Mathematical Logic
Lecture 4: Normal forms
Ashutosh Gupta
TIFR, India
Compile date: 2015-08-19
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Where are we and where are we going?
We have seen
I
propositional logic syntax and semantics
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truth tables as methods for deciding SAT
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some common equivalences
We will learn
I Various normal forms
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I
I
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NNF
CNF
DNF
k-SAT
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Topic 4.1
Normal forms
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Some terminology
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Propositional variables are also referred as atoms
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Some terminology
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Propositional variables are also referred as atoms
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A literal is either an atom or its negation
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Some terminology
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Propositional variables are also referred as atoms
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A literal is either an atom or its negation
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A clause is a disjunction of literals.
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Some terminology
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Propositional variables are also referred as atoms
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A literal is either an atom or its negation
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A clause is a disjunction of literals.
Since ∨ is associative, commutative and absorbs multiple occurrences, a
clause may be referred as a set of literals
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Negation normal form(NNF)
Definition 4.1
A formula is in NNF if ¬ appears only in front of the propositional variables.
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Negation normal form(NNF)
Definition 4.1
A formula is in NNF if ¬ appears only in front of the propositional variables.
Theorem 4.1
For every formula F there is another formula F 0 in NNF s.t. F ≡ F 0 .
Proof.
Recall from the tautologies that we can always push negation inside the
operators.
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Negation normal form(NNF)
Definition 4.1
A formula is in NNF if ¬ appears only in front of the propositional variables.
Theorem 4.1
For every formula F there is another formula F 0 in NNF s.t. F ≡ F 0 .
Proof.
Recall from the tautologies that we can always push negation inside the
operators.
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Often we assume that the formulas are in negation normal form.
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However, there are negations hidden inside ⊕, ⇒, and ⇔. In practice,
these symbols are also expected to be removed while producing NNF
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Negation normal form(NNF)
Definition 4.1
A formula is in NNF if ¬ appears only in front of the propositional variables.
Theorem 4.1
For every formula F there is another formula F 0 in NNF s.t. F ≡ F 0 .
Proof.
Recall from the tautologies that we can always push negation inside the
operators.
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Often we assume that the formulas are in negation normal form.
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However, there are negations hidden inside ⊕, ⇒, and ⇔. In practice,
these symbols are also expected to be removed while producing NNF
Exercise 4.1
a. Write an efficient algorithm to covert a propositional formula to NNF?
b. Are there any added difficulties if the formula is given as a DAG not as a
tree?
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Example :NNF
Example 4.1
Consider ¬(((p ∨ ¬s) ⊕ r ) ⇒ q)
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Example :NNF
Example 4.1
Consider ¬(((p ∨ ¬s) ⊕ r ) ⇒ q)
≡ ¬q ⇒ ¬((p ∨ ¬s) ⊕ r )
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Example :NNF
Example 4.1
Consider ¬(((p ∨ ¬s) ⊕ r ) ⇒ q)
≡ ¬q ⇒ ¬((p ∨ ¬s) ⊕ r )
≡ ¬q ⇒ ¬(p ∨ ¬s) ⊕ r
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Example :NNF
Example 4.1
Consider ¬(((p ∨ ¬s) ⊕ r ) ⇒ q)
≡ ¬q ⇒ ¬((p ∨ ¬s) ⊕ r )
≡ ¬q ⇒ ¬(p ∨ ¬s) ⊕ r
≡ ¬q ⇒ (¬p ∧ ¬¬s) ⊕ r
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Example :NNF
Example 4.1
Consider ¬(((p ∨ ¬s) ⊕ r ) ⇒ q)
≡ ¬q ⇒ ¬((p ∨ ¬s) ⊕ r )
≡ ¬q ⇒ ¬(p ∨ ¬s) ⊕ r
≡ ¬q ⇒ (¬p ∧ ¬¬s) ⊕ r
≡ ¬q ⇒ (¬p ∧ s) ⊕ r
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Example :NNF
Example 4.1
Consider ¬(((p ∨ ¬s) ⊕ r ) ⇒ q)
≡ ¬q ⇒ ¬((p ∨ ¬s) ⊕ r )
≡ ¬q ⇒ ¬(p ∨ ¬s) ⊕ r
≡ ¬q ⇒ (¬p ∧ ¬¬s) ⊕ r
≡ ¬q ⇒ (¬p ∧ s) ⊕ r
Exercise 4.2
Convert the following formulas into NNF
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¬(p ⇒ q)
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¬(¬((s ⇒ ¬(p ⇔ q))) ⊕ (¬q ∨ r ))
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Conjunctive normal form(CNF)
Definition 4.2
A formula is in CNF if it is a conjunction of clauses.
