On Undecidability of Subset Theory for Some Monoids

Early we (with B. N. Karlov) have proved the following claim for the infinite cyclic monoid ℋ. Let exp ℋ be an algebra of finite subsets of ℋ with the same operation, exp ℋ must be a monoid again. So the theory of exp ℋ is equivalent to elementary arithmetic. Thus, the theory of the monoid exp ℋ is undecidable. Here we consider an arbitrary commutative cancellative monoid ℋ with an element of infinite order, and generalize the previous claims to the corresponding monoid exp ℋ.


Introduction
Monoids (semigroups with the neutral element) are the most general kind of associative algebras. A lot of traditional universes are monoids: numbers, polynomials, matrices with addition or multiplication, words with concatenation, functions with composition etc. Monoids are tightly connected with various concepts of computation theory (see [1,2]), they can be used to describe algorithms or automata. So algorithmic problems of monoid theory are actively investigated (see [3,4]).
Investigations of formal theories are among the central problems of mathematical logic. These problems are not only of great theoretical value but have practical significance. One of its applications is database query languages. Usually such languages are variants of logic languages, while databases themselves are finite structures. This concept of relational databases was introduced by E. F. Codd in [5]. Since in real systems database items are encoded by some mathematical objects (numbers, words and so on), there are some "natural" operations or relations on such objects. For example, those can be arithmetical operations, word concatenation, or comparisons. Hence, a real database is a finite structure embedded into some infinite universe (see [6]). Therefore, a database query planner must operate with some logic language on a given universe.
Modern database management systems allow to construct aggregate types and use database items of such types. In such a manner we can define arrays, maps, sets consisting of numbers, words, Boolean values etc. These aggregates are finite due to "natural" restrictions of storage capacity. So we have a complex structure, which can contain not only atomic objects but aggregates, for example, finite sets of such objects.
Another way, which leads to the same concept, is a determinization process of automata (see [1]). The universal method is to consider sets of non-deterministic automaton states as  [7]), so its deterministic version corresponds to semigroup of subsets.
We have to note that such construction of the subset algebra is not equivalent to the monadic second order logic for the original algebra. The second order language (see [8]) admits first order variables to denote first order objects (elements of the original algebra). But in the first order theory of the subset algebra we don't have such variables generally. The described difference is important because there exist examples (constructed in [9]), where the first order theory of the subset algebra is algorithmically simpler than the theory of the original algebra. For second order logic it is impossible.
Algorithmic properties of subset algebras and its theories are very different. As we mentioned in the previous paragraph, the theory of the finite subset algebra can be decidable, while the theory of the original one is not (see [9]). And vice versa, there are algebras with decidable theories such that the corresponding subset algebras have undecidable theories. Such algebras appear when we investigate languages theory [1]: every language is a set of words, which are elements of free monoid (with concatenation). So the concatenation of entire languages can be considered as a generalization of the word concatenation onto sets of word. In this way, some result were established in [10][11][12].
In particular, one of these results is the following: if F is the free cyclic monoid (such as the set of natural numbers with addition), then the theory of finite subsets of F is algorithmically equivalent to elementary arithmetic. It was proved in [11], Corollary 1. The natural question appears: can we generalize this result to some wider class of monoids? Or, may be, the free cyclic monoid is a rare exception?
In this paper we show how to generalize the mentioned result from [11]. We establish such result for monoids those are simultaneously commutative, cancellative and have elements of infinite order. Instances of such monoids are the set of natural numbers with multiplication, the set of polynomials with multiplication, arbitrary linear space with addition over a field of characteristic zero, and many other classic algebras. In particular, every Abelian group that is not a torsion group satisfies these conditions.

Definitions and Basic Properties
The following basic concepts of monoid theory can be found in the book [13].
A monoid A = (A, * , e) is a non-empty set A with a binary operation * on it. Elements of the monoid A are elements of the set A. The operation * is called usually "multiplication", it must be associative: a * (b * c) = (a * b) * c for every a, b, c ∈ A, and e must be the (necessarily unique) neutral element: a * e = e * a = a for every a ∈ A.
