Propositional Logic: Contents

Propositional Logic Propositional Logic: Contents • Syntax and Semantics of Propositional Logic • Satisfiability (SAT) • Tableau Algorithm for SAT •...
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Propositional Logic

Propositional Logic: Contents • Syntax and Semantics of Propositional Logic • Satisfiability (SAT) • Tableau Algorithm for SAT • Structural induction • Semantic consequence and logical equivalence • Conjunctive and disjunctive normal form (CNF and DNF)

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Formulas of propositional Logic The alphabet of propositional logic consists of • an infinite set p1 , p2 , . . . of atomic formulas; • the logical connectives: – ¬ (‘not’), called negation; – ∧ (‘and’), called conjunction; – ∨ (‘or’), called disjunction; • brackets: ( and ). Remarks: • atomic formulas are also called propositional variables; • we use letters p, q, r and indexed letters q1 , q2 , . . . to denote atomic formulas. Logic in Computer Science

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Formulas of propositional logic The set P of all formulas of propositional logic is defined inductively: • all atomic formulas are formulas; • if P is a formula, then ¬P is a formula; • if P and Q are formulas, then (P ∧ Q) is a formula; • if P and Q are formulas, then (P ∨ Q) is a formula; • Nothing else is a formula. Remarks: • So, formulas are just strings over a certain alphabet without truth values or meaning. • We use P , Q, R and indexed letters such as P1 , P2 , . . . , Q1 , Q2 , . . . to denote formulas of propositional logic. Logic in Computer Science

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Truth Values An interpretation I is a function which assigns to any atomic formula pi a truth value I(pi ) ∈ {0, 1}.

• If I(pi ) = 1, then pi is called true under the interpretation I. • If I(pi ) = 0, then pi is called false under the interpretation I. Given an assignment I we can compute the truth value of compound formulas step by step using so-called truth tables.

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Truth tables: negation The negation ¬P of a formula P is true when P is false and false otherwise:

Definition Suppose an interpretation I is given and we know the value I(P ). Then the value I(¬P ) is computed by  I(¬P ) =

0 1

if if

I(P ) = 1 I(P ) = 0

Corresponding truth table: P 1 0

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¬P 0 1

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Truth tables: conjunction The conjunction (P ∧ Q) is true if and only if both P and Q are true.

Definition Suppose an interpretation I is given and we know I(P ) and I(Q). Then  I(P ∧ Q) =

1 0

if if

I(P ) = 1 and I(Q) = 1 I(P ) = 0 or I(Q) = 0

Corresponding truth table: P 1 1 0 0

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Q 1 0 1 0

(P ∧ Q) 1 0 0 0

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Truth tables: disjunction The disjunction (P ∨ Q) is true if and only if P is true or Q is true.

Definition Suppose an interpretation I is given and we know I(P ) and I(Q). Then  I(P ∨ Q) =

1 0

if if

I(P ) = 1 or I(Q) = 1 I(P ) = 0 and I(Q) = 0

Corresponding truth table: P 1 1 0 0

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Q 1 0 1 0

(P ∨ Q) 1 1 1 0

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Truth under an interpretation So, given an interpretation I, we can compute the truth value I(P ) of any formula P under I. • If I(P ) = 1, then P is called true under the interpretation I. • If I(P ) = 0, then P is called false under the interpretation I.

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Example List the Interpretations I such that P = ((p1 ∨ ¬p2 ) ∧ p3 ) is true under I. p1 1 1 1 1 0 0 0 0

p2 1 1 0 0 1 1 0 0

p3 1 0 1 0 1 0 1 0

¬p2 0 0 1 1 0 0 1 1

(p1 ∨ ¬p2 ) 1 1 1 1 0 0 1 1

P 1 0 1 0 0 0 1 0

P is true under I1 , I2 , and I3 , where • I1 (p1 ) = I1 (p2 ) = I1 (p3 ) = 1, • I2 (p1 ) = I2 (p3 ) = 1 and I2 (p2 ) = 0, • I3 (p1 ) = I3 (p2 ) = 0 and I3 (p3 ) = 1.

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Truth table for (¬P ∨ Q)

P 1 1 0 0

Q 1 0 1 0

¬P 0 0 1 1

(¬P ∨ Q) 1 0 1 1

(¬P ∨ Q) represents the assertion ‘if P is true, then Q is true’. Define a ‘new’ connective → by: (P → Q) = (¬P ∨ Q).

In what follows we use (P → Q) as an abbreviation for (¬P ∨ Q).

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Truth table for ((P → Q) ∧ (Q → P )) P 1 1 0 0

Q 1 0 1 0

(P → Q) 1 0 1 1

(Q → P ) 1 1 0 1

((P → Q) ∧ (Q → P )) 1 0 0 1

((P → Q) ∧ (Q → P )) represents the assertion ‘P is true if and only if Q is true’. Define a ‘new’ connective ↔ by: (P ↔ Q) := ((P → Q) ∧ (Q → P )).

In what follows we use (P ↔ Q) as an abbreviation for ((P → Q) ∧ (Q → P )).

