binary decision diagrams for first order predicate logic by: jan friso groote afsaneh shirazi

Binary Decision Diagrams for First Order Predicate Logic

By: Jan Friso Groote

Afsaneh Shirazi

2

BDDs for FOL

First order predicate logic Binary Decision Diagrams Simple operations on BDDs Advanced operations on BDDs Algorithm Example

3

First Order Predicate Logic

V : variables , F : functions , Pr : predicate symbols Terms:

x V is a term If f F is a function symbol of arity r and t1,…, tr are terms,

then f(t1,…, tr) is a term Set of all terms over F and V is denoted by T(F,V) Set of all predicates of the form P(t1,…,tr) is denoted

by P(Pr, F, V)

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Substitution

Substitution is a mapping :V T(F,V) [x1:=t]: maps each variable x to (x), except

x1 to t Composition: o (t) = ((t)) We assume that is extended : term to term,

predicates to predicates

5

Formulas

Formulas are defined by: t, f P(t1, …, tr) P(Pr, F, V) is a formula is a formula and are formulas is a formula is a formula, and x V is a variable x. and x. are formulas

Set of all formulas: F (Pr, F, V)

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Structure and Interpretation

A Structure is a ,multi-tuple = <A; R1, R2,…;F1, F2,…> where A is a non empty set R1, R2,… are relations on A F1, F2,… are functions on A

Herbrand structures have the form H = <T(F,); R1, R2,…;f1, f2,…>

Let be a structure and : V A be a valuation. The interpretation : T(F, V) A of a term t is defined: if x V

)(][ xx

][t

)][,...,]([)],...,([ 11

rjrj ttFttf

7

Interpretation of a formula

F (Pr, F, V) {0,1}:][

)]([max].[

)]([min].[

)][,]min([][

][1][

otherwise 0

][,...,][ if 1)],...,([

1][

0][

]:[

]:[

11

axAa

axAa

irri

x

x

Rttttp

t

f

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FOL

, ╞ iff ╞ iff , ╞ unsatisifiable iff for each there is a

valuation , , ╞

1][

9

equivalency

strongly (logically) equivalent (,╞ iff ,╞ )

logically equivalent (╞ iff ╞ ) weakly (logically) equivalent

,╞ iff ,╞ No free variables: strong logical equivalence

and logical equivalence coincide

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Binary Decision Diagrams

A BDD, is an acyclic, node labeled graph where Q : finite set of nodes l : Q {0, 1} P(Pr, F, V) {0, 1} is a node

labeling, that l(q) 0, 1 for all qQ is the false continuation of a node is the true continuation of a node s Q {0, 1} is a start node 0 is a symbol representing false, and 1 representing

truth Each sequence q0 q1 … is bounded

f

t

0

1

)1,0,,,,,( slQBtf

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Interpretation of a BDD

Let B be a BDD, be a structure and be a valuation. A , -path of a node q0 Q is the sequence

where qn {0,1} and for each i, i = f if , ╞ l(qi) and i = t if , ╞ l(qi)

If q0 ends in 1 we say that q0 holds, , ╞ q0

So, A BDD yields a truth value.

nqqqn 110

...10

12

0 Bt = 1Bf = P(t1,…,tr)BP(t1, …, tr) =

0 1

0 1 1 0

0 1 0 1 0 1

0 1

If B =

If B =

then B =

and B = then B =

13

Example

)()( xPxP

P(x)

0

0

P(x)

1

P(x)

Q(y)

1

R(z)

0

)())()(( zRyQxP

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BDD

Let be a (quantifier free) formula and B its corresponding BDD. For each structure and each valuation we find that

, ╞ iff , ╞ B Proof: Straightforward on the structure of

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BDDs for FOL

First order predicate logic Binary Decision Diagrams Simple operations on BDDs Advanced operations on BDDs Algorithm Example

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Neglect Operator

Let be a BDD. Neglect operator is defined for some q,p q, p q by:

)1,0,,,,,( slQBtf

t

}|,{},|,{'

}{\'

if

if '

112121 prqrprprrr

pQQ

psq

psss

where

f

)1,0,',',',,'()( slQBN

tf

p

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p,q-join Operator

If then

}|,{}|,{'

}{\'

if

if '

11221 qrprqrrr

qQQ

qsp

qsss

where

)1,0,',',',,'()(, slQBJtf

qp

',,',),()( rqrqrprpqlpltftf

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f-merge and f-sort operators

f-merge and f-sort operators, sort the BDD such that labels occur in ascending order.

