Lecture 09 -_davis_putnam
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The Davis-Putnam procedure is an efficient method for checking the satisfiability of propositional formulae by refutation. It works by first negating the formula and converting it to conjunctive normal form. Rules are then iterated to derive contradictory literals, showing the negated formula is unsatisfiable and the original formula is valid. The rules include deleting clauses if a literal appears alone, removing subsumed clauses, and splitting the clause set based on choosing a literal. Examples demonstrate applying the rules to derive contradictions.
![5. ( (p ∧ q) ∨ (r ⇒ s) ) ⇒ ( (p ∨ (r ⇒ s)) ∧ (q ∨ (r ⇒ s) ) )
Negate and Convert to CNF
( (p ∧ q) ∨ (¬r ∨ s) ) ∧ ¬ [ (r ⇒ s) ∨ (p ∧ q) ]
(p ∨ ¬r ∨ s) ∧ (q ∨ ¬r ∨ s) ∧ ¬ [(¬ r ∨ s) ∨ (p ∧ q)]
(p ∨ ¬r ∨ s) ∧ (q ∨ ¬r ∨ s) ∧ r ∧¬ s ∧ (¬p ∨ ¬q)(p ∨ ¬r ∨ s) ∧ (q ∨ ¬r ∨ s) ∧ r ∧¬ s ∧ (¬p ∨ ¬q)
p ∨ ¬r ∨ s
q ∨ ¬r ∨ s
r
¬s
¬p ∨ ¬q
17
By Rule (c) using 3.
p ∨ s
q ∨ s
¬s
By Rule (c) using 1.
q
¬q
5. ( (p ∧ q) ∨ (r ⇒ s) ) ⇒ ( (p ∨ (r ⇒ s)) ∧ (q ∨ (r ⇒ s) ) )
¬s
¬p ∨ ¬q
By Rule (c) using 3.
p
q
¬p ∨ ¬q
Since complementary
literals have been
derived, the negated
formula is
unsatisifiable, thus the
original formula is
valid.
18
[((p ∨ ¬q) ⇒ r) ∧ (s ⇒ (t ∧ u)) ∧ (s ∧ p)] ⇒ (r ∧ t)
Negate and Convert to CNF
[( ¬¬¬¬ (p ∨∨∨∨ ¬¬¬¬q) ∨∨∨∨ r) ∧∧∧∧ (¬¬¬¬s ∨∨∨∨ (t ∧∧∧∧ u)) ∧∧∧∧ (s ∧∧∧∧ p)] ∧∧∧∧ ¬¬¬¬ (r ∧∧∧∧ t)
[ ((¬¬¬¬ p ∧∧∧∧ q) ∨∨∨∨ r) ∧∧∧∧ (¬¬¬¬s ∨∨∨∨ (t ∧∧∧∧ u)) ∧∧∧∧ (s ∧∧∧∧ p)] ∧∧∧∧ ¬¬¬¬ (r ∧∧∧∧ t)
(¬¬¬¬ p ∨∨∨∨ r) ∧∧∧∧ (q ∨∨∨∨ r) ∧∧∧∧ (¬¬¬¬s ∨∨∨∨ t) ∧∧∧∧ (¬¬¬¬s ∨∨∨∨ u) ∧∧∧∧ (s ∧∧∧∧ p) ∧∧∧∧ (¬¬¬¬ r ∨∨∨∨ ¬¬¬¬ t)(¬¬¬¬ p ∨∨∨∨ r) ∧∧∧∧ (q ∨∨∨∨ r) ∧∧∧∧ (¬¬¬¬s ∨∨∨∨ t) ∧∧∧∧ (¬¬¬¬s ∨∨∨∨ u) ∧∧∧∧ (s ∧∧∧∧ p) ∧∧∧∧ (¬¬¬¬ r ∨∨∨∨ ¬¬¬¬ t)
¬¬¬¬ p ∨∨∨∨ r
q ∨∨∨∨ r
¬¬¬¬s ∨∨∨∨ t
¬¬¬¬s ∨∨∨∨ u
s
p
¬¬¬¬r ∨∨∨∨ ¬¬¬¬ t 19](https://image.slidesharecdn.com/lecture09-davisputnam-150424103353-conversion-gate01/85/Lecture-09-_davis_putnam-5-320.jpg)
Lecture 09 -_davis_putnam