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Conjunctive normal form(CNF)
Definition 4.2
A formula is in CNF if it is a conjunction of clauses.
Since ∧ is associative, commutative and absorbs multiple occurrences, a CNF
formula may be referred as a set of clauses
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Conjunctive normal form(CNF)
Definition 4.2
A formula is in CNF if it is a conjunction of clauses.
Since ∧ is associative, commutative and absorbs multiple occurrences, a CNF
formula may be referred as a set of clauses
Theorem 4.2
For every formula F there is another formula F 0 in CNF s.t. F ≡ F 0 .
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Conjunctive normal form(CNF)
Definition 4.2
A formula is in CNF if it is a conjunction of clauses.
Since ∧ is associative, commutative and absorbs multiple occurrences, a CNF
formula may be referred as a set of clauses
Theorem 4.2
For every formula F there is another formula F 0 in CNF s.t. F ≡ F 0 .
Proof.
Let us suppose we have removed ⊕, ⇒, ⇔ using equivalences presented so
far and converted the formula in NNF. Since ∨ distributes over ∧, we can
always push ∨ inside ∧. Eventually, we will obtain a formula that is CNF.
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Conjunctive normal form(CNF)
Definition 4.2
A formula is in CNF if it is a conjunction of clauses.
Since ∧ is associative, commutative and absorbs multiple occurrences, a CNF
formula may be referred as a set of clauses
Theorem 4.2
For every formula F there is another formula F 0 in CNF s.t. F ≡ F 0 .
Proof.
Let us suppose we have removed ⊕, ⇒, ⇔ using equivalences presented so
far and converted the formula in NNF. Since ∨ distributes over ∧, we can
always push ∨ inside ∧. Eventually, we will obtain a formula that is CNF.
Exercise 4.3
Write a formal grammar for CNF
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Conjunctive normal form(CNF)
Definition 4.2
A formula is in CNF if it is a conjunction of clauses.
Since ∧ is associative, commutative and absorbs multiple occurrences, a CNF
formula may be referred as a set of clauses
Theorem 4.2
For every formula F there is another formula F 0 in CNF s.t. F ≡ F 0 .
Proof.
Let us suppose we have removed ⊕, ⇒, ⇔ using equivalences presented so
far and converted the formula in NNF. Since ∨ distributes over ∧, we can
always push ∨ inside ∧. Eventually, we will obtain a formula that is CNF.
After the push formula size grows!
Why should this procedure terminate?
Exercise 4.3
Write a formal grammar for CNF
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CNF conversion terminates
Theorem 4.3
The procedure of converting a formula in CNF terminates.
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CNF conversion terminates
Theorem 4.3
The procedure of converting a formula in CNF terminates.
Proof.
For a formula F let ν(F ) := maximum height of nested ∧s
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CNF conversion terminates
Theorem 4.3
The procedure of converting a formula in CNF terminates.
Proof.
For a formula F let ν(F ) := maximum height of nested ∧s
Consider a formula F (G ) such that
G=
ni
m ^
_
Gij
i=0 j=0
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CNF conversion terminates
Theorem 4.3
The procedure of converting a formula in CNF terminates.
Proof.
For a formula F let ν(F ) := maximum height of nested ∧s
Consider a formula F (G ) such that
G=
ni
m ^
_
Gij
i=0 j=0
After the push we obtain F (G 0 ), where
G0 =
n1
^
j1 =0
...
nm _
m
^
Giji
jm =0 i=0
ν(
| {z }
)<ν(G )
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CNF conversion terminates
Theorem 4.3
The procedure of converting a formula in CNF terminates.
Proof.
For a formula F let ν(F ) := maximum height of nested ∧s
Consider a formula F (G ) such that
G=
ni
m ^
_
Gij
i=0 j=0
After the push we obtain F (G 0 ), where
G0 =
n1
^
j1 =0
...
nm _
m
^
Giji
jm =0 i=0
ν(
| {z }
)<ν(G )
The finitely many substituted ∨-formulas with have lower ν values.
