proportional logic and first order logic

7/24/2019 Proportional Logic and first order Logic

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Unit-I Proposional Logic & First Order Logic

October 25, 2015

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Satisability, Validity and Consequence

DenitionLet A F .

A is satisable iff vI (A) = T for some interpretation I . A

satisfying interpretation is a model for A.A is valid, denoted |= A, iff vI = T for all interpretations I . Avalid propositional formula is also called a tautology.A is unsatisable iff it is not satisable, that is, if vI (A) = Tfor all interpretations I .A is falsiable, denoted A, iff it is not valid, that is, if vI (A)= F for some interpretation v.

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Satisability and Validity of Formulas

Figure: Satisability and validity of formulas



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Satisability and Validity of Formulas

TheoremLet A F . A is valid if and only if A is unsatisable. A is satisable if and only if A is falsiable



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Proof.Let I be an arbitrary interpretation. vI (A) = T if and only if vI( A) = F by the denition of the truth value of a negation.Since I was arbitrary, A is true in all interpretations if and only if

A is false in all interpretations, that is, iff A is unsatisable.If A is satisable then for some interpretation I , vI (A) = T.By denition of the truth value of a negation, vI ( A) = F so that

A is falsiable.Conversely, if vI ( A) = F then vI (A) = T.



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Decision Procedures in Propositional Logic

Denition (Decision Procedure)

Let U F be a set of formulas. An algorithm is a decisionprocedure for U if given an arbitrary formula A F , it terminatesand returns the answer yes if A U and the answer no if A U .If U is the set of satisable formulas, a decision procedure for U iscalled a decision procedure for satisability, and similarly for

validity.



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Note

1 To decide if A is valid, apply the decision procedure forsatisability to A

2 If it reports that A is satisable, then A is not valid; if itreports that A is not satisable, then A is valid. Such andecision procedure is called a refutation procedure, becausewe prove the validity of a formula by refuting its negation

3 Refutation procedures can be efficient algorithms for decidingvalidity, because instead of checking that the formula is always

true, we need only search for a falsifying counterexample4 The existence of a decision procedure for satisability in

propositional logic is trivial, because we can build a truthtable for any formula



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Example for Satisability

Truth table for the formula p q

p q pq

T T TT F FF T TF F T

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Example for Validity and Unsatisable

Validity of (p q)( q p)Proof Refer BBProve that (p q) p q is unsatisable.Proof Refer BB

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Satisability of Set of Formulas

DenitionA set of formulas U={A1,...}s (simultaneously) satisable iff thereexists an interpretation I such that vI (Ai ) = T for all i. The

satisfying interpretation is a model of U . U is unsatisable iff forevery interpretation I , there exists an i such that vI (Ai ) = F

Example:The set U1 = {p, p q, q r} is simultaneously satisable by

the interpretation which assigns T to each atom, while the setU2 = {p, p q, p} is unsatisable. Each formula in U2 issatisable by itself, but the set is not simultaneously satisable

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Logical Consequence

Denition (Logical Consequence)

Let U be a set of formulas and A a formula. A is a logicalconsequence of U , denoted U A, iff every model of U is a model

of A

Example:Let A = (pr) ( q r). Then A is a logical consequenceof {p, q}, denoted {p, q} A, since A is true in all

interpretations I such that I (p) = T and I (q) = F . However,A is not valid, since it is not true in the interpretation I whereI (p) = F , I (q)=T , I (r) = T

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Theorem

TheoremU A if and only if

i A

i A.