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Satisfiability Definition A formula P is satisfiable if and only if there exists an interpretation I such that I(P ) = 1. Examples • Every atomic formula p is satisfiable: given p, take the interpretation I with I(p) = 1. • ¬p is satisfiable: take the interpretation I with I(p) = 0. Then I(¬p) = 1. • (p ∧ ¬p) is not satisfiable: for any interpretation I, I(p ∧ ¬p) = 0. • (p ∧ ¬q) is satisfiable: take the interpretation I with I(p) = 1 and I(q) = 0.

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Satisfiability and Puzzles (1) Isaac and Albert were excitedly describing the result of the Third Annual International Science Fair Extravaganza in Sweden. There were three contestants, Louis, Rene, and Johannes. Isaac reported that Louis won the fair, while Rene came in second. Albert, on the other hand, reported that Johannes won the fair, while Louis came in second. In fact, neither Isaac nor Albert had given a correct report of the results of the science fair. Each of them had given one true statement and one false statement. What was the actual placing of the three contestants? (Credits: based on slides by Andrei Voronkov)

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Encoding into SAT We take atomic formulas L1, L2, L3, R1, R2, R3, J 1, J 2, J 3 with the intuitive meaning: • L1: Louis came in first, L2: Louis came in second, L3: Louis came in third. • R1: Rene came in first, R2: Rene came in second, R3: Rene came in third. • J 1: Johannes came in first, J 2: Johannes came in second, J 3: Johannes came in third. We represent the information about Isaac’s report using the formula J : • J = ((L1 ∧ ¬R2) ∨ (¬L1 ∧ R2)) We represent the information about Albert’s report using the formula A: • A = ((J 1 ∧ ¬L2) ∨ (¬J 1 ∧ L2))

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Encoding into SAT We have to encode additional information. Namely, • everybody comes in at exactly one place: represent this using (P1 ∧ P2 ), where P1 = ((L1 ∨ L2 ∨ L3) ∧ (R1 ∨ R2 ∨ R3) ∧ (J 1 ∨ J 2 ∨ J 3)) and P2 = (¬(L1 ∧ L2) ∧ ¬(L1 ∧ L3) ∧ ¬(L2 ∧ L3) ∧ ¬(R1 ∧ R2) · · · ) • Only one person can come in first, etc: represent this using Q, where Q = (¬(L1 ∧ R1) ∧ ¬(L2 ∧ R2) ∧ ¬(L3 ∧ R3) ∧ (R1 ∧ J 1) · · · ) Any interpretation I with I(J ∧ A ∧ P1 ∧ P2 ∧ Q) = 1 corresponds to a possible placing of the three contestants. Logic in Computer Science

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Note on Conjunctions and Disjunctions On the previous slide, we have used formulas (P1 ∧ · · · ∧ Pn ) and (P1 ∨ · · · ∨ Pn ) which we have not defined yet. Note that according to the syntax of propositional formulas we use brackets whenever we form the conjunction/disjunction of two formulas. (P1 ∧ · · · ∧ Pn ) is defined by induction over n as follows: • (P1 ) = P1 ; • (P1 ∧ · · · ∧ Pn+1 ) = ((P1 ∧ · · · ∧ Pn ) ∧ Pn+1 ); (P1 ∨ · · · ∨ Pn ) is defined by induction over n as follows: • (P1 ) = P1 ; • (P1 ∨ · · · ∨ Pn+1 ) = ((P1 ∨ · · · ∨ Pn ) ∨ Pn+1 );

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Checking Satisfiability (SAT) We want an algorithm that checks whether a given propositional formula is satisfiable. In other words, for a given P , we search for an interpretation I such that I(P ) = 1. If this search is successful, then the output of the algorithm should be • “yes, P is satisfiable”. If no such interpretation can be found, then the output of the algorithm should be • “no, P is not satisfiable”.

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SAT applications SAT has numerous applications in computer and information science. Here are some: • Circuit design: e.g., when are two circuits equivalent? • Model checking: does a program represented as a graph structure satisfy its specification? • Planning in artificial intelligence; • Haplotyping in bioinformatics: derive haplotype data from genotype data.

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Satisfiability checking based on Truth Tables Here is an algorithm checking satisfiability that is directly based on truth tables: 1. Let P be the input formula; 2. Using truth tables, compute the value I(P ) for all interpretations I; 3. if an I is found such that I(P ) = 1, then output “P is satisfiable”; 4. If no such I is found, output “P is not satisfiable”. If P is not satisfiable, then this algorithms requires the computation of I(P ) for 2n many interpretations I, where n is the number of atomic formulas in P . Thus the running time of this algorithm is exponential. Major open problem in computer science: does there exist an algorithm checking satisfiability that runs in polynomial time? This problem is also known as the P=NP problem.

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Tableau Method Intuition: to check satisfiability of P , we apply tableau rules to P that make explicit the constraints that P imposes on formulas occuring in P (subformulas). If all sequences of rule applications lead to an “obviously unsatisfiable” constraint, then P is unsatisfiable. If at least one sequence of rule applications leads to a constraint that cannot be decomposed further and does not contain an obviously unsatisfiable set of constraints, then P is satisfiable. A constraint S is a finite set of propositional formulas. S is satisfiable if there exists an interpretation I such that I(P ) = 1 for all P ∈ S.

