1. CS 621 Artificial Intelligence Lecture 19 - 30/09/05 Prof
CS 621 Artificial Intelligence Lecture /09/05 Prof. Pushpak Bhattacharyya
Completion of “Completeness” Proof
Prof. Pushpak Bhattacharyya, IIT Bombay

2. Prof. Pushpak Bhattacharyya, IIT Bombay
Syntax & Semantics
Axioms,
Inference
Rules,
theorems
(correctness)
soundness
Reductions,
Truth table,
tautology
Completeness
(power)
syntax
semantics
Prof. Pushpak Bhattacharyya, IIT Bombay

3. Completeness of Hilbert’s Formalism of Prop. Calculus
Motivating example
p  (p ν q) is a tautology.
To show that
p  (p ν q) is a theorem.
i.e ├ (p  (p ν q))
Prof. Pushpak Bhattacharyya, IIT Bombay

4. Prof. Pushpak Bhattacharyya, IIT Bombay
Truth Table
v(p) v(q) v(p  (p ν q)) {say A}
T T T
T F T
F T T
F F T
Prof. Pushpak Bhattacharyya, IIT Bombay

5. Prof. Pushpak Bhattacharyya, IIT Bombay
Truth Table (Contd.)
For every row of the truth table, prove that
p', q'  A'
where A = p  (p ν q)
p' = p if v(p) = T
= ~p if v(p) = F
A' = A if v(A) = T
= ~A if v(A) = F
Prof. Pushpak Bhattacharyya, IIT Bombay

6. Prof. Pushpak Bhattacharyya, IIT Bombay
1st Row
v(p) = T, v(q) = T
v(A) = T
A = p  (p ν q)
To show
p, q ├ p  (p ν q)
p ν q is (p  ℱ)  q
Prof. Pushpak Bhattacharyya, IIT Bombay

7. Prof. Pushpak Bhattacharyya, IIT Bombay
To Show that
To show
p, q ├ p  ((p  ℱ)  q))
p, q, p, p  ℱ, q  ℱ ├ ℱ
By repeated application of D.T.
Prof. Pushpak Bhattacharyya, IIT Bombay

8. Prof. Pushpak Bhattacharyya, IIT Bombay
2nd Row
v(p) = T, v(q) = F
v(A) = T
To show
p, (q  ℱ) ├ p  ((p  ℱ)  q))
i.e. p, (q  ℱ), p, p  ℱ ├ ℱ
├ ℱ  q
├ q
By repeated D.T proved
Prof. Pushpak Bhattacharyya, IIT Bombay

9. Prof. Pushpak Bhattacharyya, IIT Bombay
3rd Row
v(p) = F, v(q) = T, v(A) = T
To show
(p  ℱ), q ├ p  ((p  ℱ)  q)
(p  ℱ), q, p, p  ℱ ├ q
Hence proved
Prof. Pushpak Bhattacharyya, IIT Bombay

10. Prof. Pushpak Bhattacharyya, IIT Bombay
4th Row
v(p) = F, v(q) = F, v(A) = T
To show
(p  ℱ), (q  ℱ)├ p  ((p  ℱ)  q)
p  ℱ, q  ℱ, p, p  ℱ ├ ℱ
├ ℱ  q
├ q
Hence proved
Prof. Pushpak Bhattacharyya, IIT Bombay

11. Prof. Pushpak Bhattacharyya, IIT Bombay
Finally
Thus,
p, q ├ A - (1)
p, ~q ├ A - (2)
~p, q ├ A - (3)
~p, ~q ├ A - (4)
where A is p  (p ν q)
Prof. Pushpak Bhattacharyya, IIT Bombay

12. Prof. Pushpak Bhattacharyya, IIT Bombay
Lemma
(A  B)  ((~A  B)  B) is a theorem
Proof
to show
├ (A  B)  ((~A  B)  B)
Prof. Pushpak Bhattacharyya, IIT Bombay

13. Prof. Pushpak Bhattacharyya, IIT Bombay
To show
To show
A B, ~A B ├ B
Or to show
A  B, ~A  B, B ℱ├ ℱ
We use the fact that
(A  B) (~B  ~A) is a theorem
Prof. Pushpak Bhattacharyya, IIT Bombay

14. Prof. Pushpak Bhattacharyya, IIT Bombay
A  B : L1
~A  B : L2
B  ℱ : L3
(A  B)  (~B  ~A) :L4
~B  ~A, MP L1,L4 : L5
~A, MP L3,L5 :L6
B, MP L2,L6 : L7
ℱ, MP L3,L7
Thus, A  B, ~A  B, ~B  ℱ
By repeated D.T.
├ (A  B)  ((~A  B)  B)
Prof. Pushpak Bhattacharyya, IIT Bombay

