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\begin{document}

\sloppy
\noindent
{\bf  {\textquotedblleft Vasile Alecsandri" University  of Bac\u au \\
Faculty of Sciences \\
Scientific Studies and Research \\
Series Mathematics and Informatics \\
Vol. 35 (2025), No. 1, 5 - 15} }

\vspace{0.5cm}

\title[Strong ideals in QI-algebras]{Strong ideals in QI-algebras}
\author[D. A. Romano]{Daniel Abraham Romano}
\maketitle

\vspace{0.5cm}

\textbf{Abstract. }The notion of QI-algebras was introduced in 2017 as a generalization of the concept of BI-algebras. In this article, the concepts of strong  ideals in QI-algebras are created and its properties are observed. Also, we study some properties of weak ideals and $\alpha$-ideals in a
QI-algebra. Moreover, we prove that the direct product of a family of QI-algebras is
a QI-algebra.

\section{\textbf{Introduction}}

\qquad Logical algebras are algebraic structures designed to model logical systems, in order
to encode inference rules as algebraic identities. Some logical algebras that model
non-classical logics are implication algebras, described by preffixes as BCI, BCK,
BCH, BI, BH, QI. In 1966, K. Is\'{e}ki introduced BCI-algebras as models for the so-
called BCI-logic, and also Y. Imai and K. Is\'{e}ki introduced BCK-algebras. In 1967, J.
C. Abbott studied implication algebras (\cite{Ab67}). Since then, many types of logical algebras
have been introduced and studied.\vspace{3pt}

\qquad In 2017, in \cite{S17}, Borumand Saeid et al. introduced BI-algebras as a generalization
of BCI-algebras and, in the same year, R. Kumar Bandaru (\cite{B17}) further generalized
BI-algebras by introducing the concept of QI-algebra. The study of QI-algebras has
be continued in \cite{B20, B25, R25, S20, W17}.
\vspace{0.5cm}

-------------------------------------- \newline
\textbf{Keywords and phrases:}  QI-algebra, ideal, strong ideal.\newline
\textbf{(2020) Mathematics Subject Classification:} 06F35, 03G25. \newpage

\qquad Some topics of interest are to examine the internal architecture of each logical
algebra, but also to investigate some sub-structures of this algebra, as sub-algebras
and ideals.\vspace{3pt}

\qquad  While in \cite{B20}, the subject of study was various types of ideals (implicative, fantastic and normal ideals) in right distributive QI-algebras, in the article \cite{W17} pseudo-valuations of those algebras are discussed by M. Wojciechowska-Rysiawa. The article \cite{R25},  written by this author, discusses the concept of positive implicative ideal in QI-algebras and establishes some of its important properties. The article \cite{B25} examines the internal architecture of (right distributive) QI-algebras.
\vspace{3pt}

\qquad S. A. Bhatti \cite{B91} (see also \cite{B90} and \cite{B94}) introduced the notion of strong ideals in
BCI-algebras and obtained some results about it. The concept of strong ideals in
BH-algebras was introduced in a slightly different way and studied in \cite{A10} by S. S.
Ahn and J. H. Lee.\vspace{3pt}

\qquad The main aim of this this article is to introduce the concept of strong ideal in
QI-algebras and to investigate some of its properties. We also introduce and discuss
the concepts of weak ideal and $\alpha$-ideal in QI-algebras.


\section{\textbf{Preliminaries}}

\qquad The notion of BI-algebras comes from the (dual) implication
algebra. An algebra $\mathfrak{A} =: (A,\ast,0)$ of type $(2,0)$ is called a BI-algebra (\cite{S17}, Definition 3.1) if the following holds:\vspace{3pt}

(Re) $(\forall x \in A)(x \ast x = 0)$, \vspace{3pt}

(Im) $(\forall x,y \in A)(x \ast (y \ast x) = x)$.\vspace{3pt}

\qquad A BI-algebra $\mathfrak{A}$ is said to be right distributive if the following \vspace{3pt}

(DR) $(\forall x,y,z\in A)((x \ast y) \ast z = (x \ast z) \ast (y \ast z))$\\ is valid.\vspace{3pt}