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Tableau method (Intuition) To check satisfiable of P , one starts with constraint {P }. Then, one applies rules that reflect the following facts: • if (P ∧ Q) is satisfiable, then {P, Q, (P ∧ Q)} is satisfiable; • if ¬¬P is satisfiable, then {P, ¬¬P } is satisfiable; • if ¬(P ∨ Q) is satisfiable, then {¬P, ¬Q, ¬(P ∨ Q)} is satisfiable. • if (P ∨ Q) is satisfiable, then {P, (P ∨ Q)} is satisfiable or {Q, (P ∨ Q)} is satisfable; • if ¬(P ∧Q) is satisfiable, then {¬P, ¬(P ∧Q)} is satisfiable or {¬Q, ¬(P ∧Q)} is satisfiable; To avoid “branching”, we first consider satisfiability of formulas not containing any (P ∨ Q) and ¬(P ∧ Q).

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Ingredients of the algorithm (partial) • A constraint S is a finite set of propositional formulas; • A constraint S contains a clash if there exists a formula P such that P ∈ S and ¬P ∈ S. • A non-branching completion rule is of the form S =⇒ S 0 , where S, S 0 are constraints. • A constraint S is complete if no completion rule is applicable to S.

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Completion Rules (partial) Assume that S does not contain a clash (if it does, no rule is applicable). (∧-rule)

S =⇒∧ S ∪ {P, Q} if (a) (P ∧ Q) ∈ S and (b) {P, Q} 6⊆ S.

(¬¬-rule)

S =⇒¬ S ∪ {P } if (a) ¬¬P ∈ S and (b) P 6∈ S.

(¬∨-rule)

S =⇒¬∨ S ∪ {¬P, ¬Q} if (a) ¬(P ∨ Q) ∈ S and (b) {¬P, ¬Q} 6⊆ S.

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Example 1 We check satisfiability of P = ((¬p ∧ q) ∧ ¬¬r). Set S0 = {((¬p ∧ q) ∧ ¬¬r)}. • An application of =⇒∧ gives S1 = S0 ∪ {(¬p ∧ q), ¬¬r}. • An application of =⇒∧ gives S2 = S1 ∪ {¬p, q} • An application of =⇒¬ gives S3 = S2 ∪ {r} Note that S3 = {P, (¬p ∧ q), ¬¬r, ¬p, q, r}.

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Example 1 (continued) S3 = {P, (¬p ∧ q), ¬¬r, ¬p, q, r}. • No completion rule is applicable to S3 ; • Thus, S3 is complete. • S3 does not contain any clash. • Thus, the output is “P is satisfiable”. S3 describes an interpretation I under which P is true. Namely, we set for any atomic formula x from P : • I(x) = 1 if, and only if, x ∈ S3 . Thus, I(q) = I(r) = 1 and I(p) = 0. Then I(P ) = 1.

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Example 2 We check satisfiability of P = ((p ∧ q) ∧ ¬¬¬p). Set S0 = {((p ∧ q) ∧ ¬¬¬p)}. • An application of =⇒∧ gives S1 = S0 ∪ {(p ∧ q), ¬¬¬p}. • An application of =⇒∧ gives S2 = S1 ∪ {p, q} • An application of =⇒¬ gives S3 = S2 ∪ {¬p} Note that S3 = {P, (p ∧ q), ¬¬¬p, p, q, ¬p}. Thus S3 contains a clash: p ∈ S3 and ¬p ∈ S3 and we output “P is not satisfiable”. Logic in Computer Science

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Tableau Path (partial) A sequence S0 , S1 , . . . , Sn of constraints is a tableau path if for any i < n at least one of the following conditions is satisfied: • Si =⇒∧ Si+1 • Si =⇒¬∨ Si+1 • Si =⇒¬ Si+1

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The tableau algorithm (partial) • A tableau path S0 , . . . , Sn is complete if Sn is complete. • A tableau path S0 , . . . , Sn contains a clash if Sn contains a clash. To check satisfiability of a formula P , do the following: • Generate a tableau path starting with the constraint {P }; • If the tableau path is complete and does not contain a clash, then output “P is satisfiable”. • If the tableau path contains a clash, then output “P is not satisfiable”.

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Example 3 We check satisfiability of P = ((p ∧ q) ∧ ¬(p ∨ q)). Set S0 = {((p ∧ q) ∧ ¬(p ∨ q))}. • An application of =⇒∧ gives S1 = S0 ∪ {(p ∧ q), ¬(p ∨ q)}. • An application of =⇒∧ gives S2 = S1 ∪ {p, q} • An application of =⇒¬∨ gives S3 = S2 ∪ {¬p, ¬q} S3 contains a clash: p ∈ S3 and ¬p ∈ S3 and we output “P is not satisfiable”.

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Analysing the Tableau Algorithm (partial) To show that the tableau algorithm does what it is supposed to do, one has to show the following. Let P be a propositional formula. • Termination: The algorithm terminates: there is no infinite tableau path S0 , S1 , . . . starting with {P }. • Soundness: If there exists a complete tableau path S0 , S1 , . . . , Sn with {P } = S0 and without clash, then P is satisfiable. • Completeness: If P is satisfiable, then no tableau path (generated by the three rules above) S0 , S1 , . . . , Sn with {P } = S0 contains a clash. For the proof, we require definitions and proofs by structural induction. This will be done later.

