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Normal forms,resolution methodLesson 4

- Marie Duží

Normal forms of PL formulas

- To each PL formula there is just one truth-value

function (truth table). - However, to each such truth-function there are

infinitely many, mutually equivalent, formulas. - Definition Formulas A, B are equivalent, denoted

A ? B, iff A and B have exactly the same models.

In other words, they express the same truth-value

function. - Thus A ? B iff A B a B A.
- Example p ? q ? ?p ? q ? ?(p ? ?q) ? (p ? q) ?

(p ? ?p) ? (p ? q) ? (p ? ?p) ? ... - Remark Do not confuse equivalence of formulas (A

? B) with a formula of the form (A ? B). - Here ? is a truth connective, whereas ? is a

metasymbol. - It holds that A ? B iff A ? B is a tautology.
- Example (p ? q) ? (p ? q) ? (q ? p) iff
- (p ? q) ? ((p ? q) ? (q ? p))

Example

p q f(p,q)

1 1 1

1 0 0

0 1 0

0 0 1

There are infinitely many formulas corresponding to this function (p ? q) ? (p ? q) ? (q ? p) ?(?p ? q) ? (?q ? p) ? (p ? q) ? (?p ? ?q) ? .

Normal forms in PL

- (p ? q) ? (p ? q) ? (q ? p) ?
- (?p ? q) ? (?q ? p) ?
- (p ? q) ? (?p ? ?q) ? .
- It is useful to find some normal forms of a

formula, that is to choose among those infinitely

many formulas some canonic ones. - The class of equivalent formulas is then

represented by these normal forms formulas. - In our example the formulas in bold (second and

third line) are in normal form.

Some definitions

- Literal is an atomic formula or its negation

p, ?q, r, ... - Elementary conjunction (EC) is a conjunction of

literals p ? ?q, r ? ?r, ... - Elementary disjunction (ED) is a disjunction of

literals p ? ?q, r ? ?r, ... - Complete elementary conjunction (CEC) is an

elementary conjunction in which each

propositional symbol occurs just ones (either

without or with negation) p ? ?q, r ? ?r - Complete elementary disjunction (CED) is an

elementary conjunction in which each

propositional symbol occurs just ones (either

without or with negation) p ? ?q, r ? ?r

Some definitions

- Disjunctive normal form (DNF) of a formula F is a

formula F such that F ? F and F is a

disjunction of elementary conjunctions DNF(p

? p) (p ? p) ? (?p ? ?p), p ? ?p - Conjunctive normal form (CNF) of a formula F is a

formula F such that F ? F and F is a

conjunction of elementary disjunctions KNF(p ?

p) (?p ? p) ? (?p ? p) - Complete disjunctive normal form (CDNF) of a

formula F is a formula F such that F ? F and F

is a disjunction of complete elementary

conjunctions CDNF(p ? q) (p ? q) ? (?p ? ?q) - Complete conjunctive normal form (CCNF) of a

formula F is a formula F such that F ? F and F

is a conjunction of complete elementary

disjunctions CCNF(p ? q) (?p ? q) ? (?q ? p) - CDNF and CCNF are canonic (standard) forms of a

formula.

Normal forms in PL

- How to find a canonical form of a formula?
- CDNF disjunction 1, if at least one CEC 1,

i.e. all its literals must be 1. - CCNF conjunction 0, if at least one CED 0,

i.e. all its literals must be 0. - Thus in order to find CDNF and CCNF we can use a

truth-table - CDNF look at the lines with value 1
- CCNF look at the lines with value 0

CDNF, CCNF ? table

- Formula ?(p?q)
- CDNF p??q
- CCNF
- (?p??q) ? (p??q) ? (p?q)

p q ?(p?q) CEC CED

1 1 0 ?p??q

1 0 1 p??q

0 1 0 p??q

0 0 0 p?q

CDNF, CCNF

- Method of equivalent transformations
- ??p ? q? ? ???p ? q? ? (p ? ?q) UDNF ? p ? (q

? ?q? ? ?q ? (p ? ?p? ? - ? ?p ? q? ? ?p ? ?q? ? ??p ? ?q? UKNF
- Remark Here we use PL tautologies as defined

above. - In the second line we apply the rule A ? F ? A

where F is a contradiction. - In the third line we apply distributive laws
- To each formula that is not a contradiction there

is a CDNF - To each formula that is not a tautology there is

a CCNF.

