# On Ideals of Implication Groupoids

Authors: Ravi Kumar Bandaru
Journal: Advances in Decision Sciences
 (2012)

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## Abstract

Ideals of implication groupoids are considered. Given a subset of a distributive implication groupoid, the smallest ideal containing it is constructed. A characterization of ideals in distributive implication groupoid using upper sets is given.


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## Body

## 1. Introduction

In 50-ties L-Henkin and T-Skolem introduced the notion of Hilbert algebra as an algebraic counterpart of intuitionistic logic. A Hilbert algebra [1] is an algebra ℋ=(H,*,1) of type (2,0) satisfying the axioms:(H1)
x*(y*x)=1,(H2)
(x*(y*z))*((x*y)*(x*z))=1,(H3)
x*y=1 and y*x=1 imply x=y.One can easily show that (H2) can be replaced by two rather simpler axioms:(LD)
x*(y*z)=(x*y)*(x*z) (left distributivity),(E)
x*(y*z)=y*(x*z) (exchange).Chajda and Halaš [2] introduced the concept of distributive implication groupoid and studied deductive systems, ideals, and congruence relations in distributive implication groupoid. In this paper we consider ideals in distributive implication groupoid. Given a subset of a distributive implication groupoid, we make the smallest ideal containing it. We provide an equivalent condition of the ideals using the notion of upper sets.


## 2. Preliminaries

Definition 2.1 (see [2]).
An algebra(A,*,1) of type (2,0) is called an implication groupoid if it satisfies the identities:(1)
x*x=1,(2)
1*x=x for all x,y∈A.Example 2.2.
LetA={1,a,b} in which * is defined by
(2.1)*1ab11abaa1bbab1
Then (A,*,1) is an implication groupoid.Example 2.3.
LetA={1,a,b,c} in which * is defined by
(2.2)*1abc11abca11bbb1a1ac1ab1
Then (A,*,1) is an implication groupoid.Definition 2.4 (see [2]).
An implication groupoid(A,*,1) of type (2,0) is called a distributive implication groupoid if it satisfies the following identity:
(2.3)(LD)x*(y*z)=(x*y)*(x*z)(leftdistributivity)
for all x,y,z∈A.Example 2.5.
LetA={1,a,b,c,d} in which * is defined by
(2.4)*1abcd11abcda11bb1b1a11dc1a11dd11cc1
Then (A,*,1) is a distributive implication groupoid.In every implication groupoid, one can introduce the so-called induced relation≤ by the setting
(2.5)x≤yiffx*y=1.Lemma 2.6 (see [2]).
Let(A,*,1) be a distributive implication groupoid. Then A satisfies the identities
(2.6)x*1=1,x*(y*x)=1.
Moreover, the induced relation ≤ is a quasiorder on A, and the following relationships are satisfied:(i)
x≤1,(ii)
x≤y*x,(iii)
x*((x*y)*y)=1,(iv)
1≤x implies x=1,(v)
y*z≤(x*y)*(x*z),(vi)
x≤y implies y*z≤x*z,(vii)
x*(y*z)≤y*(x*z),(viii)
x*y≤(y*z)*(x*z).


