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\title{\bf On interval-valued fuzzy prime ideals of a semiring}
\author{\bf \bf T. K. Dutta.\footnote{duttatapankumar@yahoo.co.in} \\ Department of
Pure Mathematics, University of Calcutta.
\\35, Ballygunge Circular Road, Kolkata --- 700019, India.
\\ \bf S. Kar.\footnote{karsukhendu@yahoo.co.in} and \bf S. Purkait.\footnote {sanuiitg@gmail.com} \footnote {The third author is grateful to CSIR, India for providing financial assistance.} \\ Department of
Mathematics, Jadavpur University.
\\ 188, Raja S. C. Mallick Road, Kolkata --- 700032, India.}


\date{}
\maketitle

\begin{abstract}
In this paper we introduce the notion of interval-valued fuzzy
prime ideal of a semiring, interval-valued fuzzy completely prime
ideal of a semiring and study some important properties of these
two ideals in the context of interval-valued fuzzy algebra.
\end{abstract}

AMS Mathematics Subject Classification (2010) : 08A72.

{\bf Keywords :} Semiring, Interval Number, Interval-valued Fuzzy
Ideal, Interval-valued Fuzzy Prime Ideal, Interval-valued Fuzzy
Completely Prime Ideal.

\section{Introduction}
In 1965, L. Zadeh first introduced the theory of fuzzy sets in his
pioneer paper [15]. After the birth of fuzzy set theory it has
achieved manifold applications than the ordinary set theory. From
the beginning of fuzzy set theory it was clear that this theory
was an extraordinary tool for representing human knowledge.
However, L. Zadeh himself established that sometimes, in
decision-making processes, knowledge is better represented by
means of some generalizations of fuzzy sets. The so-called
extensions of fuzzy set theory arise in this way. Interval-valued
fuzzy sets were introduced independently by Zadeh [14],
Grattan-Guiness [5], Jahn [7], Sambuc [12] in the same year 1975
as a generalization of ordinary fuzzy set. In the field of
applications, the success of the use of fuzzy set theory depends
on the choice of the membership function that we take. However,
there are applications in which experts do not have precise
knowledge of the function that should be taken. In these cases, it
is appropriate to represent the membership degree of each element
to the fuzzy set by means of an interval. From these
considerations arises the extension of fuzzy sets called theory of
Interval-valued Fuzzy Sets (IVFSs)  that is, fuzzy sets such that
the membership degree of each element of the fuzzy set is given by
a closed subinterval of the interval [0, 1]. Thus an
interval-valued fuzzy set (IVFS) is defined by an interval-valued
membership function. It is important to note that not only
vagueness (lack of sharp class boundaries), but also a feature of
uncertainty (lack of information) can be addressed intuitively by
interval valued fuzzy set. After the introduction of the concept
of interval-valued fuzzy set some authors [2, 6, 8, 10, 13]
investigated the topics related to interval-valued fuzzy set  and
obtained many meaningful conclusions. There are natural ways to
fuzzify various algebraic structures and it has been done
successfully by many mathematicians. A. Rosenfeld [11] is the
father of fuzzy abstract algebra. He first studied the notion of
fuzzy subgroup in 1971. After that in 1979, N. Kuroki [9]
introduced the concept of fuzzy semigroup. In 1993, J. Ahsan, K.
Saifullah and M. Farid Khan [1] introduced the notion of fuzzy
semirings. In 1994, T.K. Dutta and B.K. Biswas [3] characterized
fuzzy prime ideals of a semiring. Recently, many results of
semiring theory are investigated by many researchers in fuzzy
context. In this paper, we initiate the study of fuzzification of
some concepts of semiring by using the concept of interval valued
fuzzy set. Our main purpose of this paper is to study prime ideals
and completely prime ideals in semirings with the help of interval
valued fuzzy sets.


\section{Preliminaries}
Firstly, we recall some definitions and results of semirings and
fuzzy algebra which we shall use in this paper.

\begin{Definition}$[4]$
A non-empty set $S$ together with two binary operations $`+$' and
$`\cdot$' is said to be a semiring if
\begin{enumerate}[(i)]
\item $(S, +)$ is an abelian semigroup; \item $(S, \cdot)$ is a
semigroup and \item $a \cdot (b + c) = a \cdot b + a \cdot c$ and
$(b + c) \cdot a = b \cdot a + c \cdot a$ for all $a, b, c \in S$.
\end{enumerate}
\end{Definition}

Let $(S, +, \cdot)$ be a semiring. If there exists an element
$`0_{S}$' $\in S$ such that $a + 0_{S} = a = 0_{S} + a$ and $a
\cdot 0_{S} = 0_{S} = 0_{S} \cdot a$ for all $a \in S$; then
$`0_{S}$' is called the zero element of $S$. If there exists an
element $`1_{S}$' $\in S$ such that $a \cdot 1_{S} = a = 1_{S} \cdot
a$ for all $a \in S$, then $`1_{S}$' is called the identity element
of $S$.

