Constructing Some Logical Algebras with Hoops

Abstract

In any logical algebraic structures, by using of different kinds of filters, one can construct various kinds of other logical algebraic structures. With this inspirations, in this paper by considering a hoop algebra or a hoop, that is introduced by Bosbach, the notion of co-filter on hoops is introduced and related properties are investigated. Then by using of co-filter, a congruence relation on hoops is defined, and the associated quotient structure is studied. Thus Brouwerian semilattices, Heyting algebras, Wajsberg hoops, Hilbert algebras and BL-algebras are obtained.

1. Introduction

Non-classical logics (or called alternative logics) are formal systems that differ in a significant way from standard logical systems such as propositional and predicate logic. Many-valued logics are non-classical logics which are similar to classical logic. Bosbach [1,2] proposed the concept of hoop which is a nice algebraic structure to research the many-valued logical system whose propositional value is given in a lattice. For various information on hoops, refer to [3,4,5,6,7,8].

In this paper, we introduce the notion of co-filter in hoops and we get some properties of it. Then we construct a congruence relation by using co-filters on hoops. Finally, we investigate under which conditions the quotient structure of this congruence relation will be Brouwerian semilattice, Heyting algebra, Wajsberg hoop, Hilbert algebra and BL-algebra.

2. Preliminaries

In this section, we recollect some definitions and results which will be used in the following and we shall not cite them every time they are used.

Definition 1

([9]). A hoop is an algebraic structure $(\mathbb{H},\odot ,\to ,1)$ of type $(2,2,0)$ such that, for all $\alpha ,\beta ,\gamma \in \mathbb{H}$ it satisfies in the following conditions:

$(HP1)$ $(\mathbb{H},\odot ,1)$ is a commutative monoid.

$(HP2)$ $\alpha \to \alpha =1$.

$(HP3)$ $(\alpha \odot \beta )\to \gamma =\alpha \to (\beta \to \gamma )$.

$(HP4)$ $\alpha \odot (\alpha \to \beta )=\beta \odot (\beta \to \alpha )$.

On hoop $\mathbb{H}$, a binary relation ≤ is defined on $\mathbb{H}$ such that $\alpha \le \beta$ iff $\alpha \to \beta =1$ and $(\mathbb{H},\le )$ is a poset. If the least element $0\in \mathbb{H}$ exists such that, for all $\alpha \in \mathbb{H}$, $0\le \alpha$, then $\mathbb{H}$ is called a bounded hoop. We let ${\alpha }^0=1$ and ${\alpha }^n={\alpha }^{n-1}\odot \alpha$, for any $n\in \mathbb{N}$. If $\mathbb{H}$ is bounded, then, for all $\alpha \in \mathbb{H}$, the operation negation “ ′ ” is defined on $\mathbb{H}$ by, ${\alpha }^{\prime }=\alpha \to 0$. If ${({\alpha }^{\prime })}^{\prime }=\alpha$, for all $\alpha \in \mathbb{H}$, then $\mathbb{H}$ is said to have (DNP) property.

Proposition 1

([1,2]). Let $(\mathbb{H},\odot ,\to ,1)$ be a hoop. Then, for all $\alpha ,\beta ,\gamma \in \mathbb{H}$, it satisfies in the following conditions:

$(i)$$(\mathbb{H},\le )$ is a meet-semilattice with $\alpha \wedge \beta =\alpha \odot (\alpha \to \beta )$.

$(ii)$$\alpha \odot \beta \le \gamma$ iff $\alpha \le \beta \to \gamma$.

$(iii)$$\alpha \odot \beta \le \alpha ,\beta$ and ${\alpha }^n\le \alpha$, for any $n\in \mathbb{N}$.

$(iv)$$\alpha \le \beta \to \alpha$.

$(v)$$1\to \alpha =\alpha$ and $\alpha \to 1=1$.

$(vi)$$\alpha \odot (\alpha \to \beta )\le \beta$.

$(vii)$$\alpha \to \beta \le (\beta \to \gamma )\to (\alpha \to \gamma )$.

$(viii)$$\alpha \le \beta$ implies $\alpha \odot \gamma \le \beta \odot \gamma$, $\gamma \to \alpha \le \gamma \to \beta$ and $\beta \to \gamma \le \alpha \to \gamma$.

Proposition 2

([1,2]). Let $\mathbb{H}$ be a bounded hoop. Then, for any $\alpha ,\beta \in \mathbb{H}$, the following conditions hold:

$(i)$$\alpha \le {\alpha }^{″}$ and $\alpha \odot {\alpha }^{\prime }=0$

$(ii)$${\alpha }^{\prime }\le \alpha \to \beta$.

$(iii)$${\alpha }^{‴}={\alpha }^{\prime }$.

$(iv)$ If $\mathbb{H}$ has (DNP), then $\alpha \to \beta ={\beta }^{\prime }\to {\alpha }^{\prime }$.

$(v)$ If $\mathbb{H}$ has (DNP), then $(\alpha \to \beta )\to \beta =(\beta \to \alpha )\to \alpha$.

Proposition 3

([10]). Let $\mathbb{H}$ be a hoop and for any $\alpha ,\beta \in \mathbb{H}$, define the operation ∨ on $\mathbb{H}$ as follows,

$$\alpha \vee \beta =((\alpha \to \beta )\to \beta )\wedge ((\beta \to \alpha )\to \alpha ).$$

Then, for all $\alpha ,\beta ,\gamma \in \mathbb{H}$, the following conditions are equivalent:

$(i)$ ∨ is associative,

$(ii)$$\alpha \le \beta$ implies $\alpha \vee \gamma \le \beta \vee \gamma$,

$(iii)$$\alpha \vee (\beta \wedge \gamma )\le (\alpha \vee \beta )\wedge (\alpha \vee \gamma )$,

$(iv)$ ∨ is the join operation on $\mathbb{H}$.

A hoop $\mathbb{H}$ is said to a ∨-hoop, if it satisfies one of the above equivalent conditions.

Proposition 4

([10]). Let $\mathbb{H}$ be a ∨-hoop and $\alpha ,\beta ,\gamma \in \mathbb{H}$. Then ∨-hoop $(\mathbb{H},\vee ,\wedge )$ is a distributive lattice and $(\alpha \vee \beta )\to \gamma =(\alpha \to \gamma )\wedge (\beta \to \gamma )$.

Definition 2

([10]). A non-empty subset F of a hoop $\mathbb{H}$ is called a filter of $\mathbb{H}$ if, for any $\alpha ,\beta \in \mathbb{H}$, the following condition hold:

$(F1)$$\alpha ,\beta \in F$ implies $\alpha \odot \beta \in F$.

$(F2)$$\alpha \le \beta$ and $\alpha \in F$ imply $\beta \in F$.

The set of all filters of $\mathbb{H}$ is denoted by $\mathcal{F}(\mathbb{H})$. Clearly, for any filter F of $\mathbb{H}$, $1\in F$. F is called a proper filter if $F\ne \mathbb{H}$. So, if $\mathbb{H}$ is a bounded hoop, then a filter is proper iff it does not contain 0. It is easy to see that $F\in \mathcal{F}(\mathbb{H})$ iff, for any $\alpha ,\beta \in \mathbb{H}$, $1\in F$ and if $\alpha ,\alpha \to \beta \in F$, then $\beta \in F$.

3. Co-Filters in Hoops

From here on, if there is no mention, $\mathbb{H}$ denotes a bounded hoop.

We introduce the notion of co-filters on hoops, and it is proved that co-filters are not filters and some properties of them are studied. Moreover, a congruence relation is defined by them and is investigated the quotient structure of this congruence relation.

Definition 3.

A subset I of $\mathbb{H}$ is said to be a co-filter of $\mathbb{H}$ if, for any $\alpha ,\beta \in \mathbb{H}$,

$(C{F}_1)$$0\in I$.

$(C{F}_2)$${(\alpha \to \beta )}^{\prime }\in I$ and $\beta \in I$ imply $\alpha \in I$.

Example 1.