Sorting a BDD is NP-hard Avoid sorting a BDD

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f-merge Operator

If then

}|,{},|,{'

removed) are parts reachablenon ('

2121 qprpqrprrr

QQ

where

fff

)()(, qlplqpf

)1,0,,,',,'()( slQBMtf

fp

20

f-sort Operator If then

}|,'{}',''{

}|'',{}|,{'

}|,{}|,'{},''{

}|'',{}or |,{'

removed) are parts reachablenon }('','{'

otherwise )(

'' if )(

' if )(

)('

221

2121

rprppp

prprprrr

rqrprqrppp

prprprprrr

ppQQ

rl

prql

prpl

rl

where

t

ttt

ft

fff

)()(, qlplqpf

)1,0,',',',','()( slQBSft

fp

21

22

Simple Operations on BDDs

Lemma: (Soundness) Let B be a BDD. In case O is applicable to B, O(B) B

(O is one of )

Proof: check that for all structures and valuations , ╞ O(B) iff , ╞ B

Definition: B is reduced iff non of the operators is applicable

fp

tp

fp

tpqpp SSMMJN and ,,,, ,

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Simple Operations on BDDs

Lemma: Let B, C be BDDs pQB and qQC .Let be a structure and a valuation such that , ╞ p and , ╞ q. P(t1,…,tn) : label not occurring in subdags p and q. Then Exists a structure and a valuation , ╞ p, , ╞ q and , ╞ P(t1,…,tn) Exists a structure and a valuation , ╞ p, , ╞ q and , ╞ P(t1,…,tn)

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Simple Operations on BDDs

Lemma: Let B and C be reduced BDDs. pQB and qQC such that p q. Then, l(p) = l(q) pf qf

pt qt

If B,C are the same, then p=q

back

25

Isomorphism

Let B and C be BDDs.

f: QB{0B,1B} QC{0C,1C} is called homomorphism iff lC(f(p))=lB(p), f(pf ) = f(p)f and f(pt) = f(p)t . In case f is bijective, f is called isomorphism.

B = C (isomorphic) if there exists an isomorphism f.

26

Example

)()( xQxP

P(x)

0

0

Q(x)

1

P(x)

Q(x)

B C

f

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Theorem

B and C are reduced BDDs, B C. Then B = C (isomorphic).

Proof

back

28

Theorem

Operators can be applied a finite number of times to B.

Proof

fp

tp

fp

tpqpp SSMMJN and ,,,, ,

back

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R(B)

Let B be a BDD, C be a reduced BDD, B C. According to theorem C is unique up to an isomorphism and It can be efficiently obtained (Thrm) R(B) for C

R(B) = Bt ( tautology)

R(B) = Bf ( contradiction) Basis for Propositional Logic

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Advanced Operations on BDDs Copying Operator C(B): puts B in conjunction

with a copy of itself (different variables) Unification Operator U(B): instantiate B

according to ( is a relevant unifier)

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Copy Operator

Let B be a BDD in which variables occur.

not occurring in B

x

]:[)( 1

xxBBBC

1

x

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Unifier

A substitution is called a unifier of P(t1, …, tn) and Q(u1, …, um) iff (P(t1,…, tn))=(Q(u1,…,um))

A unifier is most general (MGU) iff for each unifier ’ of P(t1,…, tn) and Q(u1,…,um) there is a substitution such that o = ’

Idempotent MGU: ((x)) = x Linear time

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Relevant Unifier

A node p of B is redundant iff pt pf A path

is allowed iff there are no i, j such that l(pi)=l(pj) and i j

A node is truth-truth capable iff there is an allowed path

lpppnn

n

110

...10

1...11

1

nn

n

t

ppp

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Relevant Unifier

A valuation is a relevant unifier iff

pi is not redundant, if i = f, pi is not truth-truth capable, and for some i, j, i = f, j = t and is an idempotent MGU of l(pi) and l(pj).