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Branching How to deal with formulas containing (P ∨ Q) or ¬(P ∧ Q)? Recall that • if (P ∨ Q) is satisfiable, then {P, (P ∨ Q)} is satisfiable or {Q, (P ∨ Q)} is satisfable; • if ¬(P ∧Q) is satisfiable, then {¬P, ¬(P ∧Q)} is satisfiable or {¬Q, ¬(P ∧Q)} is satisfiable; Thus, we have to explore different ways to satisfy a formula.

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Example 4 We check satisfiability of P = ((p ∧ ¬p) ∨ (q ∧ q)). Set S0 = {P }. • An application of =⇒∨ gives S1 = S0 ∪ {p ∧ ¬p}

or

S2 = S0 ∪ {(q ∧ q)}.

• We first try to satisfy S1 . An application of =⇒∧ to S1 gives S3 = S1 ∪ {p, ¬p} which contains a clash. We have been unsuccessful. • We now try to satisfy S2 . An application of =⇒∧ to S2 gives S4 = S2 ∪ {q, q} S4 does not contain a clash and is complete. Logic in Computer Science

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Example 4 (continued) We have two tableau paths S0 , S1 , S3

S0 , S2 , S4

The second path is complete and does not contain a clash. Thus, the output is “P is satisfiable”.

Recall that S4 = {P, (q ∧ q), q}. S4 also describes an interpretation I under which P is true, namely I(q) = 1.

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Ingredients of the algorithm (complete) • A constraint S is a finite set of propositional formulas; • A constraint S contains a clash if there exists a formula P such that P ∈ S and ¬P ∈ S. • A non-branching completion rule is of the form S =⇒ S 0 , where S, S 0 are constraints. • A branching completion rule is of the form S =⇒ S1 or S2 , where S, S1 , S2 are constraints. • A constraint S is complete if no completion rule is applicable to S.

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Completion Rules Assume that S does not contain a clash (if it does, no rule is applicable). (∧-rule)

S =⇒∧ S ∪ {P, Q} if (a) (P ∧ Q) ∈ S and (b) {P, Q} 6⊆ S.

(¬¬-rule)

S =⇒¬ S ∪ {P } if (a) ¬¬P ∈ S and (b) P 6∈ S.

(¬∨-rule)

S =⇒¬∨ S ∪ {¬P, ¬Q} if (a) ¬(P ∨ Q) ∈ S and (b) {¬P, ¬Q} 6⊆ S.

(∨-rule)

S =⇒∨ S ∪ {P } or S ∪ {Q} if (a) (P ∨ Q) ∈ S and (b) P 6∈ S and Q 6∈ S.

(¬∧-rule)

S =⇒¬∧ S ∪ {¬P } or S ∪ {¬Q} if (a) ¬(P ∧ Q) ∈ S and (b) ¬P 6∈ S and ¬Q 6∈ S.

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Tableau Path A sequence S0 , S1 , . . . , Sn of constraints is a tableau path if for any i < n at least one of the following conditions is satisfied: • Si =⇒∧ Si+1 • Si =⇒¬∨ Si+1 • Si =⇒¬ Si+1 • for some S: Si =⇒¬∧ Si+1 or S

or

Si =⇒¬∧ S or Si+1

Si =⇒∨ Si+1 or S

or

Si =⇒∨ S or Si+1

• for some S:

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The tableau algorithm • A tableau path S0 , . . . , Sn is complete if Sn is complete. • A tableau path S0 , . . . , Sn contains a clash if Sn contains a clash. To check satisfiability of a formula P , do the following: • Generate tableau paths starting with the constraint {P }; • If there is a tableau path that is complete and does not contain a clash, then output “P is satisfiable”. • If no such tableau path can be found (i.e., all complete tableau paths starting with P contain a clash), then output “P is not satisfiable”.

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Example 5 We check satisfiability of P = (((¬p ∨ q) ∧ p) ∧ ¬¬q). Set S0 = {P }. • An application of =⇒∧ gives S1 = S0 ∪ {((¬p ∨ q) ∧ p), ¬¬q}. • An application of =⇒∧ gives S2 = S1 ∪ {(¬p ∨ q), p}. • An application of =⇒∨ gives S3 = S2 ∪ {¬p} S3 contains a clash: ¬p ∈ S3 and p ∈ S3 . • The other possible application of =⇒∨ to S2 gives S4 = S2 ∪ {q} S4 is complete and does not contain a clash. Logic in Computer Science

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Example 5 (continued) We have two tableau paths S0 , S1 , S2 , S3

S0 , S1 , S2 , S4

The second path is complete and does not contain a clash. Thus, the output is “P is satisfiable”.

Recall that S4 = {P, ((¬p ∨ q) ∧ p), p, ¬¬q, (¬p ∨ q), q}. S4 also describes an interpretation I under which P is true. Namely, we set for any atomic formula x from P : • I(x) = 1 if, and only if, x ∈ S4 . Thus, I(p) = I(q) = 1. Then I(P ) = 1.

