1. CS344 : Introduction to Artificial Intelligence
Pushpak Bhattacharyya CSE Dept., IIT Bombay
Lecture 9- Completeness

2. Formalization of propositional logic (review)
Axioms : A1 A2 A3
Inference rule:
Given and A, write B
A Proof is:
A sequence of
i) Hypotheses
ii) Axioms
iii) Results of MP
A Theorem is an
Expression proved from axioms and inference rules

3. Soundness, Completeness & Consistency
Semantic
World
Valuation,
Tautology
Syntactic
World
Theorems,
Proofs
Completeness * *

4. Soundness
Provability Truth
Completeness
Truth Provability

5. Soundness: Correctness of the System
Proved entities are indeed true/valid
Completeness: Power of the System
True things are indeed provable

6. Tautology An expression ‘E’ is a tautology if V(E) = T
for all valuations of constituent propositions
Each ‘valuation’ is called a ‘model’.

7. Problem (P Q) (P Q) Semantic Proof A B P Q P Q P Q A B T F F T T
T T T T T
F F F F T
F T F T T

8. To show syntactically
(P Q) (P Q)
i.e.
[(P (Q F )) F ]
[(P F ) Q]

9. If we can establish
(P (Q F )) F ,
(P F ), Q F ⊢ F
This is shown as
Q F hypothesis
(Q F ) (P (Q F )) A1

10. QF; hypothesis
(QF)(P(QF)); A1
P(QF); MP
F; MP
Thus we have a proof of the line we started with

11. Completeness

12. Necessary results Statement: (pq)((~pq)q) Proof:
If we can show that
(pq), (~pq) |- q
Or,
(pq), (~pq), qF |- F
Then we are done.

13. Proof continued 1. (pq) H1 2. (~pq) H2 3. qF H3
4. (~pq) (~qp) theorem of contraposition
5. ~qp MP, 2, 4
6. P MP, 3,5
7. q MP, 6, 1
8. F MP,7,3
QED

14. How to prove contraposition
To show
(pq)(~q~p)
Proof:
pq, ~q, p |- F
Very obvious!

15. An example to illustrate the completeness proofq
p(p V q) T F

16. 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)

17. p, ~q |- p  (p V q)
i.e. p, ~q, p |- p V q
p, ~q, p, ~p |- q
p, ~q, p, ~p |- F
|- F  q
|- q

18. p, q |- p  (p V q)
i.e. p, q, p, ~p |- q
same as 1

19. ~p, q |- p  (p V q)
~p, q, p, ~p |- q
Same as 1, since F is derived
4. ~p, ~q |- p  (p V q)

20. Why all this? If we have shown p, q |- A and p, ~q |- A
then we can show that
p |- A

21. p |- (q  A)
also p |- (~q  A)
But (q  A)  ((~q  A)  A)
is a theorem
by MP twice
p |- A

22. General Statement of the completeness proof
If V(A) = T for all models
then
|- A

23. Elaborating,
If P1, P2, …, Pn are constituent propositions of A
and if V(A) = T for every model V(Pi) = T/F
then
|- A

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

25. 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

26. Lemma
If row has
P1’, P2’, …, Pn’, A’
Then
P1’, P2’, …, Pn’ |- A’

27. Check p q p q F F F F T F T F F T T F It is true that p’, q’ |- (p q)’
i.e., ~p, ~q |- ~(p q)

28. Proof of Lemma By Induction on the no. of ‘’ symbols in A Base Case:
A is either Pi or F
A is
V(A) |- A’
F |- A  F
|- F  F

29. A is Pi
V(Pi) V(A)
T T
F F
Pi |- A But |- A A is a theorem
and ~Pi |- ~A which is also fine