DECIDABILITY OF PRESBURGER ARITHMETIC USING FINITE AUTOMATA Presented by : Shubha Jain Reference : Paper by Alexandre Boudet and Hubert Comon

TOPICS TO BE COVERED  Decidability  Presburger Arithmetic  Construction of Finite Automata

DECIDABILITY  Some definitions :  Formula : A formula is a well formed string over the alphabet : {^, ˇ, ¬, (, ), , , , ,,x, y, z, …} that is, binary operators, parenthesis, quantifiers, relational operators and variables. For ex,  x  y (y=x+1).  An atomic formula contains just addition and relational operators.  Sentence : A sentence is a formula with no free variables (variable not within the scope of any quantifier).The above example is a sentence. A sentence can consist of some atomic formulae along with the Boolean operators ^,ˇ and ¬ and the quantifiers  and .  Theory : It is a set of true sentences in a formal language.  Decidability : The term “decidable” (in logic) refers to the existence of an effective method using which we determine whether a sentence is true or not.

PRESBURGER ARITHMETIC  Presburger arithmetic is named in honor of Mojzesz Presburger, who introduced it in  It is the theory of the natural numbers with addition denoted as (N,+,<).  Presburger arithmetic is a decidable theory. This means it is possible to effectively determine, for any sentence in the language of Presburger arithmetic, whether that sentence is provable from the axioms of Presburger arithmetic.  Limitation : Presburger Arithmetic cannot formalize concepts such as divisibility or prime number. Generally, any number concept leading to multiplication cannot be defined in Presburger Arithmetic.

PRESBURGER ARITHMETIC The axioms of Presburger arithmetic are the universal closures of the following: 1)¬(0 = x+ 1) 2)x + 1 = y+ 1 → x= y 3)x + 0 = x 4)(x + y) + 1 = x + (y + 1) 5)Let P(x) be a first-order formula in the language of Presburger arithmetic with a free variable x (and possibly other free variables). Then the following formula is an axiom: (P(0) ∧ ∀x(P(x) → P(x + 1))) → P(y). (5) is an axiom schema of induction.

Some examples :  We can express the fact that every number is either even or odd :  x  y ((y+y=x)ˇ(y+y+1=x))  We can express that there are infinitely many natural numbers :  x  y (x+1=y)  (x+x = y+1+1) is a formula which can be written as (2x+3=y+2).

PRESBURGER ARITHMETIC  Mojzesz Presburger proved Presburger arithmetic to be:  Consistent : There is no statement in Presburger arithmetic which can be deduced from the axioms such that its negation can also be deduced.  Complete : For each statement in Presburger arithmetic, either it is possible to deduce it from the axioms or it is possible to deduce its negation.  Decidable : There exists an algorithm which decides whether any given statement in Presburger arithmetic is true or false.  Fischer and Rabin (1974) proved that every algorithm which decides the truth of Presburger statements has a running time of at least 2 2^(cn) for some constant c, where n is the length of the Presburger statement. Therefore, the problem is one of the few currently known that provably requires more than polynomial run time.  The decidability of Presburger arithmetic can be shown using FINITE AUTOMATA.

CONSTRUCTION OF FINITE AUTOMATA FOR DECIDABILITY OF PRESBURGER ARITHMETIC  Natural numbers can be encoded as finite words over the alphabet {0,1} : it is sufficient to consider their binary representation, which we write from right to left.  The representation is not unique as we may add some zeroes on the right. For example, we may represent the number thirteen as 1011,10110,…  More generally, tuples of natural numbers can be represented in binary notation as words over the alphabet {0,1} n, simply by stacking them using an equal length representation for each number. For example, the pair (thirteen, four) can be represented as or

 If we assume given, for all atomic formulae φ,π, automata A φ, A π which accept the solutions of φ and π respectively, then for every formula ψ, the automaton accepting ψ can be defined inductively as follows :  If ψ = φ ^ π, then A ψ = A φ ∩ A π i.e. the automaton accepting the intersection language.  If ψ = φ ˇ π, then A ψ = A φ U A π i.e. the automaton accepting the union language.  If ψ = ¬φ, then A ψ = A ¬φ is the automaton accepting the complement of the language accepted by A φ.  If ψ =  x φ, then A ψ =A  x φ is computed by projection.  If ψ =  x φ, then A ψ = A  x φ = A ¬  x ¬φ  If we want to decide the validity of the sentence ψ, it is sufficient to compute A ψ and check whether it contains an accessible final state.

CONSTRUCTION OF FINITE AUTOMATA FOR THE ATOMIC FORMULAE INVOLVING EQUATIONS Given an atomic formula (e) a 1 x a n x n = k We construct an automaton A = (Q, , T, [e], F). We have  = {0,1} n. We start from a set of states containing [e] where e is the equation to be solved and the junk state [-]. [e] is the start state. The transition rules T only contain initially the transition from [-] to itself by any letter of the alphabet .Then we saturate Q and T according to the following rules: If [a 1 x a n x n = k]  Q and  =   b1b1... bnbn

Then if k - (a 1 b a n b n ) is even then add [a 1 x a n x n = k] [a 1 x a n x n = (k – (a 1 b a n b n ))/2] to T and [a 1 x a n x n = (k – (a 1 b a n b n ))/2] to Q. If k - (a 1 b a n b n ) is odd then add [a 1 x a n x n = k] [ - ] to T.  

