The Order of Hypersubstitutions of Type (2,1)

Abstract

Hypersubstitutions are mappings which map operation symbols to terms of the corresponding arities. They were introduced as a way of making precise the concept of a hyperidentity and generalizations to 𝑀-hyperidentities. A variety in which every identity is satisfied as a hyperidentity is called solid. If every identity is an 𝑀-hyperidentity for a subset 𝑀 of the set of all hypersubstitutions, the variety is called 𝑀-solid. There is a Galois connection between monoids of hypersubstitutions and sublattices of the lattice of all varieties of algebras of a given type. Therefore, it is interesting and useful to know how semigroup or monoid properties of monoids of hypersubstitutions transfer under this Galois connection to properties of the corresponding lattices of 𝑀-solid varieties. In this paper, we study the order of each hypersubstitution of type (2,1), that is, the order of the cyclic subsemigroup of the monoid of all hypersubstitutions of type (2,1) generated by that hypersubstitution.

1. Preliminaries

Let ℕ denote the set of all positive integers. Let 𝜏={(𝑓_𝑖,𝑛_𝑖)∣𝑖∈𝐼} be a type. Let 𝑋={𝑥₁,𝑥₂,𝑥₃,…} be a countably infinite alphabet of variables such that the sequence of the operation symbols (𝑓_𝑖)_𝑖∈𝐼 is disjoint with 𝑋, and let 𝑋_𝑛={𝑥₁,𝑥₂,…,𝑥_𝑛} be an 𝑛-element alphabet where 𝑛∈ℕ. Here, 𝑓_𝑖 is 𝑛_𝑖-ary for a natural number 𝑛_𝑖≥1. An 𝑛-ary (𝑛≥1) term of type 𝜏 is inductively defined as follows: (i)every variable 𝑥_𝑗∈𝑋_𝑛 is an 𝑛-ary term,(ii)if 𝑡₁,…,𝑡_𝑛𝑖 are 𝑛-ary terms and 𝑓_𝑖 is an 𝑛_𝑖-ary operation symbol, then 𝑓_𝑖(𝑡₁,…,𝑡_𝑛𝑖) is an 𝑛-ary term.

Let 𝑊_𝜏(𝑋_𝑛) be the smallest set containing 𝑥₁,…,𝑥_𝑛 and being closed under finite application of (ii). The set of all terms of type 𝜏 over the alphabet 𝑋 is defined as the disjoint union 𝑊_𝜏(𝑋)∶=⋃^∞_𝑛=1𝑊_𝜏(𝑋_𝑛).

Any mapping 𝜎∶{𝑓_𝑖∶𝑖∈I}→𝑊_𝜏(𝑋) is called a hypersubstitution of type 𝜏 if 𝜎(𝑓_𝑖) is an 𝑛_𝑖-ary term of type 𝜏 for every 𝑖∈𝐼. Any hypersubstitution 𝜎 of type 𝜏 can be uniquely extended to a map 𝜎 on 𝑊_𝜏(𝑋) as follows: (i)𝜎[𝑡]∶=𝑡 if 𝑡∈𝑋,(ii)𝜎[𝑡]∶=𝜎(𝑓_𝑖)(𝜎[𝑡₁],…,𝜎[𝑡_𝑛𝑖]) if 𝑡=𝑓_𝑖(𝑡₁,…,𝑡_𝑛𝑖).

A binary operation ∘_ℎ is defined on the set Hyp(𝜏) of all hypersubstitutions of type 𝜏 by 𝜎₁∘_ℎ𝜎₂𝑓_𝑖∶=𝜎₁𝜎₂𝑓_𝑖, (1.1) for all 𝑛_𝑖-ary operation symbols 𝑓_𝑖. This binary associative operation makes Hyp(𝜏) into a monoid, with the identity hypersubstitution 𝜎_id which maps every 𝑓_𝑖 to 𝑓_𝑖(𝑥₁,…,𝑥_𝑛𝑖) as an identity element. For a submonoid 𝑀 of Hyp(𝜏) an identity 𝑠≈𝑡 of a variety 𝑉 of type (𝜏) is called an 𝑀-hyperidentity of 𝑉 if for every hypersubstitution 𝜎∈𝑀, the equation 𝜎[𝑠]≈𝜎[𝑡] holds in 𝑉. A variety 𝑉 is called 𝑀-solid if every identity of 𝑉 is an 𝑀-hyperidentity of 𝑉. If 𝑀 is a submonoid of Hyp(𝜏), then the collection of all 𝑀-solid varieties of type 𝜏 is a complete sublattice of the lattice of all varieties of type 𝜏 [1].

Let 𝜎∈Hyp(𝜏) and let ⟨𝜎⟩∶={𝜎^𝑛∣𝑛∈ℕ} be the cyclic subsemigroup of Hyp(𝜏) generated by 𝜎. The order of 𝜎∈Hyp(𝜏) is defined as the order of the semigroup ⟨𝜎⟩. If ⟨𝜎⟩ is finite, then the order of the hypersubstition 𝜎 is finite, otherwise the order of 𝜎 is infinite. The hypersubstitution 𝜎 is idempotent if and only if the order of 𝜎 is 1.

The order of a hypersubstitution of type (2) is 1, 2, or infinite [2]. The order of a hypersubstitution of type (3) is 1, 2, 3, or infinite [3]. The order of a hypersubstitution of type (2,2) is 1, 2, 3, 4, or infinite [4]. We are interested in type (2,1). The main result is as follows

Main Theorem. Any hypersubstitution of type (2,1) has order either infinite or less than or equal to 3.

Throughout this paper, let 𝑓 and 𝑔 be the binary operation symbols and the unary operation symbols of type 𝜏=(2,1), respectively. For a binary term 𝑎 and an unary term 𝑏 of type 𝜏, the hypersubstitution which maps the operation symbol 𝑓 to the term 𝑎 and the operation symbol 𝑔 to the term 𝑏 will be denoted by 𝜎_𝑎,𝑏.

For a binary term 𝑡∈𝑊_(2,1)(𝑋₂), we introduce the following notations:leftmost(𝑡)—the first variable (from the left) occurring in 𝑡,rightmost(𝑡)—the last variable occurring in 𝑡,var(𝑡)—the set of all variable occurring in 𝑡,op(𝑡)—the total number of all operation symbols occurring in 𝑡,ops(𝑡)—the set of all operation symbols occurring in 𝑡,firstops(𝑡)—the first operation symbol (from the left) occurring in 𝑡.

For 𝑡∈𝑊_(2,1)(𝑋₂), let Lp(𝑡) denote the left path from the root to the leaf which is labelled by the leftmost variable in 𝑡 and Rp(𝑡) denote the right path from the root to the leaf which is labelled by the rightmost variable in 𝑡. The operation symbols occurring in Lp(𝑡) and Rp(𝑡) will be denoted by ops(Lp(𝑡)) and ops(Rp(𝑡)), respectively. If 𝑡∈𝑊_(2,1)(𝑋₂) such that var(𝑡)={𝑥₁} or var(𝑡)={𝑥₂}, we define 𝑡¹=𝑡 and 𝑡^𝑛=𝑡^𝑛−1(𝑡,𝑡),if𝑛≥1. (1.2)

For 𝑎 and 𝑏 of the hypersubstitution 𝜎_𝑎,𝑏, we break our analysis into the following six cases, which cover all possibilities for 𝑎 and 𝑏: (I)op(𝑎)≤1 and op(𝑏)≤1,(II)op(𝑎)>1 and op(𝑏)>1,(III)op(𝑎)=1 and op(𝑏)>1,(IV)op(𝑎)>1 and op(𝑏)=1,(V)op(𝑎)>1 and op(𝑏)=0,(VI)op(𝑎)=0 and op(𝑏)>1.

In case (I), we have term 𝑎∈{𝑥₁,𝑥₂,𝑓(𝑥₁,𝑥₁),𝑓(𝑥₁,𝑥₂),𝑓(𝑥₂,𝑥₁),𝑓(𝑥₂,𝑥₂),𝑔(𝑥₁), 𝑔(𝑥₂)} and 𝑏∈{𝑥₁,𝑔(𝑥₁),𝑓(𝑥₁,𝑥₁)}. For 𝜎_𝑎,𝑏 we have twenty-four cases, for which the orders can be verified by simple calculations. The case 𝑎=𝑔(𝑥₂) and 𝑏=𝑓(𝑥₁,𝑥₁) gives order 3, six other cases give order 2, and the remaining 17 cases give order 1.

2. Case II: op(𝑎)>1 and op(𝑏)>1

In this section, we consider the order of a hypersubstitution 𝜎_𝑎,𝑏 where op(𝑎)>1 and op(𝑏)>1. We consider three subcases of Case II: (II-1)var(𝑎)={𝑥₁,𝑥₂}, var(𝑏)={𝑥₁}, (II-2)var(𝑎)={𝑥₁}, var(𝑏)={𝑥₁}, (II-3)var(𝑎)={𝑥₂}, var(𝑏)={𝑥₁}.

The following formula for the operation symbol count of the compound term 𝑠(𝑡₁,…,𝑡_𝑛) for some 𝑠,𝑡₁,…,𝑡_𝑛∈𝑊_𝜏(𝑋) was proved in [5]: op𝑠𝑡₁,…,𝑡_𝑛=^𝑛_𝑗=1𝑣𝑏_𝑗(𝑠)op𝑡_𝑗+op(𝑠),𝑠,𝑡₁,…,𝑡_𝑛∈𝑊_(𝜏)(𝑋), (2.1) where 𝑣𝑏_𝑗(𝑠) = the number of occurrences of variable 𝑥_𝑗 in the term  𝑠.

