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Logic as a Tool
Chapter 2: Deductive Reasoning in Propositional Logic
2.5 Normal forms of propositional formulae
Propositional Resolution
Valentin Goranko
Stockholm University
October 2016
Goranko
Disjunctive and conjunctive normal forms: basic definitions
1. A literal is a propositional constant or variable or its negation.
2. An elementary disjunction (resp., elementary conjunction) is a
disjunction (resp., conjunction) of one or more literals. Examples:
p, ¬q, p ∨ ¬q, p ∨ ¬p ∨ q ∨ ¬r are elementary disjunctions;
p, ¬q, ¬p ∧ q, ¬p ∧ q ∧ ¬r ∧ ¬p are elementary conjunctions.
3. A disjunctive normal form (DNF) is a disjunction of elementary
conjunctions. Examples:
p, ¬q, p ∧ ¬q, p ∨ ¬q, (p ∧ ¬p) ∨ ¬q,
(r ∧ q ∧ ¬p) ∨ (¬q ∧ p) ∨ (¬r ∧ p).
4. A conjunctive normal form (CNF) is a conjunction of elementary
disjunctions. Examples:
p, ¬q, p ∧ ¬q, p ∨ ¬q, p ∧ (¬p ∨ ¬q),
(r ∨ q ∨ ¬r ) ∧ ¬q ∧ (¬p ∨ r ).
Goranko
Algorithm for equivalent transformation to CNF/DNF
Theorem (Conjunctive normal form)
Every propositional formula is equivalent to a disjunctive normal form
and to a conjunctive normal form.
Algorithm transforming a formula into a DNF, respectively CNF:
1. Eliminate all occurrences of ↔ and → using the equivalences
A ↔ B ≡ (A → B) ∧ (B → A)
and A → B ≡ ¬A ∨ B.
2. Transform to negation normal form by using the relevant
equivalences.
3. For a DNF: distribute all conjunctions over disjunctions using
p ∧ (q ∨ r ) ≡ (p ∧ q) ∨ (p ∧ r ).
4. For a CNF: distribute all disjunctions over conjunctions using
p ∨ (q ∧ r ) ≡ (p ∨ q) ∧ (p ∨ r ).
Goranko
Some useful simplifications
Throughout this process the formulae can be simplified by using
commutativity, associativity, and idempotency of ∨ and, as well as:
I p ∨ ¬p ≡ >; p ∧ ¬p ≡ ⊥;
I p ∧ > ≡ p; p ∧ ⊥ ≡ ⊥;
I p ∨ > ≡ >; p ∨ ⊥ ≡ p.
Goranko
Transformation to DNF and CNF: example
(p ∧ ¬r ) → (p ↔ ¬q)
≡ (p ∧ ¬r ) → ((p → ¬q) ∧ (¬q → p)) (eliminating ↔)
≡ ¬(p ∧ ¬r ) ∨ ((¬p ∨ ¬q) ∧ (¬¬q ∨ p)) (eliminating →)
≡ (¬p ∨ ¬¬r ) ∨ ((¬p ∨ ¬q) ∧ (q ∨ p)) (driving ¬ inside)
≡ ¬p ∨ r ∨ ((¬p ∨ ¬q) ∧ (q ∨ p))
For a DNF we further distribute ∧ over ∨ and simplify:
≡ ¬p ∨ r ∨ (((¬p ∨ ¬q) ∧ q) ∨ ((¬p ∨ ¬q) ∧ p))
≡ ¬p ∨ r ∨ ((¬p ∧ q) ∨ (¬q ∧ q)) ∨ ((¬p ∧ p) ∨ (¬q ∧ p))
≡ ¬p ∨ r ∨ ((¬p ∧ q) ∨ ⊥) ∨ (⊥ ∨ (¬q ∧ p))
≡ ¬p ∨ r ∨ (¬p ∧ q) ∨ (¬q ∧ p).
For a CNF we distribute ∨ over ∧ and simplify:
≡ (¬p ∨ r ∨ ¬p ∨ ¬q) ∧ (¬p ∨ r ∨ q ∨ p)
≡ (¬p ∨ r ∨ ¬q) ∧ (> ∨ r ∨ q)
≡ (¬p ∨ r ∨ ¬q) ∧ >
≡ ¬p ∨ r ∨ ¬q. (Note that this is a DNF, too.)
Goranko
The rule of Propositional Resolution
RES
A ∨ C , B ∨ ¬C
A∨B
The formula A ∨ B is called a resolvent of A ∨ C and B ∨ ¬C ,
denoted Res(A ∨ C , B ∨ ¬C ).
Exercise: Show that the Resolution rule is logically sound.
That is, if both premises are valid then the conclusion is valid.
Moreover, RES preserves satisfiability:
Exercise: Show that if the set of premises is (simultaneously)
satisfiable, then it is satisfiable together with the conclusion.