Let $\mathbb{H}=\{0,a,b,c,d,1\}$. Define the operations ⊙ and → on $\mathbb{H}$ as below,

$$\begin{array}{ccccccc}\to \hfill & 0& a& b& c& d& 1\\ 0\hfill & 1& 1& 1& 1& 1& 1\\ a\hfill & d& 1& d& 1& d& 1\\ b\hfill & c& c& 1& 1& 1& 1\\ c\hfill & b& c& d& 1& d& 1\\ d\hfill & a& a& b& c& 1& 1\\ 1\hfill & 0& a& b& c& d& 1\end{array}\begin{array}{ccccccc}\odot \hfill & 0& a& b& c& d& 1\\ 0\hfill & 0& 0& 0& 0& 0& 0\\ a\hfill & 0& a& 0& a& 0& a\\ b\hfill & 0& 0& 0& 0& b& b\\ c\hfill & 0& a& 0& a& b& c\\ d\hfill & 0& 0& b& b& d& d\\ 1\hfill & 0& a& b& c& d& 1\end{array}$$

Then $(\mathbb{H},\odot ,\to ,0,1)$ is a hoop and $I=\{0,b,d\}$ is a co-filter of $\mathbb{H}$, which is not a filter of $\mathbb{H}$ because $1\notin I$.

Note. For $S\subseteq \mathbb{H}$, define $S^{\prime }=\{\alpha \in \mathbb{H}\mid {\alpha }^{\prime }\in S\}$.

Proposition 5.

If $\mathbb{H}$ has (DNP) and $\mathsf{\varnothing }\ne I\subseteq \mathbb{H}$, then I is a filter of $\mathbb{H}$ iff $I^{\prime }$ is a co-filter of $\mathbb{H}$.

Proof. 

$(\Rightarrow )$ Suppose $I\in \mathcal{F}(\mathbb{H})$. Then $1\in I$, and so $0=1{}^{\prime }\in I{}^{\prime }$. Let $\alpha ,\beta \in \mathbb{H}$ such that $(\alpha \to \beta ){}^{\prime }\in I^{\prime }$ and $\beta \in I{}^{\prime }$. Since $\mathbb{H}$ has (DNP), by Proposition 2(iv), $\beta {}^{\prime }\to \alpha {}^{\prime }=\alpha \to \beta \in I$, $\beta {}^{\prime }\in I$ and since $I\in \mathcal{F}(\mathbb{H})$, by Definition 2, $\alpha {}^{\prime }\in I$, and so ${\alpha }^{″}=\alpha \in I{}^{\prime }$. Hence, $I{}^{\prime }$ is a co-filter of $\mathbb{H}$.

$(\Leftarrow )$ Let $I{}^{\prime }$ be a co-filter of $\mathbb{H}$. Then $0\in I{}^{\prime }$, and so $1\in I$. Now, suppose $\alpha ,\beta \in A$ such that $\alpha ,\alpha \to \beta \in I$. Thus $(\alpha \to \beta ){}^{\prime }\in I^{\prime }$ and $\alpha {}^{\prime }\in I{}^{\prime }$. Since $\mathbb{H}$ has (DNP), by Proposition 2(iv), $(\beta {}^{\prime }\to \alpha {}^{\prime }){}^{\prime }\in I^{\prime }$, $\alpha {}^{\prime }\in I^{\prime }$ and since $I{}^{\prime }$ is a co-filter of $\mathbb{H}$, by definition, $\beta {}^{\prime }\in I{}^{\prime }$. Hemce, by (DNP), $\beta \in I$. Therefore, $I\in \mathcal{F}(\mathbb{H})$. □

If $\mathbb{H}$ does not have (DNP), then Proposition 5 is not true, in general. We show this in the following example.

Example 2.

Let $\mathbb{H}=\{0,a,b,1\}$ be a chain such that $0\le a\le b\le 1$ and two binary operations ⊙ and → which are given below,

$$\begin{array}{ccccc}\begin{array}{c}\to \hfill \end{array}\hfill & 0& a& b& 1\\ 0\hfill & 1& 1& 1& 1\\ a\hfill & a& 1& 1& 1\\ b\hfill & 0& a& 1& 1\\ 1\hfill & 0& a& b& 1\end{array}\begin{array}{ccccc}\begin{array}{c}\odot \hfill \end{array}\hfill & 0& a& b& 1\\ 0\hfill & 0& 0& 0& 0\\ a\hfill & 0& 0& a& a\\ b\hfill & 0& a& b& b\\ 1\hfill & 0& a& b& 1\end{array}$$

By routine calculations, $(\mathbb{H},\odot ,\to ,0,1)$ is a hoop that does not have (DNP). It is clear that $\mathbb{H}$ is a co-filter of $\mathbb{H}$ but ${\mathbb{H}}^{\prime }=\{0,a,1\}$ is not a filter of $\mathbb{H}$.

Note. If F is a proper filter of $\mathbb{H}$, then by Definition 2, $0\notin F$. Thus, F is not a co-filter of $\mathbb{H}$. On the other hand, for any proper co-filter I of $\mathbb{H}$, if $1\notin I$, then $I\notin \mathcal{F}(\mathbb{H})$.

Proposition 6.

Let I be a co-filter of $\mathbb{H}$. Then the following statements hold:

$(i)$ If $\alpha \le \beta$ and $\beta \in I$, then $\alpha \in I$, for any $\alpha ,\beta \in \mathbb{H}$.

$(ii)$ If $\alpha \in I$, then $\alpha \odot \beta \in I$, for any $\beta \in \mathbb{H}$.

$(iii)$ If $\mathbb{H}$ is a ∨-hoop with (DNP), then $\alpha \vee \beta \in I$, for any $\alpha ,\beta \in I$.

Proof. 

$(i)$ Let $\alpha ,\beta \in \mathbb{H}$ such that $\alpha \le \beta$ and $\beta \in I$. Then $\alpha \to \beta =1$, and so $(\alpha \to \beta ){}^{\prime }=0\in I$. Since I is a co-filter of $\mathbb{H}$, $(\alpha \to \beta ){}^{\prime }\in I$ and $\beta \in I$, we have $\alpha \in I$.

$(ii)$ Let $\alpha ,\beta \in \mathbb{H}$ and $\alpha \in I$. By Proposition 1(iii), $\alpha \odot \beta \le \alpha$. Since $\alpha \in I$, by (i), $\alpha \odot \beta \in I$.

$(iii)$ Suppose $\alpha ,\beta \in I$. By Proposition 4,

$$(\alpha \vee \beta )\to \beta =(\alpha \to \beta )\wedge (\beta \to \beta )=\alpha \to \beta ,$$

then $((\alpha \vee \beta )\to \beta ){}^{\prime }=(\alpha \to \beta ){}^{\prime }$. By Proposition 2(ii), $\alpha {}^{\prime }\le \alpha \to \beta$, and so, by Proposition 1(viii) and (DNP), $(\alpha \to \beta ){}^{\prime }\le \alpha {}^{″}=\alpha$. Hence, $((\alpha \vee \beta )\to \beta ){}^{\prime }\le \alpha$. From I is a co-filter of $\mathbb{H}$ and $\alpha \in I$, by (i) $((\alpha \vee \beta )\to \beta ){}^{\prime }\in I$. Moreover, by assumption, $\beta \in I$ and I is a co-filter of $\mathbb{H}$. Therefore, $\alpha \vee \beta \in I$. □

Corollary 1.

If I is a co-filter of $\mathbb{H}$ and $1\in I$, then $I=\mathbb{H}$.

Proof. 

By Proposition 6(i), the proof is straightforward. □

We provide conditions for a nonempty subset to be a co-filter.

Proposition 7.

Let $\alpha ,\beta \in \mathbb{H}$ and $\varnothing \ne I\subseteq \mathbb{H}$ such that I has the following properties,

$(i)$ if $\alpha ,\beta \in I$, then $\alpha {}^{\prime }\to \beta \in I$,

$(ii)$ if $\alpha \le \beta$ and $\beta \in I$, then $\alpha \in I$.

Then I is a co-filter of $\mathbb{H}$.

Proof. 

Let $\alpha \in I$. Since, for all $\alpha \in \mathbb{H}$, $0\le \alpha$, by (ii), $0\in I$. Suppose $\alpha ,\beta \in \mathbb{H}$ such that $(\alpha \to \beta ){}^{\prime }\in I$ and $\beta \in I$. Then by (i),

$$\beta {}^{\prime }\to (\alpha \to \beta ){}^{\prime }=\beta {}^{\prime }\to ((\alpha \to \beta )\to 0)\in I.$$

By (HP3), $(\alpha \to \beta )\to \beta {}^{″}\in I$. Moreover, by Propositions 2(i) and 1(viii), $\beta \le \beta {}^{″}$ and so $(\alpha \to \beta )\to \beta \le (\alpha \to \beta )\to \beta {}^{″}$. Since $(\alpha \to \beta )\to \beta {}^{″}\in I$, by (ii), $(\alpha \to \beta )\to \beta \in I$. Also, by Proposition 1(vi), $\alpha \le (\alpha \to \beta )\to \beta$, and by (ii), $\alpha \in I$. Hence I is a co-filter of $\mathbb{H}$. □

By below example, we show that the converse of Proposition 7, is not true.

