boolean algebra with presburger arithmetic

Lavdrim Selimi

Winter 2013

Boolean Algebra with Presburger Arithmetic

Introduction

1

Define BAPA

Explain the decision procedure and quantifier elimination for the BAPA logic

NP completeness of the QFBAPA

Define BAPA

2

BA – a variant of ordinary algebra

– instead of the usual algebra of numbers, BA is the algebra of truth values (0 and 1) or equivalently of subsets of a given set

PA – first order theory of natural numbers

– contains only addition and equality operations

– is a decidable theory

Define BAPA

3

BAPA combines:

1. boolean algebras of sets of uninterpreted integers

2. Presburger Arithmetic operations

It can express the relationship between int vars and cardinalities of unbounded finite sets and verification of data structure consistency properties.

Was found to be of interest in program verification and constraint DB (ex. Revesz 2004)

Define BAPA

4

BAPA formulas with quantifier alternations arise when

- verifying programs with annotations containing quantifiers

- Proving simulation relation conditions for refinement and equivalence of program fragments

BAPA constraints can be used for proving the termination of programs that manipulate data structures

Contributions

5

1. BAPA constraints can be used for prog. Analysis and verification by expressing:

DS invariants and the correctness of procedures with respect to their specification

Simulation relation between prog. Fragments

Termination conditions for programs that manipulate DS

2. Presenting an alg. α that translates BAPA sentences into PA sentences

3. Analyze the alg. α and show that its complexity matches the lower bound for PA and is therefore optimal

First–Order theory of BAPA

6

Our initial motivation for BAPA was the use of BA to reason about data structures in terms of sets.


7

BA language allows cardinality constraints of the form |A| = K, where K constant integer

Such constant cardinality constraints are useful and enable quantifier elimination for the resulting language

But not constraints such as |A| = |B|

Also cannot represent constraints on relationships between sizes of sets and integer variables


8

If we consider the equicardinality A ~ B that holds iff |A| = |B|, and consider BA extended with the relation

A ~ B

Define the relation:

plus(A, B, C) ⇔ (|A|+|B|=|C|)

by the formula:

∃x1. ∃x2. x1 ∩ x2 = ∅ ˄ A ~ x1 ˄ B ~ x2 ˄ x1∪ x2 = C


9

The relation plus(A, B, C) allows us to:

Express addition using arbitrary sets as representatives for

natural numbers

∅ can represent the natural number 0 and

Any singleton set can represent the natural number 1

Therefore a natural closure under definable operations leads to our formulation of the language BAPA which contains both sets and integers.


10

In the BAPA language we distinguish two kinds of quantifiers; over sets and over integers

Decision procedure and quantifier elimination

11

The algorithm α which

Decides the validity of BAPA sentences

Reduces a BAPA sentence to an equivalent PA sentence with the same number of quantifier alternations and an exponential increase in the total size of the formula

Algorithm procedures

12

This algorithm describes the steps to transform a BAPA sentence F0 into a PA sentence

First step: Put F0 Into Prenex form

Example:

∃A. ∃B. |A| = |B| ˄ A ∪ B = C ˄ |A ∩ B| = ∅


13

The next step: seperate BA from PA by replacing each b1 = b2 formula with b1 ⊆ b2 ˄ b2 ⊆ b1 and every b1 ⊆ b2 with |b1 ∩ b2

c|= 0

∃A. ∃B. |A| = |B| ˄ |(A ∪ B) ∩ Cc| = 0 ˄ |C ∩ (A ∪ B)c| = 0 ˄ |A ∩ B| = 0


14

Next step: write each set expression as a union of cubes (regions in the Venn Diagram)- -

and for each Si introduce a new variable li

The resulting formula will be of the form:


15

In our example:

|A| = |S1|+|S2|+|S4|+|S5|

|B| = |S2|+|S3|+|S5|+|S6|

|(A ∪ B) ∩ Cc| = |S1|+ |S2| + |S3|

|C ∩ (A ∪ B)c| = |S7| + |S8|

|A ∩ B| = |S2|+| S5|


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|A| = |B| => l1+ l2+ l4+ l5 = l2+ l3+ l5+ l6 => l1+ l4 = l3+ l6 |(A ∪ B) ∩ Cc| = l1+ l2+ l3 = 0 => l1= l2= l3 = 0 |C ∩ (A ∪ B)c| = l7 + l8 = 0 => l7 = l8 = 0 |A ∩ B| = l2+ l5= 0 => l2= l5= 0


17

l1 = |A ∩ Bc ∩ Cc|

l2 = |A ∩ B ∩ Cc|

l3 = |Ac ∩ B ∩ Cc|

l4 = |A ∩ Bc ∩ C|

l5 = |A ∩ B ∩ C|

l6 = |Ac ∩ B ∩ C|

l7 = |Ac ∩ Bc ∩ C|

l8 = |Ac ∩ Bc ∩ Cc|


18

∃A. ∃ B. ∃ l1 . ∃ l2 . ∃ l3 . ∃ l4 . ∃ l5 . ∃ l6 . ∃ l7 . ∃ l8

l1 = |A ∩ Bc ∩ Cc|

˄ l2 = |A ∩ B ∩ Cc|

˄ l3 = |Ac ∩ B ∩ Cc|

˄ l4 = |A ∩ Bc ∩ C|

˄ l5 = |A ∩ B ∩ C|

˄ l6 = |Ac ∩ B ∩ C|

˄ l7 = |Ac ∩ Bc ∩ C|

˄ l8 = |Ac ∩ Bc ∩ Cc|

˄ l1+ l4 = l3+ l6 ˄ l1= l2= l3 = 0 ˄ l7 = l8 = 0 ˄ l2= l5= 0


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Since: l1= l2= l3 = 0 ; l7 = l8 = 0 and l2= l5= 0

We simplify our formula to: ∃A. ∃ B. ∃ l4 . ∃ l6 . l1 = |A ∩ Bc ∩ Cc| ˄ l2 = |A ∩ B ∩ Cc| ˄ l3 = |Ac ∩ B ∩ Cc| ˄ l4 = |A ∩ Bc ∩ C| ˄ l5 = |A ∩ B ∩ C| ˄ l6 = |Ac ∩ B ∩ C| ˄ l7 = |Ac ∩ Bc ∩ C| ˄ l8 = |Ac ∩ Bc ∩ Cc| ˄ l4 = l6

Algorithm procedures Removing quantifiers

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Consider first the case ∃k . Because k does not occur in , simply move the existential quantifier to Gr and let Gr+1 = ∃k.Gr , which completes the step

Case ∀k , it suffices to let Gr+1 = ∀k.Gr ,

again because k does not occur in

Algorithm procedures Removing set quantifiers

21

To remove ∃y :

LEMMA 1: Let b1, . . . , bn be finite disjoint sets, and l1, . . . , ln, k1, . . . , kn be natural numbers. Then the following two statements are equivalent:


22

To remove ∃ B we have to consider following cases:

A ∩ C A ∩ Cc Ac ∩ C Ac ∩ Cc

Regarding the statements (1) and (2) of the previous Lemma:

|A ∩ C| = |A ∩ Bc ∩ C| ˄ |A ∩ B ∩ C| = l4 + l5

|A ∩ Cc| = |A ∩ Bc ∩ Cc| ˄ |A ∩ B ∩ Cc| = l1 + l2

|Ac ∩ C| = |Ac ∩ B ∩ C| ˄ |Ac ∩ Bc ∩ C| = l6 + l7

|Ac ∩ Cc|= |Ac ∩ B ∩ Cc| ˄ |Ac ∩ Bc ∩ Cc| = l3 + l8


23

After removing ∃B our example takes this form:

∃A. ∃ l4 . ∃ l6 . |A ∩ C| = l4

˄ |A ∩ Cc| = 0

˄ |Ac ∩ C| = l6

˄ |Ac ∩ Cc| = 0

˄ l4 = l6


24

To remove ∃A we consider the following two cases:

C and Cc

Again regarding to Lemma1:

|C| = |A ∩ C| ˄ |Ac ∩ C| = l4 + l6

|Cc|= |A ∩ Cc| ˄ |Ac ∩ Cc| = 0

We achieve the final step providing the PA formula:

∃ l4 . ∃ l6 . |C| = l4 + l6

|Cc|= 0

˄ l4 = l6


25

In the cases when we have only one set variable

y and its negation yc we can write |y| and | yc| as

|1 ∩ y| and |1 ∩ yc| and apply the algorithm one more time

Finally we obtain a formula of the form :

which is the last step and the algorithm terminates


26

Case ∀y:

is equivalent to

because existential quantifiers over li together with the conjuncts |Si| = li act as definitions for li

So by expressing ∀y as ∃y we show that elimination of ∀y is analogous to elimination of ∃k

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1) arbitrary BA expressions denoting sets

2) arbitrary quantifier-free PA expressions

3) cardinality operator

The constant MAXC denotes the size of the universal set, so |1| = MAXC.

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To be NP – complete:

NP – hard (QFBAPA is since it has propositional operators on formulas)

NP membership (the challenge)

QFPA formula is in NP because it has a small model property implying that satisfiable formulas have solutions whose binary representation is polynomial

29

We show that:

It is not necessary to generate all Venn region variables (drop them which are empty)

Instead of using exponentially many Venn region cardinality vars to encode sets, we can use polinomially many generic variables

New QFPA formula is equisatisfiable with the original one

30

Starting with an example

From a QFBAPA formula construct a polynomially large QFPA formula which is equisatisfiable with the original one

Again separate BA from PA by replacements:

b1 = b2 ⇒ b1 ⊆ b2 ˄ b2 ⊆ b1

and each |b1 ∩ b2c|= 0

ex: |A| = |B| ˄ |A ∩ B| = ∅

31

G ˄ F where : G – a quantifier – free PA formula

F – of the form

|A| = |S1|+|S2|

|B| = |S2|+|S3|

|A ∩ B| = |S2|

32

k1 = k2 ˄ |A| = k1

k3 = 0 |B| = k2

|A ∩ B| = k3

for β = (p1, … , pe)

where pi ∊ {0, 1} let [bi]β ∊ {0, 1}

express each |bi| = ∑ β⊨bi |Sβ |

and for each β ∊ {0, 1}e introduce a non-negative integer variable lβ denoting |Sβ|

33

|b1| = l(1, 0) + l(1, 1)

|b2| = l(0, 1) + l(1, 1)

|b3| = l(1, 1)

Finally:

k1 = k2 ˄ |b1| = l(1, 0) + l(1, 1)

k3 = 0 |b2| = l(0, 1) + l(1, 1)

|b3| = l(1, 1)

34

Our formula is of the form:

But instead of using this exponentially large formula

where β ranges over all 2e propositional assignments

We will check the satisfiability of a smaller formula

where β ranges over N assignemnts

35

β ranges over a set of N assignments β1, . . . , βN for

βi = (pi1, . . . , pie) and

pij are fresh free variables ranging over {0, 1}.

We are interested in the best upper bound N(d) on the number of non-zero Venn regions over all possible systems of equations

37

|b1| = β(1, 0) + β(1, 1)

|b2| = β(0, 1) + β(1, 1)

|b3| = β(1, 1)

(k1, k2, k3) = (β(1, 0), 0, 0)

+ (0, β(0, 1), 0)

+ (β(1, 1), β(1, 1), β(1, 1))

38

We will use a set of integer vectors con(X)

where X = {Xβ | β ∊ {0, 1}e } and

Xβ ∊ {0, 1}e is given by Xβ = ([b0]β, [b1]β, … , [be]β)

A direct application of Fact 1 yelds N = 2d log (4d)

Which is sufficient to show that QFBAPA is in NP

boolean algebra with presburger arithmetic

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