Using the facts (see [5]) that op(𝜎[𝑡])>op(𝑡) for all 𝑡∈𝑊_𝜏(𝑋) if 𝜎∈Hyp(𝜏) is regular, that is, var(𝜎(𝑓_𝑖))=𝑋_𝑛, for all 𝑖∈𝐼, and the formula above, we obtain the following theorem.

Theorem 2.1. Let 𝑎∈𝑊_(2,1)(𝑋₂), 𝑏∈𝑊_(2,1)(𝑋₁). If op(𝑎)>1, op(𝑏)>1, var(𝑎)={𝑥₁,𝑥₂}, var(𝑏)={𝑥₁}, then the order of 𝜎_𝑎,𝑏 is infinite.

Proof. Since 𝜎_𝑎,𝑏 is regular, by induction we obtain op(𝜎^𝑘_𝑎,𝑏(𝑓))<op(𝜎^𝑘+1_𝑎,𝑏(𝑓)). Then, the order of 𝜎_𝑎,𝑏 is infinite.

The following lemmas are easy to prove.

Lemma 2.2. Let 𝑎,𝑡∈𝑊_(2,1)(𝑋₂), 𝑏∈𝑊_(2,1)(𝑋₁) be such that op(𝑎)≥1, op(𝑏)≥1. If 𝑡∉𝑋₂, then 𝜎^𝑛_𝑎,𝑏[𝑡] is not a variable for all 𝑛∈ℕ.

Lemma 2.3. Let 𝑎,𝑡∈𝑊_(2,1)(𝑋₂), 𝑏∈𝑊_(2,1)(𝑋₁). If var(𝑡)={𝑥_𝑖} for some 𝑖∈{1,2}, then var(𝜎^𝑛_𝑎,𝑏[𝑡])={𝑥_𝑖} for all 𝑛∈ℕ.

Lemma 2.4. Let 𝑎∈𝑊_(2,1)(𝑋₂), 𝑏∈𝑊_(2,1)(𝑋₁) be such that op(𝑎)≥1, op(𝑏)≥1, and 𝑎=𝑓(𝑎₁,𝑎₂), 𝑏=𝑔(𝑏₁) for some 𝑎₁,𝑎₂∈𝑊_(2,1)(𝑋₂), 𝑏₁∈𝑊_(2,1)(𝑋₁). Then, the following hold: (i)if var(𝑎)={𝑥₁} and 𝑎₁≠𝑥₁, then the order of 𝜎_𝑎,𝑏 is infinite;(ii)if var(𝑏)={𝑥₁} and 𝑏₁≠𝑥₁, then the order of 𝜎_𝑎,𝑏 is infinite.

Proof. (i) Assume var(𝑎)={𝑥₁} and 𝑎₁≠𝑥₁. Then, 𝑎₁∉𝑋₂. Let 𝑛∈ℕ. By Lemma 2.2, 𝜎^𝑛+1_𝑎,𝑏[𝑎₁]∉𝑋₂. Since var(𝑎)={𝑥₁}, by Lemma 2.3, var(𝜎^𝑛_𝑎,𝑏[𝑎])={𝑥₁}. Now, op𝜎^𝑛+1_𝑎,𝑏[𝑎]=op𝜎^𝑛_𝑎,𝑏𝜎_𝑎,𝑏𝑓𝑎₁,𝑎₂=op𝜎^𝑛_𝑎,𝑏𝜎_𝑎,𝑏𝑓𝜎_𝑎,𝑏𝑎₁,𝜎_𝑎,𝑏𝑎₂=op𝜎^𝑛_𝑎,𝑏𝑎𝜎_𝑎,𝑏𝑎₁,𝜎_𝑎,𝑏𝑎₂=op𝜎^𝑛_𝑎,𝑏[𝑎]𝜎^𝑛+1_𝑎,𝑏𝑎₁,𝜎^𝑛+1_𝑎,𝑏𝑎₂>op𝜎^𝑛_𝑎,𝑏[𝑎]. (2.2) This shows that op(𝜎^𝑛+1_𝑎,𝑏[𝑎])>op(𝜎^𝑛_𝑎,𝑏[𝑎]) for all 𝑛∈ℕ. Hence, the order of 𝜎_𝑎,𝑏 is infinite.
(ii) Assume var(𝑏)={𝑥₁} and 𝑏₁≠𝑥₁, then 𝑏₁∉𝑋₁. Let 𝑛∈ℕ. By Lemma 2.2, 𝜎^𝑛+1_𝑎,𝑏[𝑏₁]∉𝑋₁. Since var(𝑏)={𝑥₁}, by Lemma 2.3  var(𝜎^𝑛_𝑎,𝑏[𝑏])={𝑥₁} for all 𝑛∈ℕ. Then, op𝜎^𝑛+1_𝑎,𝑏[𝑏]=op𝜎^𝑛_𝑎,𝑏𝜎_𝑎,𝑏𝑔𝑏₁=op𝜎^𝑛_𝑎,𝑏[𝑏]𝜎^𝑛+1_𝑎,𝑏𝑏₁>op𝜎^𝑛_𝑎,𝑏[𝑏]. (2.3) Therefore, op(𝜎^𝑛+1_𝑎,𝑏[𝑏])>op(𝜎^𝑛_𝑎,𝑏[𝑏]) for all 𝑛∈ℕ. Hence we have the claim.

Throughout the rest of this paper, we assume that when 𝑎 is not a variable term it has the form 𝑓(𝑎₁,𝑎₂) or 𝑔(𝑎₁), for some terms 𝑎₁, 𝑎₂ and that when 𝑏 is not a variable term it has the form 𝑓(𝑏₁,𝑏₂) or 𝑔(𝑏₁), for some terms 𝑏₁, 𝑏₂.

In Case (II-2), we have var(𝑎)={𝑥₁}, var(𝑏)={𝑥₁}. We consider the following four subcases: (2.1)firstops(𝑎)=𝑓, firstops(𝑏)=𝑓; (2.2)firstops(𝑎)=𝑔, firstops(𝑏)=𝑔; (2.3)firstops(𝑎)=𝑓, firstops(𝑏)=𝑔; (2.4)firstops(𝑎)=𝑔, firstops(𝑏)=𝑓.

In Case (2.1), we can separate into four subcases: (2.1.1)𝑎₁∉𝑋₂; (2.1.2)𝑎₁=𝑥₁, 𝑏₁=𝑥₁; (2.1.3)𝑎₁=𝑥₁, 𝑏₁∉𝑋₁, ops(Lp(𝑏))={𝑓}; (2.1.4)𝑎₁=𝑥₁, 𝑏₁∉𝑋₁, ops(Lp(𝑏))≠{𝑓}.

Theorem 2.5. Let 𝑎∈𝑊_(2,1)(𝑋₂), 𝑏∈𝑊_(2,1)(𝑋₁) be such that op(𝑎)>1, op(𝑏)>1, var(𝑎)={𝑥₁}=var(𝑏), and 𝑎=𝑓(𝑎₁,𝑎₂), 𝑏=𝑓(𝑏₁,𝑏₂) for some 𝑎₁,𝑎₂∈𝑊_(2,1)(𝑋₂), 𝑏₁,𝑏₂∈𝑊_(2,1)(𝑋₁). Then, the following hold: (i)if 𝑎 and 𝑏 satisfy (2.1.1) or (2.1.4), then 𝜎_𝑎,𝑏 has infinite order;(ii)if 𝑎 and 𝑏 satisfy (2.1.2) or (2.1.3), then the order of 𝜎_𝑎,𝑏 is less than or equal to 3.