15. Prof. Pushpak Bhattacharyya, IIT Bombay
Wanted to Prove
Wanted to prove
├ A, if A is a tautology
Already shown
p, q ├ A - (1)
p, ~q ├ A - (2)
~ p, q ├ A - (3)
~ p, ~ q ├ A - (4)
Prof. Pushpak Bhattacharyya, IIT Bombay

16. Prof. Pushpak Bhattacharyya, IIT Bombay
Proof - 1
Use (1) & (2)
p, q ├ A
p, ~q ├ A
i.e.
p ├ (q  A)
p ├ (~ q  A)
Now (q  A)  ((~ q  A)  A)
So, p ├ A
Lemma
Prof. Pushpak Bhattacharyya, IIT Bombay

17. Prof. Pushpak Bhattacharyya, IIT Bombay
Proof - 2
Thus q has been eliminated from (1) & (2) to give
p ├ A -(5)
Using (3) & (4)
~p ├ A -(6)
From (5) & (6)
├ A
Prof. Pushpak Bhattacharyya, IIT Bombay

18. Prof. Pushpak Bhattacharyya, IIT Bombay
Proof - 3
Thus
p  (p ν q) which is a tautology
is shown to have a proof
i.e.
├ p  (p ν q)
Prof. Pushpak Bhattacharyya, IIT Bombay

19. Prof. Pushpak Bhattacharyya, IIT Bombay
Completeness Theorem
If a wff A is a tautology then ├ A
Prof. Pushpak Bhattacharyya, IIT Bombay

20. Prof. Pushpak Bhattacharyya, IIT Bombay
Truth Table of A
p1 p2 p3 …. pn A
T T T T T T
T T T T F T
F F F F F T
Total of 2n rows.
Prof. Pushpak Bhattacharyya, IIT Bombay

21. Prof. Pushpak Bhattacharyya, IIT Bombay
Proof Sketch
p1' , p2' , p3' …. pn' ├ A'
Where pi' = pi if v(Pi) = T
= ~pi if v(Pi) = F
A' = A if v(A) = T
= ~A if v(A) = F
Prof. Pushpak Bhattacharyya, IIT Bombay

22. Prof. Pushpak Bhattacharyya, IIT Bombay
Proof By Induction
Proof by Induction on the number of ‘’ symbols in A (say k)
Basis – k = 0
A is ℱ
A is p (a literal)
Prof. Pushpak Bhattacharyya, IIT Bombay

23. Prof. Pushpak Bhattacharyya, IIT Bombay
Proof (Contd 1)
To show
├ A' , V(A) = V(ℱ) = F
i.e. ├ ~A
├ ~ℱ
├ ℱℱ
Proved since a theorem
Prof. Pushpak Bhattacharyya, IIT Bombay

24. Prof. Pushpak Bhattacharyya, IIT Bombay
Proof (Contd 2)
A = p
i.e. p ├ A'
i.e. p ├ p
But p  p is a theorem
So, ├ A'
Prof. Pushpak Bhattacharyya, IIT Bombay

25. Prof. Pushpak Bhattacharyya, IIT Bombay
Hypothesis
Suppose the statement is true for number of ‘’ less than or equal to k.
Prof. Pushpak Bhattacharyya, IIT Bombay

26. Prof. Pushpak Bhattacharyya, IIT Bombay
Induction Step
Suppose A has (k+1) ‘’ .
The form of A must be (B C)
where both B & C have <= k ‘’ symbols.
We can show that
B', C'  (B C) '
Where B' = B if v(B) = T
= ~B if v(B) = F
Prof. Pushpak Bhattacharyya, IIT Bombay

27. Induction Step (Contd)
Similarly for C' & (B C) '
Thus
├ A is shown
Completeness of propositional calculus is established.
Prof. Pushpak Bhattacharyya, IIT Bombay

28. Prof. Pushpak Bhattacharyya, IIT Bombay
Discussions
Typically ‘completeness’ proofs are more difficult than ‘soundness’ proofs.
Completeness proof is really an algorithm for proof.
This shows that proof of theorems in propositional calculus can be mechanised.
There is a decision procedure.
Prof. Pushpak Bhattacharyya, IIT Bombay

29. Prof. Pushpak Bhattacharyya, IIT Bombay
Discussions (Contd.)
Heart of all this is the deduction theorem.
D.T. is the product of a higher level intelligence.
Meta statement & meta theorem are distinct from statement & theorem.
‘Aboutness’ & ‘Withiness’
Prof. Pushpak Bhattacharyya, IIT Bombay