\qquad The concept of QI-algebras, as a generalization of BI-algebras, was determined in 2017 by R. Kumar Bandaru.
An algebra $\mathfrak{A} =: (A,\ast,0)$ of type $(2,0)$ is called a QI-algebra (\cite{B17}, Definition 3.1) if the following holds:\vspace{3pt}

(Re) \, $(\forall x \in A)(x \ast x = 0)$, \vspace{3pt}

(MR) $(\forall x \in A)(x \ast 0 = x)$ \vspace{3pt}

(QI) \, $(\forall x,y \in A)( x \ast (y \ast (x \ast y)) = x \ast y)$.\vspace{3pt}

\qquad A QI-algebra $\mathfrak{A}$ is said to be right distributive if, additionally, the formula (DR) is satisfied.\vspace{3pt}

\qquad Note that every BI-algebra is a QI-algebra but the converse need not be true (\cite{B17}, Example 3.2). \vspace{3pt}

\qquad The concept of sub-algebras in a QI-algebra $\mathfrak{A} =: (A,\cdot,0)$ is introduced by a standard way (\cite{S20}, Definition 3.1). A nonempty subset $S$ of $A$ is a sub-algebra in $\mathfrak{A}$ if it satisfies the condition\vspace{3pt}

(S1) $(\forall x,y \in A)((x \in S \, \wedge\, y \in S) \,\Longrightarrow \, x\ast y \in S)$.\\[3pt]
It can immediately be concluded that the sub-algebra $S$ in a QI-algebra $\mathfrak{A}$ satisfies the condition \vspace{3pt}

(S0) $0 \in S$.\\[3pt]
Indeed, since $S$ is not empty there exists an element $x \in S$. Then, according to (S1) and (Re), we have $x \in S \, \Longrightarrow 0 = x\ast x \in S$.
We denote the family of all sub-algebras of one QI-algebra $\mathfrak{A} =: (A,\ast,0)$ by $\mathfrak{S}(A)$.\vspace{3pt}

\qquad The concept of ideal in QI-algebras is determined by the following definition:

\begin{definition}\label{D3.1} \emph{(\cite{B17}, Definition 4.1)} A subset $J$ of a QI-algebra $\mathfrak{A} =: (A,\cdot,0)$ is called an ideal of $\mathfrak{A}$ if the following holds:\vspace{3pt}

\emph{(J0)} $0 \in J$,\vspace{3pt}

\emph{(J1)} $(\forall x,y \in A)((x\ast y \in J \, \wedge \, y \in J) \, \Longrightarrow \, x \in J)$.\\[3pt]
We denote the family of all ideals of a QI-algebra $\mathfrak{A} =: (A,\ast,0)$ by $\mathfrak{J}(A)$.
\end{definition}

\qquad The following example illustrates the relationship between the concept of sub-algebra and the concept of ideal in QI-algebras: The families $\mathfrak{S}(A)$ and $\mathfrak{J}(A)$ are mutually distinct (\cite{B25}, Remark 3.1). Additionally, these families are complete lattices by \cite{B20}, Theorem 4.1 and \cite{B25}, Theorem 3.1.

\begin{example}\label{E2.1} Let $A = \{0,a,b,c\}$ be a set with the operation given by the table\begin{align*}
	\begin{array}{c|cccc}
		\ast  & 0 & a & b & c\\
		\hline
   		0  & 0 & b & a & 0\\
		a  & a & 0 & a & 0\\
		b  & b & b & 0 & b\\
        c  & c & c & b & 0\\
\end{array} \end{align*}\\[3pt] Then $\mathfrak{A}=: (A,\ast,0)$ is a QI-algebra \emph{(}\cite{B17}, Example 3.2\emph{)}.

The subsets $S_{0} = \{0\}$, $S_{3} = \{0,c\}$ and $S_{4} = \{0,a,b\}$ are sub-algebras of the QI-algebra $\mathfrak{A}$. However, by direct checking it can be determined that the subsets $S_{1} = \{0,a\}$, $S_{2} = \{0,b\}$,  $S_{5} = \{0,a,c\}$  and $S_{6} = \{0,b,c\}$ are not sub-algebras of the QI-algebra $\mathfrak{A}$.
For illustration, for example, for the elements $0 \in S_{6}$ and $ b \in S_{6}$ we have $ 0\ast b = a \notin S_{6}$.