Proof. (i) Assume 𝑎 and 𝑏 satisfy (2.1.1). By Lemma 2.2, we have 𝜎^𝑛_𝑎,𝑏[𝑎₁]∉𝑋₂ for all 𝑛∈ℕ. Since var(𝜎_𝑎,𝑏[𝑎])={𝑥₁}, by Lemma 2.3  var(𝜎^𝑛_𝑎,𝑏[𝑎])={𝑥₁} for all 𝑛∈ℕ. Therefore, op𝜎^𝑘+1_𝑎,𝑏[𝑎]=op𝜎^𝑘_𝑎,𝑏𝑎𝜎_𝑎,𝑏𝑎₁,𝜎_𝑎,𝑏𝑎₂=op𝜎^𝑘_𝑎,𝑏[𝑎]𝜎^𝑘+1_𝑎,𝑏𝑎₁,𝜎^𝑘+1_𝑎,𝑏𝑎₂>op𝜎^𝑘_𝑎,𝑏[𝑎]. (2.4) This shows that op(𝜎^𝑛+1_𝑎,𝑏[𝑎])>op(𝜎^𝑛_𝑎,𝑏[𝑎]) for all 𝑛∈ℕ. Hence, 𝜎_𝑎,𝑏 has infinite order.
Assume 𝑎 and 𝑏 satisfy (2.1.4). Since var(𝜎_𝑎,𝑏[𝑎])={𝑥₁}, 𝜎^𝑛_𝑎,𝑏[𝑎]=𝜎_𝑎,𝑏[𝑎] for all 𝑛∈ℕ. Since 𝑔∈ops(Lp(𝑏)), we have op(𝜎^𝑛+1_𝑎,𝑏[𝑏])>op(𝜎^𝑛_𝑎,𝑏[𝑏]) for all 𝑛∈ℕ. Hence, the order of 𝜎_𝑎,𝑏 is infinite.
(ii) Assume 𝑎 and 𝑏 satisfy (2.1.2). Since var(𝜎_𝑎,𝑏[𝑎])={𝑥₁}, 𝑎₁=𝑥₁ and firstops(𝑎)=𝑓, we obtain 𝜎_𝑎,𝑏[𝑎]=𝑎. This gives 𝜎³_𝑎,𝑏𝑓=𝜎²_𝑎,𝑏[𝑎]=𝜎_𝑎,𝑏[𝑎]=𝜎²_𝑎,𝑏𝑓. (2.5) Since 𝑏₁=𝑥₁, var(𝜎_𝑎,𝑏[𝑎])={𝑥₁} and firstops(𝑏)=𝑓, we have 𝜎_𝑎,𝑏[𝑏])=𝑎. So, 𝜎³_𝑎,𝑏𝑔=𝜎²_𝑎,𝑏[𝑏]=𝜎_𝑎,𝑏[𝑎]=𝑎=𝜎_𝑎,𝑏[𝑏]=𝜎²_𝑎,𝑏𝑔. (2.6) Hence, 𝜎²_𝑎,𝑏=𝜎³_𝑎,𝑏. This shows that the order of 𝜎_𝑎,𝑏 is less than or equal to 2.
Assume 𝑎 and 𝑏 satisfy (2.1.3). Because of (2.1.2), we have 𝜎_𝑎,𝑏[𝑎]=𝑎, which implies 𝜎³_𝑎,𝑏[𝑓])=𝜎²_𝑎,𝑏[𝑓]. Since ops(Lp(𝑏))={𝑓}, we have 𝜎_𝑎,𝑏[𝑏]=𝑎^′₁𝑎^′₂⋯𝑎^′_𝑚−1𝑎^′_𝑚𝑥₁,𝑥₁,𝜎_𝑎,𝑏𝑡^′_𝑚−1,…,𝜎_𝑎,𝑏𝑡^′₂,𝜎_𝑎,𝑏𝑡^′₁, (2.7) for some 𝑡^′₁,…,𝑡^′_𝑚−1∈𝑊_(2,1)(𝑋₁) with 𝑚=op(Lp(𝑏)) and 𝑎^′₁=𝑎^′₂=⋯=𝑎^′_𝑚=𝑎. It follows that 𝜎²_𝑎,𝑏[𝑏]=(𝜎_𝑎,𝑏[𝑎])^𝑚. Since 𝜎_𝑎,𝑏[𝑎]=𝑎, we obtain 𝜎⁴_𝑎,𝑏𝑔=𝜎³_𝑎,𝑏[𝑏]=𝜎_𝑎,𝑏𝜎_𝑎,𝑏[𝑎]^𝑚=𝜎²_𝑎,𝑏[𝑎]^𝑚=𝜎_𝑎,𝑏𝜎_𝑎,𝑏[𝑎]^𝑚=𝜎_𝑎,𝑏[𝑎]^𝑚=𝜎²_𝑎,𝑏[𝑏]=𝜎³_𝑎,𝑏𝑔. (2.8) Thus, 𝜎³_𝑎,𝑏=𝜎⁴_𝑎,𝑏. Hence, the order of 𝜎_𝑎,𝑏 is less than or equal to 3.

In Case (2.2), we have firstops(𝑎)=𝑔 and firstops(𝑏)=𝑔. Since op(𝑎)>1 and op(𝑏)>1, 𝑎₁∉𝑋₂ and 𝑏₁∉𝑋₁. Then, we obtain the following.

Theorem 2.6. Let 𝑎∈𝑊_(2,1)(𝑋₂), 𝑏∈𝑊_(2,1)(𝑋₁) be such that op(𝑎)>1, op(𝑏)>1, var(𝑎)={𝑥₁}, 𝑎=𝑔(𝑎₁) and 𝑏=𝑔(𝑏₁) for some 𝑎₁∈𝑊_(2,1)(𝑋₂), 𝑏₁∈𝑊_(2,1)(𝑋₁). Then, 𝜎_𝑎,𝑏 has infinite order.

Proof. Assume that 𝑎=𝑔(𝑎₁), 𝑏=𝑔(𝑏₁), op(𝑎)>1, op(𝑏)>1, var(𝑎)={𝑥₁}. Then, 𝑎₁∉𝑋₂, 𝑏₁∉𝑋₁. It can be proved, as in the proof of Theorem 2.5(i), that the order of 𝜎_𝑎,𝑏 is infinite.

In Case (2.3), we have firstops(𝑎)=𝑓 and firstops(𝑏)=𝑔. Since op(𝑏)>1, 𝑏₁∉𝑋₁. Then, we obtain the following.

Theorem 2.7. Let 𝑎∈𝑊_(2,1)(𝑋₂), 𝑏∈𝑊_(2,1)(𝑋₁) be such that op(𝑎)>1, op(𝑏)>1, var(𝑎)={𝑥₁}, 𝑎=𝑓(𝑎₁,𝑎₂) and 𝑏=𝑔(𝑏₁) for some 𝑎₁,𝑎₂∈𝑊_(2,1)(𝑋₂), 𝑏₁∈𝑊_(2,1)(𝑋₁)⧵𝑋₁. Then, 𝜎_𝑎,𝑏 has infinite order.

Proof. By op(𝑏)>1, 𝑏₁∉𝑋₁. Since firstops(𝑏)=𝑔, var(𝑏)={𝑥₁} and Lemma 2.4(ii), we have 𝜎_𝑎,𝑏 has infinite order.

In Case (2.4), we have firstops(𝑎)=𝑔 and firstops(𝑏)=𝑓. Since op(𝑎)>1 and op(𝑏)>1, 𝑎₁∉𝑋₂ and 𝑏₁∉𝑋₁ or 𝑏₂∉𝑋₁. We consider the following three subcases: (2.4.1)𝑏₁∉𝑋₁; (2.4.2)𝑏₁=𝑥₁, op(𝑎₁)={𝑔}; (2.4.3)𝑏₁=𝑥₁, op(𝑎₁)≠{𝑔}.

Theorem 2.8. Let 𝑎∈𝑊_(2,1)(𝑋₂), 𝑏∈𝑊_(2,1)(𝑋₁) be such that op(𝑎)>1, op(𝑏)>1, var(𝑎)={𝑥₁}, var(𝑏)={𝑥₁}, and 𝑎=𝑔(𝑎₁), 𝑏=𝑓(𝑏₁,𝑏₂) for some 𝑎₁∈𝑊_(2,1)(𝑋₂), 𝑏₁,𝑏₂∈𝑊_(2,1)(𝑋₁). Then, the following hold: (i)if 𝑎 and 𝑏 satisfy (2.4.1) or (2.4.3), then 𝜎_𝑎,𝑏 has infinite order;(ii)if 𝑎 and 𝑏 satisfy (2.4.2), then the order of 𝜎_𝑎,𝑏 is less than or equal to 2.

Proof. (i) Assume that 𝑎 and 𝑏 satisfy (2.4.1). By Lemma 2.2, we have 𝜎^𝑛_𝑎,𝑏[𝑎₁]∉𝑋₂ for all 𝑛∈ℕ. Since var(𝜎_𝑎,𝑏[𝑎])={𝑥₁}, by Lemma 2.3  var(𝜎^𝑛_𝑎,𝑏[𝑎])={𝑥₁} for all 𝑛∈ℕ. Therefore, op𝜎^𝑘+1_𝑎,𝑏[𝑎]=op𝜎^𝑘_𝑎,𝑏𝑎𝜎_𝑎,𝑏𝑎₁,𝜎_𝑎,𝑏𝑎₂=op𝜎^𝑘_𝑎,𝑏[𝑎]𝜎^𝑘+1_𝑎,𝑏𝑎₁,𝜎^𝑘+1_𝑎,𝑏𝑎₂>op𝜎^𝑘_𝑎,𝑏[𝑎]. (2.9) This shows that op(𝜎^𝑛+1_𝑎,𝑏[𝑎])>op(𝜎^𝑛_𝑎,𝑏[𝑎]) for all 𝑛∈ℕ. Hence, 𝜎_𝑎,𝑏 has infinite order.
Assume that 𝑎 and 𝑏 satisfy (2.4.3). This case can be proved as in (2.4.1).
(ii) Assume that 𝑎 and 𝑏 satisfy (2.4.2). Then, 𝜎_𝑎,𝑏[𝑎]=𝑏 and 𝜎_𝑎,𝑏[𝑏]=𝑎. We have 𝜎³_𝑎,𝑏=𝜎_𝑎,𝑏 and 𝜎⁴_𝑎,𝑏=𝜎²_𝑎,𝑏. Hence, the order of 𝜎_𝑎,𝑏 is less than or equal to 3.

In Case (II-3), we have var(𝑎)={𝑥₂}, var(𝑏)={𝑥₁}. We consider the following four subcases: (3.1)firstops(𝑎)=𝑓, firstops(𝑏)=𝑓; (3.2)firstops(𝑎)=𝑔, firstops(𝑏)=𝑔; (3.3)firstops(𝑎)=𝑓, firstops(𝑏)=𝑔; (3.4)firstops(𝑎)=𝑔, firstops(𝑏)=𝑓.