Example 3.

Let $\mathbb{H}=\{0,a,b,c,1\}$ be a set with the following Cayley tabels:

$$\begin{array}{cccccc}\to \hfill & 0& a& b& c& 1\\ 0\hfill & 1& 1& 1& 1& 1\\ a\hfill & b& 1& 0& 0& 1\\ b\hfill & c& 0& 1& 0& 1\\ c\hfill & c& 0& 0& 1& 1\\ 1\hfill & 0& a& b& c& 1\end{array}\begin{array}{cccccc}\odot \hfill & 0& a& b& c& 1\\ 0\hfill & 0& 0& 0& 0& 0\\ a\hfill & 0& a& 0& 0& a\\ b\hfill & 0& 0& b& 0& b\\ c\hfill & 0& 0& 0& c& c\\ 1\hfill & 0& a& b& c& 1\end{array}$$

Then $(\mathbb{H},\odot ,\to ,0,1)$ is a hoop and $I=\{0,a\}$ is a co-filter of $\mathbb{H}$ but $a^{\prime }\to 0=b\to 0=c\notin I$.

Proposition 8.

Let $\mathbb{H}$ has (DNP). Then I is a co-filter of $\mathbb{H}$ iff for any $\alpha ,\beta \in \mathbb{H}$, I has the following properties,

$(i)$ if $\alpha ,\beta \in I$, then $\alpha {}^{\prime }\to \beta \in I$.

$(ii)$ if $\alpha \le \beta$ and $\beta \in I$, then $\alpha \in I$.

Proof. 

(⇒) Let I be a co-filter of $\mathbb{H}$. Then by Proposition 6(i), item (ii) is clear. Suppose $\alpha ,\beta \in I$. By Proposition 1(vi) and (viii), ${((\alpha {}^{\prime }\to \beta )\to \beta )}^{\prime }\le {\alpha }^{″}$. Since $\mathbb{H}$ has (DNP), $((\alpha {}^{\prime }\to \beta )\to \beta ){}^{\prime }\le \alpha$. By assumption, $\alpha \in I$, and so by Proposition 6(i), $((\alpha {}^{\prime }\to \beta )\to \beta ){}^{\prime }\in I$. Moreover, since $\beta \in I$ and I is a co-filter of $\mathbb{H}$, $\alpha {}^{\prime }\to \beta \in I$.

(⇐) The proof is similar to the proof of Proposition 7. □

Theorem 1.

Let I be a co-filter of $\mathbb{H}$. Then, for all $\alpha ,\beta ,\gamma \in \mathbb{H}$, the following statements hold:

$(i)$$((\alpha \to \beta )\to \alpha ){}^{\prime }\in I$ and $\alpha \in I$ imply $\beta \in I$.

$(ii)$ If $\alpha \to (\alpha \to \beta )\in I$, then $\alpha \to \beta \in I$.

$(iii)$$((\beta \to (\beta \to \alpha ))\to \gamma ){}^{\prime }\in I$ and $\gamma \in I$ imply $\beta \to \alpha \in I$.

$(iv)$ If $(\alpha \to \beta ){}^{\prime }\in I$, then $((\alpha \to \beta ){}^{\prime }\to \beta ){}^{\prime }\in I$.

Proof. 

$(i)$ Let $\alpha ,\beta \in \mathbb{H}$ such that $((\alpha \to \beta )\to \alpha ){}^{\prime }\in I$ and $\alpha \in I$. Since I is a co-filter of $\mathbb{H}$, $\alpha \to \beta \in I$. By Proposition 1(iv), $\beta \le \alpha \to \beta \in I$. From Proposition 6(i), $\beta \in I$.

$(ii)$ By Proposition 1(iii) and (viii), ${\alpha }^2\le \alpha$ and $\alpha \to \beta \le {\alpha }^2\to \beta =\alpha \to (\alpha \to \beta )$. Since $\alpha \to (\alpha \to \beta )\in I$, by Proposition 6(i), $\alpha \to \beta \in I$.

$(iii)$ Suppose $\alpha ,\beta ,\gamma \in \mathbb{H}$ such that $((\beta \to (\beta \to \alpha ))\to \gamma ){}^{\prime }\in I$ and $\gamma \in I$. Since I is a co-filter of $\mathbb{H}$, $\beta \to (\beta \to \alpha )\in I$, and so by (ii), $\beta \to \alpha \in I$.

$(iv)$ Let $\alpha ,\beta \in \mathbb{H}$ such that $(\alpha \to \beta ){}^{\prime }\in I$. Then by (HP3), we have

$$\begin{array}{ccc}\hfill ((\alpha \to \beta ){}^{\prime }\to \beta ){}^{\prime }\to (\alpha \to \beta ){}^{\prime }& =& (\alpha \to \beta )\to ((\alpha \to \beta ){}^{\prime }\to \beta ){}^{″}\mathrm{by}\mathrm{Propositions}1(\mathrm{viii})\mathrm{and}2(\mathrm{i})\hfill \\ & \ge & (\alpha \to \beta )\to ((\alpha \to \beta ){}^{\prime }\to \beta )\mathrm{by}(\mathrm{HP}3)\hfill \\ & =& [(\alpha \to \beta )\odot (\alpha \to \beta ){}^{\prime }]\to \beta \mathrm{by}\mathrm{Propositions}2(\mathrm{i})\hfill \\ & =& 0\to \beta \hfill \\ & =& 1\hfill \end{array}$$

Thus, $((\alpha \to \beta ){}^{\prime }\to \beta ){}^{\prime }\to (\alpha \to \beta ){}^{\prime }=1$, and so $((\alpha \to \beta ){}^{\prime }\to \beta ){}^{\prime }\le (\alpha \to \beta ){}^{\prime }$. Since $(\alpha \to \beta ){}^{\prime }\in I$ and I is a co-filter of $\mathbb{H}$, by Proposition 6(i), $((\alpha \to \beta ){}^{\prime }\to \beta ){}^{\prime }\in I$. □

If $X\subseteq \mathbb{H}$, then the least co-filter of $\mathbb{H}$ contains X is called the co-filter generated by X of $\mathbb{H}$ and we show it by $[X)$.

Theorem 2.

If $\mathbb{H}$ has (DNP), then, for any $a\in \mathbb{H}$,

$$[a)=\{\alpha \in A\mid \exists n\in \mathbb{N}suchthat{(a{}^{\prime })}^n\le \alpha {}^{\prime }\}.$$

Proof. 

Let $B=\{\alpha \in A\mid \exists n\in \mathbb{N}\mathrm{such}\mathrm{that}{(a{}^{\prime })}^n\le \alpha {}^{\prime }\}$. Since ${(a{}^{\prime })}^n\le 1=0{}^{\prime }$, for all $n\in \mathbb{N}$, we have $0\in B$, and so $B\ne \varnothing$. Now, suppose $\alpha ,\beta \in \mathbb{H}$ such that $(\alpha \to \beta ){}^{\prime }\in B$ and $\beta \in B$. Then there exist $n,m\in \mathbb{N}$, such that ${(a{}^{\prime })}^n\le (\alpha \to \beta ){}^{″}$ and ${(a{}^{\prime })}^m\le \beta {}^{\prime }$. By Proposition 1(viii),

$${(a{}^{\prime })}^n\odot {(a{}^{\prime })}^m\le (\alpha \to \beta ){}^{″}\odot {(a{}^{\prime })}^m\le (\alpha \to \beta ){}^{″}\odot \beta {}^{\prime }$$

By (HP3), we get

$$\begin{array}{ccc}\hfill ((\alpha \to \beta ){}^{″}\odot \beta {}^{\prime })\to \alpha {}^{\prime }& =& \beta {}^{\prime }\to ((\alpha \to \beta ){}^{″}\to \alpha {}^{\prime })\hfill \\ & =& {\beta }^{\prime }\to (\alpha \to {(\alpha \to \beta )}^{‴})\mathrm{by}\mathrm{Proposition}2(\mathrm{iii})\hfill \\ & =& \beta {}^{\prime }\to (\alpha \to (\alpha \to \beta ){}^{\prime })\hfill \\ & =& \alpha \to (\beta {}^{\prime }\to (\alpha \to \beta ){}^{\prime })\hfill \\ & =& \alpha \to ((\alpha \to \beta )\to \beta {}^{″})\hfill \\ & =& (\alpha \to \beta )\to (\alpha \to \beta {}^{″})\mathrm{by}\mathrm{Propositions}2(\mathrm{i})\mathrm{and}1(\mathrm{viii})\hfill \\ & =& 1\hfill \end{array}$$