Subsets $J_{0} = \{0\}$, $J_{1} = \{0,a\}$, $J_{3} = \{0,c\}$ and $J_{5} = \{0,a,c\}$ of the set $A$ are ideals of the QI-algebra $\mathfrak{A}$. The subset $J_{2} = \{0,b\}$ is not an ideal in $\mathfrak{A}$ because, for example, for $b\in J_{2}$ we have $c\ast b = b \in J_{2}$ but $c \notin J_{2}$.
The subsets $J_{4} = \{0,a,b\}$ and $J_{6} = \{0,b,c\}$ are not ideals in $\mathfrak{A}$ because, for example, for $c\in J_{6}$ we have $a\ast c = 0 \in J_{6}$ but $a \notin J_{6}$.
Similarly, we have $b \in J_{4}$ and $c\ast b = b\in J_{4}$ but $c \notin J_{4}$.\qed
\end{example}

\begin{remark} As shown in the previous example, the sub-algebra $S_{4} = \{0,a,b\}$ in $\mathfrak{A}$ is not an ideal in $\mathfrak{A}$, while the ideal $J_{5} = \{0,a,c\}$ in $\mathfrak{A}$ is not a sub-algebra in $\mathfrak{A}$. Therefore, the families $\mathfrak{S}(A)$ and $\mathfrak{J}(A)$ are not
comparable by inclusion. This reconfirms the observation made in \cite{B25}, Remark 3.1.
\end{remark}

\qquad The following proposition gives some of the basic properties of QI-algebras.

\begin{proposition}[\cite{B17}, Proposition 3.5]\label{P2.1} Let $\mathfrak{A}=: (A, \ast, 0)$ be a QI-algebra. Then:\vspace{3pt}

\emph{(1)} $(\forall x \in A)(x \ast (0 \ast x) = x)$,\vspace{3pt}

\emph{(2)} $(\forall x,y \in A)(x\ast y = y \, \Longrightarrow \, x = y)$, ,\vspace{3pt}

\emph{(3)} $(\forall x \in A)(x \ast 0 = 0 \, \Longrightarrow \, x = 0)$, \vspace{3pt}

\emph{(4)} $(\forall x,y \in A)(x\ast y = x \, \Longrightarrow \, x\ast(y\ast x) = x)$
\end{proposition}

\section{\textbf{The main results: Strong ideals in QI-algebras}}

\qquad This section is the central part of this article. First it was shown (Theorem \ref{T3.1}) that the direct product of a family of QI-algebras is a QI-algebra again.
This is followed by the design of the concept of strong ideals in QI-algebras and the study of its properties.\vspace{3pt}

\qquad In what follows, we introduce the direct product of a family of  QI-algebras.
Let $\{(A_{i},\ast_{i},0_{i}): i \in I\}$ be a family of QI-algebras. On the cartesian product
\[\prod_{i\in I}A_{i} =: \{f : I \longrightarrow \cup_{i \in I}A_{i}\mid (\forall i \in I)(f(i) \in A_{i})\},\] we define the operation $\odot$ as follows
\[(\forall f,g \in \prod_{i \in I}A_{i})(\forall \in I)((f\odot g)(i) =: f(i) \ast_{i} g(i)),\]
we created the structure $(\prod_{i \in I}A_{i},\odot, f_{0})$, where $f_{0}$ was chosen as follows \[(\forall i \in I)(f_{0}(i) =: 0_{i}).\]

\qquad Before we start working with direct products of QI-algebras, we say that the operation determined in this way is well-defined.
If a priori we accept conditions that ensure the existence of non-empty direct product, we can prove the following theorem.

\begin{theorem}\label{T3.1} The direct product of any family of QI-algebras, determined as above, is a QI-algebra.\end{theorem}\begin{proof} By direct verification, it can be proved that this structure satisfies the axioms of QI-algebra:\vspace{3pt}

Let $f,g \in \prod_{i\in I}A_{i}$ be arbitrary elements and $i \in I$. Then, we have:\vspace{3pt}

(Re) \, $(f \odot f)(i) = f(i) \ast_{i} f(i) =  0_{i}.$ \vspace{3pt}

(M) \,\, $(f\odot f_{0})(i) = f(i)\ast_{i} f_{0}(i) = f(i) \ast_{i} 0_{i} = f(i)$.\vspace{3pt}