## 3. On Ideals of Implication Groupoids

In this section, we study some properties of ideals in a distributive implication groupoid and give the smallest ideal containing a subset of a distributive implication groupoid. We characterize ideals in terms of upper sets.Definition 3.1 (see [2]).
Let𝒜=(A,*,1) be an implication groupoid. A subset I⊆A is called an ideal of 𝒜 if(I1)
1∈I,(I2)
x∈A,y∈I imply x*y∈I,(I3)
x∈A,y1,y2∈I imply (y2*(y1*x))*x∈I.Remark 3.2.
IfI is an ideal of an implication groupoid 𝒜=(A,*,1) and a∈I,x∈A, then (a*x)*x∈I.Definition 3.3 (see [2]).
Let𝒜=(A,*,1) be an implication groupoid. A subset D⊆A is called a deductive system of 𝒜 if(D1)
1∈D,(D2)
x∈D and x*y∈D imply y∈D.Lemma 3.4 (see [2]).
Let𝒜 be an implication groupoid. Then every ideal of 𝒜 is a deductive system of 𝒜.Converse of the above lemma does not hold in general.Example 3.5.
From Example2.2, we can see that {1,a} is its deductive system which is not an ideal since b*a=b∉{1,a}.Theorem 3.6 (see [2]).
A nonempty subsetI of a distributive implication groupoid 𝒜 is an ideal if and only if it is a deductive system of 𝒜.For anyx1,x2,…,xn,a∈A, we define
(3.1)∏i=1n‍x1*a=xn*(⋯*(x1*a)⋯).Lemma 3.7.
LetA be a distributive implication groupoid and x,y,z∈A such that x≤y. Then z*x≤z*y.Proof.
Letx,y,z∈A and x≤y. Then x*y=1 and hence (z*x)*(z*y)=z*(x*y)=z*1=1. Therefore z*x≤z*y.Lemma 3.8.
LetA be a distributive implication groupoid and x,y∈A such that x*y=1. Then for all a1,a2,…,an∈A,∏i=1nai*x=1 implies ∏i=1nai*y=1.Proof.
We havex*y=1;thatis,x≤y, and from Lemma 3.7, we can see that
(3.2)1=∏i=1n‍ai*x≤∏i=1n‍ai*y.
Therefore, from Lemma 2.6(iv), ∏i=1nai*y=1.We denote the set of all ideals ofA by ℐ(A). It is obvious that {1},A∈ℐ(A).Example 3.9.
From Example2.2, we can see that ℐ(A)={{1},A}.Example 3.10.
From Example2.5, we can see that ℐ(A)={{1},{1,a,d},{1,b,c},A}.Example 3.11.
LetA={1,a,b,c,d} in which * is defined by
(3.3)*1abcd11abcdaa1cddbaa1cccaaa1cdaaaa1
Then (A,*,1) is an implication groupoid. We can see that ℐ(A)={{1},{1,a},{1,a,c,d},A}.The following theorem is straightforward.Theorem 3.12.
IfIi(i∈Δ) are ideals of an implication groupoid A, then ⋂i∈Δ‍Ii is an ideal of A.Note 1.
In an implication groupoid, union of two ideals need not be an ideal. From Example2.3, we can see that I={1,a} and J={1,b} are ideals of A but I∪J={1,a,b} is not an ideal of A.The following is a characterization of idealsTheorem 3.13.
LetI be a subset of a distributive implication groupoid A containing 1. Then I∈ℐ(A) if and only if for any a,b∈I and x∈A, a*(b*x)=1 implies x∈I.Proof.
LetI∈ℐ(A). Assume a,b∈I and x∈A such that a*(b*x)=1. Since I is an ideal of A, we have a*(b*x)∈I. Since every ideal of A is deductive system, by applying (D2) twice, we conclude that x∈I. Conversely, assume that the condition holds. Since ideals and deductive systems coincide in distributive implication groupoid, it is enough to show that I satisfies (D1) and (D2). Since 1∈I, the condition (D1) holds. Suppose x∈I and x*a∈I. Then x*((x*a)*a)=(x*(x*a))*(x*a)=((x*x)*(x*a))*(x*a)=(1*(x*a))*(x*a)=(x*a)*(x*a)=1. Therefore x*((x*a)*a)∈I and hence a∈I. Thus I∈ℐ(A).Corollary 3.14.