Then $(\alpha \to \beta ){}^{″}\odot \beta {}^{\prime }\le \alpha {}^{\prime }$, and so ${(a{}^{\prime })}^n\odot {(a{}^{\prime })}^m\le {\alpha }^{\prime }$. Hence, $n+m\in \mathbb{N}$ exists such that ${(a{}^{\prime })}^{n+m}\le \alpha {}^{\prime }$. Therefore, $\alpha \in B$, and so B is a co-filter of $\mathbb{H}$. Also, by Proposition 1(iii), ${(a{}^{\prime })}^n\le a{}^{\prime }$. Thus, $a\in B$ and B is a co-filter of $\mathbb{H}$ which containing a. Now, it is enough to prove that B is the least co-filter of $\mathbb{H}$ which containing a. Suppose C is a co-filter of $\mathbb{H}$ that contains a. We show that $B\subseteq C$. Let $\alpha \in B$. Then there exists $n\in \mathbb{N}$ such that ${(a{}^{\prime })}^n\le \alpha {}^{\prime }$. Thus ${(a{}^{\prime })}^n\to \alpha {}^{\prime }=1$. Since $\mathbb{H}$ has (DNP), by (HP3) and Proposition 2(iv), we get

$$\begin{array}{ccc}\hfill 1& =& {(a{}^{\prime })}^n\to \alpha {}^{\prime }\hfill \\ & =& ({(a{}^{\prime })}^{n-1}\odot (a{}^{\prime }))\to \alpha {}^{\prime }\hfill \\ & =& {(a{}^{\prime })}^{n-1}\to (a{}^{\prime }\to \alpha {}^{\prime })\hfill \\ & =& {(a{}^{\prime })}^{n-1}\to (\alpha \to a{}^{″})\hfill \\ & =& {(a{}^{\prime })}^{n-1}\to (\alpha \to a)\hfill \\ & =& ({(a{}^{\prime })}^{n-2}\odot (a{}^{\prime }))\to (\alpha \to a)\hfill \\ & =& {(a{}^{\prime })}^{n-2}\to (a{}^{\prime }\to (\alpha \to a))\hfill \\ & =& {(a{}^{\prime })}^{n-2}\to ((\alpha \to a){}^{\prime }\to a{}^{″})\hfill \\ & =& {(a{}^{\prime })}^{n-2}\to ((\alpha \to a){}^{\prime }\to a)\hfill \end{array}$$

By continuing this method, we have

$$1=((((\alpha \to a){}^{\prime }\to a){}^{\prime }\to a){}^{\prime }\to \dots \to a){}^{\prime }\to a$$

Hence,

$$[(((((\alpha \to a){}^{\prime }\to a){}^{\prime }\to a){}^{\prime })\to )\dots )\to a){}^{\prime }\to a]{}^{\prime }=1^{\prime }=0\in C.$$

Since C is a co-filter of $\mathbb{H}$ and $a\in C$, we obtain,

$$(((((\alpha \to a){}^{\prime }\to a){}^{\prime }\to a){}^{\prime })\to )\dots )\to a){}^{\prime }\in C.$$

By continuing this method, we can see that $(\alpha \to a){}^{\prime }\in C$. Since $(\alpha \to a){}^{\prime }\in C$, $a\in C$ and C is a co-filter of $\mathbb{H}$, we have $\alpha \in C$. Hence, $B\subseteq C$. Therefore, $B=[a)$. □

Corollary 2.

Let $\mathbb{H}$ has (DNP), $X\subseteq \mathbb{H}$ and $a\in \mathbb{H}$. Then the following statements hold:

$(i)$$[X)=\{\alpha \in A\mid \exists n\in \mathbb{N}and{a}_1,\dots ,a_n\in Xs.t,a_1{}^{\prime }\odot a_2{}^{\prime }\odot \dots \odot a_n{}^{\prime }\le \alpha {}^{\prime }\}$.

$(ii)$$[I\cup \{a\})=\{\beta \in A\mid \exists n,m\in \mathbb{N}and{\alpha }_1,\dots ,{\alpha }_m\in Is.t,({\alpha }_1^{\prime }\odot \dots \odot {\alpha }_m^{\prime })\odot {(a^{\prime })}^n\le {\beta }^{\prime }\}$.

Proof. 

The proof is similar to the proof of Theorem 2. □

Example 4.

Let A be the hoop as in Example 3. It is clear that A has (DNP). Since $a^{\prime }=d$ and $d\le d=a^{\prime }$ and $d\le 1=0^{\prime }$, we get $[a)=\{0,a\}$. Also, since $d^{\prime }=a$ and $a\le 1,a,c$, we have $[d)=\{0,b,d\}$.

Theorem 3.

Let I be a co-filter of $\mathbb{H}$. We define the relation ${\equiv }_I$ on $\mathbb{H}$ as follows,

$$\alpha {\equiv }_I\beta iff(\alpha \to \beta ){}^{\prime }\in Iand(\beta \to \alpha ){}^{\prime }\in I,forall\alpha ,\beta \in \mathbb{H}.$$

Then ${\equiv }_I$ is a congruence relation on $\mathbb{H}$.

Proof. 

At first, we prove that ${\equiv }_I$ is an equivalence relation on $\mathbb{H}$. Since, for all $\alpha \in \mathbb{H}$, $\alpha \to \alpha =1$ and I is a co-filter of $\mathbb{H}$, $(\alpha \to \alpha ){}^{\prime }=0\in I$. Thus, $\alpha {\equiv }_I\alpha$, and so ${\equiv }_I$ is reflexive. It is obvious that ${\equiv }_I$ is symmetric. For proving transitivity of ${\equiv }_I$, suppose $\alpha ,\beta ,\gamma \in \mathbb{H}$ such that $\alpha {\equiv }_I\beta$ and $\beta {\equiv }_I\gamma$. Hence, $(\alpha \to \beta ){}^{\prime },(\beta \to \alpha ){}^{\prime },(\beta \to \gamma ){}^{\prime }\mathrm{and}(\gamma \to \beta ){}^{\prime }\in I$ and by Proposition 1(vii) and (viii),

$$(\alpha \to \gamma ){}^{\prime }\le {((\alpha \to \beta )\odot (\beta \to \gamma ))}^{{}^{\prime }}.$$

By (HP3) and Propositions 1(vii),(viii) and 2(iii), we have,

$$\begin{array}{ccc}\hfill (((\alpha \to \beta )\to (\beta \to \gamma ){}^{\prime })\to (\alpha \to \beta ){}^{\prime }){}^{\prime }& =& (((\alpha \to \beta )\to (\beta \to \gamma ){}^{\prime })\to ((\alpha \to \beta )\to 0)){}^{\prime }\hfill \\ & =& ([(\alpha \to \beta )\odot ((\alpha \to \beta )\to (\beta \to \gamma ){}^{\prime })]\to 0){}^{\prime }\hfill \\ & \le & (\beta \to \gamma ){}^{‴}\hfill \\ & =& (\beta \to \gamma ){}^{\prime }\in I.\hfill \end{array}$$

Thus, by Proposition 6(i),

$$(((\alpha \to \beta )\to (\beta \to \gamma ){}^{\prime })\to (\alpha \to \beta ){}^{\prime }){}^{\prime }\in I.$$

Since $(\alpha \to \beta ){}^{\prime }\in I$ and I is a co-filter of $\mathbb{H}$, $(\alpha \to \beta )\to (\beta \to \gamma ){}^{\prime }\in I$. Moreover, $(\alpha \to \gamma ){}^{\prime }\le (\alpha \to \beta )\to (\beta \to \gamma ){}^{\prime }$ and $(\alpha \to \beta )\to (\beta \to \gamma ){}^{\prime }\in I$, by Proposition 6(i), $(\alpha \to \gamma ){}^{\prime }\in I$. By the similar way, $(\gamma \to \alpha ){}^{\prime }\in I$. Hence, $\alpha {\equiv }_I\gamma$. Therefore, ${\equiv }_I$ is an equivalence relation on $\mathbb{H}$. Now, let $\alpha {\equiv }_I\beta$, for some $\alpha ,\beta \in \mathbb{H}$. Then $(\alpha \to \beta ){}^{\prime },(\beta \to \alpha ){}^{\prime }\in I$. Thus, by Proposition 1(vi), $\alpha \le (\alpha \to \gamma )\to \gamma$, for all $\gamma \in \mathbb{H}$. So, by Proposition 1(viii), $\beta \to \alpha \le \beta \to ((\alpha \to \gamma )\to \gamma )$. Then by Proposition 1(viii) and (HP3),

$$((\alpha \to \gamma )\to (\beta \to \gamma )){}^{\prime }\le (\beta \to \alpha ){}^{\prime }.$$