Reverse task having CDNF, CCNF find a simpler

formula

- Let p, q, r be
- p You will manage to turn lead to gold
- q On April 1st your brother in law will become

an Attorney General - r The verdict will come after April 1st.
- An alchemist in prison obtained 5 messages
- First message p ? q ? r
- Second message p ? q ? ?r
- Third message ?p ? ?q ? r
- Fourth message ?p ? ?q ? ?r
- Fifth message At least one of the previous

messages is true. - Question Which piece of information did the

alchemist obtain? - Solution (p ? q ? r) ? (p ? q ? ?r) ? (?p ? ?q ?

r) ? (?p ? ?q ? ?r). Applying equivalent

transformations we simplify the formula - (p ? q ? r) ? (p ? q ? ?r) ? (?p ? ?q ? r) ? (?p

? ?q ? ?r) ? - (p ? q) ? (r ? ?r) ? (?p ? ?q) ? (r ? ?r) ? (p ?

q) ? (?p ? ?q) ? (p ? q) - Answer You will manage to turn lead to gold if

and only if on April 1st your brother in law

becomes an Attorney General.

How many binary truth functions are there?

p q 0 1 2 3 4 5 6 7 8 9 A B C D E F

1 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1

1 0 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1

0 1 0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1

0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1

? ? ? ? ? ?

NOR

NAND

Which is the minimal number of logical

connectives?

- Functionally complete systems
- 1. ?, ?, ?,
- 2. ?, ? or ?, ?,
- 3. ?, ?,
- ? or ?.
- Thus to express any truth-value function just one

connective is sufficient! - Either Scheffers NAND ? or Pierces NOR ?

Functionally complete systems

- ?, ?, ? is sufficient due to the statement on

normal forms - Transformation to ?, ? or ?, ?A ? B ? ?(?A

? ?B),A ? B ? ?(?A ? ?B) - Transformation to ?, ? A ? B ? ?A ? B,A ? B

? ?(A ? ?B) - Transformation to ? or ?
- ?A ? A?A, A?B ? (A?B)?(A?B),
- ?A ? A?A, A?B ? (A?B)?(A?B).

Semantic tableau a method to create CNF, DNF

- Disjunctive tableau a tree whose leaves are

conjunctions of literals - Conjunctive tableau a tree whose leaves are

disjunctions of literals - A (tautology) ? all the leaves of a

conjunctive tableau must be closed, i.e., contain

opposite literals p, ?p (p ? ?p) tautology - A (contradiction) ? all the leaves of a

disjunctive tableau must be closed, i.e., contain

opposite literals p, ?p (p ? ?p) contradiction

Semantic tableau

- Execute negations move the negation inside to

particular atomic formulas - Transform formulas of an implicative or

equivalence form applying the laws - (p ? q) ? (?p ? q),
- (p ? q) ? (?p ? q) ? (?q ? p) ? (p ? q) ? (?p ?

?q) - Apply distributive laws in order to obtain a

tableau.

Disjunctive normal form

- Create a disjunctive tableau
- branching ? disjunction
- comma ? conjunction
- Proof of a contradiction
- If all the branches are closed, i.e. contain a

pair of opposite literals like p, ?p, which means

p ? ?p, the proven formula is a contradiction.

(?p ? q ? r) ? (s ? ?q) ? (t ? ?r) ? p ? ?s ? ?t

- ?p,(s ? ?q),(t ? ?r),p,?s,?t q,(s ? ?q),(t ?