(QI) Considering that\vspace{3pt}

$(f\odot(g\odot(x \odot (f \odot g)))(i) = f(i)\ast_{i}(g(i)\ast_{i}(f(i) \ast_{i} g(i))) = f(i) \ast_{i} g(i)$ \\[3pt]
\hspace*{48mm}$= (f \odot g)(i)$,\\[3pt]
we have  that (QI) is a valid formula for the observed structure. \vspace{3pt}

Therefore, the structure $(\prod_{i \in I}A_{i},\odot, f_{0})$ is a QI-algebra.
\end{proof}

\qquad This result is strongly related to Theorem \ref{T3.7}.\vspace{3pt}

\qquad In what follows, we will deal with the creation of the concept of a strong ideals in QI-algebras and an examination of its properties.
The design of the concept of strong ideals in QI-algebras introduces the following concept of strong ideals in BH-algebras (\cite{A10}).

\begin{definition}\label{D3.1} A non-empty subset $J$ of a QI-algebra $\mathfrak{A}=: (A,\ast,0)$ is called
a strong ideal in $\mathfrak{A}$ if it satisfies (J0) and the following condition:\vspace{3pt}

\emph{(StJ)} $(\forall x,y,z \in A)(((x \ast y) \ast z \in J \, \wedge \, y \in J) \, \Longrightarrow \, x\ast z \in J)$.\\[3pt]
The family of all strong ideals
in a QI-algebra $\mathfrak{A} =: (A,\ast,0)$ is denoted by $\mathfrak{J}_{s}(A)$.
\end{definition}

\begin{remark} Sometimes, this class of ideals in logical algebras is called a T-ideal as, for example, in \cite{J16}, Definition 2.5.
\end{remark}

\qquad The set $A$ is a trivial strong ideal in a QI-algebra $\mathfrak{A} =: (A,\ast,0)$. So, $A \in \mathfrak{J}_{s}(A)$.

\begin{proposition}\label{P3.1} Any strong ideal in a QI-algebra $\mathfrak{A}$ is an ideal in $\mathfrak{A}$. This means $\mathfrak{J}_{s}(A) \, \subseteq \, \mathfrak{J}(A)$.\end{proposition}
\begin{proof} It is clear that $J$ satisfies the condition (J0). Putting $z = 0$ in (StJ), we obtain (J1).
\end{proof}

\begin{proposition}\label{P3.2} In every QI-algebra $\mathfrak{A} =: (A,\ast,0)$, the subset $\{0\}$ is a strong ideal in $\mathfrak{A}$. So, $\{0\} \in \mathfrak{J}_{s}(A)$.
\end{proposition} \begin{proof}
Let $x,y,z \in A$ be arbitrary elements such that $(x\ast y) \ast z = 0$ and $y = 0$. Then $x \ast z = (x \ast 0) \ast z = 0$ in accordance with (M).
So, the subset $\{0\}$ is a strong ideal in $\mathfrak{A}$.
\end{proof}

\qquad Since the family $\mathfrak{J}_{s}(A)$ is not empty, the following theorem can be proved.

\begin{theorem}\label{T3.2} The family $\mathfrak{J}_{s}(A)$ for an arbitrary QI-algebra $\mathfrak{A} =: (A,\ast,0)$ is a complete lattice.
\end{theorem} \begin{proof} Let $\{J_{k}\}_{k \in I}$ be a family of string ideals in $\mathfrak{A}$.

It is clear that $0 \in \bigcap_{k\in I}J_{k}$ holds. Let $x,y,z \in A$ be  arbitrary elements such that $(x\ast y)\ast z \in \bigcap_{k\in I}J_{k}$ and $y \in \bigcap_{k\in I}J_{k}$.
Then, for each $k \in I$, $(x\ast y)\ast z \in J_{k}$ and $y \in J_{k}$ hold. Thus $x\ast z \in J_{k} $ since $J_{k}$ is a strong ideal in $\mathfrak{A}$.
This means $x\ast z \in \bigcap_{k\in I}J_{k}$. Therefore, $\bigcap_{k\in I}J_{k}$ is a strong ideal in $\mathfrak{A}$.