- 1. Davis-Putnam Procedure 1 Lecture 9 Davis-Putnam Procedure The Davis-Putnam method is the most efficient method for deciding the validity of propositional formulae. This procedure provides a mechanism for checking the satisfiability of a set of formulae. 2 the satisfiability of a set of formulae. It is a proof by refutation method. The first step of the procedure is to negate the formula to be proved valid. Then it is converted into conjunctive normal form. Next some rules are iterated through until all the elements of the set are contradictory which means that the negated formula is unsatisfiable. Rules a) If literal l is in a clause but no occurrence of lc appears in any clause, delete all clauses containing l. b) Remove any clause which is subsumed by any other clause. 3 other clause. c) If a one literal clause, say l is present, delete all clauses containing l and remove each occurrence of lc in any other clause. d) If (a) or (c) cannot be applied, choose a literal l, such that l, lc appear in some clauses and replace the clause set by 2 sets of clauses. i. The first set formed as in (c ). ii. The set of clauses with all clauses containing lc removed and l removed from all remaining. Example 1 1. p ∨∨∨∨ ¬¬¬¬ q ∨∨∨∨ r 2. ¬¬¬¬ p 3. ¬¬¬¬ s ∨∨∨∨ ¬¬¬¬ r 4. p ∨∨∨∨ q 5. t ∨∨∨∨ q s By (c) using (2) obtain 1. ¬¬¬¬ q ∨∨∨∨ r 2. ¬¬¬¬ s ∨∨∨∨ ¬¬¬¬ r 3. q 4. s 4 6. s By (a) remove 5. 1. p ∨∨∨∨ ¬¬¬¬ q ∨∨∨∨ r 2. ¬¬¬¬ p 3. ¬¬¬¬ s ∨∨∨∨ ¬¬¬¬ r 4. p ∨∨∨∨ q 5. s By (c) using (3) obtain 1. r 2. ¬¬¬¬ s ∨∨∨∨ ¬¬¬¬ r 3. s By (c ) using (3) obtain 1. r 2. ¬¬¬¬ r
- 2. Example 2 1. p ∨∨∨∨ q 2. p ∨∨∨∨ ¬¬¬¬q 3. ¬¬¬¬ p ∨∨∨∨ r 4. ¬¬¬¬ r ∨∨∨∨ ¬¬¬¬ p By (d) choosing p. Instead we could have choose q 2nd rule 1. p 2. ¬¬¬¬ p ∨∨∨∨ r 1st rule 1. p 2. ¬¬¬¬ p ∨∨∨∨ r 5 By (d) choosing p. 1st rule 1. r 2. ¬¬¬¬ r 2nd rule 1. q 2. ¬¬¬¬ q 2. ¬¬¬¬ p ∨∨∨∨ r 3. ¬¬¬¬ r ∨∨∨∨ ¬¬¬¬ p By (c) using (1) 1. r 2. ¬¬¬¬ r 2. ¬¬¬¬ p ∨∨∨∨ r 3. ¬¬¬¬ r ∨∨∨∨ ¬¬¬¬ p By (c) using (1) 1. r 2. ¬¬¬¬ r Exercise Apply the Davis and Putnam procedure to the following set of clauses 6 clauses 1. s 2. ¬p 3. p ∨ q 4. ¬ s ∨ ¬ r 5. p ∨ ¬q ∨ r Exercise 1. s 2. ¬¬¬¬p 3. p ∨∨∨∨ q 4. ¬¬¬¬ s ∨∨∨∨ ¬¬¬¬ r 5. p ∨∨∨∨ ¬¬¬¬q ∨∨∨∨ r By (c) using (1) 1. q 2. ¬¬¬¬ r 3. ¬¬¬¬q ∨∨∨∨ r 7 5. p ∨∨∨∨ ¬¬¬¬q ∨∨∨∨ r By (c) using (1) 1. ¬¬¬¬p 2. p ∨∨∨∨ q 3. ¬¬¬¬ r 4. p ∨∨∨∨ ¬¬¬¬q ∨∨∨∨ r 3. ¬¬¬¬q ∨∨∨∨ r By (c) using (1) 1. ¬¬¬¬ r 2. r Exercise Use the Davis-Putnam procedure to establish the validity of: 8