An element a of a monoid has an infinite order if all powers a 0 = e, a 1 = a, a 2 , . . . are pairwise distinct. Otherwise, the element a is of finite order.
An element a of a commutative monoid A is An element a of a commutative monoid is invertible if a * b = e for some b. This element b is unique and is denoted by a −1 .
On subsets of a monoid A we can also define the binary operation: if x, y ⊆ A, then If the set x contains a unique element a, then we write a * y instead of x * y or {a} * y. For instance, we get x * ∅ = ∅, {e} * y = e * y = y. The last equality means that the set E = {e} is the neutral element for the operation * on subsets of A. The operation * on subsets is associative (commutative) whenever the operation * has corresponding properties. Also, it is clear that the 3 product x * y is a finite set whenever the sets x and y are finite. Therefore, for any commutative monoid A = (A, * , e) there is the commutative monoid exp A = (P f in (A), * , E), where P f in (A) is the set of all finite subsets of A.
If a monoid A is cancellative, then the function b → a * b on A is injective for any fixed element a. So for cardinalities we have |x| = |a * x| for each x ∈ exp A and a ∈ A. Also, for every x, y ∈ exp A, y = ∅, the equality x * y = a∈y x * a holds, so |x * y| ≥ |x|.
In any commutative monoid the divisibility relation is definable: We can define the set Q of invertible elements: The divisibility relation is transitive and reflexive, so the mutual divisibility relation is an equivalency. Also, the last relation can be defined as x and y are distinguished by some invertible multiplier: x The empty set ∅ is a zero of the monoid exp A and has corresponding definition: because ∅ * y = ∅ for all y, but x * y = x if x = ∅ and y = ∅.
Everywhere in the following we consider an arbitrary commutative cancellative monoid A with at least one element of infinite order. Hence, there is the commutative monoid exp A. We consider only non-zero elements of the monoid exp A, i. e. non-empty sets. These elements form a submonoid of exp A because x * y = ∅ whenever the sets x and y are not empty.
Further, we write xy everywhere instead of x * y. Also, we use letters a, . . . , h to denote elements of an original monoid A whereas letters t, . . . , z are used to denote elements of the corresponding monoid exp A.
It is easy to prove the next claim. Therefore, each invertible element of exp A is {a} for a suitable invertible a.
Another simple claim is Proposition 2. If u, v ∈ exp A and e ∈ u, then v ⊆ uv.
that follows the claim.

Definability of Powers
The main goal of our paper is to interpret elementary arithmetic in the monoid exp A, where A is a commutative cancellative monoid with an element of infinite order. To do that we sequentially define some relations and establish its properties.
The first ones are used to distinguish sets of one or two elements: Proposition 3. The formula R 1 (x) is true in the monoid exp A if and only if the set x consists of one element exactly.
Let us suppose the set x contains at least two different elements a and b. Then, consider the Now we must show the inclusion x 3 ⊆ xu. Let us consider an arbitrary item of x 3 . This item is of the form f cd, where f, c, d ∈ x. There are three possible cases.
• In the case cd = ab we have cd ∈ u and f cd = f (cd) ∈ xu.
• The second case is cd = ab and f = a. Then, we can transform f cd = f ab = a(f b). Since f = a, we can conclude f b = ab using cancellation. The last inequality follows f b ∈ u and f cd = a(f b) ∈ xu. • The last case is cd = ab and f = a. Thus, we have the equality f cd = aab = ba 2 . According to cancellation we have a 2 = ab and f cd = ba 2 ∈ xu.
Hence, we have proved f cd ∈ xu in any case, so x 3 ⊆ xu. Consequently, xu = x 3 . Therefore, there exists u such that u = x 2 and xu = x 3 , so the formula R 1 (x) is false.
Proposition 4. Let the formula R 2 (x) be true in the monoid exp A. Then, the set x consists of two elements exactly.
