The decidability of Presburger Arithmetic By Guillermo Guillen 04/13/05 Dr. Smith COT 6421 FIU Spring 2005

Theorem (Decidability of Presburger Arithmetic, 1929 Warsaw Pol.) : Th(N, +) is decidable, where Th(N, +) is the set of true sentences in the language of the model (N, +). Note: (N, +) is the model with predicate symbols +, ≤ and N is the set of natural numbers.

Let Σ = { 00, 01, 10, 11} where i j represents the 2 x 1 matrix (i, j) T for i, j є {0,1}. Note that every symbol in Σ represents two numbers in binary form, where top and bottom entries of each matrix is either a 0 or a 1. Consider A = { w є Σ * | the first row is equal to the second row }. Ex: є A and ¬( є A ) 00, 11

Let Σ = { 000, 001, 010, 011, 100, 101, 110, 111}. Consider the following language : B = { w є Σ* | the sum of the top two rows is equal to the third row } Ex: є B top row 010 ≡ 2, middle row 011 ≡ 3, bottom row 101 ≡ 5. ¬ ( є B ) top row 0100 ≡ 4, middle row 0110 ≡ 6, bottom row 1001 ≡ 9.

є B and ¬ ( є B )

Automata Theory Results: The DFAs for the reverse of A,B implies that A,B are regular. A problem of addition can be decided by some DFA if the addition can be encoded and represented in a particular way so that a DFA may read and decide such a problem. Application of DFAs in Theorem : For each sentence φ, we’ll produce finite automatons that will compute each atomic formula of φ. By combining all of φ’s atomic formulas with the use of connectives and quantifiers, we’ll produce a DFA for φ.

proof of Theorem (Decidability of Presburger Arithmetic): GOAL: To construct an algorithm that determines on a given input φ, where φ is a sentence in the language of (N, +), is true in (N,+). Consider the following algorithm : On input φ, construct the following sequence : φ 0 = φ = Q 1 x 1 … Q n x n ( ψ), where Q i is either and ψ has no quantifiers with x i as its variables. φ 1 = Q 2 x 2... Q n x n (ψ), …, φ n = ψ.

φ = ( x 1 + 5x 2 = x ) Next, construct a DFA for φ. First, set the left side of φ equal to 0. Then 0 = 3 – x 1 – 5x 2 + x 3 and set b = 3. Idea: Consider all the 0(mod 2) solutions for the above eq. when b = 3, 0, 1, 2, -1, -2, (any other necessary integers).

b a є Σ □1 □ □ □-2 □ 001 □1 □ 20 □ □ □ □ -3 □-4 □ □ □ □-3 □-4 □ 100 □0 □ 1 □-2 □ □1 □ □ □-2 □ □-2 □ □-4 □-5 □ 111 □-2 □ -3 □-4 □-5 For each b, create a new state q b and a transition from q b to q c under some a when c lies under the column of b for some a in Σ. Set q 3 to be the initial state and q 0 to be the final state.

q1q1 q3 q2 q0 q-2 q-3 q-4 q-5 q-1

Now construct a NDFA for A i from A i + 1 : If φ i = Эx i φ i + 1, then for each state in A i + 1, A i will contain that state and a new start state. A i will take as input, say an i x n matrix, and nondeterministically guesses the (i + 1)th row by simulating A i + 1 on the new (i + 1) x n matrix. Hence, A i accepts the i x n matrix iff A i + 1 accepts the (i + 1) x n matrix for some (i + 1)th row. If φ i = for all x i (φ i + 1 ), then it is equivalent to ¬ there does exists an x i ( ¬ φ i + 1 ). Then construct the NDFA for the complement of L (A i + 1 ).

Then apply the construction used in the first “if” to get a NDFA which recognizes the “there exists” quantifier. Finally, construct an NDFA for the language under the “there exists” quantifier. So, we get our desired A i. Last Step of Algorithm on input φ: Note first that A 0 accepts any input string iff φ is true. If A 0 accepts ε then φ is true and accept. Otherwise, reject. □