Example:{p, q} (p r) ( q r), so by Theorem, (p q) (p r) ( q r)

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Semantic Tableaux

The method of semantic tableaux is an efficient decisionprocedure for satisability (and by duality validity) inpropositional logic

The principle behind semantic tableaux is very simple: searchfor a model (satisfying interpretation) by decomposing theformula into sets of atoms and negations of atomsIt is easy to check if there is an interpretation for each set: aset of atoms and negations of atoms is satisable iff the setdoes not contain an atom p and its negation pThe formula is satisable iff one of these sets is satisable

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Decomposing Formulas into Sets of Literals

Denition (Literals)

A literal is an atom or the negation of an atom. An atom is a

positive literal and the negation of an atom is a negative literal.For any atom p, {p, p} is a complementary pair of literals

For any formula A, {A, A} is a complementary pair of formulas. A is the complement of A and A is thecomplement of A

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Problems

1 Analyze satisability of the formula:A = p ( q p).Solution Refer BB

2 Prove that B = (p q) ( p q)is unsatisable

TheoremA set of literals is satisable if and only if it does not contain a

complementary pair of literals - Proof: Refer BB

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SAT Solvers

Denition (SAT Solver)A computer program that searches for a model for a propositionalformula is called a SAT Solver

Properties of Clausal FormDenitionLet S, S be sets of clauses. S S denotes that S is satisable if and only if S is satisable

It is important to understand that S S (S is satisable if and only if S is satisable) does not imply that S S (S islogically equivalent to S)

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Pure Literals and Renaming

Denition (Pure Literals)

Let S be a set of clauses. A Pure literal in S is a literal l thatappears in atleast one clause of S, but its complement lc does notappear in any clause of S

Denition (Renaming)

Let S be a set of clauses and U a set of atomic propositions RU (S)

the renaming of S by U is obtained from S by replacing each literall on an atomic proposition in U by lc

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Example

If S={pqr, p q, q r ,r}, nd Rp ,q (S)

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Example

If S={pqr, p q, q r ,r}, nd Rp ,q (S)Solution: Rp ,q (S)= {p q ,pq ,qr ,r},



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Davis Putnam Algorithm

Input : A formula A in Clausal FormOutput : Report A is Satisable or UnsatisablePerform the following rules repeatedly but the third rule is usedonly if the rst two do not apply.

1 Unit Literal rule: If there is a unit clause (l), delete all clausescontaining l and delete all occurrences of lc from other clauses

2 Pure Literal rule:If there is a pure literal l, delete all clausescontaing l

3 Eliminate a variable by resolution:Choose an atom p and

perform all possible resolutions on clauses that clash on p andp . Add these resolvents to the set of clauses and then deleteall clauses containing p or p

Algorithm 1: Davis Putnam Algorithm

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Termination condition for the Algorithm

1 If empty clause is produced, report the formula isunsatisable

2 If no more rules are applicable, report that formula issatisable

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Example

Consider the set of clauses {p, p q,q r, r st}. Apply DPalgorithm.

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Example

Consider the set of clauses {p, p q,q r, r st}. Apply DPalgorithm.Step(i) According to step(i) of D.P algorithm, p is a unitliteral. Hence it must be deleted in all clauses containing p

and delete all occurrences of pc

from other clauses. thegiven set becomes {q,q r, r st}

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Example



from other clauses. thegiven set becomes {q,q r, r st}Step(ii) Like the previous step, remove q and qc . the setbecomes {r, r st}

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Example




Step(iii) Like the previous step, remove r and its complement. the set becomes {st}

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Example




Step(iii) Like the previous step, remove r and its complement. the set becomes {st}No more rules are applicable for the set {st}, the set of clauses is satisable



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Hilbert System (H)

Gentzen System vs Hilbert SystemGentzen Hilbert

One Axiom Several AxiomsMany Inference Rules Only one rule of inferenceNotations used

A,B,C denote arbitrary formulas in propositional logicAA means AA can be proved



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Hilbert System(H)

Denition (Axioms of H & Rules of Inference)

1 (A(BA))2 (A(BC))((AB)(AC))

3 (B A)(AB)4 Rule of Inference: (Modus Ponens)(MP) A, AB B

Theorem

A A

Proof.Refer BB



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Assignment Due Date:27.10.15

1 Prove (AB)(B A)2 Prove (A A) A in H



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First Order Logic: Deductive Systems

Denition (First Order Logic)

An extension of propositional logic that includes predicatesinterpreted as relations on domains

Notations usedP,A,V denotes the countable sets of predicate symbols,constant symbols and variablespn P n-ary predicate

n=1 or 2 unary or binary Universal Quantier Existential Quantier



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Example

x

y

p(x,y)

y

p(y,x)