Since $(\beta \to \alpha ){}^{\prime }\in I$ and I is a co-filter of $\mathbb{H}$, by Proposition 6(i), $((\alpha \to \gamma )\to (\beta \to \gamma )){}^{\prime }\in I$. By the similar way, ${((\beta \to \gamma )\to (\alpha \to \gamma ))}^{{}^{\prime }}\in I$. Hence, $\alpha \to \gamma {\equiv }_I\beta \to \gamma$. Suppose $\alpha {\equiv }_I\beta$, for some $\alpha ,\beta \in \mathbb{H}$. Then ${(\alpha \to \beta )}^{\prime },{(\beta \to \alpha )}^{\prime }\in I$ and by Proposition 1(vii) and (HP3), $\alpha \to \beta \le (\gamma \to \alpha )\to (\gamma \to \beta )$, for all $\gamma \in \mathbb{H}$. Also, by Proposition 1(viii),

$${((\gamma \to \alpha )\to (\gamma \to \beta ))}^{\prime }\le {(\alpha \to \beta )}^{\prime }.$$

From ${(\alpha \to \beta )}^{\prime }\in I$ and I is a co-filter of $\mathbb{H}$, by Proposition 6(i), ${((\gamma \to \alpha )\to (\gamma \to \beta ))}^{\prime }\in I$. By the similar way, ${((\gamma \to \beta )\to (\gamma \to \alpha ))}^{\prime }\in I$. Hence, $\gamma \to \alpha {\equiv }_I\gamma \to \beta$. Finally, if $\alpha {\equiv }_I\beta$, for some $\alpha ,\beta \in \mathbb{H}$, then $(\alpha \to \beta ){}^{\prime },(\beta \to \alpha ){}^{\prime }\in I$. From $\alpha \odot \gamma \le \alpha \odot \gamma$, by Proposition 1(ii),(viii) and (HP3), $\alpha \le \gamma \to (\alpha \odot \gamma )$, and so

$$\beta \to \alpha \le \beta \to (\gamma \to (\alpha \odot \gamma ))=(\beta \odot \gamma )\to (\alpha \odot \gamma ).$$

Then by Proposition 1(viii),

$$((\beta \odot \gamma )\to (\alpha \odot \gamma )){}^{\prime }\le (\beta \to \alpha ){}^{\prime }.$$

Since $(\beta \to \alpha ){}^{\prime }\in I$ and I is a co-filter of $\mathbb{H}$, by Proposition 6(i), $((\beta \odot \gamma )\to (\alpha \odot \gamma )){}^{\prime }\in I$. Similarly, $((\alpha \odot \gamma )\to (\beta \odot \gamma )){}^{\prime }\in I$. Hence, $\alpha \odot \gamma {\equiv }_I\beta \odot \gamma$. Therefore, ${\equiv }_I$ is a congruence relation on $\mathbb{H}$. □

For any $\alpha \in \mathbb{H}$, $I_{\alpha }$ will denote the equivalence class of α with respect to ${\equiv }_I$. It is clear that

$$I_{\alpha }=\{\beta \in \mathbb{H}\mid \alpha {\equiv }_I\beta \}=\{\beta \in \mathbb{H}\mid {(\alpha \to \beta )}^{\prime }\in I\mathrm{and}{(\beta \to \alpha )}^{\prime }\in I\}.$$

Easily we can see that $I_0=I$ and $I_1=\{\beta \in \mathbb{H}\mid {\beta }^{\prime }\in I\}$.

Theorem 4.

Let $\mathbb{H}/I=\{I_{\alpha }\mid \alpha \in \mathbb{H}\}$. Define the operations ⊗ and ⇝ on $\mathbb{H}/I$ as follows:

$$I_{\alpha }\otimes I_{\beta }=I_{\alpha \odot \beta }and{I}_{\alpha }⇝I_{\beta }=I_{\alpha \to \beta }.$$

Then $(\mathbb{H}/I,\otimes ,⇝,I_0,I_1)$ is a bounded hoop.

Proof. 

The proof is straightforward.

Note. Let $\alpha ,\beta \in \mathbb{H}$. Then the binary relation “${\le }_I$” is defined on $\mathbb{H}/I$ as follows,

$$I_{\alpha }{\le }_I{I}_{\beta }\mathrm{iff}(\alpha \to \beta ){}^{\prime }\in I.$$

Then ${\le }_I$ is a partially order relation on $\mathbb{H}/I$. Since $(\alpha \to \alpha ){}^{\prime }=0\in I$, for any $\alpha \in \mathbb{H}$, $I_{\alpha }{\le }_I{I}_{\alpha }$. Suppose $I_{\alpha }{\le }_I{I}_{\beta }$ and $I_{\beta }{\le }_I{I}_{\alpha }$, for any $\alpha ,\beta \in \mathbb{H}$. Then $(\alpha \to \beta ){}^{\prime }\in I$ and $(\beta \to \alpha ){}^{\prime }\in I$. Thus, $\alpha {\equiv }_I\beta$, and so $I_{\alpha }=I_{\beta }$. Now, let $I_{\alpha }{\le }_I{I}_{\beta }$ and $I_{\beta }{\le }_I{I}_{\gamma }$. Then $(\alpha \to \beta ){}^{\prime }\in I$ and $(\beta \to \gamma ){}^{\prime }\in I$. By Proposition 1(vii), for $\alpha ,\beta ,\gamma \in \mathbb{H}$, we have $(\alpha \to \beta )\odot (\beta \to \gamma )\le \alpha \to \gamma$. Thus, by Proposition 1(viii) and (HP3),

$$(\alpha \to \gamma ){}^{\prime }\le ((\alpha \to \beta )\odot (\beta \to \gamma )){}^{\prime },$$

and so

$$(\alpha \to \gamma ){}^{\prime }\le [((\alpha \to \beta )\odot (\beta \to \gamma ))\to 0].$$

Thus, $(\alpha \to \gamma ){}^{\prime }\le (\alpha \to \beta )\to ((\beta \to \gamma )\to 0)$. Moreover, by Proposition 1(ii),

$$\alpha \to \beta \le (\alpha \to \gamma ){}^{\prime }\to (\beta \to \gamma ){}^{\prime }.$$

Hence, by Proposition 1(viii), we obtain,

$$((\alpha \to \gamma ){}^{\prime }\to (\beta \to \gamma ){}^{\prime }){}^{\prime }\le (\alpha \to \beta ){}^{\prime }\in I.$$

Since ${(\alpha \to \beta )}^{\prime }\in I$ and I is a co-filter of $\mathbb{H}$, by Proposition 6(i), $((\alpha \to \gamma ){}^{\prime }\to (\beta \to \gamma ){}^{\prime }){}^{\prime }\in I$. Also, $(\beta \to \gamma ){}^{\prime }\in I$, then $(\alpha \to \gamma ){}^{\prime }\in I$. Hence, $I_{\alpha }\le I_{\gamma }$. Therefore, ${\le }_I$ is a partially order relation on $\mathbb{H}/I$.

For proving $(\mathbb{H}/I,\otimes ,⇝,I_0,I_1)$ is a bounded hoop, we have $I_{\alpha }=I_{\beta }$ and $I_{\gamma }=I_{\delta }$ iff $\alpha {\equiv }_I\beta$ and $\gamma {\equiv }_I\delta$. Since ${\equiv }_I$ is a congruence relation on $\mathbb{H}$, so all operations are well-defined. Thus, by routine calculations, we can see that $(\mathbb{H}/I,\otimes ,I_1)$ is a commutative monoid and (HP2) holds. Let $I_{\alpha },I_{\beta },I_{\gamma }\in \mathbb{H}/I$, for any $\alpha ,\beta ,\gamma \in \mathbb{H}$. Since $\mathbb{H}$ is a hoop, by (HP3) and (HP4) we have,

$$(I_{\alpha }\otimes I_{\beta })⇝I_{\gamma }=I_{(\alpha \odot \beta )}⇝I_{\gamma }=I_{(\alpha \odot \beta )\to \gamma }=I_{\alpha \to (\beta \to \gamma )}=I_{\alpha }⇝(I_{\beta \to \gamma })=I_{\alpha }⇝(I_{\beta }⇝I_{\gamma }).$$

Also, for any $I_{\alpha },I_{\beta }\in \mathbb{H}/I$, we get

$$I_{\alpha }\otimes (I_{\alpha }⇝I_{\beta })=I_{\alpha }\otimes (I_{\alpha \to \beta })=I_{\alpha \odot (\alpha \to \beta )}=I_{\beta \odot (\beta \to \alpha )}=I_{\beta }\otimes (I_{\beta \to \alpha })=I_{\beta }\otimes (I_{\beta }⇝I_{\alpha }).$$

Therefore, $(\mathbb{H}/I,\otimes ,⇝,I_0,I_1)$ is a bounded hoop. □

Example 5.

