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T-Ordering in ternary semiring

Abstract

We investigate the concept of minus ordering in Ternary Semiring. We derive some equivalent conditions for minus ordering by using generalized inverses and prove that T-ordering is a partial ordering. 2020 mathematical science classification: 16Y60, 16U90.

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Introduction

In 1932, Lehmer [1] explored ternary algebraic system theory. He studied "triplexes", which are ternary algebraic systems that finally turn out to be ternary groups. A generalization of ternary rings is called ternary semirings, as presented by Dutta et al. [2] . Ternary semiring is a superset of semiring. Negative integers form a ternary semiring but it does not form a semiring. In 1920, E.H. Moore established the concept of the generalized inverse of a matrix, defining the unique generalized inverse through the use of matrices projectors. "Penrose demonstrated in 1955 that Moore's inverse is the unique matrix that can satisfy four equations [3] . The notion of regular element was first introduced by Von Neumann in 1936 [4] . Jian Miao Cen [5] introduced the concept of t-ordering in fuzzy matrices and he has discussed the relationships between the T-ordering and the generalized inverses. In this paper we studied the concept of T-ordering for elements in ternary semiring as a generalization of the T-ordering for regular fuzzy matrices.

Preliminaries

In this work, T represents the Ternary semiring; it is abbreviated as T.S.R. throughout this paper.

Definition 2.1

The term "ternary semiring" refers to 𝑇 ≠ , a binary addition, and a ternary multiplication called juxtaposition if (𝑇, +) is a semigroup that commutes that fulfilling the requirements listed below.  (abc)de = a (bcd)e = ab (cde),  (a +b)cd = acd+bcd,  a (b +c)d = abd+acd,  ab (c + d) = abc+abd, for all a, b, d, d, e in 𝑇 Remark z -is an natural example of a ternary semiring, which has ternary multiplication and regular addition.  Definition 2.2: If an element 𝑎 exists in T such that 𝑎𝑏𝑎𝑏𝑎 = 𝑎, then 𝑎 is considered regular. The element 𝑏 is represented as 𝑎 -and is also known as the generalized inverse or 1-inverse of the element ′𝑎′and the set of all 1-inverses is denoted by 𝑎{1}. If every element in 𝑎 T.S.R is regular, then we say T.S.R is regular.  Definition 2.3: If there is an element 𝑎 in 𝑇 ∃ 𝑏𝑎𝑏𝑎𝑏 = 𝑏, then 𝑎 is considered antiregular. The name for element 𝑏 is 2-inverse of 𝑎 and set of all 2-inverses of 𝑎 is denoted by 𝑎{2}.

The both conditions in Definition (2.2) and Definition (2.3) satisfied by a single element 𝑏 is called 1-2 inverse and is denoted by 𝑎 ̂. In Definition (2.2), if the ′𝑏′ is unit then ′𝑎′ is called unit regular.

Definition 2.4:

The Moore-Penrose inverse of an element 𝑎 ∈ 𝑇 if ∃ 𝑏 ∈ 𝑇 such that the following equations hold:

Where 𝑒 is identity element, here 𝑏 is unique and denoted by 𝑎 + In Definition (2.4), the element 𝑏 is satisfy the equations ( 1) and ( 3), then we say that element is {1,3}-inverse of ′𝑎 ′ and 𝑏 satisfy the equation ( 1) and ( 4) then we say that 𝑏 is {1,4}-inverse of ′𝑎′.

The transpose of 𝑎 is written as 𝑎 𝑇 and the following properties of the transpose will be used:

𝑎 𝑇 𝑇 = 𝑎, (𝑎 + 𝑏) 𝑇 = 𝑎 𝑇 + 𝑏 𝑇 , (𝑎𝑏𝑐) 𝑇 = 𝑐 𝑇 𝑏 𝑇 𝑎 𝑇 .

T-ordering in ternary semiring Definition 4.1:

The T-ordering 𝑎 ≤ 𝑇 𝑏in 𝑇 is defined as 𝑎 ≤ 𝑇 𝑏 iff 𝑎𝑎 𝑇 𝑎 = 𝑎𝑎 𝑇 𝑏 = 𝑏𝑎 𝑇 𝑎; 𝑎𝑎𝑎 𝑇 = 𝑎𝑏𝑎 𝑇 = 𝑏𝑎𝑎 𝑇 ; 𝑎 𝑇 𝑎𝑎 = 𝑎 𝑇 𝑏𝑎 = 𝑎 𝑇 𝑎𝑏.