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Example 6 We check satisfiability of P = (((¬p ∨ q) ∧ p) ∧ ¬q). Set S0 = {P }. • An application of =⇒∧ gives S1 = S0 ∪ {((¬p ∨ q) ∧ p), ¬q}. • An application of =⇒∧ gives S2 = S1 ∪ {(¬p ∨ q), p}. • An application of =⇒∨ gives S3 = S2 ∪ {¬p} S3 contains a clash: ¬p ∈ S3 and p ∈ S3 . • The other possible application of =⇒∨ gives S4 = S2 ∪ {q} S4 contains a clash: q ∈ S4 and ¬q ∈ S4 . Logic in Computer Science

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Example 6 (continued) We have two tableau paths S0 , S1 , S2 , S3

S0 , S1 , S2 , S4

Both contain a clash. Thus, the output is “P is not satisfiable”.

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Analysing the Tableau Algorithm To show that the tableau algorithm does what it is supposed to do, one has to show the following. Let P be a propositional formula. • Termination: The algorithm terminates: there are only finitely many tableau paths starting with {P }. • Soundness: If there exists a complete tableau path S0 , S1 , . . . , Sn with S0 = {P } without clash, then P is satisfiable. • Completeness: If P is satisfiable, then there exists a complete tableau path S0 , S1 , . . . , Sn with S0 = {P } without clash. For the proof, we require definitions and proofs by structural induction. We first introduce this important concept.

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Definitions by structural induction Many important functions F which have as domain the set of all propositional formulas are defined by specifying the values • F (pi ), for all propositional variables pi , • F (P ∧ Q), given the values F (P ) and F (Q), • F (P ∨ Q), given the values F (P ) and F (Q), • F (¬P ), given the value F (P ). Such a definition is called a definition by structural induction. (The idea should be familar from proofs by induction for natural numbers.)

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Example 1: Interpretations The definition of interpretations I was given by structural induction. To define an interpretation I it is sufficient to define • I(pi ) for all atomic formulas pi . The values I(P ), P an arbitrary propositional formula, are then given by means of truth tables. In other words, truth tables define the values • I(P ∧ Q), given the values I(P ) and I(Q), • I(P ∨ Q), given the values I(P ) and I(Q), • I(¬P ), given the value I(P ).

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Example 2: Subformulas The function sub(P ) giving the subformulas of a formula P is defined by structural induction as follows: • sub(pi ) = {pi }, for all atomic formulas pi , • sub(P ∧ Q) = {(P ∧ Q)} ∪ sub(P ) ∪ sub(Q) • sub(P ∨ Q) = {(P ∨ Q)} ∪ sub(P ) ∪ sub(Q) • sub(¬P ) = {¬P } ∪ sub(P ). The set sub(P ) is called the set of subformulas of P .

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Subformulas Compute sub(P ) for P = ((p1 ∧ ¬p2 ) ∨ ¬p3 ). sub(P ) = {P } ∪ sub(p1 ∧ ¬p2 ) ∪ sub(¬p3 ) = {P } ∪ {(p1 ∧ ¬p2 )} ∪ sub(p1 ) ∪ sub(¬p2 ) ∪ sub(¬p3 ) = {P, (p1 ∧ ¬p2 )} ∪ {p1 } ∪ {¬p2 } ∪ sub(p2 ) ∪ {¬p3 } ∪ sub(p3 ) = {P, (p1 ∧ ¬p2 ), p1 , ¬p2 , p2 , ¬p3 , p3 }

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Example 3: Length of a formula The function L(P ) giving the length of a formula is defined by structural induction as follows: • L(pi ) = 1, for all atomic formulas pi , • L(P ∧ Q) = 1 + L(P ) + L(Q), • L(P ∨ Q) = 1 + L(P ) + L(Q), • L(¬P ) = 1 + L(P ). L(P ) is called the length of formula P .

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Length of a formula Compute L(P ) for P = ¬(p0 ∧ ¬p1 ). L(P ) = 1 + L(p0 ∧ ¬p1 ) = 1 + 1 + L(p0 ) + L(¬p1 ) = 1 + 1 + 1 + 1 + L(p1 ) = 5.

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Proofs by structural induction Statements about objects defined by structural induction can often be proved by structural induction. We illustrate this proof method by means of the following example. (|sub(P )| denotes the number of subformulas of P .) Theorem For every formula P : |sub(P )| ≤ L(P ). Proof The proof is by structural induction. In other words, we show: 1. for all atomic formulas pi : |sub(pi )| ≤ L(pi ); 2. for all formulas P and Q: if |sub(P )| ≤ L(P ) and |sub(Q)| ≤ L(Q), then |sub(P ∧ Q)| ≤ L(P ∧ Q); 3. for all formulas P and Q: if |sub(P )| ≤ L(P ) and |sub(Q)| ≤ L(Q), then |sub(P ∨ Q)| ≤ L(P ∨ Q); 4. for every formula P : if |sub(P )| ≤ L(P ), then |sub(¬P )| ≤ L(¬P ). Logic in Computer Science

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We now check (1.)-(4.): 1. |sub(pi )| = 1 ≤ 1 = L(pi ). 2. Suppose |sub(P )| ≤ L(P ) and |sub(Q)| ≤ L(Q). Then |sub(P ∧ Q)| = |{P ∧ Q} ∪ sub(P ) ∪ sub(Q)| ≤ 1 + |sub(P )| + |sub(Q)| ≤ 1 + L(P ) + L(Q) = L(P ∧ Q). 3. Exercise. 4. Exercise.