If we denote by $\mathcal{Z}$ the family of all strong ideals in $\mathfrak{A}$ that contain $\bigcup_{k \in I}J_{k}$, then $\cap \mathcal{Z}$ is a strong ideal in $\mathfrak{A}$ according to the first part of this proof.

If we put $\sqcap_{k\in I}J_{k} = \bigcap_{k\in I}J_{k}$ and $\sqcup_{k \in I}J_{k} = \cap \mathcal{Z}$, then $(\mathfrak{J}_{s}(A),\sqcap,\sqcup)$ is a complete lattice.
\end{proof}

\begin{corollary}\label{C3.1} Let $\mathfrak{A} =: (A,\ast,0)$ be a QI-algebra. For each $x \in A$, there exists a minimal strong ideal $J_{x}$ in $\mathfrak{A}$ that contains $x$.\end{corollary}
\begin{proof} If $\mathcal{Z}$ is the family of all strong ideals in $\mathfrak{A}$ that contain $x$, then $J_{x} =: \cap \mathcal{Z}$ is a strong ideal in $\mathfrak{A}$ that contains $x$. If $J$ is an strong ideal in $\mathfrak{A}$ that contains $x$, then $J \in \mathcal{Z}$. Thus $J_{x} \, \subseteq\, J$. Therefore, $J_{x}$ is a minimal strong ideal in $\mathfrak{A}$ that contains $x$.\end{proof}

\qquad The following theorem gives another determination of the concept of strong ideal in QI-algebras.

\begin{theorem}\label{T3.3}Let $J$ be an ideal in a QI-algebra $\mathfrak{A} =: (A,\ast,0)$. Then $J$ is a strong ideal in $\mathfrak{A}$  if and only if the following holds\vspace{3pt}

\emph{(StJ1)} $(\forall x,y,z \in A)((x\ast z \in A \setminus J \, \wedge \, y \in J) \, \Longrightarrow \, (x\ast y)\ast z \in A \setminus J)$.
\end{theorem}\begin{proof} Let $J$ be a strong ideal in $\mathfrak{A}$ and let $x,y,z \in A$ be arbitrary elements such that $x\ast z \notin J$ and $y \in J$.
If we assume that $(x\ast y)\ast z \in J$, then it would be $x\ast z \in J$ since $J$ is a strong ideal in $\mathfrak{A}$. We got a contradiction. So, it must be $(x\ast y) \ast z \notin J$.

Conversely, let (StJ1) be a valid formula for the ideal $J$ and let $x,y,z \in A$ be such that $(x\ast y)\ast z \in J$ and $y \in J$.
Assume that $x\ast z \notin J$. Then, according to (StJ1), there would be $(x\ast y)\ast z \notin J$. We got a contradiction. Therefore, $x\ast z \in J$. This proves that $J$ is a strong ideal in $\mathfrak{A}$.
\end{proof}

\qquad Analogously to the previous one, the following theorem can be proven:

\begin{theorem}\label{T3.4}Let $J$ be an ideal in a QI-algebra $\mathfrak{A} =: (A,\ast,0)$. Then $J$ is a strong ideal in $\mathfrak{A}$  if and only if the following holds\vspace{3pt}

\emph{(StJ2)} $(\forall x,y,z \in A)(((x\ast y)\ast z  \in  J \, \wedge \, x\ast z \in A \setminus J) \, \Longrightarrow \, y \in A \setminus J)$.
\end{theorem} \begin{proof} Let $J$ be a strong ideal in $\mathfrak{A}$ and let $x,y,z \in A$ be arbitrary elements such that $(x\ast y)\ast z  \in  J$ and  $x\ast z \notin J$.
If we assume that $y \in J$, then it would be $x\ast z \in J$ since $J$ is a strong ideal in $\mathfrak{A}$. We got a contradiction. So, it must be $y \notin J$.