1...110

1

nn

npps

35

Lemma

Let B be a reduced BDD. No redundant nodes Every path allowed If pi is not truth-truth capable pit = 0

is a relevant unifier iff for some i, j, i=f, j=t and is the MGU of l(pi) and l(pj) on the rightmost path of B

1...110

10

nn

nppp

36

MGU

y = a

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Unification Operator

If is a relevant unifier of B

Lemma: (Soundness) B C(B) B B U(B)

Proof: easy logical consequence

)()( BBU

back

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BDDs for FOL

First order predicate logic Binary Decision Diagrams Simple operations on BDDs Advanced operations on BDDs Algorithm Example

39

Algorithm

)))((e(TryToReduc of unifiersrelevant allFor

stop and ble'unsatisfia'report , If

)e(TryToReduc

Endrepeat

))((:

)e(TryToReduc

Repeat

)(:

)Solve(

BURB

BB

B

BCRB

B

BRB

f

Avoid expensive sorting operator

All pairs of pi, pj in the rightmost path, with pit=0 and pjt0 Linearly in size of terms

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Algorithm

)))((e(TryToReduc of unifiersrelevant allFor

stop and ble'unsatisfia'report , If

)e(TryToReduc

Endrepeat

))((:

)e(TryToReduc

Repeat

)(:

)Solve(

BURB

BB

B

BCRB

B

BRB

f

Avoid sorting by grouping predicates with the same predicate symbol

The depth of recursive calls is limited by the number of free variables

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Algorithm

)))((e(TryToReduc of unifiersrelevant allFor

stop and ble'unsatisfia'report , If

)e(TryToReduc

Endrepeat

))((:

)e(TryToReduc

Repeat

)(:

)Solve(

BURB

BB

B

BCRB

B

BRB

f

))))))))(((...((...(((21 BCRCRURUR

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Algorithm

The program is browsing through larger and larger BDDs of the form

We stop: Bf Lemma

Bt

))))))))(((...((...(((21 BCRCRURUR

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Russel’s Paradox

A set which contains those sets that are not members of themselves

F(x,y): x is a member of y Problem: negated and Skolemised formula

)),(),(( yyFxyFyx

),(),( yyFayF

44

Russel’s Paradox

(y) = a

),(),( yyFayF

Example

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Thank You

46

Appendix A

Thrm: B and C are reduced BDDs, B C. Then B = C (isomorphic).

Proof: Define f: QBQC and g: QCQB

f(p) = q such that p q

g(q) = p such that p q

f is a homomorphism, g is the inverse of f

f is an isomorphism

We assumed f and g are well defined.

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Proof

B C and all nodes in B are reachable from the root each node in B is related via to 1 node in C

p related to q1,q2 q1 q2

q1 = q2

(using thrm)

back

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Appendix B

Proof: The transformation operators can be formulated as rewrite rules.l1 and l2 are predicates. l1 > l2

tp

fp

p

tp

fp

Mzxlzylxl

Mzxlzyxll

Nxxxl

Szylzxllzylxl

Szylzxllzyxll

),()),(,(

),()),,((

),(

)),(),,(()),(,(

)),(),,(()),,((

111

111

1

11221

11221

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Proof

To each DAG we can obtain its canonical tree by undoing the sharing of subdags. Application of these rules must terminate on these trees.

Each rewrite of the DAG corresponds to one or more rewrite of canonical tree.

In Join operator the number of nodes are strictly decreasing It should terminate

back

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Appendix C

A set is circular if it is a member of another set, which in turn is a member of the original.

There is no set containing all non circular sets.

Problem: negated and Skolemised formula (a, f are

skolem functions)

))),(),((),(( xzFzxFzyxFxy

)),())),(())(,(((

))),(),((),((

axFxxfFxfxF

xzFzxFaxF

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There are no circular sets

Two relevant unifier:1. Mapping z and x to a2. Mapping x to z

)),())),(())(,(((

))),(),((),((

axFxxfFxfxF

xzFzxFaxF

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There are no circular sets

We must apply copying operator.6 relevant unifier:1. x:=a z:=a2. u:=a v:=a3. x:=v u:=a4. u:=z x:=a5. z:=x6. v:=u

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There are no circular sets

We apply first unifier.4 relevant unifier:1. u:=f(a) v:=a2. u:=a v:=a3. u:=a4. v:=u

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