Due to Köing lemma, the procedure terminates.
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CNF examples
Example 4.2
Consider (p ⇒ (¬q ∧ r )) ∧ (p ⇒ ¬q)
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CNF examples
Example 4.2
Consider (p ⇒ (¬q ∧ r )) ∧ (p ⇒ ¬q)
≡ (¬p ∨ (¬q ∧ r )) ∧ (¬p ∨ ¬q)
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CNF examples
Example 4.2
Consider (p ⇒ (¬q ∧ r )) ∧ (p ⇒ ¬q)
≡ (¬p ∨ (¬q ∧ r )) ∧ (¬p ∨ ¬q)
≡ ((¬p ∨ ¬q) ∧ (¬p ∨ r )) ∧ (¬p ∨ ¬q)
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CNF examples
Example 4.2
Consider (p ⇒ (¬q ∧ r )) ∧ (p ⇒ ¬q)
≡ (¬p ∨ (¬q ∧ r )) ∧ (¬p ∨ ¬q)
≡ ((¬p ∨ ¬q) ∧ (¬p ∨ r )) ∧ (¬p ∨ ¬q)
≡ (¬p ∨ ¬q) ∧ (¬p ∨ r ) ∧ (¬p ∨ ¬q)
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CNF examples
Example 4.2
Consider (p ⇒ (¬q ∧ r )) ∧ (p ⇒ ¬q)
≡ (¬p ∨ (¬q ∧ r )) ∧ (¬p ∨ ¬q)
≡ ((¬p ∨ ¬q) ∧ (¬p ∨ r )) ∧ (¬p ∨ ¬q)
≡ (¬p ∨ ¬q) ∧ (¬p ∨ r ) ∧ (¬p ∨ ¬q)
Exercise 4.4
Convert the following formulas into CNF
1. ¬((p ⇒ q) ⇒ ((q ⇒ r ) ⇒ (p ⇒ r )))
2. (p ⇒ (¬q ⇒ r )) ∧ (p ⇒ ¬q) ⇒ (p ⇒ r )
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Conjunctive normal form(CNF) (2)
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A unit clause contains only one single literal.
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A binary clause contains two literals.
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A ternary clause contains three literals.
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We naturally extend definition of the clauses to empty set of literals. We
refer to ⊥ as empty clause.
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Conjunctive normal form(CNF) (2)
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A unit clause contains only one single literal.
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A binary clause contains two literals.
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A ternary clause contains three literals.
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We naturally extend definition of the clauses to empty set of literals. We
refer to ⊥ as empty clause.
Example 4.3
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(p ∧ q ∧ ¬r ) has three unit clauses
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Conjunctive normal form(CNF) (2)
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A unit clause contains only one single literal.
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A binary clause contains two literals.
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A ternary clause contains three literals.
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We naturally extend definition of the clauses to empty set of literals. We
refer to ⊥ as empty clause.
Example 4.3
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(p ∧ q ∧ ¬r ) has three unit clauses
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(p ∨ ¬q ∨ ¬s) ∧ (p ∨ q) ∧ ¬r has a ternary, a binary and a unit clause
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Conjunctive normal form(CNF) (2)
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A unit clause contains only one single literal.
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A binary clause contains two literals.
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A ternary clause contains three literals.
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We naturally extend definition of the clauses to empty set of literals. We
refer to ⊥ as empty clause.
Example 4.3
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(p ∧ q ∧ ¬r ) has three unit clauses
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(p ∨ ¬q ∨ ¬s) ∧ (p ∨ q) ∧ ¬r has a ternary, a binary and a unit clause
Exercise 4.5
a. Show F 0 obtained from the procedure may be exponentially larger than F
b. Give a linear time algorithm to prove validity of a CNF formula
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Tseitin’s encoding
We can translate every formula into CNF without exponential explosion using
Tseitin’s encoding by introducing fresh variables.
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Tseitin’s encoding
We can translate every formula into CNF without exponential explosion using
Tseitin’s encoding by introducing fresh variables.
1. Assume input formula F is NNF without ⊕, ⇒, and ⇔.
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Tseitin’s encoding
We can translate every formula into CNF without exponential explosion using
Tseitin’s encoding by introducing fresh variables.