A semiring may or may not have a zero and an identity element.

\vspace{.2in} 

$\bullet$ Throughout this paper we consider a semiring $(S, +,
\cdot)$ with zero element $`0_{S}$' and identity element
$`1_{S}$'. Unless otherwise stated a semiring $(S, +, \cdot)$ will
be denoted simply by $S$ and multiplication $`\cdot$' will be
denoted by juxtaposition.

\begin{Definition}$[4]$
A semiring $(S, +, \cdot)$ is said to be commutative
if $(S, \cdot)$ is commutative.
\end{Definition}

\begin{Definition}$[4]$
Let $I$ be a nonempty subset of a semiring $S$. Then
\begin{enumerate}[(i)]
\item $I$ is said to be a left ideal of
$S$ if $(I, +)$ is a subsemigroup of $(S, +)$ and $sa \in I$ for
all $s \in S$ and for all $a \in I$.
\item $I$ is said to be a right
ideal of $S$ if $(I, +)$ is a subsemigroup of $(S, +)$
and $as \in I$ for all $s \in S$ and for all $a \in I$.
\item $I$ is said to be an ideal of $S$ if it is both a left ideal and a right
ideal of $S$.
\end{enumerate}
\end{Definition}

\begin{Definition}$[4]$
A proper ideal $P$ of a semiring $S$ is said to be prime if $AB
\subseteq P$ for any two ideals $A, B$ of $S$ implies that either
$A \subseteq P$ or $B \subseteq P$.
\end{Definition}

\begin{Proposition}$[4]$
The following conditions in a semiring $S$ are equivalent :
\begin{enumerate}[(i)]
\item $P$ is a prime ideal of $S$. \item $\{arb : r \in S\}
\subseteq P$ if and only if $a \in P$ or $b \in P$.
\end{enumerate}
\end{Proposition}

\begin{Definition}$[4]$
A proper ideal $I$ of a semiring $S$ is said to be completely prime if
$ab \in I$ for $a, b$ in $S$ implies that $a \in I$ or $b \in I$.
\end{Definition}

\begin{Proposition}$[4]$
A prime ideal $P$ of a semiring $S$ is completely prime if and
only if $ab \in P$ implies that $ba \in P$ for any $a, b \in S$.
\end{Proposition}

\begin{Definition}$[8]$
An interval number on $[0,1]$, denoted by $\widetilde{a}$, is
defined as the closed subinterval of $[0,1]$, where $\widetilde{a}
= [a^{-},a^{+}]$ satisfying $0 \leq a^{-} \leq a^{+} \leq 1$.
\end{Definition}

\vspace{.2in}

$\bullet$ For any two interval numbers $\widetilde{a} =
[a^{-},a^{+}]$ and $\widetilde{b} = [b^{-},b^{+}]$ we define :
\begin{enumerate}[(i)]
\item $\widetilde{a} \leq \widetilde{b}$ if and only if $a^{-} \leq
b^{-}$ and $a^{+} \leq b^{+}$. \item $\widetilde{a} = \widetilde{b}$
if and only if $a^{-} = b^{-}$ and $a^{+} = b^{+}$.  \item
$\widetilde{a} < \widetilde{b}$ if and only if $\widetilde{a} \neq
\widetilde{b}$ and $\widetilde{a} \leq \widetilde{b}$.
\end{enumerate}

\begin{Note} We write $\widetilde{a} \geq \widetilde{b}$ whenever
$\widetilde{b} \leq \widetilde{a}$ and $\widetilde{a} > \widetilde{b}$
whenever $\widetilde{b} < \widetilde{a}$. We denote the interval
number $[0,0]$ by $\widetilde{0}$ and $[1,1]$ by $\widetilde{1}$.
\end{Note}

\begin{Definition}$[2]$
Let $\{\widetilde{a_{i}}: i \in \Lambda\}$ be a family of interval
numbers, where $\widetilde{a_{i}} = [a_{i}^{-},a_{i}^{+}]$. Then we
define $\displaystyle\sup_{i \in \Lambda}\{\widetilde{a_{i}}\} =
[\displaystyle\sup_{i \in \Lambda}a_{i}^{-},\displaystyle\sup_{i
\in \Lambda}a_{i}^{+}]$ and $\displaystyle\inf_{i \in
\Lambda}\{\widetilde{a_{i}}\} = [\displaystyle\inf_{i \in
\Lambda}a_{i}^{-},\displaystyle\inf_{i \in \Lambda}a_{i}^{+}]$.
\end{Definition}