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Termination of the tableau algorithm Assume P is given. We have to show that there are only finitely many tableau paths {P } = S0 , S1 , . . . , Sn . Let sub¬ (P ) = sub(P ) ∪ {¬Q | Q ∈ sub(P )}. Now observe for any tableau path {P } = S0 , S1 , . . . , Sn : • S0 ⊂ S1 ⊂ · · · ⊂ Sn ⊆ sub¬ (P ). Hence • the length of any tableau path {P } = S0 , S1 , . . . , Sn is not greater than |sub¬ (P )| ¬

• and the number of such tableau paths is not greater that |sub¬ (P )||sub

(P )|

.

(Here, by X ⊂ Y we denote that X is a proper subset of Y .)

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Soundness of the tableau algorithm Let {P } = S0 , S1 , . . . , Sn be a complete tableau path such that Sn does not contain a clash. We define an interpretation I by  I(pi ) =

1 0

if if

p i ∈ Sn pi 6∈ Sn

We show the following claim by structural induction: Claim 1 • I(Q) = 1 for all Q ∈ Sn ; • I(Q) = 0 for all ¬Q ∈ Sn . Since P ∈ Sn , we obtain I(P ) = 1. Thus P is satisfiable.

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The steps of the structural induction We have to show: 1. Claim 1 holds for all atomic formulas pi ; 2. if Claim 1 holds for P1 and P2 , then Claim 1 holds for (P1 ∧ P2 ); 3. if Claim 1 holds for P1 and P2 , then Claim 1 holds for (P1 ∨ P2 ); 4. if Claim 1 holds for Q, then it holds for ¬Q.

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Proof of Point 1 Let pi be an atomic formula. We have to show (a) I(pi ) = 1 if pi ∈ Sn ; (b) I(pi ) = 0 if ¬pi ∈ Sn . Point (a) follows by definition of I. For Point (b), • assume that ¬pi ∈ Sn . • Since Sn does not contain a clash, pi 6∈ Sn . • Hence, by definition of I, I(pi ) = 0.

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Proof for Point 2 Assume Claim 1 holds for P1 and P2 . Suppose (P1 ∧ P2 ) ∈ Sn . • Then, by non-applicability of =⇒∧ to Sn , P1 ∈ Sn and P2 ∈ Sn ; • By induction hypothesis, I(P1 ) = 1 and I(P2 ) = 1; • Hence I(P1 ∧ P2 ) = 1. Suppose ¬(P1 ∧ P2 ) ∈ Sn . • Then, by non-applicability of =⇒¬∧ to Sn , ¬P1 ∈ Sn or ¬P2 ∈ Sn ; • By induction hypothesis, I(P1 ) = 0 or I(P2 ) = 0; • Hence I(P1 ∧ P2 ) = 0.

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Proof for Point 3 Assume Claim 1 holds for P1 and P2 . Suppose (P1 ∨ P2 ) ∈ Sn . • Then, by non-applicability of =⇒∨ to Sn , P1 ∈ Sn or P2 ∈ Sn ; • By induction hypothesis, I(P1 ) = 1 or I(P2 ) = 1; • Hence I(P1 ∨ P2 ) = 1. Suppose ¬(P1 ∨ P2 ) ∈ Sn . • Then, by non-applicability of =⇒¬∨ to Sn , ¬P1 ∈ Sn and ¬P2 ∈ Sn ; • By induction hypothesis, I(P1 ) = 0 and I(P2 ) = 0; • Hence I(P1 ∨ P2 ) = 0.

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Proof for Point 4 Assume Claim 1 holds for Q. We show Claim 1 for ¬Q. Suppose ¬Q ∈ Sn . • By induction hypothesis, I(Q) = 0. • Hence I(¬Q) = 1. Suppose ¬¬Q ∈ Sn . • Then, by non-applicability of =⇒¬¬ to Sn , Q ∈ Sn . • By induction hypothesis, I(Q) = 1; • Hence I(¬Q) = 0.

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Completeness of the tableau algorithm Assume that P is satisfiable. We have to construct a complete tableau path {P } = S0 , S1 , . . . , Sn such that Sn does not contain a clash. Let I be an interpretation with I(P ) = 1. We construct the tableau path as follows: Let S0 = {P } and assume that S0 ⊂ S1 ⊂ · · · ⊂ Si have already been defined such that I(Q) = 1 for all Q ∈ Si . Then Si does not contain a clash. If no completion rule is applicable, then the path is complete and we are done. Now assume that a completion rule is applicable. We show that we can apply the rule in such a way that Si =⇒ Si+1 and I(Q) = 1 for all Q ∈ Si+1 .

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Construction of Si+1 1. If P1 ∧ P2 ∈ Si and =⇒∧ is applicable, then set Si+1 = Si ∪ {P1 , P2 }. Then I(P1 ) = I(P2 ) = 1 since I(P1 ∧ P2 ) = 1. Thus I(Q) = 1 for all Q ∈ Si+1 . 2. Otherwise, if ¬(P1 ∨ P2 ) ∈ Si and =⇒¬∨ is applicable, then set Si+1 = Si ∪ {¬P1 , ¬P2 }. Then I(¬P1 ) = I(¬P2 ) = 1 since I(¬(P1 ∨ P2 )) = 1. Thus I(Q) = 1 for all Q ∈ Si+1 . 3. Otherwise, if ¬¬P1 ∈ Si and =⇒¬ is applicable, then set Si+1 = Si ∪ {P1 }. Then I(P1 ) = 1 since I(¬¬P1 ) = 1. Thus I(Q) = 1 for all Q ∈ Si+1 .