1. Assume input formula F is NNF without ⊕, ⇒, and ⇔.
2. find a G1 ∧ · · · ∧ Gn that is below a ∨ in F (G1 ∧ · · · ∧ Gn )
3. F (G1 ∧ · · · ∧ Gn ) := F (p) ∧ (¬p ∨ G1 ) ∧ · · · ∧ (¬p ∨ Gn ), where p is a
fresh variable
4. goto 2
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Tseitin’s encoding
We can translate every formula into CNF without exponential explosion using
Tseitin’s encoding by introducing fresh variables.
1. Assume input formula F is NNF without ⊕, ⇒, and ⇔.
2. find a G1 ∧ · · · ∧ Gn that is below a ∨ in F (G1 ∧ · · · ∧ Gn )
3. F (G1 ∧ · · · ∧ Gn ) := F (p) ∧ (¬p ∨ G1 ) ∧ · · · ∧ (¬p ∨ Gn ), where p is a
fresh variable
4. goto 2
Exercise 4.6
Convert the following formulas into CNF using Tseitin’s encoding
1. (p ⇒ (¬q ∧ r )) ∧ (p ⇒ ¬q)
2. (p ⇒ q) ∨ (q ⇒ ¬r ) ∨ (r ⇒ q) ⇒ ¬(¬(q ⇒ p) ⇒ (q ⇔ r ))
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Tseitin’ encoding preserves satisfiability
Theorem 4.4
if m |= F (p) ∧ (¬p ∨ G1 ) ∧ · · · ∧ (¬p ∨ Gn ) then m |= F (G1 ∧ · · · ∧ Gn )
Proof.
Case m(p) = 1:
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Therefore, m |= Gi for all i ∈ 1..n.
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Tseitin’ encoding preserves satisfiability
Theorem 4.4
if m |= F (p) ∧ (¬p ∨ G1 ) ∧ · · · ∧ (¬p ∨ Gn ) then m |= F (G1 ∧ · · · ∧ Gn )
Proof.
Case m(p) = 1:
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Therefore, m |= Gi for all i ∈ 1..n.
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Therefore, m |= G1 ∧ · · · ∧ Gn .
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Tseitin’ encoding preserves satisfiability
Theorem 4.4
if m |= F (p) ∧ (¬p ∨ G1 ) ∧ · · · ∧ (¬p ∨ Gn ) then m |= F (G1 ∧ · · · ∧ Gn )
Proof.
Case m(p) = 1:
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Therefore, m |= Gi for all i ∈ 1..n.
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Therefore, m |= G1 ∧ · · · ∧ Gn .
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Therefore, m |= F (G1 ∧ · · · ∧ Gn ).
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Tseitin’ encoding preserves satisfiability
Theorem 4.4
if m |= F (p) ∧ (¬p ∨ G1 ) ∧ · · · ∧ (¬p ∨ Gn ) then m |= F (G1 ∧ · · · ∧ Gn )
Proof.
Case m(p) = 1:
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Therefore, m |= Gi for all i ∈ 1..n.
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Therefore, m |= G1 ∧ · · · ∧ Gn .
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Therefore, m |= F (G1 ∧ · · · ∧ Gn ).
Case m(p) = 0:
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Tseitin’ encoding preserves satisfiability
Theorem 4.4
if m |= F (p) ∧ (¬p ∨ G1 ) ∧ · · · ∧ (¬p ∨ Gn ) then m |= F (G1 ∧ · · · ∧ Gn )
Proof.
Case m(p) = 1:
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Therefore, m |= Gi for all i ∈ 1..n.
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Therefore, m |= G1 ∧ · · · ∧ Gn .
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Therefore, m |= F (G1 ∧ · · · ∧ Gn ).
Case m(p) = 0:
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Since F (G1 ∧ · · · ∧ Gn ) is in NNF, p occurs only positively in F .
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Tseitin’ encoding preserves satisfiability
Theorem 4.4
if m |= F (p) ∧ (¬p ∨ G1 ) ∧ · · · ∧ (¬p ∨ Gn ) then m |= F (G1 ∧ · · · ∧ Gn )
Proof.
Case m(p) = 1:
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Therefore, m |= Gi for all i ∈ 1..n.
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Therefore, m |= G1 ∧ · · · ∧ Gn .
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Therefore, m |= F (G1 ∧ · · · ∧ Gn ).