Let A be the hoop as in Example 3. Then $I=\{0,b,d\}$ is a co-filter of A. Thus, by routine calculations, we can see that $\left[b\right]=\left[d\right]=\left[0\right]=\{0,b,d\}$ and $\left[a\right]=\left[c\right]=\left[1\right]=\{a,c,1\}$. Hence, $\frac{A}{{\equiv }_I}=\{\left[0\right],\left[1\right]\}$. Therefore, $\frac{A}{{\equiv }_I}$ is a bounded hoop.

Example 6.

Let A be the hoop as in Example 2. We can see that A does not have (DNP) property, in general. So by Proposition 5 and Example 2, filter and co-filter are different notions. Then A is a co-filter of A and the quotient is $\frac{A}{A}=\{\left[1\right]\}$ that is a hoop algebra. But $F=\{b,1\}$ is a filter of A and the quotient $\frac{A}{{\equiv }_F}=\{\left[0\right],\left[a\right],\left[1\right]\}$ that is a hoop with (DNP).

4. Some Applications of Co-Filters

In this section, we try to investigate under which conditions the quotient structure of this congruence relation will be Brouwerian semilattice, Heyting algebra, Wajsberg hoop, Hilbert algebra and BL-algebra.

Definition 4

([11]). A Brouwerian lattice is an algebra $(\mathbb{H},\wedge ,\vee ,\to ,{}^{\prime })$ with the lattice infimum ($\wedge$) and the lattice supremum ($\vee$) in which two operations “$\prime$” and “$\to$” are defined by ${\alpha }^{\prime }=\alpha \to 0$ and

$$\alpha \wedge \beta \le \gamma iff\alpha \le \beta \to \gamma$$

respectively.

Theorem 5.

Let I be a co-filter of $\mathbb{H}$ and for all $\alpha \in \mathbb{H}$, ${\alpha }^2=\alpha$. Then $\mathbb{H}/I$ is a Brouwerian semilattice.

Proof. 

Let I be a co-filter of $\mathbb{H}$. By Theorem 4, $\mathbb{H}/I$ is a hoop. Thus, by Proposition 1(i), $(\mathbb{H}/I,{\le }_I)$ is a meet-semilattice with $I_{\alpha }{\wedge }_I{I}_{\beta }=I_{\alpha }\otimes (I_{\alpha }⇝I_{\beta })$, for all $I_{\alpha },I_{\beta }\in \mathbb{H}/I$. Now, we prove that, for all $I_{\alpha },I_{\beta }\in \mathbb{H}/I$,

$$I_{\alpha }{\wedge }_I{I}_{\beta }\le I_{\gamma }\mathrm{iff}{I}_{\alpha }\le I_{\beta }⇝I_{\gamma }.$$

Since $\mathbb{H}/I$ is a hoop, by Proposition 1(iii), $I_{\alpha }\otimes I_{\beta }\le I_{\alpha }{\wedge }_I{I}_{\beta }\le I_{\gamma }$. Thus, $I_{\alpha }\otimes I_{\beta }\le I_{\gamma }$, and so by Proposition 1(ii), $I_{\alpha }\le I_{\beta }⇝I_{\gamma }$. Conversely, suppose $I_{\alpha }\le I_{\beta }⇝I_{\gamma }$, for all $I_{\alpha },I_{\beta },I_{\gamma }\in \mathbb{H}/I$. According to definition of ${\le }_I$, ${(\alpha \to (\beta \to \gamma ))}^{\prime }\in I$. By Proposition 1(vii), $\beta \to \gamma \le (\alpha \to \beta )\to (\alpha \to \gamma )$ and by (HP3), $\beta \to \gamma \le (\alpha \odot (\alpha \to \beta ))\to \gamma$. Also, by Proposition 1(viii) and (HP3), we get

$$\alpha \to (\beta \to \gamma )\le \alpha \to ((\alpha \odot (\alpha \to \beta ))\to \gamma ),$$

and so

$$\alpha \to (\beta \to \gamma )\le (\alpha \odot (\alpha \odot (\alpha \to \beta ))\to \gamma ).$$

Thus,

$$\alpha \to (\beta \to \gamma )\le (((\alpha \odot \alpha )\odot (\alpha \to \beta ))\to \gamma ).$$

Since for any $\alpha \in \mathbb{H}$, ${\alpha }^2=\alpha$, we obtain, $\alpha \to (\beta \to \gamma )\le ((\alpha \odot (\alpha \to \beta ))\to \gamma )$ and so,

$$\alpha \to (\beta \to \gamma )\le (\alpha \to \beta )\to (\alpha \to \gamma ).$$

Hence, by Proposition 1(viii), we get ${((\alpha \to \beta )\to (\alpha \to \gamma ))}^{\prime }\le {(\alpha \to (\beta \to \gamma ))}^{\prime }$. Since I is a co-filter of $\mathbb{H}$ and ${(\alpha \to (\beta \to \gamma ))}^{\prime }\in I$, by Proposition 6(i), ${((\alpha \to \beta )\to (\alpha \to \gamma ))}^{\prime }\in I$, so $I_{\alpha }⇝I_{\beta }\le I_{\alpha }⇝I_{\gamma }$. Thus, by Proposition 1(ii),(i) and (viii),

$$I_{\alpha }{\wedge }_I{I}_{\beta }=I_{\alpha }\otimes (I_{\alpha }⇝I_{\beta })\le I_{\alpha }\otimes (I_{\alpha }⇝I_{\gamma })=I_{\alpha }\wedge I_{\gamma }\le I_{\gamma }.$$

Hence, $I_{\alpha }{\wedge }_I{I}_{\beta }\le I_{\gamma }$. Therefore, $\mathbb{H}/I$ is a Brouwerian semilattice. □

Example 7.

Let $\mathbb{H}=\{0,a,b,c,1\}$ be a set with two operations which are given below:

$$\begin{array}{cccccc}\to \hfill & 0\hfill & a& b& c& 1\\ 0\hfill & 1\hfill & 1& 1& 1& 1\\ a\hfill & b\hfill & 1& b& 1& 1\\ b\hfill & a\hfill & a& 1& 1& 1\\ c\hfill & 0\hfill & a& b& 1& 1\\ 1\hfill & 0\hfill & a& b& c& 1\end{array}\begin{array}{cccccc}\odot \hfill & 0\hfill & a& b& c& 1\\ 0\hfill & 0\hfill & 0& 0& 0& 0\\ a\hfill & 0\hfill & a& 0& a& a\\ b\hfill & 0\hfill & 0& b& b& b\\ c\hfill & 0\hfill & a& b& c& c\\ 1\hfill & 0\hfill & a& b& c& 1\end{array}$$

Thus, $(\mathbb{H},\odot ,\to ,0,1)$ is a hoop and ${\alpha }^2=\alpha$, for all $\alpha \in \mathbb{H}$. Then $I=\{0,a\}$ is a co-filter of $\mathbb{H}$, $I_0=I_a=I$ and $I_b=I_c=I_1=\{b,c,1\}$. Hence, by Theorem 5, $\mathbb{H}/I=\{I_0,I_1\}$ is a Brouwerian semilattice.

Theorem 6.

Let $\mathbb{H}$ has (DNP) and $\mathbb{H}/I$ be a Brouwerian semilattice. Then I is a co-filter of $\mathbb{H}$.

Proof. 