Proof. By the definition of the relations R 1 and R 2 , it follows immediately that if R 2 (x) is true then R 1 (x) is false. Hence, if the formula R 2 (x) is true, then x can't contain one element exactly due to Proposition 3. Now let us assume that a set x contain at least three different items: Now if we prove the inclusion x 3 ⊆ xv 2 , then we obtain xv 2 = xv 1 = x 3 . The last equalities follow the falsehood of the formula R 2 (x) because the sets v 1 , v 2 , and x 3 are pairwise distinct. Therefore, the formula R 2 (x) is true only if x contains two elements exactly.
At last, let us prove the inclusion x 3 ⊆ xv 2 . The set v 2 must contain all products aγ (where γ ∈ x) from x 2 except for ab and ac. Also, according to cancellation the set v 2 must contain bc: if bc = ab then a = c, if bc = ac then a = b, both cases are impossible. Let us consider an arbitrary triple product cdf ∈ x 3 , where c, d, f ∈ x. If cdf is not equal to any triple product with a, then cdf ∈ xv 2 . Otherwise, we consider all the possible triple products with a: • abc = a(bc) ∈ xv 2 ; • if b 2 = ac, then abb = ab 2 = a(ac) = ca 2 ∈ xv 2 ; • if b 2 = ac, then abb = ab 2 ∈ xv 2 , because b 2 = ab due to cancellation; • abγ = b(aγ) ∈ xv 2 for any γ / ∈ {b, c}; • products of the form acγ are considered analogously to the products abγ; Thus, we have proved the needed inclusion x 3 ⊆ xv 2 .
Further, consider the binary relation P : and the unary relation L: Proposition 5. Let the formula L(x) be true in the monoid exp A. By Proposition 4, it follows that x = {a, b} for some different a, b ∈ A. Then, (ii) at least one item of the set x has infinite order in the monoid A; (iii) for all natural numbers l > 0 we have a l = b l ; (iv) for all natural numbers m > 0 the set x m contains exactly m + 1 elements of the form Proof. The truth of y ∼ x m means y = x m w for some invertible element w of exp A. We prove Claim (i) by induction. For m = 1 consider the relation P (x, x 1 w). Since x = {a, b}, by Proposition 1, it follows that x is not invertible. Thus, the element x 1 w is not invertible too. Further, let uv = x 1 w, and v be not invertible. Then, vw −1 is not invertible also, hence, u(vw −1 ) = x and (1) follow u to be invertible. Thus, we obtain v = xu −1 w and xw = ux(u −1 w). Therefore, we have proved the truth of P (x, x 1 w). For every m > 1 the truth of P (x, x m w) can be easy concluded from (2) by induction.
To prove Claim (ii) let us suppose both elements of x are of finite order. Thus, there exist finitely many non-equal products of the form a k b n . Let M be the amount of all such products. For all natural m > 0 the power x m consists of such products only, so there exist at most 2 M different sets of the form x m . But m itself can have infinitely many values, so there are natural numbers m 1 and m 2 such that m 1 < m 2 and x m 1 = x m 2 . It follows that xx m 1 | x m 2 . By Claim (i), we have the truth of P (x, x m 1 ). Then, by (3), we can conclude that x m 2 | x m 1 is false. Hence, we have got a contradiction with x m 1 = x m 2 . Therefore, our assumption is false, at least one item of x must have infinite order. Now we have to prove Claim (iii). Let us assume that a l = b l for some natural number l > 0. Then, for any natural number q ≥ l we have a p b q = a p+l b q−l . So an arbitrary product a m−s b s must be equal to some product a m−rs b rs , where r s < l. For every natural m ≥ l the power x m consists of products a m−s b s = a m−rs b rs . Since r s < l, we have m − r s > m − l and x m = a m−l x m , where the set x m contains only products a p b q with p + q < l. There exist finitely many products of the last form, hence, there exist finitely many different x m . Thus, there must exist x * such that x m = x * infinitely often. In particular, we can select x m = a m−l x * and x m+n = a m+n−l x * for some n > 0, then, we get x m+n = {a n }x m . By Claim (i), it follows that the formulas P (x, x m ) and P (x, x m+n ) are true. Since R 1 ({a n }) is true, we can use (4), so {a n } is invertible. The inequality n > 0 follows xx m | x m+n . Now we can apply (3) divisible by x m+n . The last follows that {a n } is not invertible. This contradiction proves the falsehood of the assumption a l = b l .
