Download Propositional Logic - Department of Computer Science

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)
Logic in Computer Science
2
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
3
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
4
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.
Logic in Computer Science
5
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
Logic in Computer Science
¬P
0
1
6
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
Logic in Computer Science
Q
1
0
1
0
(P ∧ Q)
1
0
0
0
7
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
Logic in Computer Science
Q
1
0
1
0
(P ∨ Q)
1
1
1
0
8
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.
Logic in Computer Science
9
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.
Logic in Computer Science
10
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).
Logic in Computer Science
11
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 )).
Logic in Computer Science
12
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.
Logic in Computer Science
13
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)
Logic in Computer Science
14
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))
Logic in Computer Science
15
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
16
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 );
Logic in Computer Science
17
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”.
Logic in Computer Science
18
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.
Logic in Computer Science
19
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.
Logic in Computer Science
20
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.
Logic in Computer Science
21
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).
Logic in Computer Science
22
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.
Logic in Computer Science
23
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.
Logic in Computer Science
24
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}.
Logic in Computer Science
25
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.
Logic in Computer Science
26
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
27
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
Logic in Computer Science
28
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”.
Logic in Computer Science
29
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”.
Logic in Computer Science
30
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.
Logic in Computer Science
31
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.
Logic in Computer Science
32
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
33
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.
Logic in Computer Science
34
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.
Logic in Computer Science
35
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.
Logic in Computer Science
36
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:
Logic in Computer Science
37
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”.
Logic in Computer Science
38
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
39
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.
Logic in Computer Science
40
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
41
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”.
Logic in Computer Science
42
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.
Logic in Computer Science
43
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.)
Logic in Computer Science
44
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 ).
Logic in Computer Science
45
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 .
Logic in Computer Science
46
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 }
Logic in Computer Science
47
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 .
Logic in Computer Science
48
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.
Logic in Computer Science
49
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
50
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.
Logic in Computer Science
51
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 .)
Logic in Computer Science
52
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.
Logic in Computer Science
53
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.
Logic in Computer Science
54
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.
Logic in Computer Science
55
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.
Logic in Computer Science
56
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.
Logic in Computer Science
57
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.
Logic in Computer Science
58
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 .
Logic in Computer Science
59
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 .
Logic in Computer Science
60
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 .
Logic in Computer Science
61
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.
Logic in Computer Science
62
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.
Logic in Computer Science
63
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
Logic in Computer Science
64
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.
Logic in Computer Science
65
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.
Logic in Computer Science
66
Example 2
Show {p1 } 6|= p2 .
Solution: Take the interpretation I with I(p1 ) = 1 and I(p2 ) = 0.
Logic in Computer Science
67
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.
Logic in Computer Science
68
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.
Logic in Computer Science
69
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.
Logic in Computer Science
70
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
71
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.
Logic in Computer Science
72
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
Logic in Computer Science
73
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.
Logic in Computer Science
74
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)}
Logic in Computer Science
75
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.
Logic in Computer Science
76
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.
Logic in Computer Science
77
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.
Logic in Computer Science
78
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.
Logic in Computer Science
79
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.
Logic in Computer Science
80
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
81