?r),p,?s,?t - r,(s ? ?q),(t ? ?r),p,?s,?t
- q,s,(t ? ?r),p,?s,?t q,?q,(t ? ?r),p,?s,?t
- r,s,(t ? ?r),p,?s,?t r,?q,(t ? ?r),p,?s,?t
- r,?q,t,p,?s,?t r,?q,?r,p,?s,?t

Conjunctive normal form (dual method)

- Create a conjunctive tableau
- branching ? conjunction
- comma ? disjunction
- Proof of a tautology
- If all the branches are closed, i.e. contain a

pair of opposite literals like p, ?p, which means

p ? ?p, the proven formula is a tautology.

Proof of a tautology

- (p ? (q ? r)) ? (?s ? ?q) ? (t ? ?r) ? (p ? (s

? t)) - (p ? ?q ? ?r) ? (?s ? q) ? (?t ? r) ? ?p ? s ? t
- p, (?s ? q), (?t ? r), ?p, s, t
- ?q, (?s ? q), (?t ? r), ?p, s, t
- ?r, (?s ? q), (?t ? r), ?p, s, t
- 2.1 ?q, ?s, (?t ? r), ?p, s, t
- 2.2 ?q, q, (?t ? r), ?p, s, t
- 3.1 ?r, ?s, (?t ? r), ?p, s, t
- 3.2 ?r, q, (?t ? r), ?p, s, t
- 3.2.1 ?r, q, ?t, ?p, s, t
- 3.2.2 ?r, q, r, ?p, s, t
- All the branches are closed, the formula is a

tautology

What is entailed by a formula?

- Using disjunctive tableau
- If the tableau is not closed, complete the all

the branches by a set of literals that close

them. - Negation of the completion formula is entailed.
- For instance, if we complete the branches by p,

?q, then we added the formula p??q hence the

formula ?p?q, i.e. p?q is entailed. - Dually we can use a conjunctive tableau.

What is entailed by the formula G ?

- G (p ? (q ? r)) ? (?s ? ?q) ? (t ? ?r)
- Solution the disjunctive semantic tableau of G

is not closed (check) in all the leaves the

literals p, ?s, ?t are missing. - Thus the set p, ?s, ?t completes the tableau.

Hence the formula (G ? p ? ?s ? ?t) is a

contradiction. - For this reason G ? ?(p ? ?s ? ?t).
- Hence G ?(p ? ?s ? ?t),
- G (?p ? s ? t), G (p ? (s ? t))

Complete normal forms

- Using the tableau method we can easily prove that
- A tautology does not have a complete conjunctive

normal form (CCNF) - A contradiction does not have a complete

disjunctive normal form (CDNF) - Tautology ? all the branches of the conjunctive

tableau are closed contain a pair (p, ?p), which

means p ? ?p, which is not allowed in the

complete form ? no CCNF. - Dually for a contradiction

CDNF of tautologies

- T1 p ? ?p
- T2 (p ? q) ? (p ? ?q) ? (?p ? q) ? (?p ? ?q)
- T3 (p ? q ? r) ? (p ? q ? ?r) ? (p ? ?q ? r) ?
- (p ? ?q ? ?r) ? (?p ? q ? r) ? (?p ? q ? ?r) ?
- (?p ? ?q ? r) ? (?p ? ?q ? ?r)
- For each possible valuation at least one CEC

must be true.

Rezolution method in PL

- Disadvantage of the semantic tableau method
- If we are proving that P1,...,Pn Z, it is

sufficient to prove (P1? ... ? Pn) ? Z, i.e.

that P1? ... ? Pn ? ?Z is a contradiction. - However, proof by semantic tableau - disjunctive

normal form. - Thus we must apply a lot of distributive

operations in order to transform the formula. - It is easier to prove directly that the formula

P1? ... ? Pn ? ?Z is a contradiction.