Example 1: Let 𝑇 = {0,1,2,3,4} be a ternary semiring with respect to modulo addition and modulo multiplication 5, the Tordering defined as follows:

Let 𝑎 = 3; 𝑎 𝑇 = 2; 𝑏 = 3 satisfy 𝑎𝑎 𝑇 𝑎 = 𝑎𝑎 𝑇 𝑏 = 𝑏𝑎 𝑇 𝑎; ; 𝑎𝑎𝑎 𝑇 = 𝑏𝑎𝑎 𝑇 = 𝑎𝑏𝑎 𝑇 ; 𝑎 𝑇 𝑎𝑎 = 𝑎 𝑇 𝑏𝑎 = 𝑎 𝑇 𝑎𝑏

Throughout in this paper, let 𝑎 + be Moore-Penrose inverse Theorem 3.1: [8] Let 𝑎, 𝑏 ∈ 𝑇 and 𝑎 + exists. The subsequent items are therefore equal.

 𝑎 ≤ 𝑇 𝑏  𝑎𝑎 + 𝑎 = 𝑎𝑎 + 𝑏 = 𝑏𝑎 + 𝑎; 𝑎 + 𝑎𝑎 = 𝑎 + 𝑏𝑎 = 𝑎 + 𝑎𝑏; 𝑎𝑎𝑎 + = 𝑎𝑏𝑎 + = 𝑏𝑎𝑎 + ;  𝑎 = 𝑎𝑎 + 𝑎𝑎 + 𝑏 = 𝑏𝑎 + 𝑎𝑎 + 𝑎 = 𝑎𝑎 + 𝑏𝑎 + 𝑎 = 𝑏𝑎𝑎 + 𝑏𝑎 + 𝑎𝑏 Proof: (i)  𝑎 𝑇 = 𝑎 𝑇 𝑏𝑏 + 𝑏𝑏 + = 𝑏 + 𝑏𝑏 + 𝑏𝑎 𝑇 Theorem 3.3: [8] For 𝑎, 𝑏 ∈ 𝑇 if 𝑎 ≤ 𝑇 𝑏 then 𝑎 = 𝑎𝑏 + 𝑏𝑏 + 𝑏 = 𝑏𝑏 + 𝑏𝑏 + 𝑎 = 𝑎𝑏 + 𝑏𝑏 + 𝑎 = 𝑏𝑎𝑏 + 𝑏𝑏 + 𝑎𝑏 Proof: we know that 𝑎 ≤ 𝑇 𝑏 ⇔ 𝑎𝑎 + 𝑎 = 𝑎𝑎 + 𝑏 = 𝑏𝑎 + 𝑎;

⇔ 𝑎 + 𝑎𝑎 = 𝑎 + 𝑏𝑎 = 𝑎 + 𝑎𝑏; From Equations (4.6), (4.7), (4.8), (4.9) we get 𝑎 = 𝑎𝑏 + 𝑏𝑏 + 𝑏 = 𝑏𝑏 + 𝑏𝑏 + 𝑎 = 𝑎𝑏 + 𝑏𝑏 + 𝑎 = 𝑏𝑎𝑏 + 𝑏𝑏 + 𝑎𝑏 Theorem 3.4: [8] Let𝑎, 𝑏 ∈ 𝑇, and𝑎 ≤ 𝑇 𝑏. If 𝑏 + exists then 𝑏 + 𝑏𝑏 + ∈ 𝑎{1,3} ∩ 𝑎{1,4}.

Proof: Using Theorem 3.2 and 3. (𝑒𝑏 + 𝑏𝑏 + 𝑎) 𝑇 = (𝑒𝑏 + 𝑏𝑏 + 𝑎𝑒𝑎 𝑇 (𝑏 + ) 𝑇 𝑏 𝑇 (𝑏 + ) 𝑇 𝒆) 𝑇 = 𝑒𝑏 + 𝑏𝑏 + 𝑎𝑎 𝑇 (𝑏 + ) 𝑇 𝑏 𝑇 (𝑏 + ) 𝑇 = 𝑒𝑏 + 𝑏𝑏 + 𝑎 ∴ 𝑏 + 𝑏𝑏 + ∈ 𝑎{1,3} ∩ 𝑎{1,4}.