- 3. 1. ((p→ q) ∧ (q→ r)) → ¬ (¬ r ∧ p) Negate & convert to CNF ((p→ q) ∧ (q→ r)) ∧ ¬¬(¬ r ∧ p) (¬p ∨ q) ∧ (¬q ∨ r) ∧ ¬r ∧ p 9 (¬p ∨ q) ∧ (¬q ∨ r) ∧ ¬r ∧ p 1. ¬p ∨ q 2. ¬q ∨ r 3. ¬r 4. p 1. ((p→ q) ∧ (q→ r)) → ¬ (¬ r ∧ p) By (c) using (3) 1. ¬p ∨ q 2. ¬q By (c) using (2) 1. ¬p 2. p 10 2. ¬q 3. p 2. p Since complementary literals have been derived, the negated formula is unsatisifiable, thus the original formula is valid. 2. ((p→ q) ∧ (r→ s) ∧ (p ∨ r)) → (q ∨ s) Negate & convert to CNF (¬¬¬¬p ∨∨∨∨ q) ∧∧∧∧ (¬¬¬¬r ∨∨∨∨ s) ∧∧∧∧ (p ∨∨∨∨ r) ∧∧∧∧ (¬¬¬¬q ∧∧∧∧ ¬¬¬¬s) 1. ¬¬¬¬p ∨∨∨∨ q 11 1. ¬¬¬¬p ∨∨∨∨ q 2. ¬¬¬¬r ∨∨∨∨ s 3. p ∨∨∨∨ r 4. ¬¬¬¬q 5. ¬¬¬¬s 2. ((p→ q) ∧ (r→ s) ∧ (p ∨ r)) → (q ∨ s) By (c) using (4) 1. ¬¬¬¬p 2. ¬¬¬¬r ∨∨∨∨ s 3. p ∨∨∨∨ r By (c) using (2) 1. s 2. ¬¬¬¬s 12 3. p ∨∨∨∨ r 4. ¬¬¬¬s By (c) using (1) 1. ¬¬¬¬r ∨∨∨∨ s 2. r 3. ¬¬¬¬s Since complementary literals have been derived, the negated formula is unsatisifiable, thus the original formula is valid.
- 4. 3. ¬(p → q) ∨ r ∨ ¬ ((p ∧ ¬ q) ∨ (p ∧ r)) Negate & convert to CNF (p → q) ∧ ¬r ∧ ¬¬((p ∧ ¬q) ∨ (p ∧ r)) (¬p ∨ q) ∧ ¬r ∧ ((p ∧ ¬q) ∨ (p ∧ r)) 13 (¬p ∨ q) ∧ ¬r ∧ ((p ∧ ¬q) ∨ (p ∧ r)) (¬p ∨ q) ∧ ¬r ∧ (p ∧ (¬q ∨ r)) 1. ¬p ∨ q 2. ¬r 3. p 4. ¬q ∨ r 3. ¬(p → q) ∨ r ∨ ¬ ((p ∧ ¬ q) ∨ (p ∧ r)) By (c) using (2) 1. ¬p ∨ q 2. p By (c) using (2) 1. q 2. ¬q 14 2. p 3. ¬q 2. ¬q Since complementary literals have been derived, the negated formula is unsatisifiable, thus the original formula is valid. 4. ((p ∧ ¬ q) ∧ (p ∨ r)) → (q ∧ r) Negate & convert to CNF ((p ∧ ¬ q) ∧ (p ∨ r)) ∧ (¬q ∨ ¬r) 1. p 15 1. p 2. ¬ q 3. p ∨ r 4. ¬q ∨ ¬r 4. ((p ∧ ¬ q) ∧ (p ∨ r)) → (q ∧ r) By (a) 1. ¬q 2. ¬q ∨ ¬r 16 2. ¬q ∨ ¬r By (a) 1. { } Since complementary literals have not been derived, the negated formula is satisifiable, thus the original formula is not valid.