Case m(p) = 0:
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Since F (G1 ∧ · · · ∧ Gn ) is in NNF, p occurs only positively in F .
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Therefore, m[p 7→ 1] |= F (p)(why?).
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Tseitin’ encoding preserves satisfiability
Theorem 4.4
if m |= F (p) ∧ (¬p ∨ G1 ) ∧ · · · ∧ (¬p ∨ Gn ) then m |= F (G1 ∧ · · · ∧ Gn )
Proof.
Case m(p) = 1:
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Therefore, m |= Gi for all i ∈ 1..n.
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Therefore, m |= G1 ∧ · · · ∧ Gn .
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Therefore, m |= F (G1 ∧ · · · ∧ Gn ).
Case m(p) = 0:
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Since F (G1 ∧ · · · ∧ Gn ) is in NNF, p occurs only positively in F .
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Therefore, m[p 7→ 1] |= F (p)(why?).
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Therefore, m |= F (G1 ∧ · · · ∧ Gn ) if m |= G1 ∧ · · · ∧ Gn or not.
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Tseitin’ encoding preserves satisfiability
Theorem 4.4
if m |= F (p) ∧ (¬p ∨ G1 ) ∧ · · · ∧ (¬p ∨ Gn ) then m |= F (G1 ∧ · · · ∧ Gn )
Proof.
Case m(p) = 1:
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Therefore, m |= Gi for all i ∈ 1..n.
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Therefore, m |= G1 ∧ · · · ∧ Gn .
I
Therefore, m |= F (G1 ∧ · · · ∧ Gn ).
Case m(p) = 0:
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Since F (G1 ∧ · · · ∧ Gn ) is in NNF, p occurs only positively in F .
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Therefore, m[p 7→ 1] |= F (p)(why?).
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Therefore, m |= F (G1 ∧ · · · ∧ Gn ) if m |= G1 ∧ · · · ∧ Gn or not.
Exercise 4.7
a. Show if 6|= F (p) ∧ (¬p ∨ G1 ) ∧ .. ∧ (¬p ∨ Gn ) then 6|= F (G1 ∧ .. ∧ Gn )
b. Modify the encoding such that it works without assumptions at step 1
Hint: look into a sat solver frontend limboole and look for function tseitin $wget http://fmv.jku.at/limboole/limboole1.1.tar.gz
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Disjunctive normal form(DNF)
Definition 4.3
A formula is in DNF if it is a disjunction of conjunctions of literals.
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Disjunctive normal form(DNF)
Definition 4.3
A formula is in DNF if it is a disjunction of conjunctions of literals.
Theorem 4.5
For every formula F there is another formula F 0 in DNF s.t. F ≡ F 0 .
Proof.
Proof is similar to CNF.
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Disjunctive normal form(DNF)
Definition 4.3
A formula is in DNF if it is a disjunction of conjunctions of literals.
Theorem 4.5
For every formula F there is another formula F 0 in DNF s.t. F ≡ F 0 .
Proof.
Proof is similar to CNF.
Exercise 4.8
a. Give the formal grammar of DNF
b. Give a linear time algorithm to prove satisfiability of a DNF formula
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k-sat
Definition 4.4
A k-sat formula is a CNF formula and has at most k literals in each of its
clauses
Example 4.4
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(p ∧ q ∧ ¬r ) is 1-SAT
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k-sat
Definition 4.4
A k-sat formula is a CNF formula and has at most k literals in each of its
clauses
Example 4.4
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(p ∧ q ∧ ¬r ) is 1-SAT
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(p ∨ ¬p) ∧ (p ∨ q) is 2-SAT
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(p ∨ ¬q ∨ ¬s) ∧ (p ∨ q) ∧ ¬r is 3-SAT
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3-SAT satisfiablity
Theorem 4.6
For each k-SAT formula F there is a 3-SAT formula F 0 with linear blow up
such that F is sat iff F 0 is sat.
Proof.
Consider F a k-SAT formula with k ≥ 4.
Consider a clause G = (`1 ∨ · · · ∨ `k ) in F , where `i are literals.
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3-SAT satisfiablity
Theorem 4.6
For each k-SAT formula F there is a 3-SAT formula F 0 with linear blow up
such that F is sat iff F 0 is sat.
Proof.
Consider F a k-SAT formula with k ≥ 4.