Corollary 4.1 [9] If 𝑎 ≤ 𝑇 𝑏and 𝑎 + exists then 𝑏 ∈ 𝑎 + {1}

Proof: 𝑎 + = 𝑎 + 𝑎𝑎 + 𝑎𝑎 + = 𝑎 + (𝑎𝑎 + 𝑎)𝑎 + = 𝑎 + (𝑏𝑎 + 𝑎)𝑎 + = 𝑎 + 𝑏𝑎 + (𝑏𝑎 + 𝑎𝑎 + 𝑎)𝑎 + = 𝑎 + 𝑏𝑎 + 𝑏𝑎 + 𝑎 + = 𝑎 + 𝑏𝑎 + 𝑏𝑎 + ∴ 𝑏 ∈ 𝑎 + {1}.

Theorem 3.5 [9] Let 𝑎, 𝑏 ∈ 𝑇 If 𝑎 + and 𝑏 + exists, then the subsequent are equivalent:

 𝑎 ≤ 𝑇 𝑏  𝑎 + 𝑎𝑎 + =𝑏 + 𝑎𝑎 + =𝑎 + 𝑎𝑏 + ; 𝑎𝑎 + 𝑎 + = 𝑎𝑎 + 𝑏 + = 𝑎𝑏 + 𝑎 + ; 𝑎 + 𝑎 + 𝑎 = 𝑎 + 𝑏 + 𝑎 = 𝑏 + 𝑎 + 𝑎  𝑎 + = 𝑏 + 𝑎𝑎 + 𝑎𝑎 + =𝑎 + 𝑎𝑎 + 𝑎𝑏 + = 𝑎 + 𝑎𝑏 + 𝑎𝑎 + = 𝑏𝑎 + 𝑎𝑏 + 𝑎𝑎 + 𝑏 Proof: (i) (ii) Since𝑎𝑎 𝑇 𝑎 = 𝑎𝑎 𝑇 𝑏 (𝑎𝑎 𝑇 𝑎) 𝑇 = (𝑎𝑎 𝑇 𝑏) 𝑇 = 𝑏 𝑇 𝑎𝑎 𝑇 = 𝑏 + 𝑎𝑎 + (∵ 𝑏 𝑇 = 𝑏 + ) And𝑎𝑎 𝑇 𝑎 = 𝑏𝑎 𝑇 𝑎(𝑎𝑎 𝑇 𝑎) 𝑇 = (𝑏𝑎 𝑇 𝑎) 𝑇 𝑎 𝑇 𝑎𝑎 𝑇 = 𝑎 𝑇 𝑎𝑏 𝑇 𝑎 + 𝑎𝑎 + = 𝑎 + 𝑎𝑏 + ∴ 𝑎 + 𝑎𝑎 + =𝑏 + 𝑎𝑎 + =𝑎 + 𝑎𝑏 + .