LetI be a subset of a distributive implication groupoid A containing 1. Then I∈ℐ(A) if and only if for any a1,a2,…,an∈I and x∈A,∏i=1nai*x=1 implies x∈I.Definition 3.15.
For every subsetX⊆A, the smallest ideal of A which contains X, that is, the intersection of all ideals I⊇X, is said to be the ideal generated by X, and will be denoted by (X]. Obviously, (∅]={1}.Lemma 3.16.
LetA be a distributive implication groupoid and x,y,z∈A. Then x*(y*z)=1 if and only if y*(x*z)=1.Proof.
Letx*(y*z)=1. Then y*(x*(y*z))=y*1=1 and hence (y*x)*(y*(y*z))=1. Therefore (y*x)*(y*z)=1. Thus y*(x*z)=1. Similarly, we can prove the converse.Theorem 3.17.
LetA be a distributive implication groupoid and X(≠∅)⊆A. Then
(3.4)(X]={x∈A:x=1or∏i=1n‍ai*x=1forsomea1,a2,…,an∈X}.Proof.
LetI={x∈A:x=1 or ∏i=1nai*x=1 for some a1,a2,…an∈X}. Since a*a=1 for all a∈X, we obtain X⊆I. Obviously 1∈I. Let x*y∈I and x∈I. To prove y∈I, we will consider three cases. Case 1: x=1. Then y=1*y∈I. Case 2: x*y=1 and x≠1. Since x∈I and x≠1, we conclude that ∏i=1nai*x=1 for some a1,a2,…,an∈X. From Lemma 3.8, ∏i=1nai*y=1. Therefore y∈I. Case 3: x*y≠1 and x≠1. Then there are a1,a2,…,an,b1,b2,…,bm∈X such that ∏i=1nai*(x*y)=1 and ∏j=1mbj*x=1. Applying Lemma 3.16, we deduce that x≤∏i=1nai*y and by Lemma 3.7, we see that
(3.5)1=∏j=1m‍bj*x≤∏j=1m‍bj*(∏i=1n‍ai*y).
By Lemma 2.6(iv), ∏j=1mbj*(∏i=1nai*y)=1. Hence I is an ideal of A.
Suppose thatU is any ideal of A containing X. Let x∈I. If x=1, then obviously x∈U. Assume that x≠1. Then there are a1,a2,…,an∈X such that ∏i=1nai*x=1. Since X⊆U, it follows that a1,a2,…,an∈U. Therefore x∈U by Corollary 3.14. Thus I⊆U and hence I=(X].LetI1,I2∈ℐ(A); we define the meet of I1 and I2 (denoted by I1∧I2) by I1∧I2=I1∩I2 and the join of I1 and I2 (denoted by I1∨I2) by I1∨I2=(I1∪I2]. We note that (ℐ(A),∧,∨) is a lattice.Theorem 3.18.
(
ℐ
(
A
)
,
∧
,
∨
) is a complete lattice.LetA be a distributive implication groupoid. For any x,y∈A, consider a set
(3.6)A(x)={z∈A∣x*z=1},A(x,y)={z∈A∣x*(y*z)=1}.
The set A(x) (resp., A(x,y)) is called an upper set of x (resp., of x and y). Obviously, 1,x∈A(x) and 1,x,y∈A(x,y). We know that A(1)={1} is always an ideal of A. But the sets A(x) and A(x,y) need not be ideals of A in an implication groupoid, since A(a)={a} and A(a,1)={a} are not ideals of A in Example 2.2. The following lemma can be proved easily.Lemma 3.19.
IfA is an implication groupoid, then A(u)=A(u,1).Theorem 3.20.
IfA is a distributive implication groupoid, then, for any x,y∈A, the set A(x,y) is an ideal of A.Proof.
LetA be a distributive implication groupoid. Clearly 1∈A(x,y). Let r∈A(x,y) and r*s∈A(x,y). Then x*(y*r)=1 and x*(y*(r*s))=1. Now x*(y*(r*s))=1 implies that (x*(y*r))*(x*(y*s))=1 which gives x*(y*s)=1. Therefore s∈A(x,y). Hence A(x,y) is an ideal of A.Corollary 3.21.
LetA be a distributive implication groupoid. Then for any x∈A, the set A(x) is an ideal of A.Lemma 3.22.
IfA is a distributive implication groupoid, then A(x)⊆A(x,y) for any x,y∈A.Theorem 3.23.