In Case (3.1), we can separate into four subcases: (3.1.1)𝑎₂∉𝑋₂; (3.1.2)𝑎₂=𝑥₂, 𝑏₂=𝑥₁; (3.1.3)𝑎₂=𝑥₂, 𝑏₂∉𝑋₁, ops(Rp(𝑏))={𝑓}; (3.1.4)𝑎₂=𝑥₂,𝑏₂∉𝑋₁, ops(Rp(𝑏))≠{𝑓}.

Theorem 2.9. Let 𝑎∈𝑊_(2,1)(𝑋₂), 𝑏∈𝑊_(2,1)(𝑋₁) be such that op(𝑎)>1, op(𝑏)>1, var(𝑎)={𝑥₂}, var(𝑏)={𝑥₁}, and 𝑎=𝑓(𝑎₁,𝑎₂), 𝑏=𝑓(𝑏₁,𝑏₂) for some 𝑎₁,𝑎₂∈𝑊_(2,1)(𝑋₂), 𝑏₁,𝑏₂∈𝑊_(2,1)(𝑋₁). Then, the following hold: (i)if 𝑎 and 𝑏 satisfy (3.1.1) or (3.1.4), then 𝜎_𝑎,𝑏 has infinite order;(ii)if 𝑎 and 𝑏 satisfy (3.1.2) or (2.1.3), then the order of 𝜎_𝑎,𝑏 is less than or equal to 2.

Proof. (i) Assume that 𝑎 and 𝑏 satisfy (3.1.1). By Lemma 2.2, we have 𝜎^𝑛_𝑎,𝑏[𝑎₁]∉𝑋₂ for all 𝑛  in ℕ. Since var(𝜎_𝑎,𝑏[𝑎])={𝑥₂}, by Lemma 2.3  var(𝜎^𝑛_𝑎,𝑏[𝑎])={𝑥₂} for all 𝑛∈ℕ. Therefore, op𝜎^𝑘+1_𝑎,𝑏[𝑎]=op𝜎^𝑘_𝑎,𝑏𝑎𝜎_𝑎,𝑏𝑎₁,𝜎_𝑎,𝑏𝑎₂=op𝜎^𝑘_𝑎,𝑏[𝑎]𝜎^𝑘+1_𝑎,𝑏𝑎₁,𝜎^𝑘+1_𝑎,𝑏𝑎₂>op𝜎^𝑘_𝑎,𝑏[𝑎]. (2.10) This shows that op(𝜎^𝑛+1_𝑎,𝑏[𝑎])>op(𝜎^𝑛_𝑎,𝑏[𝑎]) for all 𝑛∈ℕ. Hence, 𝜎_𝑎,𝑏 has infinite order.
Assume that 𝑎 and 𝑏 satisfy (3.1.4). Then, 𝜎^𝑛_𝑎,𝑏[𝑎]=𝜎_𝑎,𝑏[𝑎] for all 𝑛∈ℕ. Since var(𝜎_𝑎,𝑏[𝑎])={𝑥₂} and 𝑏₂∉𝑋₁, we have op𝜎^𝑘+1_𝑎,𝑏[𝑏]=op𝜎^𝑘_𝑎,𝑏𝑏𝜎_𝑎,𝑏𝑏₁,𝜎_𝑎,𝑏𝑏₂=op𝜎^𝑘_𝑎,𝑏[𝑏]𝜎^𝑘+1_𝑎,𝑏𝑏₁,𝜎^𝑘+1_𝑎,𝑏𝑏₂>op𝜎^𝑘_𝑎,𝑏[𝑏]. (2.11) This implies that op(𝜎^𝑛+1_𝑎,𝑏[𝑏])>op(𝜎^𝑛_𝑎,𝑏[𝑏]) for all 𝑛∈ℕ. Hence, the order of 𝜎_𝑎,𝑏 is infinite.
(ii) Assume that 𝑎 and 𝑏 satisfy (3.1.2). Then, we get 𝜎_𝑎,𝑏[𝑎]=𝑎, 𝜎²_𝑎,𝑏[𝑏]=𝜎_𝑎,𝑏[𝑏] or 𝜎_𝑎,𝑏[𝑎]=𝑎, 𝜎_𝑎,𝑏[𝑏]=𝑏. In the first, the order of 𝜎_𝑎,𝑏 is equal to 2. For the latter, 𝜎_𝑎,𝑏 is idempotent.
Assume that 𝑎 and 𝑏 satisfy (3.1.3). Then, 𝜎_𝑎,𝑏[𝑎]=𝑎, 𝜎²_𝑎,𝑏[𝑏]=𝜎_𝑎,𝑏[𝑏], which can be proved the same way as the first case of (ii).

In Case (3.2), we have firstops(𝑎)=𝑔 and firstops(𝑏)=𝑔. Since op(𝑎)>1 and op(𝑏)>1, 𝑎₁∉𝑋₂ and 𝑏₁∉𝑋₁. Then, we obtain the following.

Theorem 2.10. Let 𝑎∈𝑊_(2,1)(𝑋₂), 𝑏∈𝑊_(2,1)(𝑋₁) be such that op(𝑎)>1, op(𝑏)>1, var(𝑎)={𝑥₂}, 𝑎=𝑔(𝑎₁) and 𝑏=𝑔(𝑏₁) for some 𝑎₁∈𝑊_(2,1)(𝑋₂)⧵𝑋₂, 𝑏₁∈𝑊_(2,1)(𝑋₁)⧵𝑋₁. Then 𝜎_𝑎,𝑏 has infinite order.

Proof. Since var(𝑎)={𝑥₂} and 𝑏₁∉𝑋₁, we have op𝜎^𝑘+1_𝑎,𝑏[𝑏]=op𝜎^𝑘_𝑎,𝑏𝑏𝜎_𝑎,𝑏𝑏₁,𝜎_𝑎,𝑏𝑏₂=op𝜎^𝑘_𝑎,𝑏[𝑏]𝜎^𝑘+1_𝑎,𝑏𝑏₁,𝜎^𝑘+1_𝑎,𝑏𝑏₂>op𝜎^𝑘_𝑎,𝑏[𝑏]. (2.12) Hence, the order of 𝜎_𝑎,𝑏 is infinite.

In Case (3.3), we have firstops(𝑎)=𝑓 and firstops(𝑏)=𝑔. Since op(𝑏)>1, 𝑏₁∉𝑋₁. Then, we obtain the following.

Theorem 2.11. Let 𝑎∈𝑊_(2,1)(𝑋₂), 𝑏∈𝑊_(2,1)(𝑋₁) be such that op(𝑎)>1, op(𝑏)>1, var(𝑎)={𝑥₂}, 𝑎=𝑓(𝑎₁,𝑎₂) and 𝑏=𝑔(𝑏₁) for some 𝑎₁,𝑎₂∈𝑊_(2,1)(𝑋₂), 𝑏₁∈𝑊_(2,1)(𝑋₁)⧵𝑋₁. Then 𝜎_𝑎,𝑏 has infinite order.

Proof. Since 𝑏₁∉𝑋₁  firstops(𝑏)=𝑔 and Lemma 2.4(ii), we have 𝜎_𝑎,𝑏 has infinite order.

In Case (3.4), we have firstops(𝑎)=𝑔 and firstops(𝑏)=𝑓. Since op(𝑎)>1 and op(𝑏)>1, 𝑎₁∉𝑋₂ and 𝑏₁∉𝑋₁. We have the following result.

Theorem 2.12. Let 𝑎∈𝑊_(2,1)(𝑋₂), 𝑏∈𝑊_(2,1)(𝑋₁) be such that op(𝑎)>1, op(𝑏)>1, var(𝑎)={𝑥₂}, var(𝑏)={𝑥₁}, and 𝑎=𝑔(𝑎₁), 𝑏=𝑓(𝑏₁,𝑏₂) for some 𝑎₁∈𝑊_(2,1)(𝑋₂), 𝑏₁,𝑏₂∈𝑊_(2,1)(𝑋₁). Then, 𝜎_𝑎,𝑏 has infinite order.

Proof. Since op(𝑎)>1, 𝑎₁∉𝑋₂. We have, by Lemma 2.2, 𝜎^𝑛_𝑎,𝑏[𝑎₁]∉𝑋₂ for all  𝑛in ℕ. Since var(𝜎_𝑎,𝑏[𝑎])={𝑥₂}, by Lemma 2.3  var(𝜎^𝑛_𝑎,𝑏[𝑎])={𝑥₂} for all 𝑛∈ℕ. We obtain op𝜎^𝑘+1_𝑎,𝑏[𝑎]=op𝜎^𝑘_𝑎,𝑏𝑎𝜎_𝑎,𝑏𝑎₁,𝜎_𝑎,𝑏𝑎₂=op𝜎^𝑘_𝑎,𝑏[𝑎]𝜎^𝑘+1_𝑎,𝑏𝑎₁,𝜎^𝑘+1_𝑎,𝑏𝑎₂>op𝜎^𝑘_𝑎,𝑏[𝑎]. (2.13) Hence, 𝜎_𝑎,𝑏 has infinite order.

3. Case III: op(𝑎)=1 and op(𝑏)>1

We consider three subcases: (III-1)var(𝑎)=𝑋₂, var(𝑏)={𝑥₁}; (III-2)var(𝑎)={𝑥₁}, var(𝑏)={𝑥₁}; (III-3)var(𝑎)={𝑥₂}, var(𝑏)={𝑥₁}.