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Construction of Si+1 1. Otherwise, if ¬(P1 ∧ P2 ) ∈ Si and =⇒¬∧ is applicable, then I(¬P1 ) = 1 or I(¬P2 ) = 1 since I(¬(P1 ∧ P2 )) = 1. In the first case let Si+1 = Si ∪ {¬P1 }. In the second case let Si+1 = Si ∪ {¬P2 }. In both cases I(Q) = 1 for all Q ∈ Si+1 . 2. Otherwise, if (P1 ∨ P2 ) ∈ Si and =⇒∨ is applicable, then I(P1 ) = 1 or I(P2 ) = 1 since I(P1 ∨ P2 ) = 1. In the first case let Si+1 = Si ∪ {P1 }. In the second case let Si+1 = Si ∪ {P2 }. In both cases I(Q) = 1 for all Q ∈ Si+1 .

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Modern SAT solvers High performance SAT solvers are not tableau based. They are based on modern versions of • the Davis-Putnam-Logemann-Loveland algorithm (DPLL) developed in the 1960s (which is based on a very general proof method called resolution); • and on stochastic local search algorithms. Many solvers are available as free and open source software.

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Tautology Definition A tautology is a formula which is true under all interpretations. Example All formulas of the form P ∨ ¬P are tautologies, because I(P ∨ ¬P ) = 1 for all interpretations I: P 1 0

¬P 0 1

P ∨ ¬P 1 1

Observation: A formula P is a tautology if, and only if, ¬P is not satisfiable.

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Contradiction Definition A contradiction is a formula which is false under all interpretations. Example All formulas of the form P ∧ ¬P are tautologies, because I(P ∧ ¬P ) = 0 for all interpretations I: P 1 0

¬P 0 1

P ∧ ¬P 0 0

Observation: A formula P is a contradiction if, and only if, P is not satisfiable

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Semantic consequence Definition Suppose X is a finite set of formulas and P is a formula. Then P follows from X (is a semantic consequence of X) if the following holds for every interpretation I: If I(Q) = 1 for all Q ∈ X, then I(P ) = 1. This is denoted by X |= P.

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Example 1 Show {p1 ∧ p2 } |= p1 ∨ p2 . Solution: p1 1 1 0 0

p2 1 0 1 0

p1 ∧ p2 1 0 0 0

p1 ∨ p2 1 1 1 0

The statement follows, because in any row where the column for p1 ∧p2 contains 1 the column for p1 ∨ p2 also contains 1.

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Example 2 Show {p1 } 6|= p2 . Solution: Take the interpretation I with I(p1 ) = 1 and I(p2 ) = 0.

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Example 3 We show that ∅ |= P if, and only if, P is a tautology. (⇒) Assume P is not a tautology. Take interpretation I with I(P ) = 0. Then I(Q) = 1 for all Q ∈ ∅, but I(P ) 6= 1, Hence ∅ 6|= P . (⇐) Assume ∅ 6|= P . Take interpretation I with I(Q) = 1 for all Q ∈ ∅ and I(P ) 6= 1. Then P is not a tautology.

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Example 4 (ex falso quodlibet) We show that {(P ∧ ¬P )} |= Q holds for all formulas Q. Let Q be arbitrary. There is no interpretation I such that I(P ∧ ¬P ) = 1. Thus, if I is an interpretation such that I(P ∧ ¬P ) = 1, then I(Q) = 1. Thus {(P ∧ ¬P )} |= Q.

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Reduction to Satisfiability Recall that we call a finite set S of formulas satisfiable if there exists an interpretation I such that I(Q) = 1 for all Q ∈ S. Note that the set S = {Q1 , . . . , Qn } is satisfiable if, and only if, the formula obtained by taking the conjunction of all Q1 , . . . , Qn , (Q1 ∧ . . . ∧ Qn ), is satisfiable.

Observation For every finite set S of formulas and every formula P : S |= P if, and only if, S ∪ {¬P } is not satisfiable.

Thus, we can use the tableau algorithm to check semantic consequence: to check whether S |= P check that S ∪ {¬P } is not satisfiable.

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Example We check {p ∧ q} |= p ∨ q. To this end, we have to show that S0 = {p ∧ q, ¬(p ∨ q)} is not satisfiable. We do this using the tableau algorithm: • an application of =⇒∧ to S0 gives S1 = S0 ∪ {p, q, ¬(p ∨ q)} • an application of =⇒¬∨ to S1 gives S2 = S1 ∪ {¬p, ¬q} S2 contains a clash: {p, ¬p} ⊆ S2 . Thus, all tableau paths starting with S0 contain a clash. Hence S0 is not satisfiable. Logic in Computer Science

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Logical equivalence Definition Two formulas P and Q are called equivalent if they have the same truth value under every possible interpretation. In other words, P and Q are equivalent if I(P ) = I(Q) for every interpretation I. This is denoted by P ≡ Q.

Observation For any two formulas P and Q: P ≡ Q if, and only if, neither (P ∧ ¬Q) nor (Q ∧ ¬P ) are satisfiable.