Let $I_{\alpha }{\otimes }_I{I}_{\beta }=I_{\alpha }{\wedge }_I{I}_{\beta }$, for all $I_{\alpha },I_{\beta }\in \mathbb{H}/I$. Then $I_{\alpha }{\otimes }_I{I}_{\alpha }=I_{\alpha }{\wedge }_I{I}_{\alpha }=I_{\alpha }$. Thus, $I_{{\alpha }^2}=I_{\alpha }$, and so ${({\alpha }^2\to \alpha )}^{\prime }\in I$. By Proposition 1(iii), $0\in I$. Now, suppose ${(\alpha \to \beta )}^{\prime }$ and $\beta \in I$, for some $\alpha ,\beta \in \mathbb{H}$. Since $I=I_0$, we have $\beta \in I_0$. It means that ${(\beta \to 0)}^{\prime }\in I$, and equivalently $I_{\beta }\le I_0$. Moreover, ${(\alpha \to \beta )}^{\prime }\in I$, then $I_{\alpha }\le I_{\beta }$, and so $I_{\alpha }\le I_0$ i.e., ${(\alpha \to 0)}^{\prime }\in I$. Hence, ${\alpha }^{″}\in I$. Since $\mathbb{H}$ has (DNP), we get $\alpha \in I$. Therefore, I is a co-filter of $\mathbb{H}$. □

Definition 5

([11]). A hoop $(\mathbb{H},\odot ,\to ,1)$ is called Wajsberg if, for any $\alpha ,\beta \in \mathbb{H}$,

$$(\alpha \to \beta )\to \beta =(\beta \to \alpha )\to \alpha .$$

Theorem 7.

Let $\mathbb{H}$ has (DNP). Then I is a co-filter of $\mathbb{H}$ iff $\mathbb{H}/I$ is a Wajsberg hoop.

Proof. 

(⇒) Since $\mathbb{H}$ has (DNP), by Proposition 2(v), $(\alpha \to \beta )\to \beta =(\beta \to \alpha )\to \alpha$, for all $\alpha ,\beta \in \mathbb{H}$. Thus,

$${(((\alpha \to \beta )\to \beta )\to ((\beta \to \alpha )\to \alpha ))}^{\prime }=0\in I,$$

and so $(I_{\alpha }⇝I_{\beta })⇝I_{\beta }\le (I_{\beta }⇝I_{\alpha })⇝I_{\alpha }$. By the similar way, $(I_{\beta }⇝I_{\alpha })⇝I_{\alpha }\le (I_{\alpha }⇝I_{\beta })⇝I_{\beta }$. Thus, $(I_{\alpha }⇝I_{\beta })⇝I_{\beta }=(I_{\beta }⇝I_{\alpha })⇝I_{\alpha }$, for all $I_{\alpha },I_{\beta }\in \mathbb{H}/I$. Therefore, $\mathbb{H}/I$ is a Wajsberg hoop.

(⇐) The proof is similar to the proof of Theorem 6. □

Example 8.

In Example 1, $\mathbb{H}$ is a hoop with (DNP). Since $I=\{0,b,d\}$ is a co-filter of $\mathbb{H}$, $I_0=I_b=I_d=I$ and $I_a=I_c=I_1=\{a,c,1\}$. Hence, by Theorem 7, $\mathbb{H}/I=\{I_0,I_1\}$ is a Wajsberg hoop.

Definition 6

([11]). A Heyting algebra is an algebra $(A,\vee ,\wedge ,\to ,1)$, where $(A,\vee ,\wedge ,1)$ is a distributive lattice with the greatest element and the binary operation → on A verifies, for any $x,y,z\in A$,

$$x\le y\to ziffx\wedge y\le z.$$

Theorem 8.

Let $\mathbb{H}$ has (DNP) and ${\alpha }^2=\alpha$, for all $\alpha \in \mathbb{H}$. Then I is a co-filter of $\mathbb{H}$ iff $\mathbb{H}/I$ is a Heyting algebra.

Proof. 

(⇒) Since I is a co-filter of $\mathbb{H}$ and ${\alpha }^2=\alpha$, for all $\alpha \in \mathbb{H}$, by Theorem 5, $\mathbb{H}/I$ is a Brouwerian semilattice. Moreover, since $\mathbb{H}$ has (DNP), by Theorem 7, $\mathbb{H}/I$ is a Wajsberg hoop. Define $I_{\alpha }{\vee }_I{I}_{\beta }=(I_{\beta }⇝I_{\alpha })⇝I_{\alpha }$, for all $I_{\alpha },I_{\beta }\in \mathbb{H}/I$. Then by Propositions 3 and 4, $(\mathbb{H}/I,{\wedge }_I,{\vee }_I)$ is a distributive lattice. Therefore, $\mathbb{H}/I$ is a Heyting algebra.

(⇐) Since $\mathbb{H}/I$ is a Heyting algebra, it is a Brouwerian semilattice. On the other side, $\mathbb{H}$ has (DNP), then by Theorem 6, I is a co-filter of $\mathbb{H}$. □

Example 9.

Let $\mathbb{H}=\{0,a,b,1\}$ be a set with the following Cayley tabels,

$$\begin{array}{ccccc}\to \hfill & 0& a& b& 1\\ 0\hfill & 1& 1& 1& 1\\ a\hfill & b& 1& b& 1\\ b\hfill & a& a& 1& 1\\ 1\hfill & 0& a& b& 1\end{array}\begin{array}{ccccc}\odot \hfill & 0& a& b& 1\\ 0\hfill & 0& 0& 0& 0\\ a\hfill & 0& a& 0& a\\ b\hfill & 0& 0& b& b\\ 1\hfill & 0& a& b& 1\end{array}$$

Then $(\mathbb{H},\odot ,\to ,0,1)$ is a hoop with (DNP) and for any $\alpha \in \mathbb{H}$, ${\alpha }^2=\alpha$. From $I=\{0,b\}$ is a co-filter of $\mathbb{H}$, $I_0=I_b=I and I_1=I_a=\{1,a\}$. Then by Theorem 8, $\mathbb{H}/I=\{I,I_1\}$ is a Heyting algebra.

Definition 7

([11]). A Hilbert algebra is a tripe $(A,\to ,1)$ of type $(2,0)$ such that, for all $x,y,z\in A$, the following three axioms are satisfied,

$(H1)$$x\to (y\to x)=1$.

$(H2)$$x\to (y\to z)=(x\to y)\to (x\to z)$.

$(H3)$ If $x\to y=y\to x=1$, then $x=y$.

The Hilbert algebra induces a partial order ≤ on A, defined by, $x\le y$ iff $x\to y=1$ and 1 is the greatest element of the induced poset $(A,\le )$. A Hilbert algebra A is bounded if there is an element $0\in A$ such that, for any $x\in A$, $0\le x$.

Lemma 1.

Let ${\alpha }^2=\alpha$, for all $\alpha \in \mathbb{H}$. Then, for all $\alpha ,\beta ,\gamma \in \mathbb{H}$,

$$\alpha \to (\beta \to \gamma )=(\alpha \to \beta )\to (\alpha \to \gamma ).$$

Proof. 

Let $\alpha \in \mathbb{H}$ such that ${\alpha }^2=\alpha$. Then by Proposition 1(iv), $\beta \le \alpha \to \beta$, for any $\alpha ,\beta \in \mathbb{H}$ and by Proposition 1(viii), $\alpha \to ((\alpha \to \beta )\to \gamma )\le \alpha \to (\beta \to \gamma )$. Then by (HP3), $(\alpha \to \beta )\to (\alpha \to \gamma )\le \alpha \to (\beta \to \gamma )$. Conversely, by (HP3), for all $\alpha ,\beta ,\gamma \in \mathbb{H}$,

$$[\alpha \to (\beta \to \gamma )]\to [(\alpha \to \beta )\to (\alpha \to \gamma )]=[(\alpha \to \beta )\odot \alpha \odot (\alpha \to (\beta \to \gamma ))]\to \gamma .$$

By Proposition 1(vii), $\alpha \odot (\alpha \to (\beta \to \gamma ))\le \beta \to \gamma$. Then by Proposition 1(viii) and (vii),

$$(\alpha \to \beta )\odot \alpha \odot (\alpha \to (\beta \to \gamma ))\le (\alpha \to \beta )\odot (\beta \to \gamma )\le \alpha \to \gamma .$$

Thus, ${\alpha }^2\odot (\alpha \to \beta )\odot (\alpha \to (\beta \to \gamma ))\le \gamma$. Since ${\alpha }^2=\alpha$, we get $\alpha \odot (\alpha \to \beta )\odot (\alpha \to (\beta \to \gamma ))\le \gamma$. Hence, by (HP3), $\alpha \to (\beta \to \gamma )\le (\alpha \to \beta )\to (\alpha \to \gamma )$. □

Theorem 9.