Similarly we get, 𝑎𝑎 + 𝑎 + = 𝑎𝑎 + 𝑏 + = 𝑎𝑏 + 𝑎 + ; 𝑎 + 𝑎 + 𝑎 = 𝑎 + 𝑏 + 𝑎 = 𝑏 + 𝑎 + 𝑎 (II)(iii) 𝑎 + 𝑎𝑎 + =𝑏 + 𝑎𝑎 + (𝑎 + 𝑎𝑎 + )𝑎𝑎 + =(𝑏 + 𝑎𝑎 + )𝑎𝑎 + 𝑎 + = 𝑏 + 𝑎𝑎 + 𝑎𝑎 + 𝑎 + 𝑎𝑎 + = 𝑎 + 𝑎𝑏 + 𝑎 + 𝑎(𝑎 + 𝑎𝑎 + ) = 𝑎 + 𝑎(𝑎 + 𝑎𝑏 + )𝑎 + = 𝑎 + 𝑎𝑎 + 𝑎𝑏 + 𝑎 + = 𝑎 + 𝑎𝑎 + 𝑎𝑎 + = (𝑎 + 𝑎𝑎 + )𝑎𝑎 + = 𝑎 + 𝑎𝑏 + 𝑎𝑎 + 𝑎 + = 𝑎 + 𝑎𝑏 + 𝑎𝑎 + 𝑎 + = 𝑎 + 𝑎𝑎 + 𝑎𝑎 + = 𝑎 + (𝑎 + 𝑎𝑎 + 𝑎𝑎 + )𝑎 + = (𝑎 + 𝑎 + 𝑎)𝑎 + 𝑎𝑎 + 𝑎 + = (𝑏 + 𝑎 + 𝑎)𝑎 + 𝑎𝑎 + 𝑎 + = 𝑏 + 𝑎 + 𝑎𝑎 + (𝑎𝑎 + 𝑎 + ) = 𝑏 + 𝑎 + 𝑎𝑎 + (𝑎𝑎 + 𝑏 + ) = 𝑏 + (𝑎 + 𝑎𝑎 + )𝑎𝑎 + 𝑏 + = 𝑏 + 𝑎 + 𝑎𝑏 + 𝑎𝑎 + 𝑏 + ∴ 𝑎 + = 𝑏 + 𝑎 + 𝑎𝑏 + 𝑎𝑎 + 𝑏 + (III)(i) 𝑎 + = 𝑏 + 𝑎𝑎 + 𝑎𝑎 + = 𝑎 + 𝑎𝑎 + 𝑎𝑏 + 𝑎 + = 𝑎 + 𝑎𝑎 + 𝑎𝑏 + (𝑎 + 𝑎𝑎 + 𝑎𝑎 + ) 𝑇 = (𝑎 + 𝑎𝑎 + 𝑎𝑏 + ) 𝑇 𝑎𝑎 𝑇 𝑎𝑎 𝑇 𝑎 = 𝑏𝑎 𝑇 𝑎𝑎 𝑇 𝑎(∵ 𝑎 𝑇 = 𝑎 + ) (𝑎𝑎 𝑇 𝑎𝑎 𝑇 𝑎)𝑎 𝑇 𝑎 = (𝑏𝑎 𝑇 𝑎𝑎 𝑇 𝑎)𝑎 𝑇 𝑎 𝑎𝑎 𝑇 𝑎 = 𝑏𝑎 𝑇 𝑎 Similarly𝑎 + = 𝑏 + 𝑎𝑎 + 𝑎𝑎 + (𝑎 + 𝑎𝑎 + 𝑎𝑎 + ) 𝑇 = (𝑏 + 𝑎𝑎 + 𝑎𝑎 + ) 𝑇 𝑎𝑎 𝑇 𝑎𝑎 𝑇 𝑎 = 𝑎𝑎 𝑇 𝑎𝑎 𝑇 𝑏𝑎𝑎 𝑇 (𝑎𝑎 𝑇 𝑎𝑎 𝑇 𝑎) = 𝑎𝑎 𝑇 (𝑎𝑎 𝑇 𝑎𝑎 𝑇 𝑏) 𝑎𝑎 𝑇 𝑎 = 𝑎𝑎 𝑇 𝑏 ∴ 𝑎𝑎 𝑇 𝑎 = 𝑏𝑎 𝑇 𝑎 = 𝑎𝑎 𝑇 𝑏 (*) 𝑎 + = 𝑏 + 𝑎𝑎 + 𝑎𝑎 + (𝑎 + 𝑎𝑎 + 𝑎𝑎 + ) 𝑇 = (𝑏 + 𝑎𝑎 + 𝑎𝑎 + ) 𝑇 𝑎𝑎 𝑇 𝑎𝑎 𝑇 𝑎 = 𝑎𝑎 𝑇 𝑎𝑎 𝑇 𝑏 𝑎 𝑇 (𝑎𝑎 𝑇 𝑎𝑎 𝑇 𝑎) = 𝑎 𝑇 (𝑎𝑎 𝑇 𝑎𝑎 𝑇 𝑏)𝑎𝑎 𝑇 𝑎𝑎 = 𝑎 𝑇 𝑏𝑎 Similarly we obtain 𝑎 𝑇 𝑎𝑎 = 𝑎 𝑇 𝑎𝑏 ∴𝑎 𝑇 𝑎𝑎 = 𝑎 𝑇 𝑏𝑎 = 𝑎 𝑇 𝑎𝑏 (**) 𝑎 + = 𝑎 + 𝑎𝑎 + 𝑎𝑏 + (𝑎 + 𝑎𝑎 + 𝑎𝑎 + ) 𝑇 = (𝑎 + 𝑎𝑎 + 𝑎𝑏 + ) 𝑇 𝑎𝑎 𝑇 𝑎𝑎 𝑇 𝑎 = 𝑏𝑎 𝑇 𝑎𝑎 𝑇 𝑎 𝑎(𝑎𝑎 𝑇 𝑎𝑎 𝑇 𝑎)𝑎 𝑇 = 𝑎(𝑏𝑎 𝑇 𝑎𝑎 𝑇 𝑎)𝑎 𝑇 𝑎𝑎𝑎 𝑇 = 𝑎𝑏𝑎 𝑇 and similarly, 𝑎𝑎𝑎 𝑇 = 𝑏𝑎𝑎 𝑇 ∴ 𝑎𝑎𝑎 𝑇 = 𝑎𝑏𝑎 𝑇 = 𝑏𝑎𝑎 𝑇 (***) From (*),(**) and (***) we get𝑎 ≤ 𝑇 𝑏 Theorem 3.6 [11] For 𝑎, 𝑏 ∈ 𝑇 + , then the following are hold