Thus, we can use the tableau algorithm to check logical equivalence: to check whether P ≡ Q check that (P ∧ ¬Q) is not satisfiable and (Q ∧ ¬P ) is not satisfiable.

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Laws for equivalences The following equivalences can be checked using the tableau algorithm or by truth tables: • Associative laws: P ∨ (Q ∨ R) ≡ (P ∨ Q) ∨ R, P ∧ (Q ∧ R) ≡ (P ∧ Q) ∧ R • Commutative laws: P ∨ Q ≡ Q ∨ P, P ∧ Q ≡ Q ∧ P

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Laws for Equivalence • Distributive laws: P ∧ (Q ∨ R) ≡ (P ∧ Q) ∨ (P ∧ R), P ∨ (Q ∧ R) ≡ (P ∨ Q) ∧ (P ∨ R) • Complement law: ¬¬P ≡ P • De Morgan’s laws: ¬(P ∨ Q) ≡ ¬P ∧ ¬Q, ¬(P ∧ Q) ≡ ¬P ∨ ¬Q.

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Proof of {(p ∧ (q ∨ r))} |= ((p ∧ q) ∨ (p ∧ r)) We have to show that S0 = {(p ∧ (q ∨ r)), ¬((p ∧ q) ∨ (p ∧ r))} is not satisfiable. • an application of =⇒∧ to S0 gives S1 = S0 ∪ {p, (q ∨ r)} • an application of =⇒¬∨ to S1 gives S2 = S1 ∪ {¬(p ∧ q), ¬(p ∧ r)}

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Continue by decomposing (q ∨ r) ∈ S2 (1) First option: • an application of =⇒∨ to (q ∨ r) ∈ S2 gives S3 = S2 ∪ {q} • an application of =⇒¬∧ to ¬(p ∧ q) ∈ S3 gives S4 = S3 ∪ {¬p} which contains the clash {p, ¬p}. • the other application of =⇒¬∧ to ¬(p ∧ q) ∈ S3 gives S5 = S3 ∪ {¬q} which contains the clash {q, ¬q}.

Thus, every complete tableau path starting with S0 , S1 , S2 , S3 contains a clash.

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Continue by decomposing (q ∨ r) ∈ S2 (2) Second option: • an application of =⇒∨ to (q ∨ r) ∈ S2 gives S6 = S2 ∪ {r}. • an application of =⇒¬∧ to ¬(p ∧ r) ∈ S6 gives S7 = S6 ∪ {¬p} which contains the clash {p, ¬p}. • the other application of =⇒¬∧ to ¬(p ∧ r) ∈ S6 gives S8 = S6 ∪ {¬r} which contains the clash {r, ¬r}.

Thus, every complete tableau path starting with S0 , S1 , S2 , S6 contains a clash. We can conclude that all complete tableau paths starting with S0 contain a clash.

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Conjunctive and disjunctive normal form • A formula (P1 ∨ P2 ∨ · · · ∨ Pn ) is called a disjunction of P1 , . . . , Pn ; • Similarly, (P1 ∧ P2 ∧ · · · ∧ Pn ) is called a conjunction of P1 , . . . , Pn ; • A formula which is either an atomic formula or its negation is called a literal; • A formula is in conjunctive normal form (CNF) if it is a conjunction of disjunctions of literals. • A formula is in disjunctive normal form (DNF) if it is a disjunction of conjunctions of literals.

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Examples • p1 , ¬p1 , ¬p5 are literals. They are also in CNF and in DNF. • (p ∨ q) is in CNF and in DNF. • ((p1 ∨ p2 ) ∧ (¬p1 ∨ p3 )) and ((p1 ∨ p2 ) ∧ ¬p1 ) are in CNF and not in DNF. • ((p1 ∧ p2 ) ∨ (¬p1 ∧ p3 )) and (p ∨ (p ∧ ¬p)) are in DNF and not in CNF.

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CNF Theorem (1) Every formula is equivalent to a formula in CNF. (2) Every formula is equivalent to a formula in DNF. Proof of (1) Suppose a formula P is given. We transform P to a formula in CNF using the Laws of equivalence: Step 1. Apply De Morgan’s laws and Complement law ¬¬P ≡ P until negation (¬) occurs in front of atomic formulas only. Step 2. Apply Distributive law P ∨ (Q ∧ R) ≡ (P ∨ Q) ∧ (P ∨ R) and Commutative laws until the formula is in CNF.

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Example Transform (¬(p0 ∨ p1 ) ∨ (p2 ∧ p1 )) into CNF. (¬(p0 ∨ p1 ) ∨ (p2 ∧ p1 )) is equivalent to (de Morgan’s Law) ((¬p0 ∧ ¬p1 ) ∨ (p2 ∧ p1 )) is equivalent to (Distributive law) (((¬p0 ∧ ¬p1 ) ∨ p2 ) ∧ ((¬p0 ∧ ¬p1 ) ∨ p1 )) is equivalent to (Distributive law) ((¬p0 ∨ p2 ) ∧ (¬p1 ∨ p2 ) ∧ (¬p0 ∧ ¬p1 ) ∨ p1 )) is equivalent to (Distributive law) ((¬p0 ∨ p2 ) ∧ (¬p1 ∨ p2 ) ∧ (¬p0 ∨ p1 ) ∧ (¬p1 ∨ p1 )). Logic in Computer Science

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