Conversely, let (StJ2) be a valid formula for the ideal $J$ and let $x,y,z \in A$ be such that $(x\ast y)\ast z \in J$ and $y \in J$.
Assume that $x\ast z \notin J$. Then, according to (StJ2), there would be $y \notin J$. We got a contradiction. Therefore, $x\ast z \in J$. This proves that $J$ is a strong ideal in $\mathfrak{A}$.
\end{proof}

\qquad In what follows we need the following lemma:
\begin{lemma} [\cite{B25}, Proposition 3.4]\label{L3.1} Let $J$ be an ideal in a right distributive QI-algebra $\mathfrak{A} =: (A,\ast,0)$. Then:\vspace{3pt}

$(\forall x, y \in A)(x \in J \, \Longrightarrow \, x \ast y \in J).$\end{lemma}


\qquad Further on, we have:

\begin{theorem}\label{T3.5} Every ideal in a right distributive QI-algebra is a strong ideal in it.
\end{theorem} \begin{proof} Let $J$ be an ideal in a right distributive QI-algebra $\mathfrak{A} =: (A,\ast,0)$ and let $x,y,z \in A$ be such that $(x\ast y)\ast z \in J$ and $y \in J$.
First, by Lemma \ref{L3.1}, we have $y \in J \, \Longrightarrow \, y\ast z \in J$ for arbitrary $z \in A$.
On the other hand, from $(x\ast z)\ast (y\ast z) = (x\ast y)\ast z \in J$ and $y\ast z \in J$ it follows $x\ast z \in J$ according to (J1) since $\mathfrak{A}$ is a right distributive QI-algebra.
Therefore, $J$ is a strong ideal in $\mathfrak{A}$.
\end{proof}

\qquad Judging by the previous theorem, it makes sense to talk about strong ideals in QI-algebras only in non right distributive QI-algebras.\vspace{3pt}

\qquad Let $\mathfrak{A} =: (A,\ast,0_{A})$ and $\mathfrak{B} =: (B,\star,0_{B})$ be QI-algebras. A QI-homomorphism is a mapping $f: A \longrightarrow B$ satisfying the condition
\[(\forall x,y \in A)(f(x\ast y) = f(x)\star f(y)).\] It is easy to prove that $f(0_{A}) = 0_{B}$ holds.


\begin{theorem}\label{T3.6} Let $f : \mathfrak{A} \longrightarrow \mathfrak{B}$ be a homomorphism of QI-algebras. If $C$ is a strong ideal of $\mathfrak{B}$, then $f^{-1}(C)$ is a strong ideal in $\mathfrak{A}$.
\end{theorem}
\begin{proof}
Since $f(0) = 0$, we have $0 \in f^{-1}(C)$.

Let $x, y, z \in A$ be such that $(x \ast y) \ast z \in f^{-1}(C)$ and $y \in f^{-1}(C)$. Then $(f(x) \ast f(y)) \ast f(z) =
f((x\ast y)\ast z) \in C$ and $f(y) \in C$. Since $C$ is a strong ideal in $\mathfrak{B}$, it follows
from (StJ) that $f(x \ast z) = f(x) \ast f(z) \in C$. So that $x \ast z \in f^{-1}(C)$. Hence
$f^{-1}(C)$ is a strong ideal in $\mathfrak{A}$.\end{proof}

\begin{corollary}\label{C3.2} Let $f : \mathfrak{A} \longrightarrow \mathfrak{B}$ be a homomorphism of QI-algebras. Then $Kerf =: \{x \in A : f(x) = 0\}$ is a strong ideal of $\mathfrak{A}$.\end{corollary}\begin{proof}
Since the subset $\{0\}$ is a strong ideal in the
QI-algebra $\mathfrak{B}$, by Proposition \ref{P3.2}, we have that the kernel $Kerf = f^{-1}(\{0\})$ of the homomorphism $f$ is a strong ideal in $\mathfrak{A}$ in accordance with the previous theorem.
\end{proof}

\begin{example}\label{E3.1} Let $\mathfrak{A} =: (A,\ast,0)$ be a QI-algebra as in Example \ref{E2.1}.

Subsets $J_{0}$, $J_{1}$ and $J_{5}$ are strong ideals in $\mathfrak{A}$.