Consider a clause G = (`1 ∨ · · · ∨ `k ) in F , where `i are literals.
Let x2 , . . . , xk−2 some variables that do not appear in F .
Let G 0 be the following set of clauses
(`1 ∨ `2 ∨ x2 ) ∧
^
(¬xi ∨ xi+1 ∨ `i+1 ) ∧ (¬xk−2 ∨ `k−1 ∨ `k ).
i∈2..k−2
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3-SAT satisfiablity
Theorem 4.6
For each k-SAT formula F there is a 3-SAT formula F 0 with linear blow up
such that F is sat iff F 0 is sat.
Proof.
Consider F a k-SAT formula with k ≥ 4.
Consider a clause G = (`1 ∨ · · · ∨ `k ) in F , where `i are literals.
Let x2 , . . . , xk−2 some variables that do not appear in F .
Let G 0 be the following set of clauses
(`1 ∨ `2 ∨ x2 ) ∧
^
(¬xi ∨ xi+1 ∨ `i+1 ) ∧ (¬xk−2 ∨ `k−1 ∨ `k ).
i∈2..k−2
We show G is sat iff G 0 is sat.
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3-SAT satisfiablity
Theorem 4.6
For each k-SAT formula F there is a 3-SAT formula F 0 with linear blow up
such that F is sat iff F 0 is sat.
Proof.
Consider F a k-SAT formula with k ≥ 4.
Consider a clause G = (`1 ∨ · · · ∨ `k ) in F , where `i are literals.
Let x2 , . . . , xk−2 some variables that do not appear in F .
Let G 0 be the following set of clauses
(`1 ∨ `2 ∨ x2 ) ∧
^
(¬xi ∨ xi+1 ∨ `i+1 ) ∧ (¬xk−2 ∨ `k−1 ∨ `k ).
i∈2..k−2
We show G is sat iff G 0 is sat.
Assume m |= G 0 :
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3-SAT satisfiablity
Theorem 4.6
For each k-SAT formula F there is a 3-SAT formula F 0 with linear blow up
such that F is sat iff F 0 is sat.
Proof.
Consider F a k-SAT formula with k ≥ 4.
Consider a clause G = (`1 ∨ · · · ∨ `k ) in F , where `i are literals.
Let x2 , . . . , xk−2 some variables that do not appear in F .
Let G 0 be the following set of clauses
(`1 ∨ `2 ∨ x2 ) ∧
^
(¬xi ∨ xi+1 ∨ `i+1 ) ∧ (¬xk−2 ∨ `k−1 ∨ `k ).
i∈2..k−2
We show G is sat iff G 0 is sat.
Assume m |= G 0 :
Assume for each i < 1..k m(`i ) = 0.
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3-SAT satisfiablity
Theorem 4.6
For each k-SAT formula F there is a 3-SAT formula F 0 with linear blow up
such that F is sat iff F 0 is sat.
Proof.
Consider F a k-SAT formula with k ≥ 4.
Consider a clause G = (`1 ∨ · · · ∨ `k ) in F , where `i are literals.
Let x2 , . . . , xk−2 some variables that do not appear in F .
Let G 0 be the following set of clauses
(`1 ∨ `2 ∨ x2 ) ∧
^
(¬xi ∨ xi+1 ∨ `i+1 ) ∧ (¬xk−2 ∨ `k−1 ∨ `k ).
i∈2..k−2
We show G is sat iff G 0 is sat.
Assume m |= G 0 :
Assume for each i < 1..k m(`i ) = 0. Due to the first clause m(x2 ) = 1 and
due to ith clause m(xi ) = 1 then m(xi+1 ) = 1. By induction, m(xk−2 ) = 1.
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3-SAT satisfiablity
Theorem 4.6
For each k-SAT formula F there is a 3-SAT formula F 0 with linear blow up
such that F is sat iff F 0 is sat.
Proof.
Consider F a k-SAT formula with k ≥ 4.
Consider a clause G = (`1 ∨ · · · ∨ `k ) in F , where `i are literals.
Let x2 , . . . , xk−2 some variables that do not appear in F .
Let G 0 be the following set of clauses
(`1 ∨ `2 ∨ x2 ) ∧
^
(¬xi ∨ xi+1 ∨ `i+1 ) ∧ (¬xk−2 ∨ `k−1 ∨ `k ).
i∈2..k−2
We show G is sat iff G 0 is sat.