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First Order Logic:Deductive System

Gentzen System:Deductive SystemOne Axiom4 rules of Inference

Two Special rules: and (a)xA(x) A(a) xA(x) A(a)

(a)xA(x) A(a) xA(x) A(a)

U {, (a )}U { }U {(a )}

U {}

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Hilbert System

Denition (Axioms of H for the First Order Logic)1 Apart from the three axioms of H for Propositional Logic we

have:1 xA(x)A(a)2 x(AB(x))(A xB(x))

Rules of Inference(Modus Ponens and Generalization)A B , A

B A(a )

xA (x )

C-Rule:(i) U xA(x) (Existential Quantier)(i+1) U A(a) (C rule)

Deduction Rule:U { A} B U A B

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Theorems

TheoremA(a) xA(x)

Proof.Refer BB

Theorem x A(x) x A(x)

Proof.Refer BB



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Skolems Algorithm

An algorithm to transform a formula A into a formula Ainclausal form x(p(x) q(x)) (xp(x) xq(x))

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Skolems Algorithm I

Input : A closed formula A of rst-order logicOutput : A formula Ain clausal form such that A A

1 Rename bound variables so that no variable appears in twoquantiers

x(p(x) q(x)) (yp(y) zq(z))2 Eliminate all binary Boolean operators other than and

x( p(x) q(x)) yp(y) zq(z)3 Push negation operators inward, collapsing double negation,

until they apply to atomic formulas only. Use the equivalences: xA(x) x A(x), xA(x) x A(x)



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Skolems Algorithm Contd..

4 The given formula is transformed to:x(p(x) q(x)) y p(y) zq(z)

5 Extract quantiers from the matrix. Choose an outermostquantier, that is, a quantier in the matrix that is not withinthe scope of another quantier still in the matrix. Extract thequantier using the following equivalences, where Q is aquantier and op is either or:A op QxB(x) Qx(A op B(x)), QxA(x) op B Qx(A(x) opB)

Repeat until all quantiers appear in the prex and the matrixis quantier-free. The equivalences are applicable becausesince no variable appears in two quantiers.

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Skolems Algorithm Contd..6 In the example, no quantier appears within the scope of

another, so we can extract them in any order, for example, x,y, z: xyz((p(x) q(x)) p(y) q(z))

7 Use the distributive laws to transform the matrix into CNF.The formula is now in PCNF

xyz((p(x) p(y) q(z)) ( q(x) p(y) q(z)))8 For every existential quantier x in A, let y1 , . . . , yn be

the universally quantied variables preceding x and let f be anew n-ary function symbol. Delete x and replace everyoccurrence of x by f(y1 , . . . , yn ). If there are no universalquantiers preceding x, replace x by a new constant (0-aryfunction). These new function symbols are Skolem functionsand the process of replacing existential quantiers byfunctions is Skolemization

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Skolem Algorithm Contd..

9 For Example, z((p(a) p(b) q(z)) ( q(a) p(b) q(z))), where a and b are the Skolem functions (constants)corresponding to the existentially quantied variables x and y,respectively

10 The formula can be written in clausal form by dropping the(universal) quantiers and writing the matrix as sets of clauses:{{p(a), p(b), q(z) }, { q(a), p(b), q(z) }}

Algorithm 2: Skolems Algorithm

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Skolems Algorithm Example

Step TransformationOriginal formula xyp(x,y) yxp(x,y)Rename bound variables xyp(x,y) wzp(z,w)

Eliminate Boolean operators xyp(x,y) wzp(z,w)Push negation inwards xy p(x,y) wzp(z, w)Extract quantiers xywz( p(x,y) p(z,w))Distribute matrix (no change)Replace existential quantiers xw( p(x,f(x)) p(g(x,w), w))Write in clausal form {{ p(x,f(x)), p(g(x,w), w) }}

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Skolems Theorem

TheoremLet A be a closed formula. Then there exists a formula A inclausal form such that A A

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