𝑏 ≤ 𝑇 𝑎 ⇔ 𝑏 = 𝑎𝑏 + 𝑏𝑏 + 𝑏 = 𝑏𝑏 + 𝑏𝑏 + 𝑎 = 𝑏𝑏 + 𝑎𝑏 + 𝑏 = 𝑎𝑏𝑏 + 𝑎𝑏 + 𝑏𝑎 𝑎 = 𝑎𝑎 + 𝑎𝑎 + 𝑎 = 𝑏𝑎 + 𝑎𝑎 + 𝑎 = 𝑏𝑎 + (𝑎𝑎 + 𝑏𝑎 + 𝑎)𝑎 + 𝑏 = (𝑏𝑏 + 𝑏𝑏 + 𝑎)𝑎 + 𝑎𝑎 + 𝑏𝑎 + 𝑎𝑎 + (𝑎𝑏 + 𝑏𝑏 + 𝑏) = 𝑏𝑏 + 𝑏𝑏 + 𝑎𝑎 + 𝑎𝑎 + 𝑎𝑏 + 𝑏𝑏 + 𝑏 (∵ 𝑎 = 𝑏𝑎 + 𝑎𝑎 + 𝑎) Theorem 3.7: [14] If 𝑎 ≤ 𝑇 𝑏, then we have Corollary 3.7: [11] If 𝑎 ≤ 𝑇 𝑏 and 𝑎 + , 𝑏 + exists then 𝑏 + ∈ 𝑎{1}

Proof: Since 𝑎 + = 𝑏 + 𝑎𝑎 + 𝑎𝑎 + = 𝑎 + 𝑎𝑎 + 𝑎𝑏 + = 𝑎 + 𝑎𝑏𝑎𝑎 + 𝑎 = 𝑎𝑎 + 𝑎𝑎 + 𝑎 = 𝑎(𝑏 + 𝑎𝑎 + 𝑎𝑎 + )𝑎𝑎 + 𝑎 = 𝑎𝑏 + 𝑎(𝑎 + 𝑎𝑎 + )𝑎𝑎 + 𝑎 = 𝑎𝑏 + 𝑎𝑏 + 𝑎𝑎 + 𝑎𝑎 + 𝑎 = 𝑎𝑏 + 𝑎𝑏 + 𝑎 ∴ 𝑎 = 𝑎𝑏 + 𝑎𝑏 + 𝑎 Hence 𝑏 + ∈ 𝑎{1} Theorem 3.8: [11] Let 𝑎, 𝑏 ∈ 𝑇 + , Then the following conditions are equivalent  𝑎 ≤ 𝑇 𝑏  𝑏 ∈ 𝑎 + {1,3,4} Proof: (i)(ii) By corollary (3.1), 𝑏 ∈ 𝑎 + {1} Since 𝑎 = 𝑎𝑎 + 𝑎𝑎 + 𝑏 = 𝑏𝑎 + 𝑎𝑎 + 𝑎 and 𝑎 + = 𝑏 + 𝑎𝑎 + 𝑎𝑎 + = 𝑎 + 𝑎𝑎 + 𝑎𝑏 + (𝑎 + 𝑏𝑒) 𝑇 = (𝑏 + 𝑎𝑎 + 𝑎𝑎 + 𝑏𝑒) 𝑇 = (𝒆𝑏 + 𝑎𝑎 + 𝑎𝑎 + 𝑒𝑏𝑒) 𝑇 = 𝑒 𝑇 𝑏 𝑇 𝑒 𝑇 𝑎 + 𝑇 𝑎 𝑇 𝑎 + 𝑇 𝑎 𝑇 𝑏 + 𝑇 𝑒 𝑇 = 𝑒𝑏 + 𝑎𝑎 + 𝑎𝑎 + 𝑒𝑏𝑒 = 𝑎 + 𝑏𝑒 (∵ 𝑎 + = 𝑎 𝑇 and 𝑒 𝑇 = 𝑒) ∴ (𝑎 + 𝑏𝑒) 𝑇 = (𝑎 + 𝑏𝑒 ) (𝑒𝑏𝑎 + ) 𝑇 = (𝑒𝑏𝑒𝑎 + 𝑎𝑎 + 𝑎𝑏 + 𝑒) 𝑇 = 𝑒 𝑇 𝑏 + 𝑇 𝑎 𝑇 𝑎 + 𝑇 𝑎 𝑇 𝑎 + 𝑇 𝑒 𝑇 𝑏 𝑇 𝑒 𝑇 = 𝑒𝑏𝑎 + 𝑎𝑎 + 𝑎𝑏 + = 𝑒𝑏𝑎 + ∴ (𝑒𝑏𝑎 + ) 𝑇 = (𝑒𝑏𝑎 + ) Hence 𝑏 ∈ 𝑎 + {1, 3, 4}.