Let I be a co-filter of $\mathbb{H}$ and ${\alpha }^2=\alpha$, for all $\alpha \in \mathbb{H}$. Then $\mathbb{H}/I$ is a Hilbert algebra.

Proof. 

Since I is a co-filter of $\mathbb{H}$, by Theorem 5, $\mathbb{H}/I$ is a hoop. Thus by Proposition 1(iv), it is clear that $I_{\alpha }⇝(I_{\beta }⇝I_{\alpha })=I_1$, for all $I_{\alpha },I_{\beta }\in \mathbb{H}/I$. Let $I_{\alpha },I_{\beta }\in \mathbb{H}/I$ such that $I_{\alpha }⇝I_{\beta }=I_{\beta }⇝I_{\alpha }=I_1$. Then ${(\alpha \to \beta )}^{\prime }\in I$ and ${(\beta \to \alpha )}^{\prime }\in I$ and so $\alpha {\equiv }_I\beta$. Hence, $I_{\alpha }=I_{\beta }$. Moreover, since ${\alpha }^2=\alpha$, for all $\alpha \in \mathbb{H}$, by Lemma 1, $\alpha \to (\beta \to \gamma )=(\alpha \to \beta )\to (\alpha \to \gamma )$, for all $\alpha ,\beta ,\gamma \in \mathbb{H}$, and so

$${[(\alpha \to (\beta \to \gamma ))\to ((\alpha \to \beta )\to (\alpha \to \gamma ))]}^{\prime }=0\in I.$$

Thus, by definition of $I_1$,

$$(I_{\alpha }⇝(I_{\beta }⇝I_{\gamma }))⇝((I_{\alpha }⇝I_{\beta })⇝(I_{\alpha }⇝I_{\gamma }))=I_1$$

Therefore, $\mathbb{H}/I$ is a Hilbert algebra. □

Definition 8

([11]). A BL-algebra is an algebra $(A,\vee ,\wedge ,\odot ,\to ,0,1)$ of type $(2,2,2,2,0,0)$ that, for any $x,y,z\in A$, it is satisfying the following axioms:

$(BL1)$$(A,\vee ,\wedge ,0,1)$ is a bounded lattice.

$(BL2)$$(A,\odot ,1)$ is a commutative monoid.

$(BL3)$$x\odot y\le z$ iff $x\le y\to z$.

$(BL4)$$(x\to y)\vee (y\to x)=1$.

$(BL5)$$x\wedge y=x\odot (x\to y)$.

Theorem 10.

Let $\mathbb{H}$ be a ∨-hoop such that, for all $\alpha \in \mathbb{H}$, ${\alpha }^2=\alpha$ and I be a co-filter of $\mathbb{H}$. Then $\mathbb{H}/I$ is a BL-algebra.

Proof. 

Let $\mathbb{H}$ be a ∨-hoop. Then $\mathbb{H}/I$ is a ${\vee }_I$-hoop. Thus, by Proposition 4, $(\mathbb{H}/I,{\wedge }_I,{\vee }_I,I_0,I_1)$ is a bounded distributive lattice. Now, we prove that $\mathbb{H}/I$ is a BL-algebra. For this, it is enough to prove that $(I_{\alpha }⇝I_{\beta }){\vee }_I(I_{\beta }⇝I_{\alpha })=I_1$, for all $I_{\alpha },I_{\beta }\in \mathbb{H}/I$. Equivalently, we show that ${((\alpha \to \beta )\vee (\beta \to \alpha ))}^{\prime }\in I$, for all $\alpha ,\beta \in \mathbb{H}$. Since for all $\alpha ,\beta \in \mathbb{H}$,

$$(\alpha \to \beta ),(\beta \to \alpha )\le (\alpha \to \beta )\vee (\beta \to \alpha ),$$

by Proposition 1(viii),

$${((\alpha \to \beta )\vee (\beta \to \alpha ))}^{\prime }\le {(\alpha \to \beta )}^{\prime }\wedge {(\beta \to \alpha )}^{\prime }.$$

On the other hand, by Proposition 1(iv), $\beta \le \alpha \to \beta$ and by Proposition 2(ii), ${\beta }^{\prime }\le \beta \to \alpha$, then by Proposition 1(viii), ${(\alpha \to \beta )}^{\prime }\le {\beta }^{\prime }$ and ${(\beta \to \alpha )}^{\prime }\le {\beta }^{″}$. Thus, by Propositions 1(i), 2(i) and ${\alpha }^2=\alpha$, for all $\alpha \in \mathbb{H}$, we have

$$\begin{array}{ccc}\hfill {((\alpha \to \beta )\vee (\beta \to \alpha ))}^{\prime }& \le & {(\alpha \to \beta )}^{\prime }\wedge {(\beta \to \alpha )}^{\prime }\hfill \\ & \le & {\beta }^{\prime }\wedge {\beta }^{″}\hfill \\ & =& {\beta }^{\prime }\odot ({\beta }^{\prime }\to {\beta }^{″})\mathrm{by}\mathrm{Proposition}1(i)\hfill \\ & =& {\beta }^{\prime }\odot ({\beta }^{\prime }\to ({\beta }^{\prime }\to 0))\mathrm{by}(\mathrm{HP}3)\hfill \\ & =& {\beta }^{\prime }\odot ({({\beta }^{\prime })}^2\to 0)\mathrm{by}{\alpha }^2=\alpha \hfill \\ & =& {\beta }^{\prime }\odot ({\beta }^{\prime }\to 0)\mathrm{by}\mathrm{Proposition}2(i)\hfill \\ & =& {\beta }^{\prime }\odot {\beta }^{″}\hfill \\ & =& 0.\hfill \end{array}$$

Then ${((\alpha \to \beta )\vee (\beta \to \alpha ))}^{\prime }=0\in I$. Therefore, $\mathbb{H}/I$ is a BL-algebra. □

Theorem 11.

Let $\mathbb{H}$ has (DNP) and ${\alpha }^2=\alpha$, for all $\alpha \in \mathbb{H}$. Then I is a co-filter of $\mathbb{H}$ iff $\mathbb{H}/I$ is a BL-algebra.

Proof. 

(⇒) Since $\mathbb{H}$ has (DNP) and I is a co-filter of $\mathbb{H}$, by Theorem 7, $\mathbb{H}/I$ is a Wajsberg hoop. Define $I_{\alpha }{\vee }_I{I}_{\beta }=(I_{\alpha }⇝I_{\beta })⇝I_{\beta }$ for all $I_{\alpha },I_{\beta }\in \mathbb{H}/I$. Then by Proposition 3, $\mathbb{H}/I$ is a ${\vee }_I$-hoop, and so by Proposition 4, $(\mathbb{H}/I,{\wedge }_I,{\vee }_I,I_0,I_1)$ is a bounded lattice. On the other side, since ${\alpha }^2=\alpha$, for all $\alpha \in \mathbb{H}$, by Theorem 10, $\mathbb{H}/I$ is a BL-algebra.

(⇐) Since $\mathbb{H}$ has (DNP) and $\mathbb{H}/I$ is a BL-algebra, $\mathbb{H}/I$ is a distributive lattice. Thus, by Theorem 6, I is a co-filter of $\mathbb{H}$. □

Remark 1.

As you see in this section, we investigated the relation among the quotient hoop $\frac{A}{I}$ that is made by a co-filter I with other logical algebras such as Brouwerian semi-lattice, Heyting algebra, Hilbert algebra, Wajsberg hoop and BL-algebra. Clearly these conditions are similar and we know that for example if A has Godel condition ($x^2=x$) then $\frac{A}{I}$ is Hilbert algebra and by adding (DNP) property to A we obtain that $\frac{A}{I}$ is Heyting algebra.

5. Conclusions and Future Works

We have introduced the notion of co-filter of hoops and a congruence relation on hoop, and then we have constructed the quotient structures by using co-filters. We have considered the relation between filters and co-filters in a hoop with (DNP) property. We have provided conditions for a subset to be a co-filter. We have discussed characterizations of a co-filter. We have studied the relation among this structure and other algebraic structures. Using the notion of co-filters, we have established the quotient Brouwerian semilattice, the quotient Hilbert algebra and the quotient BL-algebra. We have induced a co-filter from a quotient Brouwerian semilattice. In our subsequent research, we will study some kinds of co-filter such as, implicative, ultra and prime one and investigate the relation between them. Also, we will discuss fuzzy co-filters and fuzzy congruence relation by them and study the quotient structure of this fuzzy congruence relation.