At last, let us prove Claim (iv). According to Claim (ii) we can suppose that a has infinite order in the monoid A. We note earlier that the power x m consists of all products of the form a k b m−k , where k = 0, 1, . . . , m. If we suppose that there are equal products for different k (i. e. a k 1 b m−k 1 = a k 2 b m−k 2 for k 1 < k 2 ), then we obtain b k 2 −k 1 = a k 2 −k 1 using cancellation. It contradicts Claim (iii). Now we can prove the inverse of Claim (i) in the previous proposition. The truth of P (x, y) implies immediately that the element y is not invertible. It is clear that y = Ey. From the definition of P we obtain y = Exw 1 = x 1 w 1 for some w 1 . If w 1 is not invertible, then we again use the equality y = x 1 w 1 to obtain y = x 1 xw 2 = x 2 w 2 for some w 2 , and so on. We can do it while w m is not invertible, so we get y = x m w m for some w m . Later we show that the number m can't grow infinitely. Hence, we can select the maximal m such that there is the presentation y = x m w m . If w m is not invertible, then analogously we can obtain y = x m+1 w m+1 , it contradicts the maximality of m. Therefore, w m is invertible, and we have proved y ∼ x m . Now we need to prove the used claim: there is an upper bound of m such that y = x m w m . Let us suppose this bound doesn't exist. So there exist infinitely many natural numbers m such that y = x m w m . The power x m consists of m + 1 elements exactly, it is followed by Claim (iv) of Proposition 5. But for cardinality of sets we have the inequality |x m | ≤ |x m w m | = |y| that means |y| ≥ m + 1. The monoid exp A consists of finite sets, so the inequality |y| ≥ m + 1 can't be true for infinitely many m. Therefore, the assumption is false.
If we combine the last two propositions, then we obtain Corollary 7. Let L(x) be true in the monoid exp A. Then, P (x, y) is true if and only if y ∼ x m for some natural number m > 0.
The previous claims are useful if the monoid exp A actually contains elements x those satisfy L(x). So we need to prove such x exists. Proposition 8. Let g, a ∈ A, the element g be invertible, and the element a be of infinite order. Then, for the set x = g{e, a} the formula L(x) is true in the monoid exp A.
Proof. At first, consider the condition R 2 (x) from (1). Let u = g 2 {e, a 2 }. Then, If we suppose g 2 {e, a 2 } = u = x 2 = g 2 {e, a, a 2 }, then multiplying by g −2 we obtain the equality {e, a 2 } = {e, a, a 2 } and a ∈ {e, a 2 }. Both cases a = e and a = a 2 mean the finite order of a. It contradicts the proposition assumption. Now let xv = x 3 that means gv ∪ gav = g 3 {e, a, a 2 , a 3 }. So gv ⊆ g 3 {e, a, a 2 , a 3 } and v ⊆ g 2 {e, a, a 2 , a 3 }. It is enough to check all subsets of g 2 {e, a, a 2 , a 3 } to prove that v must be equal to x 2 or u.
Consider the second part of line (1). Let uv = x or (g −1 u)v = g −1 x for suitable u and v. The neutral element e of A belongs to the set g −1 x, hence, for some invertible c ∈ A we have c ∈ g −1 u and c −1 ∈ v. Then, e belongs to the sets u = c −1 g −1 u and v = cv. By Proposition  7 2, it follows that u , v ⊆ u v = g −1 uv = g −1 x. Let us assume that both sets u and v contain non-neutral elements b and d correspondingly. As u , v ⊆ g −1 x = {e, a}, so we have b, d ∈ {e, a} and b = d = a. Thus, we obtain bd ∈ u v = g −1 x = {e, a}. If bd = e, then a 2 = e and a has finite order. If bd = a, then a 2 = a and again a has finite order. Both cases lead to the contradiction, hence, our assumption is wrong, u or v must be equal to {e}, and elements u = cgu or v = c −1 v is invertible.
