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.

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