Assume m |= G 0 :
Assume for each i < 1..k m(`i ) = 0. Due to the first clause m(x2 ) = 1 and
due to ith clause m(xi ) = 1 then m(xi+1 ) = 1. By induction, m(xk−2 ) = 1.
Due to the last clause of G 0 , m(xk−2 ) = 0. Contradiction.
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3-SAT satisfiablity
Theorem 4.6
For each k-SAT formula F there is a 3-SAT formula F 0 with linear blow up
such that F is sat iff F 0 is sat.
Proof.
Consider F a k-SAT formula with k ≥ 4.
Consider a clause G = (`1 ∨ · · · ∨ `k ) in F , where `i are literals.
Let x2 , . . . , xk−2 some variables that do not appear in F .
Let G 0 be the following set of clauses
(`1 ∨ `2 ∨ x2 ) ∧
^
(¬xi ∨ xi+1 ∨ `i+1 ) ∧ (¬xk−2 ∨ `k−1 ∨ `k ).
i∈2..k−2
We show G is sat iff G 0 is sat.
Assume m |= G 0 :
Assume for each i < 1..k m(`i ) = 0. Due to the first clause m(x2 ) = 1 and
due to ith clause m(xi ) = 1 then m(xi+1 ) = 1. By induction, m(xk−2 ) = 1.
Due to the last clause of G 0 , m(xk−2 ) = 0. Contradiction.
Therefore, exists i ∈ 1..k m(`i ) = 1. Therefore m |= G .
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3-SAT satisfiability(cont.)
Proof(contd. from last slide).
Recall
G 0 = (`1 ∨ `2 ∨ x2 ) ∧
^
(¬xi ∨ xi+1 ∨ `i+1 ) ∧ (¬xk−2 ∨ `k−1 ∨ `k ).
i∈2..k−2
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3-SAT satisfiability(cont.)
Proof(contd. from last slide).
Recall
G 0 = (`1 ∨ `2 ∨ x2 ) ∧
^
(¬xi ∨ xi+1 ∨ `i+1 ) ∧ (¬xk−2 ∨ `k−1 ∨ `k ).
i∈2..k−2
Assume m |= G :
There is a m(`i ) = 1.
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3-SAT satisfiability(cont.)
Proof(contd. from last slide).
Recall
G 0 = (`1 ∨ `2 ∨ x2 ) ∧
^
(¬xi ∨ xi+1 ∨ `i+1 ) ∧ (¬xk−2 ∨ `k−1 ∨ `k ).
i∈2..k−2
Assume m |= G :
There is a m(`i ) = 1.
Let m0 = m[x2 7→ 1, .., xi−1 7→ 1, xi 7→ 0, . . . , xk−2 7→ 0]
Therefore, m0 |= G 0 (why?).
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3-SAT satisfiability(cont.)
Proof(contd. from last slide).
Recall
G 0 = (`1 ∨ `2 ∨ x2 ) ∧
^
(¬xi ∨ xi+1 ∨ `i+1 ) ∧ (¬xk−2 ∨ `k−1 ∨ `k ).
i∈2..k−2
Assume m |= G :
There is a m(`i ) = 1.
Let m0 = m[x2 7→ 1, .., xi−1 7→ 1, xi 7→ 0, . . . , xk−2 7→ 0]
Therefore, m0 |= G 0 (why?).
G 0 contains 3(k − 2) literals.
In the worst case, the formula size will increase by 3.
16
3-SAT satisfiability(cont.)
Proof(contd. from last slide).
Recall
G 0 = (`1 ∨ `2 ∨ x2 ) ∧
^
(¬xi ∨ xi+1 ∨ `i+1 ) ∧ (¬xk−2 ∨ `k−1 ∨ `k ).
i∈2..k−2
Assume m |= G :
There is a m(`i ) = 1.
Let m0 = m[x2 7→ 1, .., xi−1 7→ 1, xi 7→ 0, . . . , xk−2 7→ 0]
Therefore, m0 |= G 0 (why?).
G 0 contains 3(k − 2) literals.
In the worst case, the formula size will increase by 3.
Exercise 4.9
a. Complete the above argument.
b. Show a 3-SAT formula cannot be converted into 2-SAT using Tseitin’s
encoding
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End of Lecture 4
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