In Case (III-1), we separate into the following subcases: (1)𝑎=𝑓(𝑥₁,𝑥₂),firstops(𝑏)=𝑓;(2)𝑎=𝑓(𝑥₁,𝑥₂),firstops(𝑏)=𝑔;(3)𝑎=𝑓(𝑥₂,𝑥₁),firstops(𝑏)=𝑓;(4)𝑎=𝑓(𝑥₂,𝑥₁),firstops(𝑏)=𝑔.

We have the following results. Subcases (2) and (4) give infinite order, by Lemma 2.4(ii). The next proposition deal with subcases (1) and (3).

Proposition 3.1. Let 𝑏=𝑓(𝑏₁,𝑏₂) for some 𝑏₁,𝑏₂∈𝑊_(2,1)(𝑋₁) be such that var(𝑏)={𝑥₁} and op(𝑏)>1. Then, the following hold: (i)if 𝑎=𝑓(𝑥₁,𝑥₂), then the order of 𝜎_𝑎,𝑏 is equal to 1 or is infinite;(ii)if 𝑎=𝑓(𝑥₂,𝑥₁), then the order of 𝜎_𝑎,𝑏 is less than or equal to 3 or infinite.

Proof. (i) Assume that 𝑎=𝑓(𝑥₁,𝑥₂). Then, 𝜎_𝑎,𝑏[𝑎]=𝑎. We consider the following.
(i′) If ops(𝑏)={𝑓}, then 𝜎_𝑎,𝑏[𝑏]=𝑏. We get that 𝜎_𝑎,𝑏 is idempotent.
(i′′) If 𝑔∈ops(𝑏), suppose that 𝑔∈ops(𝑏₁). Let 𝑘∈ℕ. Then op(𝜎^𝑘+1_𝑎,𝑏[𝑏₁])≥op(𝜎^𝑘_𝑎,𝑏[𝑏₁]).
Now, op𝜎^𝑘+1_𝑎,𝑏[𝑏]=op𝜎^𝑘_𝑎,𝑏𝜎_𝑎,𝑏[𝑏𝜎]=op^𝑘_a,𝑏𝜎_𝑎,𝑏𝑓𝜎_𝑎,𝑏𝑏₁,𝜎_𝑎,𝑏𝑏₂=op𝜎^𝑘_𝑎,𝑏𝜎_𝑎,𝑏𝑓𝜎^𝑘+1_𝑎,𝑏𝑏₁,𝜎^𝑘+1_𝑎,𝑏𝑏₂>op𝜎^𝑘_𝑎,𝑏𝑓𝜎^𝑘_𝑎,𝑏𝑏₁,𝜎^𝑘_𝑎,𝑏𝑏₂=op𝜎^𝑘_𝑎,𝑏[𝑏]. (3.1) A similar argument works for 𝑔∈ops(𝑏₂). This shows that the order of 𝜎_𝑎,𝑏 is infinite.
(ii) If ops(𝑏)={𝑓}, then 𝜎¹_𝑎,𝑏[𝑎]=𝜎³_𝑎,𝑏[𝑎] and 𝜎_𝑎,𝑏[𝑏]=𝜎³_𝑎,𝑏[𝑏]. This gives 𝜎⁴_𝑎,𝑏=𝜎²_𝑎,𝑏. Then, the order of 𝜎_𝑎,𝑏 is less than or equal to 3. If 𝑔∈ops(𝑏), then it can be proved as in (i′′) that the order is infinite.

In Case (III-2), we have var(𝑎)={𝑥₁}, var(𝑏)={𝑥₁}. We consider the following subcases: (3.2.1)𝑎=𝑓(𝑥₁,𝑥₁), firstops(𝑏)=𝑓; (3.2.1.1)𝑏₁=𝑥₁;(3.2.1.2)𝑏₁≠𝑥₁, 𝑔∈ops(Lp(𝑏));(3.2.1.3)𝑏₁≠𝑥₁, 𝑔∉ops(Lp(𝑏)); (3.2.2)𝑎=𝑔(𝑥₁), firstops(𝑏)=𝑔, then 𝑏₁∉𝑋₁;(3.2.3)𝑎=𝑓(𝑥₁,𝑥₁), firstops(𝑏)=𝑔, then 𝑏₁∉𝑋₁;(3.2.4)𝑎=𝑔(𝑥₁), firstops(𝑏)=𝑓;(3.2.4.1)𝑏₁=𝑥₁;(3.2.4.2)𝑏₁≠𝑥₁.

Proposition 3.2. Let 𝑎=𝑓(𝑥₁,𝑥₁), firstops(𝑏)=𝑓, op(𝑏)>1, and var(𝑏)={𝑥₁}. Then, the following hold: (i)if 𝑏 satisfies (3.2.1.1) or (3.2.1.3), then the order of 𝜎_𝑎,𝑏 is less than or equal to 2;(ii)if 𝑏 satisfies (3.2.1.2), then the order of 𝜎_𝑎,𝑏 is infinite.

Proof. (i) If 𝑎 and 𝑏 satisfy (3.2.1.1), then 𝜎_𝑎,𝑏[𝑎]=𝑎=𝜎_𝑎,𝑏[𝑏]. Assume that 𝑎 and 𝑏 satisfy (3.2.1.3). Since 𝑎=𝑓(𝑥₁,𝑥₁) and 𝑔∉𝜎_𝑎,𝑏[𝑏], we have 𝜎²_𝑎,𝑏[𝑏]=𝜎_𝑎,𝑏[𝑏]. Clearly, 𝜎²_𝑎,𝑏[𝑎]=𝑎. Then, 𝜎³_𝑎,𝑏=𝜎²_𝑎,𝑏. Hence, the order of 𝜎_𝑎,𝑏 is less than or equal to 2.
(ii) Assume that 𝑎 and 𝑏 satisfy (3.2.1.2). Since 𝑎=𝑓(𝑥₁,𝑥₁), firstops(𝑏)=𝑓 and 𝑔∈ops(Lp(𝑏)), it follows that 𝜎_𝑎,𝑏[𝑏]=𝑓(𝑥₁,𝑥₁)(𝑏(𝑣),𝑏(𝑣^′)) for some 𝑣,𝑣^′∈𝑊_(2,1)(𝑋₁). For 𝑛∈ℕ, we have op𝜎^𝑛+1_𝑎,𝑏[𝑏]=op𝜎^𝑛𝜎_𝑎,𝑏[𝑏𝜎]=op^𝑛_𝑎,𝑏𝑓𝑥₁,𝑥₁𝑏(𝑣),𝑏𝑣^′=op𝑓𝑥₁,𝑥₁𝜎^𝑛_𝑎,𝑏[𝑏]𝜎^𝑛_𝑎,𝑏[𝑣],𝜎^𝑛_𝑎,𝑏[𝑏]𝜎^𝑛_𝑎,𝑏𝑣^′>op𝜎^𝑛_𝑎,𝑏[𝑏]. (3.2) Hence, the order of 𝜎_𝑎,𝑏 is infinite.

Proposition 3.3. Let 𝑎=𝑔(𝑥₁) or 𝑎=𝑓(𝑥₁,𝑥₁), firstops(𝑏)=𝑔, op(𝑏)>1, and var(𝑏)={𝑥₁}. Then, the order of 𝜎_𝑎,𝑏 is infinite.

Proof. Assume that 𝑎=𝑔(𝑥₁). Since firstops(𝑏)=𝑔, 𝑎=𝑔(𝑥₁), 𝜎_𝑎,𝑏[𝑏]=𝑏(𝑔(𝑥₁)). For any natural number 𝑛, op(𝜎^𝑛+1_𝑎,𝑏[𝑏])>op(𝜎^𝑛_𝑎,𝑏[𝑏]) makes 𝜎_𝑎,𝑏 have infinite order.
Assume that 𝑎=𝑓(𝑥₁,𝑥₁). Since firstops(𝑏)=𝑔, 𝑎=𝑓(𝑥₁,𝑥₁), 𝜎_𝑎,𝑏[𝑏]=𝑏(𝑓(𝑥₁,𝑥₁)). For 𝑛∈ℕ, a similar proof to those above gives op(𝜎^𝑛+1_𝑎,𝑏[𝑏])>op(𝜎^𝑛_𝑎,𝑏[𝑏]). Hence, the order of 𝜎_𝑎,𝑏 is infinite.

Proposition 3.4. Let 𝑎=𝑔(𝑥₁), firstops(𝑏)=𝑓, op(𝑏)>1 and var(𝑏)={𝑥₁}. Then the following hold: (i)if 𝑏 satisfies (3.2.4.1), then the order of 𝜎_𝑎,𝑏 is equal to 2;(ii)if 𝑏 satisfies (3.2.4.2), then the order of 𝜎_𝑎,𝑏 is infinite.

Proof. (i) If 𝑎 and 𝑏 satisfy (3.2.4.1), then 𝜎_𝑎,𝑏[𝑎]=𝑏 and 𝜎_𝑎,𝑏[𝑏]=𝑎. Hence, the order of 𝜎_𝑎,𝑏 is equal to 2.
(ii) Assume that 𝑎 and 𝑏 satisfy (3.2.4.2). Since 𝑎=𝑔(𝑥₁), firstops(𝑏)=𝑓, it follows that 𝜎_𝑎,𝑏[𝑏]=𝑔(𝑥₁)𝑏. For 𝑛∈ℕ. As in the proof of Proposition 3.3, we have op(𝜎^𝑛+1_𝑎,𝑏[𝑏])>op(𝜎^𝑛_𝑎,𝑏[𝑏]). Hence, the order of 𝜎_𝑎,𝑏 is infinite.