The final state is [a 1 x a n x n = 0] : Every word in the set of solution might be followed eventually by a sequence of. The automaton constructed is deterministic and complete. Example : Consider an atomic formula x + 2y = 3z + 2. We can rewrite it as : (e) x + 2y - 3z = 2. Here k=2 and our input alphabet is  = {0,1} 3, thus it contains :

Initially, set of states Q = { [x + 2y - 3z = 2], [-] }. For  =, k - (a 1 b a n b n ) = 2 – ( ) = 2 is even so, we add [x + 2y - 3z = 2] → [x + 2y - 3z = (2-(0+0-0))/2]  [x + 2y - 3z = 2] [x + 2y - 3z = 1] to T and [x + 2y – 3z = 1] to Q

Similarly for  =, k - (a 1 b a n b n ) = 2 – ( 0 + 2*1 - 0) = 0 is even so, we add [x + 2y - 3z = 2] → [x + 2y - 3z = (2-(0+2-0))/2]  [x + 2y - 3z = 2] [x + 2y - 3z = 0] to T and [x + 2y – 3z = 0] to Q

Similarly for  =, k - (a 1 b a n b n ) = 2 – ( 1 + 2*0 – 3*1) = 4 is even so, we add [x + 2y - 3z = 2] → [x + 2y - 3z = (2-(1+0-3))/2]  [x + 2y - 3z = 2] [x + 2y - 3z = 2] to T

Similarly for  =, k - (a 1 b a n b n ) = 2 – ( 1 + 2*1 – 3*1) = 2 is even so, we add [x + 2y - 3z = 2] → [x + 2y - 3z = (2-(1+2-3))/2]  [x + 2y - 3z = 2] [x + 2y - 3z = 1] to T

Now for  =, k - (a 1 b a n b n ) = 2 – ( – 3*1) = 5 is odd so, we add Y [x + 2y - 3z = 2] [-] to T. Similarly for  =,andwe have transitions to dead state

This can be shown as : Transitions to the dead state are not shown. x+2y-3z=2 x +2y-3z=1 x+2y-3z=

Now we need to saturate the sets Q and T. We need to find the transitions for the state [x+2y-3z=1].Here k=1. For  =, k - (a 1 b a n b n ) = 1 – ( – 3*1) = 4 is even so, we add [x + 2y - 3z = 1] → [x + 2y - 3z = (1-(0+0-3))/2]  [x + 2y - 3z = 1] [x + 2y - 3z = 2] to T

Continuing in this manner, the set of states Q now contains 5 states : [x+2y-3z=2] [x+2y-3z=1] [x+2y-3z=0] [x+2y-3z=-1] [x+2y-3z=-2] The start state is [x+2y-3z=2] and the final state is [x+2y-3z=0] And the transition diagram can be shown as :

1 0 0 x+2y-3z= -2 x+2y-3z=2 x+2y-3z=-1 x+2y-3z=1 x+2y-3z=

CONSTRUCTION OF FINITE AUTOMATA FOR THE ATOMIC FORMULAE INVOLVING DISEQUATIONS Turning the automaton A recognizing the solutions of an equation e into one recognizing the solutions of the disequation ¬ e is straightforward : A being complete and deterministic, we just have to say that the accepting state is no longer accepting and that all the other states are accepting.

CONSTRUCTION OF FINITE AUTOMATA FOR THE ATOMIC FORMULAE INVOLVING INEQUATIONS Given an atomic formula (e) a 1 x a n x n  k The only case where this equation has no solution is when k is negative and a 1... a n are all positive. In this case, the automaton is a trivial one, with only one non-accepting state. Otherwise we start from a set of states containing [e] where e is the equation to be solved and the junk state [-]. [e] is the start state. Then the transitions are generated as follows :

If [a 1 x a n x n  k]  Q and  =   Then if k  0 or a i <0 for some i, then add [a 1 x a n x n  k] [a 1 x a n x n  (k – (a 1 b a n b n ))/2  ] to T and [a 1 x a n x n  (k – (a 1 b a n b n ))/2  ] to Q b1b1... bnbn 

if k 0,...,a n >0, then add [a 1 x a n x n  k] [-] to T. The accepting states are all those labeled by an inequation for which is a solution, that is, the states [a 1 x a n x n  k] for which k  0. The automata computed is deterministic and complete. 

PROJECTION CONSTRUCTION FOR THE PRESENCE OF EXISTENTIAL QUANTIFIER This can be done simply by removing the corresponding component in the transition rules. Consider the equation  z x+2y-3z=2 The construction of an automaton for this formula can be done simply by constructing the automaton for the equation x+2y-3z=2 on the alphabet {0,1} 3 and then removing the row corresponding to z from the inputs on the transitions. Here we remove the last row from the tuple. The resulting automaton will then be non- deterministic and can be shown as :

1 0 x+2y-3z= -2 x+2y-3z=2 x+2y-3z=-1 x+2y-3z=1 x+2y-3z=

CONCLUSION We saw an introduction to Presburger Arithmetic, and also how to construct a finite automata for its decidability by checking its emptiness..

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