Resolution rule

- Let l be a literal. Then the rule of resolution

is - (A ? l) ? (B ? ?l)
- (A ? B)
- The rule is truth preserving.
- Proof Let (A ? l) ? (B ? ?l) be true in a

valuation v. Then both the clauses (A ? l) and (B

? ?l) must be true. - Let v(l) 0. Then w(A) 1 hence w(A ? B) 1.
- Let v(l) 1. Then w(?l) 0 and w(B) 1, hence

w(A ? B) 1. - In both cases A ? B is true in a model v of (A ?

l) ? (B ? ?l).

Clausal form

- Conjunctive normal form is called here a clausal

form. - Particular elementary disjunctions are clauses.
- Example Transformation into clausal form
- ?((p ? q) ? (r ? ?q) ? ?r) ? ?p ?
- ((p ? q) ? (r ? ?q) ? ?r) ? p ?
- (?p ? q) ? (r ? ?q) ? ?r ? p
- The formula consists of four clauses. Proof of

the contradiction - (?p ? q) ? (r ? ?q) ? (?p ? r) ? ?r ? ?p ? p

Proofs by resolution method

- A is a contradiction apply the rule of

resolution until obtaining an empty clause - A is a tautology ?A is a contradiction
- Prove that the set of formulas A1,,An does not

have a model prove that the formula A1 ?... ? An

is a contradiction - Find the logical consequences of A1,,An

derive particular resolvent clauses - Proof of the validity A1,,An Z
- direct keep applying the rule of resolution on

A1,...,An till you obtain Z - indirect prove that (A1 ?...? An ? Z) is a

tautology by proving that (A1 ?...? An ? ?Z) is

a contradiction, that is the set A1,..., An,,?Z

does not have a model

Examples

- Write down the clauses and derive resolvents till

you obtain an empty clause - (?q ? p) ? (p ? r) ? (q ? ?r) ? ?p
- q ? p
- p ? r
- ?q ? ?r
- ?p
- p ? ?r rezolution of 1, 3
- p rezolution of 2, 5
- contradiction 4, 6
- Question What is logically entailed by (?q ? p)

? (p ? r) ? (q ? ?r) ? - Answer Formula p

Examples

- Direct proof of the validityp ? q ? r, ?s ? ?q,

t ? ?r p ? (s ? t) - ?p ? q ? r
- s ? ?q
- t ? ?r
- ?p ? s ? r rezolution 1, 2
- ?p ? s ? t rezolution 3, 4
- ?p ? s ? t ? p ? (s ? t) QED

examples

- Indirect proof of the validityp ? q ? r, ?s ?

?q, t ? ?r p ? (s ? t) - ?p ? q ? r
- s ? ?q
- t ? ?r
- ?p ? s ? r rezolution 1, 2
- ?p ? s ? t rezolution 3, 4
- p negated
- ?s con-
- ?t clusion
- ?p ? s ? t, p, ?s, ?t s ? t, ?s, ?t t, ?t

Summary

- Typical tasks
- Prove the validity of an argument
- What is logically entailed by given premises?
- Add missing premises so that an argument be valid
- Proof of a tautology, contradiction
- Find the models of a formula, of a set of

formulas - Methods
- Semantic tableaus
- Equivalent transformations
- Direct / indirect proofs
- Resolution method

Proof of the tautology

- (p ? q) ? (?p ? r) ? (?q ? r)
- Table A

p q r (p ? q) (?p ? r) A (?q ? r) A ? (?q ? r)

1 1 1 1 1 1 1 1

1 1 0 1 1 1 1 1

1 0 1 0 1 0 1 1

1 0 0 0 1 0 0 1

0 1 1 1 1 1 1 1

0 1 0 1 0 0 1 1

0 0 1 1 1 1 1 1

0 0 0 1 0 0 0 1

Indirect proof of the tautology

- (p ? q) ? (?p ? r) ? (?q ? r)
- A is a tautology, if and only if ?A is a

contradiction - Assume that ?A is not a contradiction, that it

has a model - Negation of implication ?(A ? B) ? (A ? ?B)
- (p ? q) ? (?p ? r) ? ?q ? ?r
- 1 1 1 0 1 0

q 0, r 0, hence p ? 0, ?p ? 0 - 0 0 0 0 p 0, ?p 0, tj.
- 1 p 1
- contradiction
- The negated formula does not have a model, hence

the original formula is a tautology.