To prove another parts of the L definition we establish the following fact: P (x, y) is true if and only if y ∼ x m for some natural number m > 0.
If P (x, y) is true, then we can use the same methods as in the proof of Proposition 6 and obtain y = x m w m for natural numbers m > 0 while w m is not invertible. But x m = g m {e, a, a 2 , . . . , a m }, a has infinite order, hence, the cardinality of x m is m + 1. This cardinality must be less or equal to the cardinality of y, so m can't grow infinitely. Then, for some m we must have y = x m w m for some invertible w m , i. e. y ∼ x m . Now let y ∼ x m and y = x m w for some natural number m > 0 and some invertible w. We have demonstrated yet that the cardinality of x m w is m + 1 > 1, so x and y are not invertible due to that means the truth of the implication in the P definition. The proved claim implies (2) immediately. Let us prove the part (4). If the formulas P (x, y) and P (x, z) are true, then we get y = x m 1 w 1 and z = x m 2 w 2 for some invertible w 1 and w 2 and some natural m 1 and m 2 . By Proposition 1, the invertible elements w 1 and w 2 of exp A are of the form {g 1 } and {g 2 } correspondingly for some invertible elements g 1 and g 2 of the monoid A. So the equality y = zu means g 1 x m 1 = g 2 x m 2 u. If the formula R 1 (u) is true, then u = {h} for some h ∈ A and we have g 1 x m 1 = g 2 hx m 2 . In particular, the set g 1 x m 1 contains g 1 g m 1 , hence, g 1 g m 1 = g 2 hg m 2 a k ∈ g 2 hx m 2 for some k ≤ m 2 and h(g −1 1 g 2 g m 2 −m 1 a k ) = e. Therefore, h is invertible in the monoid A, and u = {h} is invertible in exp A.
To prove (3) let again P (x, y) be true, so y = x m w for some invertible w. If we suppose the truth of both xy | z and z | y, then xx m wu = z and x m w = zv for some u and v. So we have x m w = xx m wuv = x m+1 wuv and x m = x m+1 uv. The last equality is impossible because the cardinality of the first set is less than of the second one.

Interpretation of Elementary Arithmetic
Now we are ready to do the last steps and to construct an interpretation of elementary arithmetic in the monoid exp A. Let us introduce the following two relations: D(x, y, z) ≡ P (x, y) ∧ R 2 (z) ∧ zy = y 2 ; C(x, z) ≡ (∃y)D(x, y, z). . Let g, a ∈ A, the element g be invertible, and the element a be of infinite order in the monoid A. Let L(x) be true in the monoid exp A for x = g{e, a}. Then, (i) the formula D(x, x n w, z) is true in the monoid exp A for every z = g n {e, a n }w, where n is an arbitrary natural number and w is an arbitrary invertible element of exp A; (ii) let z ∈ exp A be such that the formula C(x, z) is true. Then, there exists exactly one natural number n such that the formula D(x, x n w, z) is true in the monoid exp A for some invertible element w.
The formula P (x, x n w) is true due to Corollary 7. Now let us establish Claim (ii). The formula C(x, z) follows D(x, y, z) for some y. The formula D(x, y, z) contains P (x, y), hence, by Corollary 7, we conclude y = x n w for some natural number n and invertible w. Let us suppose there exists at least two different natural numbers n 1 and n 2 satisfying D(x, x n i w i , z), i = 1, 2. We can assume n 1 < n 2 , then, x n 1 w 1 z = x 2n 1 w 2 1 and x n 2 w 2 z = x 2n 2 w 2 2 . Futher, Now we can cancel the equality by w 2 and obtain x 2n 2 w 2 = x n 2 +n 1 w 1 . Remember, that both w 1 and w 2 have exactly one element (Proposition 1), the power x m has exactly m + 1 elements (Claim (iv) of Proposition 5). So the last equality is possible only for 2n 2 = n 2 + n 1 that means n 1 = n 2 . It contradicts our assumption n 1 < n 2 .
