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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
  • 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
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