In Case (III-3), we have var(𝑎)={𝑥₂}, var(𝑏)={𝑥₁}. We consider the following subcases: (3.3.1)𝑎=𝑓(𝑥₂,𝑥₂), firstops(𝑏)=𝑓;(3.3.1.1)𝑏₂=𝑥₁;(3.3.1.2)𝑏₂≠𝑥₁; (3.3.2)𝑎=𝑔(𝑥₂), firstops(𝑏)=𝑔.(3.3.3)𝑎=𝑓(𝑥₂,𝑥₂),firstops(𝑏)=𝑔.(3.3.4)𝑎=𝑔(𝑥₂),firstops(𝑏)=𝑓;(3.3.4.1)𝑏₂=𝑥₁;(3.3.4.2)𝑏₂≠𝑥₁.

Proposition 3.5. Let 𝑎=𝑓(𝑥₂,𝑥₂), firstops(𝑏)=𝑓, op(𝑏)>1, and var(𝑏)={𝑥₁}. Then, the following hold: (i)if 𝑏 satisfies (3.3.1.1), then the order of 𝜎_𝑎,𝑏 is equal to 2;(ii)if 𝑏 satisfies (3.3.1.2), then the order of 𝜎_𝑎,𝑏 is infinite.

Proof. (i) If 𝑎 and 𝑏 satisfy (3.3.1.1), then 𝜎_𝑎,𝑏[𝑎]=𝑎 and 𝜎²_𝑎,𝑏[𝑏]=𝜎_𝑎,𝑏[𝑏]. Hence the order of 𝜎_𝑎,𝑏 is equal to 2.
(ii) This can be proved in the same way as Proposition 3.2 (ii).

Proposition 3.6. Let 𝑎=𝑔(𝑥₂) or 𝑎=𝑓(𝑥₂,𝑥₂), firstops(𝑏)=𝑔, op(𝑏)>1, and var(𝑏)={𝑥₁}. Then, the order of 𝜎_𝑎,𝑏 is infinite.

Proof. Assume that 𝑎=𝑔(𝑥₂). Since firstops(𝑏)=𝑔, 𝑎=𝑔(𝑥₂), 𝜎_𝑎,𝑏[𝑏]=𝑏(𝑔(𝑥₂)). The same argument works as in Proposition 3.3. Hence, the order of 𝜎_𝑎,𝑏 is infinite. The case that 𝑎=𝑓(𝑥₂,𝑥₂) can be proved similarly.

Proposition 3.7. Let 𝑎=𝑔(𝑥₂), firstops(𝑏)=𝑓, op(𝑏)>1, and var(𝑏)={𝑥₁}. Then, the following hold: (i)if 𝑏 satisfies (3.3.4.1), then the order of 𝜎_𝑎,𝑏 is equal to 2;(ii)if 𝑏 satisfies (3.3.4.2), then the order of 𝜎_𝑎,𝑏 is infinite.

Proof. (i) If 𝑎 and 𝑏 satisfy (3.3.4.1), then 𝜎²_𝑎,𝑏[𝑎]=𝑎 and 𝜎²_𝑎,𝑏[𝑏]=𝑏. Hence, the order of 𝜎_𝑎,𝑏 is equal to 2.
(ii) Assume 𝑎 and 𝑏 satisfy (3.3.4.2), Since 𝑏=𝑔(𝑥₂), firstops(𝑏)=𝑓, 𝑏₂∉𝑋₂. By Lemma 2.2, we have 𝜎^𝑛_𝑎,𝑏[𝑏₁]∉𝑋₂. Then, op𝜎_𝑎,𝑏[𝑏]=op𝜎_𝑎,𝑏𝑓𝑏₁,𝑏₂=op𝜎_𝑎,𝑏𝑓𝜎_𝑎,𝑏𝑏₁,𝜎_𝑎,𝑏𝑏₂>op𝑓𝑏₁,𝑏₂=op(𝑏). (3.3) This shows that, op(𝜎_𝑎,𝑏[𝑏])>op(𝑏) for all 𝑎∈𝑊_(2,1)(𝑋₁)⧵𝑋₁. Consequently, op(𝜎^𝑘+1_𝑎,𝑏[𝑏])=op(𝜎_𝑎,𝑏[𝜎^𝑘_𝑎,𝑏[𝑏]]) for all 𝑘∈ℕ. Then, order of 𝜎_𝑎,𝑏 is infinite.

4. Case IV: op(𝑎)>1 and op(𝑏)=1

We consider three subcases: (IV-1)var(𝑎)=𝑋₂, var(𝑏)={𝑥₁}; (IV-2)var(𝑎)={𝑥₁}, var(𝑏)={𝑥₁};(IV-3)var(𝑎)={𝑥₂}, var(𝑏)={𝑥₁}.

In Case (IV-1), we separate into four cases: (4.1.1)𝑏=𝑓(𝑥₁,𝑥₁), firstops(𝑎)=𝑓; (4.1.2)𝑏=𝑓(𝑥₁,𝑥₁), firstops(𝑎)=𝑔; (4.1.3)𝑏=𝑔(𝑥₁), firstops(𝑎)=𝑓; (4.1.4)𝑏=𝑔(𝑥₁), firstops(𝑎)=𝑔.

Theorem 4.1. Let 𝑎∈𝑊_(2,1)(𝑋₂), 𝑏∈𝑊_(2,1)(𝑋₁) be such that op(𝑎)>1, op(𝑏)=1, var(𝑎)={𝑥₁,𝑥₂}, var(𝑏)={𝑥₁}. Then, the following hold: (i)if 𝑏 satisfies (4.1.1) or (4.1.2) or (4.1.3), then the order of 𝜎_𝑎,𝑏 is infinite;(ii)if 𝑏 satisfies (4.1.4), then the order of 𝜎_𝑎,𝑏 is equal to 1 or is infinite.

Proof. (i) If 𝑎 and 𝑏 satisfy (4.1.1), then we let 𝑎=𝑓(𝑎₁,𝑎₂) for some 𝑎₁,𝑎₂∈𝑊_(2,1)(𝑋₂). Since op(𝑎)>1, we may assume that 𝑎₁∉𝑋₂. By Lemma 2.2, we have 𝜎^𝑛_𝑎,𝑏[𝑎₁]∉𝑋₂. For any natural number 𝑛, op(𝜎^𝑛+1_𝑎,𝑏[𝑎])>op(𝜎^𝑛_𝑎,𝑏[𝑎]) makes 𝜎_𝑎,𝑏 have infinite order. Cases (4.1.2) and (4.1.3) can be proved similarly.
(ii) Assume that 𝑎 and 𝑏 satisfy (4.1.4). If ops(𝑎₁)={𝑔}, then 𝜎_𝑎,𝑏[𝑎]=𝑎 and 𝜎_𝑎,𝑏[𝑏]=𝑏. Hence, 𝜎_𝑎,𝑏 is idempotent. The case that op(𝑎₁)≠{𝑔} can be proved as in (i).

In Case (IV-2), we have var(𝑎)={𝑥₁}, var(𝑏)={𝑥₁}. If 𝑎=𝑔(𝑎₁) for some term 𝑎₁, then op(𝑎)>1 means that 𝑎₁ is not variable.

Proposition 4.2. Let 𝑏=𝑓(𝑥₁,𝑥₁) or 𝑏=𝑔(𝑥₁), op(𝑎)>1, 𝑎=𝑔(𝑎₁) for some 𝑎₁ and var(𝑎)={𝑥₁}=var(𝑏). Then, the order of 𝜎_𝑎,𝑏 is infinite.

Proof. Assume op(𝑎)>1, 𝑎=𝑔(𝑎₁) for some 𝑎₁ and var(𝑎)={𝑥₁}=var(𝑏). Since 𝑎₁∉𝑋₂. By Lemma 2.2 we have 𝜎^𝑛_𝑎,𝑏[𝑎₁]∉𝑋₂. It can be proved, as in the proof of Theorem 4.1(i), that the order of 𝜎_𝑎,𝑏 is infinite.

In Case (IV-3), we have var(𝑎)={𝑥₂}, var(𝑏)={𝑥₁}. If firstops(𝑎)=𝑔, then 𝑎₂∉𝑋₂. We consider the following cases: (4.3.1)𝑎₂∉𝑋₂, 𝑏=𝑓(𝑥₁,𝑥₁); (4.3.2)𝑎₂∉𝑋₂, 𝑏=𝑔(𝑥₁).

Proposition 4.3. Let 𝑏=𝑓(𝑥₁,𝑥₁), op(𝑎)>1, and var(𝑎)={𝑥₂}, var(𝑏)={𝑥₁}. If 𝑎 and 𝑏 satisfy (4.3.1), then the order of 𝜎_𝑎,𝑏 is infinite.

Proof. The case (4.3.1) can be verified as in Proposition 4.2.

Proposition 4.4. Let 𝑏=𝑔(𝑥₁), op(𝑎)>1, and var(𝑎)={𝑥₂}, var(𝑏)={𝑥₁}. If 𝑎 and 𝑏 satisfy (4.3.2), then the order of 𝜎_𝑎,𝑏 is equal to 1 or is infinite.

Proof. Assume that 𝑎 and 𝑏 satisfy (4.3.2). If 𝑏=𝑔(𝑥₁), firstops(𝑎)=𝑔, and ops(𝑎₂)={𝑔}, then 𝜎_𝑎,𝑏[𝑎]=𝑎 and 𝜎_𝑎,𝑏[𝑏]=𝑏. Hence, 𝜎_𝑎,𝑏 is idempotent. Assume that 𝑎₂∉𝑋₂. This case can be proved as in Proposition 4.3.