The ideal $J_{3} = \{0,c\}$ is not a strong ideal in $\mathfrak{A}$ because, for example, for $x = a$, $y = c$ and $z = 0$ we have  $(a\ast c)\ast 0 = 0 \ast 0 = 0\in J_{3}$ and $c \in J_{3}$ but $a \ast 0 = a \notin J_{3}$.
\qed\end{example}

\begin{remark} The previous example shows that an ideal in a QI-algebra, in the general case, does not have to be a strong ideal in that algebra. So, $\mathfrak{J}_{s}(A) \, \subsetneqq \mathfrak{J}(A)$.
\end{remark}

 \qquad Further on, we have:

\begin{theorem}\label{T3.7} Let $\{(A_{i},\ast_{i},0_{i}): i \in I\}$ be a family of QI-algebras, $K$ be a subset of $I$ and let $J_{i}$ be a strong ideal in $(A_{i},\ast_{i},0_{i})$ for each $i \in K$.
Then $\prod_{i\in I}T_{i}$, where $T_{i} = J_{i}$ for $i \in K$ and $T_{i} = A_{i}$ for $i \in I \backslash K$, is a strong ideal in the QI-algebra $\prod_{i\in I}A_{i}$.
\end{theorem}\begin{proof} First, it is clear that $f_{0} \in \prod_{i\in I}T_{i}$.\vspace{3pt}

If $K = \emptyset$, then $\prod_{i\in I}T_{i} = \prod_{i\in I}A_{i}$, so $\prod_{i\in I}T_{i}$ is certainly an ideal in $\prod_{i\in I}A_{i}$. Assume, therefore, that $K \neq \emptyset$.
\vspace{3pt}

Let $x,y,z \in \prod_{i\in I}A_{i}$ be such that $(x \odot y)\odot z \in \prod_{i\in I}T_{i}$ and $y \in \prod_{i\in I}T_{i}$. This means $(x(i)\ast_{i} y(i))\ast_{i} z(i) \in J_{i}$ and $y(i) \in J_{i}$ for each $i \in K$. Then $(x\odot z)(i) = x(i)\ast_{i} z(i)\in J_{i}$ since $J_{i}$ is a strong ideal in $(A_{i},\ast_{i},0_{i})$ for each $i \in K$. Hence $x \odot z \in \prod_{i\in I}T_{i}$.\vspace{3pt}

As shown, $\prod_{i\in I}T_{i}$ is a strong ideal in $\prod_{i\in I}A_{i}$.
\end{proof}

\begin{example} Let $\mathfrak{A} =: (A,\ast,0)$ be a QI-algebra as in Example \ref{E2.1}.
Then according to the Theorem \ref{T3.1}, $\mathfrak{A}\times \mathfrak{A} =: (A\times A,\otimes,(0,0))$ is  a QI-algebra also, where the operation $\otimes$ is defined as follows
\[(\forall x,y,u,v \in A)((x,y)\otimes(u,v) =: (x\ast u,y\ast v)).\] The subset $J_{1} = \{0,a\}$ is a strong ideal in $\mathfrak{A}$ as shown in Example \ref{E3.1}. The subsets $J_{1} \times A$, $A \times J_{1}$ and $J_{1} \times J_{1}$ are strong ideals in $\mathfrak{A}\times \mathfrak{A}$
according to the Theorem \ref{T3.7}.\qed
\end{example}

\qquad At the end of this section, we prove another specific property of strong ideals in QI-algebras.

\begin{proposition} Let $J$ be a strong ideal in a QI-algebra $\mathfrak{A} =: (A,\ast,0)$. Then holds
\[(\forall x,y \in A)(y \in J \, \Longrightarrow \, x \ast (x \ast y) \in J).\]\end{proposition}\begin{proof}
If we put $z = x\ast y$ in (WJ), we get \[0 = (x\ast y)\ast (x\ast y) \in J \, \wedge \, y \in J \, \Longrightarrow \, x\ast (x\ast y) \in J.\]
This gives $y \in J \, \Longrightarrow \, x\ast (x\ast y) \in J$ since $0 \in J$.
\end{proof}


\section{\textbf{Conclusion:  Weak ideals and $\alpha$-−ideals in QI-algebras}}

\qquad  In this paper, the concept of strong ideals in QI-algebras is introduced and its important properties are reviewed.
In this way, the spectrum of different ideals in this class of logical algebras, so far introduced and analyzed, is supplemented by the concept of a strong ideals.
As a continuation of previous research on the spectrum of ideals in QI-algebras, that spectrum can be supplemented, for example, by introducing the so-called weak ideals, p-ideals and $\alpha$-ideals by recognizing their properties as well as their mutual relations. For the sake of illustration, we present the first and third possibilities, since the formula for determining of $p$-ideals is generally known (See, for example, \cite{Z94} or \cite{H98}, Definition 1).

