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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
• 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)
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
• 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
• 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.

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
• (?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
• 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
• Question What is logically entailed by (?q ? p)
? (p ? r) ? (q ? ?r) ?

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
• 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
• 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
• The negated formula does not have a model, hence
the original formula is a tautology.

34
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)

35
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

36
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

37
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

38
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

39
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.

40
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