5. Case V: op(𝑎)>1 and op(𝑏)=0

In this case, we have 𝑎=𝑓(𝑎₁,𝑎₂) or 𝑎=𝑔(𝑎₁), 𝑏=𝑥₁. We consider three subcases: (V-1)var(𝑎)=𝑋₂, 𝑏=𝑥₁; (V-2)var(𝑎)={𝑥₁}, 𝑏=𝑥₁; (V-3)var(𝑎)={𝑥₂}, 𝑏=𝑥₁.

In Case (V-1), we separate into the following subcases: (5.1.1)firstops(𝑎)=𝑓, 𝑎₁∉𝑋₂, ops(Lp(𝑎))≠{𝑔}; (5.1.2)firstops(𝑎)=𝑓, 𝑎₁∉𝑋₂, ops(Lp(𝑎))={𝑔}; (5.1.3)firstops(𝑎)=𝑓, 𝑎₁∈𝑋₂, ops(Rp(𝑎))≠{𝑔}; (5.1.4)firstops(𝑎)=𝑓, 𝑎₁∈𝑋₂, ops(Rp(𝑎))={𝑔}; (5.1.5)firstops(𝑎)=𝑔, ops(𝑎₁)={𝑔}; (5.1.6)firstops(𝑎)=𝑔, ops(𝑎₁)≠{𝑔}.

Proposition 5.1. Let 𝑎=𝑓(𝑎₁,𝑎₂), op(𝑎)>1, op(𝑏)=0, var(𝑎)={𝑥₁,𝑥₂}, and 𝑏=𝑥₁. Then, the following hold: (i)if 𝑎 satisfies (5.1.1) or (5.1.3), then the order of 𝜎_𝑎,𝑏 is infinite;(ii)if 𝑎 satisfies (5.1.2) or (5.1.4), then the order of 𝜎_𝑎,𝑏 is less than or equal to 2.

Proof. (i) Assume 𝑎 and 𝑏 satisfy (5.1.1). Then, firstops(𝑎)=𝑓, var(𝑎)={𝑥₁,𝑥₂}, and 𝑓∈ops(Lp(𝑎)). Then, op(𝑎)<op(𝜎_𝑎,𝑏[𝑎]). For any natural number 𝑛, op(𝜎^𝑛+1_𝑎,𝑏[𝑎])>op(𝜎^𝑛_𝑎,𝑏[𝑎]) makes 𝜎_𝑎,𝑏 have infinite order. A similar argument works for (5.1.3).
(ii) Assume 𝑎 and 𝑏 satisfy (5.1.2). Then, firstops(𝑎)=𝑓, var(𝑎)={𝑥₁,𝑥₂}, and ops(Lp(𝑎))={𝑔}. If leftmost(𝑎₁)=𝑥₁, then 𝜎_𝑎,𝑏[𝑎]=𝑎 and 𝜎_𝑎,𝑏[𝑏]=𝑏. Hence, 𝜎_𝑎,𝑏 is idempotent. If leftmost(𝑎₁)=𝑥₂, then 𝜎²_𝑎,𝑏[𝑎]=𝜎_𝑎,𝑏[𝑎] and 𝜎_𝑎,𝑏[𝑏]=𝑏. Hence, the order of 𝜎_𝑎,𝑏 is equal to 2.
Assume 𝑎 and 𝑏 satisfy (5.1.4). Then, firstops(𝑎)=𝑓, var(𝑎)={𝑥₁,𝑥₂} and ops(Rp(𝑎))={𝑔}. If rightmost(𝑎₂)=𝑥₂, then 𝜎_𝑎,𝑏[𝑎]=𝑎 and 𝜎_𝑎,𝑏[𝑏]=𝑏. Hence, 𝜎_𝑎,𝑏 is idempotent. If leftmost(𝑎₂)=𝑥₁, then 𝜎²_𝑎,𝑏[𝑎]=𝜎_𝑎,𝑏[𝑎] and 𝜎_𝑎,𝑏[𝑏]=𝑏. Hence, the order of 𝜎_𝑎,𝑏 is equal to 2.

Proposition 5.2. Let 𝑎=𝑔(𝑎₁), op(𝑎)>1, op(𝑏)=0, var(𝑎)={𝑥₁,𝑥₂}, and 𝑏=𝑥₁. If 𝑎 satisfies (5.1.5) or (5.1.6), then, the order of 𝜎_𝑎,𝑏 is less than or equal to 2.

Proof. Assume that 𝑎 and 𝑏 satisfy (5.1.5). Then, firstops(𝑎₁)=𝑔 and ops(𝑎₁)={𝑔}. If leftmost(𝑎₁)=𝑥₁, then 𝜎_𝑎,𝑏[𝑎]=𝑏 and 𝜎_𝑎,𝑏[𝑏]=𝑏. Hence, the order of 𝜎_𝑎,𝑏 is equal to 2. If leftmost(𝑎₁)=𝑥₂, then 𝜎²_𝑎,𝑏[𝑎]=𝜎_𝑎,𝑏[𝑎] and 𝜎_𝑎,𝑏[𝑏]=𝑏. Again, the order of 𝜎_𝑎,𝑏 is equal to 2.
Assume that 𝑎 and 𝑏 satisfy (5.1.6). Then firstops(𝑎₁)=𝑓 and 𝑓∈ops(𝑎₁). If leftmost(𝑎₁)=𝑥₁, then 𝜎_𝑎,𝑏[𝑎]=𝑎 and 𝜎_𝑎,𝑏[𝑏]=𝑏. Hence, 𝜎_𝑎,𝑏 is idempotent. If leftmost(𝑎₁)=𝑥₂, then 𝜎²_𝑎,𝑏[𝑎]=𝜎_𝑎,𝑏[𝑎] and 𝜎_𝑎,𝑏[𝑏]=𝑏. Hence, the order of 𝜎_𝑎,𝑏 is equal to 2.

In Case (V-2), we have var(𝑎)={𝑥₁}, var(𝑏)={𝑥₁}. We separate this case into following subcases: (5.2.1)firstops(𝑎)=𝑓, 𝑎₁=𝑥₁; (5.2.2)firstops(𝑎)=𝑓, 𝑎₁∉𝑋₂, ops(Lp(𝑎))={𝑔}; (5.2.3)firstops(𝑎)=𝑓, 𝑎₁∉𝑋₂, ops(Lp(𝑎))≠{𝑔}; (5.2.4)firstops(𝑎)=𝑔, ops(𝑎₁)={𝑔}; (5.2.5)firstops(𝑎)=𝑔, ops(𝑎₁)≠{𝑔}.

Proposition 5.3. Let 𝑎=𝑓(𝑎₁,𝑎₂), op(𝑎)>1, op(𝑏)=0, var(𝑎)={𝑥₁}, and 𝑏=𝑥₁. Then, the following hold: (i)if 𝑎 satisfies (5.2.1) or (5.2.2), then the order of 𝜎_𝑎,𝑏 is equal to 1;(ii)if 𝑎 satisfies (5.2.3), then the order of 𝜎_𝑎,𝑏 is infinite.

Proof. (i) Assume that 𝑎 and 𝑏 satisfy (5.2.1). Since firstops(𝑎₁)=𝑓 and var(𝑎)={𝑥₁}, we have 𝜎_𝑎,𝑏[𝑎]=𝑎. Since 𝑏=𝑥₁, 𝜎_𝑎,𝑏[𝑏]=𝑏. Hence, 𝜎_𝑎,𝑏 is idempotent.
Assume 𝑎 and 𝑏 satisfy (5.2.2). Since ops(𝑎₁)=𝑔 and var(𝑎)={𝑥₁}, we have 𝜎_𝑎,𝑏[𝑎]=𝑎 and 𝜎_𝑎,𝑏[𝑏]=𝑏. Hence, 𝜎_𝑎,𝑏 is idempotent.
(ii) Assume 𝑎 and 𝑏 satisfy (5.2.3). Since firstops(𝑎₁)=𝑓 and var(𝑎)={𝑥₁}, we have op(𝜎^𝑘+1_𝑎,𝑏[𝑎₁])≥op(𝜎^𝑘_𝑎,𝑏[𝑎₁]) for all 𝑘∈ℕ. For any natural number 𝑛, op(𝜎^𝑛+1_𝑎,𝑏[𝑎])>op(𝜎^𝑛_𝑎,𝑏[𝑎]) makes 𝜎_𝑎,𝑏 have infinite order.

Proposition 5.4. Let 𝑎=𝑔(𝑎₁), op(𝑎)>1, op(𝑏)=0, var(𝑎)={𝑥₁}, and 𝑏=𝑥₁. Then, the following hold: (i)if 𝑎 satisfies (5.2.4), then the order of 𝜎_𝑎,𝑏 is equal to 2;(ii)if 𝑎 satisfies (5.2.5), then the order of 𝜎_𝑎,𝑏 is equal to 1 or is infinite.