- 5. 5. ( (p ∧ q) ∨ (r ⇒ s) ) ⇒ ( (p ∨ (r ⇒ s)) ∧ (q ∨ (r ⇒ s) ) ) Negate and Convert to CNF ( (p ∧ q) ∨ (¬r ∨ s) ) ∧ ¬ [ (r ⇒ s) ∨ (p ∧ q) ] (p ∨ ¬r ∨ s) ∧ (q ∨ ¬r ∨ s) ∧ ¬ [(¬ r ∨ s) ∨ (p ∧ q)] (p ∨ ¬r ∨ s) ∧ (q ∨ ¬r ∨ s) ∧ r ∧¬ s ∧ (¬p ∨ ¬q)(p ∨ ¬r ∨ s) ∧ (q ∨ ¬r ∨ s) ∧ r ∧¬ s ∧ (¬p ∨ ¬q) p ∨ ¬r ∨ s q ∨ ¬r ∨ s r ¬s ¬p ∨ ¬q 17 By Rule (c) using 3. p ∨ s q ∨ s ¬s By Rule (c) using 1. q ¬q 5. ( (p ∧ q) ∨ (r ⇒ s) ) ⇒ ( (p ∨ (r ⇒ s)) ∧ (q ∨ (r ⇒ s) ) ) ¬s ¬p ∨ ¬q By Rule (c) using 3. p q ¬p ∨ ¬q Since complementary literals have been derived, the negated formula is unsatisifiable, thus the original formula is valid. 18 [((p ∨ ¬q) ⇒ r) ∧ (s ⇒ (t ∧ u)) ∧ (s ∧ p)] ⇒ (r ∧ t) Negate and Convert to CNF [( ¬¬¬¬ (p ∨∨∨∨ ¬¬¬¬q) ∨∨∨∨ r) ∧∧∧∧ (¬¬¬¬s ∨∨∨∨ (t ∧∧∧∧ u)) ∧∧∧∧ (s ∧∧∧∧ p)] ∧∧∧∧ ¬¬¬¬ (r ∧∧∧∧ t) [ ((¬¬¬¬ p ∧∧∧∧ q) ∨∨∨∨ r) ∧∧∧∧ (¬¬¬¬s ∨∨∨∨ (t ∧∧∧∧ u)) ∧∧∧∧ (s ∧∧∧∧ p)] ∧∧∧∧ ¬¬¬¬ (r ∧∧∧∧ t) (¬¬¬¬ p ∨∨∨∨ r) ∧∧∧∧ (q ∨∨∨∨ r) ∧∧∧∧ (¬¬¬¬s ∨∨∨∨ t) ∧∧∧∧ (¬¬¬¬s ∨∨∨∨ u) ∧∧∧∧ (s ∧∧∧∧ p) ∧∧∧∧ (¬¬¬¬ r ∨∨∨∨ ¬¬¬¬ t)(¬¬¬¬ p ∨∨∨∨ r) ∧∧∧∧ (q ∨∨∨∨ r) ∧∧∧∧ (¬¬¬¬s ∨∨∨∨ t) ∧∧∧∧ (¬¬¬¬s ∨∨∨∨ u) ∧∧∧∧ (s ∧∧∧∧ p) ∧∧∧∧ (¬¬¬¬ r ∨∨∨∨ ¬¬¬¬ t) ¬¬¬¬ p ∨∨∨∨ r q ∨∨∨∨ r ¬¬¬¬s ∨∨∨∨ t ¬¬¬¬s ∨∨∨∨ u s p ¬¬¬¬r ∨∨∨∨ ¬¬¬¬ t 19