Dept., IIT Bombay
• Lecture 15, 16, 17- Completeness Proof Self
  References and Paradoxes
• Week of 9/2/09 (with a quiz on 9th)

2
Soundness, Completeness Consistency
Soundness
Semantic World ---------- Valuation, Tautology
Syntactic World ---------- Theorems, Proofs
Completeness


3
An example to illustrate the completeness proof
4
Running the completeness proof

• For every row of the truth table set up a proof
• p, q - p?(p V q)
• p, q - p?(p V q)
• p, q - p?(p V q)
• p, q - p?(p V q)

5
Completeness Proof
6

• We have a truth table with 2n rows
• P1 P2 P3 . . . Pn A
• F F F . . . F T
• F F F . . . T T
• .
• .
• .
• T T T . . . T T

7

• If we can 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

8
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 P1, P2, , Pn -- A, where
• A A if V(A) true
• A otherwise
• Similarly for each Pi

9
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 n1 ?

10
Proof of Lemma (contd.)

• Induction
• By hypothesis,
• P1, P2, , Pn -- B
• P1, P2, , Pn -- 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.

11
Proof of Theorem

• A is a tautology.
• There are 2n models corresponding to P1, P2, ,
  Pn propositions.
• Consider,
• P1, P2, , Pn -- A
• and P1, P2, , Pn -- A
• P1, P2, , Pn-1 -- Pn ? A
• and 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

12
Self Reference and Paradoxes
13
Paradox -1

• This statement is false
• The truth of this cannot be decided

14
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

15
Paradox -3 (Richardian Paradox)

• Order the statements about properties of number
  in same order. E.g.,
• A prime no. is one that is divisible by itself
  of 1.
• A square no. is one that is product of 2
  identical numbers.
• .
• .

16
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.

17
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.

18

• All these paradoxes came because of
• Self reference
• Confusion between what is inside a system and
  what is outside

19
Answer to Quiz-1 (9/2/09)
20
Question proof of MR theorem

• Prove that if monotone restriction is satisfied,
• the path to a node taken up for expansion is
• optimal.
• i.e., for a node n being expanded,
• g(n) g(n)

Arpits Maheswaris contribution in the proof is
acknowledged
21
Proof

• Lemma If monotone restriction is satisfied, f
  values of
• nodes expanded is non decreasing.
• i.e., if nq is expanded after np, f(np) f(nq)
• Proof of Lemma
• Case 1 nq is child of np
• g(nq) h(nq) g(np) c(np, nq) h(nq)
• but, c(np,
  nq) h(nq) h(np)
• hence, f(nq) g(np) h(np)
• 
  f(np)

np
h(np)
nq
g
h(nq)
22
Proof of Lemma

• Case 2 np and nq are both in open list, nq
  expanded after np
• by definition of A,
• f(np) f(nq)
• Case 3 nq did not exist on open list when np
  was expanded. But there was another node nr which
  led to nq later
• i.e., f(np) f(nr) . . . f(nq)
• Thus, the lemma is proved.

23
Actual Proof of the Theorem

• Proof by induction on the length of the path to
  an expanded node n. nq is in OL.
• Base Case k
  0
• g(s) g(s)
• statement is true
• Hypothesis Suppose the statement is
  true
• for length k.
• i.e., g(np) g(nk) g(np)

s
All in closed list
k
n1
nq
np
n
24
Actual Proof of the Theorem

• Induction To examine g(n) g(nk1)
• By lemma,
• f(n) f(nq)
• i.e., g(n) h(n) g(nq) h(nq)
• c(np, n) g(np) h(n) g(nq) h(nq)
• Add c(nq, n) to both sides,
• c(np, n) g(np) h(n) c(nq, n) g(nq)
  c(nq, n) h(nq)
• But, h(n) c(nq, n) h(nq)
• hence, c(np, n) g(np) g(nq) c(nq, n)
• Thus, g(n) g(n)

g(n) through np
g(n) through nq
nq
h(nq)
n
g
h(n)
25
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.