LetA be a distributive implication groupoid and a∈A. Then the following are equivalent:(i)
a≤x for any x∈A,(ii)
A=A(a),(iii)
A=A(a,x)=A(x,a) for any x∈A.Proof.
(i)
⇔
(ii): straightforward.
(ii)
⇒
(iii): by Lemma 3.22, A=A(a)⊆A(a,x)⊆A.
(iii)
⇒
(ii):
A
=
A
(
a
,
1
)
=
A
(
a
).Theorem 3.24.
LetA be a distributive implication groupoid and a∈A. Then A(a)=⋂b∈AA(a,b).Proof.
By Lemma3.22, A(a)⊆A(a,b) for any a,b∈A. Therefore A(a)⊆⋂b∈AA(a,b). If c∈⋂b∈AA(a,b), then c∈A(a,b) for all b∈A and so c∈A(a,1). Hence 1=a*(1*c)=a*c, which proves c∈A(a). This means that ⋂b∈AA(a,b)⊆A(a).Corollary 3.25.
LetA be a distributive implication groupoid. Then for any a∈A,A(a)=A(a,1)=⋂b∈AA(a,b).Theorem 3.26.
LetA be a distributive implication groupoid. Then A(a,b)=A(b,a) for any a,b∈A.Proof.
It follows from Lemma3.16.The following is a characterization of ideals.Theorem 3.27.
LetI be a nonempty subset of a distributive implication groupoid A. Then I is an ideal of A if and only if A(a,b)⊆I for all a,b∈I.Proof.
LetI be an ideal of A and a,b∈I. If c∈A(a,b), then a*(b*c)∈I and so z∈I. Hence A(a,b)⊆I. Conversely, assume that A(a,b)⊆I for all a,b∈I. Note that 1∈A(a,b)⊆I. Let x∈I and x*y∈I. Since (x*y)*(x*y)=1, we have y∈A(x*y,x)⊆I. We conclude that I is an ideal of A.Corollary 3.28.
LetA be a distributive implication groupoid. If I is an ideal of A, then A(a)⊆I for any a∈I.The converse of the above corollary need not be true in general. Consider the following example.Example 3.29.
LetA={1,a,b,c,d,e,f,g} in which * is defined by
(3.7)*abcdefg1a11111111bc1cg11g1cff1f1f11dcec1e111eaffd1fg1fcecge1g1gabcfef111abcdefg1
Then (A,*,1) is a distributive implication groupoid. Here I={1,b,e,f,g} contains A(1),A(b),A(e),A(f),A(g) but I is not an ideal of A.Theorem 3.30.
LetA be a distributive implication groupoid and x,y∈A. Then y∈A(x) if and only if A(x)=A(x,y).Proof.
Assume thaty∈A(x). Then x*y=1. We know that A(x)⊆A(x,y). For any z∈A(x,y), we have 1=x*(y*z)=(x*y)*(x*z)=x*z and so z∈A(x). Hence A(x)=A(x,y). Conversely, if A(x)=A(x,y), then y∈A(x,y)=A(x).Theorem 3.31.
LetA be a distributive implication groupoid and x,y∈A. Then x≤y if and only if A(y)⊆A(x).Proof.
Letx≤y. Then x*y=1. For any z∈A(y), we have y*z=1. Also x*z=1*(x*z)=(x*y)*(x*z)=x*(y*z)=x*1=1 and so z∈A(x). Hence A(y)⊆A(x). Conversely, if A(y)⊆A(x), then y∈A(x) and hence x≤y.Corollary 3.32.
LetA be a distributive implication groupoid and x,y∈A. Then x≤y and y≤x if and only if A(x)=A(y).Example 3.33.
LetA={1,a,b,c} be a set with the following table:
(3.8)*1abc11abca11b1b1c1cc11b1
Then (A,*,1) is a distributive implication groupoid. We can see that a≤c,c≤a and A(a)=A(c)={1,a,c}.Theorem 3.34.
LetI be an ideal of A. Then I=⋃x,y∈IA(x,y).Proof.
We know thatA(x,y)⊆I for all x,y∈I. Therefore ⋃x,y∈IA(x,y)⊆I. Let z∈I. Then z∈A(z)=A(z,1)⊆⋃x,y∈IA(x,y). Then I⊆⋃x,y∈IA(x,y).Corollary 3.35.
IfI is an ideal of A,  I=⋃x∈IA(x,1).Finally we conclude this paper with the following theorem.Theorem 3.36.
LetI be an ideal of A. Then I=⋃x∈IA(x).Proof.
SinceA(x,1)=A(x), we have, by Corollary 3.35, I=⋃x∈IA(x).



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