Introduce the following two relations: Let us select an arbitrary x ∈ exp A satisfying the formula L(x). Then, we denote by C x the following set: Proposition 10. Let g, a ∈ A, the element g be invertible, the element a be of infinite order, and x = g{e, a}. Then, the relation ≤ x is reflexive and transitive on the set C x .
Proof. If z ∈ C x , then the formula D(x, y, z) is true for some y by definition, so z ≤ x z is true for y 1 = y 2 = y.
To prove the transitivity of the relation ≤ x let us suppose z 1 ≤ x z 2 and z 2 ≤ x z 3 for elements z 1 , z 2 , and z 3 of C x . Then, we have the truth of the next four formulas: D(x, y 1 , z 1 ), D(x, y 2 , z 2 ), D(x, y 2 , z 2 ), and D(x, y 3 , z 3 ) for some y 1 , y 2 , y 2 , y 3 such that y 1 | y 2 and y 2 | y 3 . By Claim (ii) of Proposition 9, we conclude y 2 = x n w and y 2 = x n w for some invertible elements w and w , it follows y 2 | y 2 . Thus, we obtain y 1 | y 2 , y 2 | y 2 , and y 2 | y 3 . Therefore, we get y 1 | y 3 that means z 1 ≤ x z 3 . Corollary 11. The relation ≈ x is an equivalency on the set C x , the factor order on C x by ≈ x is isomorphic to natural numbers.

Immediately we have
Let us denote byn the equivalency class consisting of elements z such that D(x, x n w, z) is true for invertible elements w. By Claim (ii) of Proposition 9, it follows that such n is unique for each z ∈ C x . Now we can define formulas, which interpret arithmetic relations: Let us prove their main properties.
Proposition 12. Let g, a ∈ A, the element g be invertible, the element a be of infinite order, and x = g{e, a}. Let L * (x) be true in the monoid exp A. Then, for any z 1 , z 2 , z 3 ∈ exp A such that z i ∈n i for i = 1, 2, 3 we have Proof. To prove Claim (i) let A x (z 1 , z 2 , z 3 ) be true. Thus, the formulas D(x, x n 1 w 1 , z 1 ), D(x, x n 2 w 2 , z 2 ) and D(x, x n 1 w 1 x n 2 w 2 , z 3 ) are true for some invertible w 1 and w 2 . But z 3 ∈n 3 means the truth of D(x, x n 3 w 3 , z 2 ). Hence, the formulas D(x, x n 1 +n 2 w 1 w 2 , z 3 ) and D(x, x n 3 w 3 , z 2 ) are true for invertible (w 1 w 2 ) and w 3 . By Claim (ii) of Proposition 9, we get n 1 + n 2 = n 3 .
On the other hand, let us suppose n 1 + n 2 = n 3 . Then, the formulas D(x, x n 1 w 1 , z 1 ), D(x, x n 2 w 2 , z 2 ), and D(x, x n 1 +n 2 w 3 , z 3 ) are true for some invertible elements w 1 , w 2 and w 3 . By Claim (i) of Proposition 5, the value of P (x, x n w) doesn't depend on invertible w. Thus, the truth of D(x, x n 1 +n 2 w 3 , z 3 ) follows the truth D(x, x n 1 +n 2 w 1 w 2 , z 3 ) and, consequently, of A x (z 1 , z 2 , z 3 ). Now consider Claim (ii). Let z 1 and z 2 satisfy the formula DV x (z 1 , z 2 ). It means D(x, x n 1 w 1 , z 1 ) and D(x, x n 2 w 2 , z 2 ) for some invertible elements w 1 and w 2 . Since the truth of the equality z 1 ≈ x z 1 , we have D(x, x n 1 w 1 , z 1 ) and C(x, z 1 ), so L * (x) follows L(z 1 ). Let us apply Claim (ii) of Proposition 9 to L(z 1 ) and C(z 1 , z 2 ), we have D(z 1 , z 1 m w, z 2 ) for some natural number m and invertible w. Then, The elements w 1 , w 2 , and w are invertible, so the sets w 2 w 4m 1 w 2 and ww 3m 1 w 2 2 consist of one element. Then, for cardinalities we get n 2 + 4n 1 m + 1 = 3n 1 m + 2n 2 + 1 that is followed by Claim (iv) of Proposition 5. Therefore, we have n 2 = n 1 m and n 1 | n 2 . Now let us suppose n 1 | n 2 , it follows the equality n 2 = n 1 m for some natural number m. Assume z 1 = g n 1 {e, a n 1 } and z 2 = g mn 1 {e, a mn 1 }. Then, we have D(x, x n 1 , z 1 ) and D(x, x n 2 , z 2 ), see Claim (i) of Proposition 9. Thus, the equalities z 1 ≈ x z 1 and