\begin{definition} A non-empty subset $J$ of a QI-algebra $\mathfrak{A}=: (A,\ast,0)$ is called
a weak ideal in $\mathfrak{A}$ if it satisfies the following condition:\vspace{3pt}

\emph{(WJ)} $(\forall x,y,z \in A)((x \ast (y \ast z )\in J \, \wedge \, y \in J) \, \Longrightarrow \, x\ast z \in J)$.\end{definition}

\qquad  It can be shown that the weak ideal in a QI-algebra also satisfies the condition (J0). Indeed, since $J$ is a nonempty subset of $A$, there exists some $x \in A$ such that $x \in J$. Then we have $J \ni x = x\ast 0 = x\ast (x\ast x)$ and $x \in J$ from which, by (WJ), it follows that $0 = x\ast x \in J$.\vspace{3pt}

\qquad  In addition, we have:
\begin{proposition} Any weak ideal in a QI-algebra is a sub-algebra in it.\end{proposition}\begin{proof}
 Let $J$ be a weak ideal in a QI-algebra $\mathfrak{A}$ and let $x,y \in A$ be arbitrary elements such that $x \in J$ and $y \in J$. Then $x \ast (y \ast y) = x \ast 0 = x \in J$ and $y \in J$. Thus $x \ast y \in J$ by (WJ). \end{proof}

 \begin{proposition} Any weak ideal in a QI-algebra is an ideal in it.\end{proposition} \begin{proof} Putting $z = 0$ in (WJ), we get (J1).\end{proof}

 \qquad  The concept of $\alpha$-ideal in QI-algebra is given by the following definition:

 \begin{definition} A non-empty subset $J$ of a QI-algebra $\mathfrak{A}$
is called an $\alpha$-ideal if, in addition to (J0), it also satisfies the following condition\vspace{3pt}

\emph{($\alpha$J)} $(\forall x,y,z \in A)((x\ast y \in J \, \wedge \, y\ast z \in J) \, \Longrightarrow \, x\ast z \in J)$.\end{definition}

\begin{proposition} Any $\alpha$-ideal in a QI-algebra is an ideal in it.\end{proposition} \begin{proof} Putting $z = 0$ in ($\alpha$J), we get (J1).\end{proof}

\begin{example} Let $\mathfrak{A} = (A,\ast,0)$ be a QI-algebra as in Example \ref{E2.1}. Ideals $J_{0}$ and $J_{1}$ are $\alpha$-ideals in $\mathfrak{A}$, but the ideal $J_{3}$ is not an $\alpha$-ideal in $\mathfrak{A}$ because we have $0 \ast c = 0 \in J_{3}$ and $c \ast a = c \in J_{3}$ but $0 \ast a = b \notin J_{3}$. The ideal $J_{5}$ is also not an $\alpha$-ideal in $\mathfrak{A}$ because, for example, for $x = 0$, $y = c$ and $z = a$ we have $0 \ast c = 0 \in J_{5}$ and $c \ast a = c \in J_{5}$ but $0 \ast a = b \notin J_{5}$.\qed
\end{example}

\qquad
Thus, the family of $\alpha$-ideals in a QI-algebra differs from the families $\mathfrak{J}(A)$ and $\mathfrak{J}_{s}(A)$.\vspace{3pt}

\qquad In one of the next texts by this researcher, material on weak ideals in QI-algebras will be presented in more detail.\\[15pt]
\textbf{Acknowledgement}. The author thanks the reviewer(s) for the useful suggestions.



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\vspace{0.5cm}
\address{D. A. Romano:  International Mathematical Virtual Institute\newline 6, Korduna\v ska Street, 78000 Banja Luka, Bosnia and Herzegovina}
\newline
e-mail: daniel.a.romano@hormail.com, bato49@hotmail.com\newline
ORCID 0000-0003-1148-3258
\newline

\end{document}