Proof. (i) Assume 𝑎 and 𝑏 satisfy (5.2.4). Then, 𝜎_𝑎,𝑏[𝑎]=𝑏=𝜎_𝑎,𝑏[𝑏]. Hence, the order of 𝜎_𝑎,𝑏 is equal to 2.
(ii) Assume 𝑎 and 𝑏 satisfy (5.2.5). If there is only one 𝑓∈ops(𝑎₁), then 𝜎_𝑎,𝑏[𝑎]=𝑥₁(𝑥₁(⋯(𝑎)))=𝑎 and 𝜎_𝑎,𝑏[𝑏]=𝑏. Hence, 𝜎_𝑎,𝑏 is idempotent.
Assume that the symbol 𝑓 occurs more than twice in term 𝑎₁. Then, op(𝜎^𝑘+1_𝑎,𝑏[𝑎₁])≥op(𝜎^𝑘_𝑎,𝑏[𝑎₁]) for all 𝑘∈ℕ. For any natural number 𝑛, op(𝜎^𝑛+1_𝑎,𝑏[𝑎])>op(𝜎^𝑛_𝑎,𝑏[𝑎]) makes 𝜎_𝑎,𝑏 have infinite order.

In Case (V-3), we have var(𝑎)={𝑥₂}, 𝑏=𝑥₁. We separate into the following subcases: (5.3.1)firstops(𝑎)=𝑓, 𝑎₂=𝑥₂; (5.3.2)firstops(𝑎)=𝑓, 𝑎₂∉𝑋₂, ops(𝑎₂)={𝑔}; (5.3.3)firstops(𝑎)=𝑓, 𝑎₂∉𝑋₂, ops(𝑎₂)≠{𝑔}; (5.3.4)firstops(𝑎)=𝑔, ops(𝑎₁)={𝑔}; (5.3.5)firstops(𝑎)=𝑔, ops(𝑎₁)≠{𝑔}.

Proposition 5.5. Let 𝑎=𝑓(𝑎₁,𝑎₂), op(𝑎)>1, op(𝑏)=0, var(𝑎)={𝑥₂}, and 𝑏=𝑥₁. Then, the following hold: (i)if 𝑎 satisfies (5.3.1) or (5.3.2), then the order of 𝜎_𝑎,𝑏 is equal to 1;(ii)if 𝑎 satisfies (5.3.3), then the order of 𝜎_𝑎,𝑏 is infinite.

Proof. This can be proved similarly to the proof of Proposition 5.3.

Proposition 5.6. Let 𝑎=𝑔(𝑎₁), op(𝑎)>1, op(𝑏)=0, var(𝑎)={𝑥₂}, and 𝑏=𝑥₁. Then, the following hold: (i)if 𝑎 satisfies (5.3.4), then the order of 𝜎_𝑎,𝑏 is equal to 2;(ii)if 𝑎 satisfies (5.3.5), then the order of 𝜎_𝑎,𝑏 is equal to 1 or is infinite.

Proof. This can be proved similarly to the proof of Proposition 5.4.

6. Case VI: op(𝑎)=0 and op(𝑏)>1

In this case, we consider the following cases: (6.1)𝑎=𝑥₁;(6.1.1)firstops(𝑏)=𝑓, 𝑏₁=𝑥₁;(6.1.2)firstops(𝑏)=𝑓, 𝑏₁∉𝑋₁, ops(Lp(𝑏))={𝑓};(6.1.3)firstops(𝑏)=𝑓, 𝑏₁∉𝑋₁, ops(Lp(𝑏))≠{𝑓};(6.1.4)firstops(𝑏)=𝑔, ops(𝑏₁)={𝑓};(6.1.5)firstops(𝑏)=𝑔, ops(𝑏₁)≠{𝑓}; (6.2)𝑎=𝑥₂;(6.2.1)firstops(𝑏)=𝑓, 𝑏₂=𝑥₁;(6.2.2)firstops(𝑏)=𝑓, 𝑏₂∉𝑋₁, ops(Rp(𝑏))={𝑓};(6.2.3)firstops(𝑏)=𝑓, 𝑏₂∉𝑋₁, ops(Rp(𝑏))≠{𝑓};(6.2.4)firstops(𝑏)=𝑔, ops(𝑏₁)={𝑓};(6.2.5)firstops(𝑏)=𝑔, ops(𝑏₁)≠{𝑓}.

Proposition 6.1. Let 𝑎=𝑥₁, 𝑏=𝑓(𝑏₁,𝑏₂), op(𝑏)>1, and var(𝑏)={𝑥₁}. Then, the following hold: (i)if 𝑏 satisfies (6.1.1) or (6.1.2), then the order of 𝜎_𝑎,𝑏 is equal to 2;(ii)if 𝑏 satisfies (6.1.3), then the order of 𝜎_𝑎,𝑏 is equal to 1 or is infinite.

Proof. (i) If 𝑎 and 𝑏 satisfy (6.1.1) or (6.1.2), then 𝜎_𝑎,𝑏[𝑎]=𝑎 and 𝜎_𝑎,𝑏[𝑏]=𝑎. Hence, the order of 𝜎_𝑎,𝑏 is equal to 2.
(ii) Assume 𝑎 and 𝑏 satisfy (6.1.3). If there is only one occurrence of 𝑔 in 𝑏₁, then 𝜎_𝑎,𝑏[𝑎]=𝑎 and 𝜎_𝑎,𝑏[𝑏]=𝑏. Hence, 𝜎_𝑎,𝑏 is idempotent.
Assume that 𝑔 occurs more than twice in 𝑏₁. Then, op(𝜎^𝑘+1_𝑎,𝑏[𝑏₁])≥op(𝜎^𝑘_𝑎,𝑏[𝑏₁]) for all 𝑘∈ℕ. Then, op𝜎^𝑘+1_𝑎,𝑏[𝑏]=op𝜎^𝑘_𝑎,𝑏𝜎_𝑎,𝑏[𝑏𝜎]=op^𝑘_𝑎,𝑏𝜎_𝑎,𝑏𝑓𝜎_𝑎,𝑏𝑏₁,𝜎_𝑎,𝑏𝑏₂=op𝜎^𝑘_𝑎,𝑏𝜎_𝑎,𝑏𝑓𝜎^𝑘+1_𝑎,𝑏𝑏₁,𝜎^𝑘+1_𝑎,𝑏𝑏₂>op𝜎^𝑘_𝑎,𝑏𝑓𝜎^𝑘_𝑎,𝑏𝑏₁,𝜎^𝑘_𝑎,𝑏𝑏₂=op𝜎^𝑘_𝑎,𝑏[𝑏]. (6.1) This shows that the order of 𝜎_𝑎,𝑏 is infinite.

Proposition 6.2. Let 𝑎=𝑥₁, 𝑏=𝑔(𝑏₁), op(𝑏)>1, and var(𝑏)={𝑥₁}. Then, the following hold: (i)if 𝑏 satisfies (6.1.4), then the order of 𝜎_𝑎,𝑏 is equal to 1;(ii)if 𝑏 satisfies (6.1.5), then the order of 𝜎_𝑎,𝑏 is infinite.

Proof. (i) If 𝑎 and 𝑏 satisfy (6.1.4), then 𝜎_𝑎,𝑏[𝑎]=𝑎 and 𝜎_𝑎,𝑏[𝑏]=𝑏. Hence, 𝜎_𝑎,𝑏 is idempotent.
(ii) Assume 𝑎 and 𝑏 satisfy (6.1.5). Then, op(𝜎^𝑘+1_𝑎,𝑏[𝑏₁])≥op(𝜎^𝑘_𝑎,𝑏[𝑏₁]) for all 𝑘∈ℕ. Then, op𝜎^𝑘+1_𝑎,𝑏[𝑏]=op𝜎^𝑘_𝑎,𝑏𝜎_𝑎,𝑏[𝑏𝜎]=op^𝑘_𝑎,𝑏𝜎_𝑎,𝑏𝑔𝜎_𝑎,𝑏𝑏₁=op𝜎^𝑘_𝑎,𝑏𝜎_𝑎,𝑏𝑔𝜎^𝑘+1_𝑎,𝑏𝑏₁>op𝜎^𝑘_𝑎,𝑏𝑔𝜎^𝑘_𝑎,𝑏𝑏₁=op𝜎^𝑘_𝑎,𝑏[𝑏]. (6.2) This shows that the order of 𝜎_𝑎,𝑏 is infinite.

Proposition 6.3. Let 𝑎=𝑥₂, 𝑏=𝑓(𝑏₁,𝑏₂), op(𝑏)>1, and var(𝑏)={𝑥₁}. Then, the following hold: (i)if 𝑏 satisfies (6.2.1) or (6.2.2), then the order of 𝜎_𝑎,𝑏 is equal to 2;(ii)if 𝑏 satisfies (6.2.3), then the order of 𝜎_𝑎,𝑏 is equal to 1 or is infinite.

Proof. (i) If 𝑎 and 𝑏 satisfy (6.2.1) or (6.2.2), then 𝜎_𝑎,𝑏[𝑎]=𝑎 and 𝜎²_𝑎,𝑏[𝑏]=𝜎_𝑎,𝑏[𝑏]. Hence the order of 𝜎_𝑎,𝑏 is equal to 2.
(ii) This case can be proved similarly to the proof of Proposition 6.1(ii).

Proposition 6.4. Let 𝑎=𝑥₂, 𝑏=𝑔(𝑏₁), op(𝑏)>1, and var(𝑏)={𝑥₁}. Then, the following hold: (i)if 𝑏 satisfies (6.2.4), then the order of 𝜎_𝑎,𝑏 is equal to 1;(ii)if 𝑏 satisfies (6.2.5), then the order of 𝜎_𝑎,𝑏 is infinite.

Proof. The proofs follow those of Proposition 6.2 (i) and (ii), respectively.