Goranko
Clausal normal forms
• A clause is essentially an elementary disjunction l1 ∨ . . . ∨ ln ,
but written as a (possibly empty) set of literals {l1 , . . . , ln }.
• The empty clause {} is a clause containing no literals.
• A unit clause is a clause containing only one literal.
• A clausal form is a (possibly empty) set of clauses, written as a list:
C1 . . . Ck . It represents the conjunction of these clauses.
Thus, every CNF can be re-written in a clausal form, and therefore every
propositional formula is equivalent to one in a clausal form.
Example: the clausal form of the CNF-formula
(p ∨ ¬q ∨ ¬r ) ∧ ¬p ∧ (¬q ∨ r ) is {p, ¬q, ¬r }{¬p}{¬q, r }.
Note that the empty clause {} is not satisfiable (being an empty
disjunction), while the empty set of clauses ∅ is satisfied by any truth
assignment (being an empty conjunction).
Goranko
Clausal Propositional Resolution rule
The Propositional Resolution rule can be rewritten for clauses:
CL − RES
{A1 , . . . , C , . . . , Am } {B1 , . . . , ¬C , . . . , Bn }
.
{A1 , . . . , Am , B1 , . . . , Bn }
The clause {A1 , . . . , Am , B1 , . . . , Bn } is called a resolvent of the clauses
{A1 , . . . , C , . . . , Am } and {B1 , . . . , ¬C , . . . , Bn }.
Example
{p, q, ¬r } {¬q, ¬r }
,
{p, ¬r , ¬r }
{¬p, q, ¬r } {r }
,
{¬p, q}
{¬p} {p}
.
{}
Goranko
Some remarks
Note that two clauses can have more than one resolvent, e.g.:
{p, ¬q}{¬p, q} {p, ¬q}{¬p, q}
,
.
{p, ¬p}
{¬q, q}
However, it is wrong to apply the Propositional Resolution rule for both
pairs of complementary literals simultaneously and obtain
{p, ¬q}{¬p, q}
.
{}
Sometimes, the resolvent can (and should) be simplified, by removing
duplicated literals on the fly:
{A1 , . . . , C , C , . . . , Am } ⇒ {A1 , . . . , C , . . . , Am }.
For instance:
{p, ¬q, ¬r }{q, ¬r }
{p, ¬q, ¬r }{q, ¬r }
instead of
{p, ¬r }
{p, ¬r , ¬r }
Goranko
Propositional resolution as a deductive system
The underlying idea of Propositional Resolution: in order to prove the
validity of a logical consequence A1 , . . . , An |= B, show that there is no
truth assignment which falsifies it, i.e., show that the formulae
A1 , . . . , An and ¬B cannot be satisfied simultaneously.
That is done by transforming the formulae A1 , . . . , An and ¬B to a
clausal form, and then using repeatedly the Propositional Resolution rule
in attempt to derive the empty clause {}.
Since {} is not satisfiable, its derivation means that A1 , . . . , An and ¬B
cannot be satisfied together.
Then, the logical consequence A1 , . . . , An |= B holds.
Alternatively, after finitely many applications of the Propositional
Resolution rule, no new applications of the rule remain possible.
If the empty clause is not derived by then, it cannot be derived at all,
and hence the A1 , . . . , An and ¬B can be satisfied together,
so the logical consequence A1 , . . . , An |= B does not hold.
Goranko
Propositional resolution:
how to construct a satisfying assignment?
Suppose the Propositional Resolution applied to a given input set of
clauses ends with a set of non-empty clauses (such that no more clauses
can be derived from it).
Question:
How to construct a satisfying assignment for that set of clauses?
Goranko
Propositional resolution derivation: Example 1
Check whether p → q, q → r |= p → r holds.
First, transform p → q, q → r , ¬(p → r ) to clausal form:
C1 = {¬p, q}, C2 = {¬q, r }, C3 = {p}, C4 = {¬r }.
Now, applying Propositional Resolution successively:
C5 = Res(C1 , C3 ) = {q};
C6 = Res(C2 , C5 ) = {r };
C7 = Res(C4 , C6 ) = {}.
The derivation of the empty clause completes the proof.
Goranko
Propositional resolution derivation: Example 2
Check whether (¬p → q), ¬r p ∨ (¬q ∧ ¬r ) holds.
First, transform (¬p → q), ¬r , ¬(p ∨ (¬q ∧ ¬r )) to clausal form:
C1 = {p, q}, C2 = {¬r }, C3 = {¬p}, C4 = {q, r }.
Now, applying Propositional Resolution successively:
C5 = Res(C1 , C3 ) = {q};
C6 = Res(C2 , C4 ) = {q};
At this stage, no new applications of the Propositional Resolution rule
are possible, hence the empty clause is not derivable.
Therefore, (¬p → q), ¬r 6 p ∨ (¬q ∧ ¬r ).
A falsifying assignment can be extracted from the unit clauses in the
final set: p : F, q : T, r : F.
Goranko