(ii)(i) 𝑎 + 𝑏𝑎 + 𝑏𝑎 + = 𝑎 + , (𝑎 + 𝑏𝑒) 𝑇 = (𝑎 + 𝑏𝑒), (𝑒𝑏𝑎 + ) 𝑇 = (𝑒𝑏𝑎 + ) 𝑎𝑎 + 𝑎 = 𝑎(𝑎 + 𝑏𝑎 + 𝑏𝑎 + )𝑎 = 𝑎𝑎 + 𝑒𝑏𝑎 + 𝑒𝑏𝑎 + 𝑎 (𝑎𝑎 + 𝑒)(𝑏𝑎 + 𝑒)(𝑏𝑎 + 𝑒)𝑎 = (𝑎𝑎 + 𝑒) 𝑇 (𝑏𝑎 + 𝑒) 𝑇 (𝑏𝑎 + 𝑒) 𝑇 = (𝑏𝑎 + 𝑒𝑏𝑎 + 𝑒𝑎𝑎 + 𝑒) 𝑇 𝑎 = (𝑏𝑎 + 𝑏𝑎 + 𝑎𝑎 + 𝑒) 𝑇 𝑎 = (𝑏𝑎 + 𝑏𝑎 + (𝑏𝑎 + 𝑎𝑎 + 𝑎)𝑎 + 𝑒) 𝑇 𝑎 = (𝑏𝑎 + 𝑒𝑏𝑎 + 𝑒𝑏𝑎 + 𝑒) 𝑇 𝑎 = (𝑏𝑎 + 𝑒) 𝑇 𝑎 = 𝑏𝑎 + 𝑎 ∴ 𝑎𝑎 + 𝑎 = 𝑏𝑎 + 𝑎 𝑎𝑎 + 𝑎 = 𝑎(𝑎 + 𝑏𝑎 + 𝑏𝑎 + )𝑎 = 𝑎(𝑎 + 𝑏𝑒) 𝑇 (𝑎 + 𝑏𝑒) 𝑇 (𝑎 + 𝑎𝑒) 𝑇 = 𝑎(𝑎 + 𝑎𝑒𝑎 + 𝑏𝑒𝑎 + 𝑏𝑒) 𝑇 = 𝑎(𝑎 + (𝑎𝑎 + 𝑎𝑎 + 𝑏)𝑎 + 𝑏𝑎 + 𝑏) 𝑇 = 𝑎(𝑎 + 𝑏𝑎 + 𝑏𝑎 + 𝑏𝑒) 𝑇 = 𝑎(𝑎 + 𝑏𝑒) 𝑇 = 𝑎𝑎 + 𝑏 ∴ 𝑎𝑎 + 𝑎 = 𝑏𝑎 + 𝑎 = 𝑎𝑎 + 𝑏 Similarly we get 𝑎𝑎𝑎 + = 𝑎𝑏𝑎 + = 𝑏𝑎𝑎 + ; 𝑎 + 𝑎𝑎 = 𝑎 + 𝑎𝑏 = 𝑎 + 𝑏𝑎 Hence 𝑎 ≤ 𝑇 𝑏

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