Normal forms, resolution method Lesson 4

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Title: Normal forms, resolution method Lesson 4

1
Normal forms,resolution methodLesson 4

Marie Duží

2
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))

3
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) ? .
4
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.

5
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

6
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.

7
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

8
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
9
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.

10
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.

11
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
12
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 ?

13
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).

14
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

15
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.

16
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.

17
(?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

18
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.

19
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

20
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.

21
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))

22
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

23
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.

24
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.

25
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).

26
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

27
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

28
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

29
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

30
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

31
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

32
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
33
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.

34
Proof by equivalent transformations