Proof by equivalent transformations

- Apply the laws
- (A ? B) ? (?A ? B) ? (?(A ? ?B))
- ?(A ? B) ? (?A ? ?B) de Morgan
- ?(A ? B) ? (?A ? ?B) de Morgan
- ?(A ? B) ? (A ? ?B)
- (A ? (B ? C)) ? ((A ? B) ? (A ? C)) distributive

laws - (A ? (B ? C)) ? ((A ? B) ? (A ? C))
- 1 ? A ? 1 1 ? tautology,
- 1 ? A ? A e.g. (p ? ?p)
- 0 ? A ? 0 0 ? contradiction
- 0 ? A ? A e.g. (p ? ?p)

Proof by equivalent transformations

- (p ? q) ? (?p ? r) ? (?q ? r)
- (p ? q) ? (?p ? r) ? (?q ? r) ? ?(p ? q) ?

(?p ? r) ? (?q ? r) ? - (p ? ?q) ? (?p ? ?r) ? q ? r ?
- p ? (?p ? ?r) ? q ? r ? ?q ? (?p ? ?r) ? q ?

r ? - (p ? ?p ? q ? r) ? (p ? ?r ? q ? r) ? (?q ? ?p ?

q ? r) ? (?q ? ?r ? q ? r) - ? 1 ? 1 ? 1 ? 1 ? 1 tautology
- Remark we obtained the conjunctive normal form

Proof by resolution

- (p ? q) ? (?p ? r) ? (?q ? r)
- The clausal form of the negated formula (indirect

proof) - (p ? q) ? (?p ? r) ? ?q ? ?r ? (?p ? q) ? (p ? r)

? ?q ? ?r - 1. ?p ? q
- 2. p ? r
- 3. ?q
- 4. ?r
- 5. q ? r rezolution 1, 2
- 6. r rezolution 3, 5
- 7. rezolution 4, 6

Proof by semantic tableau

- (p ? q) ? (?p ? r) ? (?q ? r)
- Direct proof CNF (? branching, ? , all

branches closed p ? ?p) - (p ? ?q) ? (?p ? ?r) ? q ? r
- p, (?p ? ?r), q, r ?q, (?p ? ?r), q, r
- p, ?p, q, r p, ?r, q, r

Proof by semantic tableau

- (p ? q) ? (?p ? r) ? (?q ? r)
- Indirect proof DNF of the negated formula
- (? branching, ? , all branches 0 p ? ?p)
- (?p ? q) ? (p ? r) ? ?q ? ?r
- ?p, (p ? r), ?q, ?r q, (p ? r), ?q, ?r
- ?p, p, ?q, ?r ?p, r, ?q, ?r

Proof of the validity of an argument

- (p ? q) ? (?p ? r) ? (?q ? r) iff
- (p ? q) ? (?p ? r) (?q ? r) iff
- (p ? q), (?p ? r) (?q ? r)
- p The program functions well
- q The system is OK
- r It is necessary to call the trouble-shooter
- If the program functions well, the system is OK.
- If the program malfunctions, it is necessary to

call the trouble-shooter. - --------------------------------------------------

-------------------------- - If the system is not OK, it is necessary to call

the trouble-shooter.

Proof of the validity of an argument

- (p ? q), (?p ? r) (?q ? r)
- Indirect proof
- (p ? q), (?p ? r), (?q ? ?r) is a contradictory

set - ?p ? q
- p ? r
- ?q
- ?r
- q ? r rezolution 1, 2
- r rezolution 3, 5
- rezolution 4, 6