z 2 ≈ x z 2 hold. Since D(x, x n 1 , z 1 ) is true, we obtain C(x, z 1 ), hence, we have L(z 1 ) due to L * (x). By Claim (i) of Proposition 9, we get the truth of D(z 1 , (z 1 ) m , z 2 ) and C(z 1 , z 2 ). Therefore, the formula DV x (z 1 , z 2 ) is true also. Proof. It follows by Proposition 8 that L(x) is true for such x. By Corollary 7, we get that the formula P (x, y) is true for y = x n w exactly, where n is a natural number and w is an invertible element. Let us select an arbitrary z ∈ exp A such that C(x, z) is true. From C(x, z) we obtain D(x, y, z), so y = x n w for some n and w, zx n w = x 2n w 2 , and (zw −1 )x n = x 2n . As e ∈ x n , so by Proposition 2, we have a = a 1 , a 2 , . . . , a 2n }.
The powers a k are pairwise different because the element a has infinite order. Hence, we can conclude that the set zw −1 must contain a n and can't contain powers greater than a n . On the other hand, a k = e for all k > 0, hence, zw −1 must contain e. Since D(x, y, z), we get R 2 (z) and z = {e, a n }. Evidently, the element a n must have infinite order, therefore, L(z) is followed by Proposition 8. Now we can prove the main result of the paper. Proof. Remember, that the following two relations are enough to interpret elementary arithmetic: addition and divisibility (see [8,14]). Let us select an arbitrary element t of the monoid exp A such that L * (t) is true. According to Proposition 13 such t exists. Then, the domain of our interpretation is the set of equivalency classes for the relation ≈ t .
Consider any formula Φ. For every subformula φ of the formula Φ we construct its interpretation φ as following: Therefore, the interpretation of the entire formula Φ is (∃t)(L * (t) ∧ Φ ).
The soundness of this interpretation is implied by Corollary 11, Propositions 12 and 13, and the L * definition.
Corollary 15. Let a commutative cancellative monoid A contains an element of infinite order. Then, the elementary theory of the monoid exp A is algorithmically undecidable at least as elementary arithmetic.
Note that the last result distinguishes the claim of Corollary 1 in [11]. In [11] the free cyclic monoid A is considered, and it was proved that the theory of exp A is algorithmically equivalent to elementary arithmetic.
In the current article we investigate more general monoids those theories themselves can have higher degree of undecidability (see [14]). In this case the theory of exp A has also higher degree At last, remember that every Abelian group is a commutative cancellative monoid. If an Abelian group is not a torsion group, then it has an element of infinite order. So the previously proved claims are applicable to these groups.
Corollary 16. Let A be an Abelian group but not a torsion group. Then, elementary arithmetic can be interpreted in the theory of the monoid exp A. Hence, this theory is undecidable.
In particular, the theory of exp A is undecidable when A is the additive or multiplicative group of (integer, rational, real, complex) numbers.

Conclusion
We have proved that an algebra of finite subsets is undecidable for the wide class of monoids. The question appears whether it is possible to generalize this result to other monoids. Are used restrictions important or they can be eliminated?
• Are the theory of the algebra exp A is undecidable when all elements of A have finite orders?
In particular, is such claim holds for infinite Abelian torsion groups? An example of such group is the multiplicative group of complex roots of unity. • Is such claim true for monoids without cancellation?
• Is such claim true for non-commutative monoids?