\vspace{.2in}

$\bullet$ Suppose $D[0,1]$ denotes the set of all interval numbers
on $[0,1]$.
\begin{Definition}$[15]$
Let $S$ be a non-empty set. A mapping $\mu : S \longrightarrow
[0,1]$ is called a fuzzy subset of $S$.
\end{Definition}


\begin{Definition}$[8]$
Let $S$ be a non-empty set. A mapping $\widetilde{\mu} : S
\longrightarrow D[0,1]$ is called an interval-valued fuzzy subset
of $S$.
\end{Definition}

\vspace{.2in}

$\bullet$ For simplicity, throughout this paper we shall use the term $`i.v.$ fuzzy subset' for $`$interval-valued fuzzy subset'.

\begin{Note} If $\widetilde{\mu}$ be an $i.v.$ fuzzy subset of a
set $S$, then $\widetilde{\mu}(x)$ is an interval number, where $x
\in S$. Suppose $\widetilde{\mu}(x) = [\alpha,\beta]$ for some $x
\in S$. Then we have $0 \leq \alpha \leq \beta \leq 1$. So we can
always define two fuzzy subsets $\mu^{-}$ and $\mu^{+}$ of $S$
such that $\mu^{-}(x) = \alpha$ and $\mu^{+}(x) = \beta$. Thus we
have $\widetilde{\mu}(x) = [\mu^{-}(x),\mu^{+}(x)]$ for all $x \in
S$. Conversely, suppose we have two fuzzy subsets $\mu^{-}$ and
$\mu^{+}$ of $S$ such that $0 \leq \mu^{-}(x) \leq \mu^{+}(x) \leq
1$ for all $x \in S$. Then we can define an $i.v.$ fuzzy subset
$\widetilde{\mu}$ of $S$ such that $\widetilde{\mu}(x) =
[\mu^{-}(x),\mu^{+}(x)]$ for all $x \in S$.
\end{Note}

\begin{Definition}
Let $X \neq \emptyset$ be a set and $A \subseteq X$. Then the
interval-valued characteristic function $\widetilde{\chi}_{A}$ of
$A$ is an $i.v.$ fuzzy subset of $X$, defined as follows :
  $$ \widetilde{\chi}_{A}(x) = \begin{cases}
          \widetilde{1} &\text{when} \quad x \in A. \\
           \hfill \widetilde{0} &\text{when} \quad x \notin A. \\
         \end{cases} $$
\end{Definition}

\begin{Definition}
Let $\widetilde{\mu_{1}}$ and $\widetilde{\mu_{2}}$ be two $i.v.$ fuzzy subsets
of a non-empty set $X$. Then $\widetilde{\mu_{1}}$ is said to be
subset of $\widetilde{\mu_{2}}$, denoted by $\widetilde{\mu_{1}}
\subseteq \widetilde{\mu_{2}}$, if
 $\widetilde{\mu_{1}}(x) \leq \widetilde{\mu_{2}}(x)$ i.e.
$\mu_{1}^{-}(x) \leq \mu_{2}^{-}(x)$ and $\mu_{1}^{+}(x) \leq
\mu_{2}^{+}(x)$ for all $x \in X$, where $\widetilde{\mu_{1}}(x) =
[\mu_{1}^{-}(x), \mu_{1}^{+}(x)]$ and $\widetilde{\mu_{2}}(x) =
[\mu_{2}^{-}(x), \mu_{2}^{+}(x)]$.
\end{Definition}

\begin{Definition}$[8]$
Let $\widetilde{\mu}$ be an $i.v.$ fuzzy subset of a non-empty set $X$ and
$[\alpha,\beta] \in D[0,1]$. Then the level subset of
$\widetilde{\mu}$, denoted by
$\overline{U}(\widetilde{\mu},[\alpha,\beta])$, is defined as :\\
$\overline{U}(\widetilde{\mu},[\alpha,\beta]) = \Big\{x \in X :
\widetilde{\mu}(x) \geq [\alpha,\beta]\Big\}$.
\end{Definition}

If we consider two interval numbers
$[\alpha_{1},\beta_{1}]$ and $[\alpha_{2},\beta_{2}]$ such that
$[\alpha_{1},\beta_{1}] > [\alpha_{2},\beta_{2}]$, then we have
$[\alpha_{1},\beta_{1}] \geq [\alpha_{2},\beta_{2}]$ and
$[\alpha_{1},\beta_{1}] \neq [\alpha_{2},\beta_{2}]$. 

In this case, we find that
$\overline{U}(\widetilde{\mu},[\alpha_{1},\beta_{1}]) \subseteq
\overline{U}(\widetilde{\mu},[\alpha_{2},\beta_{2}])$, since for
any $x \in \overline{U}(\widetilde{\mu},[\alpha_{1},\beta_{1}])
\Longrightarrow \widetilde{\mu}(x) \geq [\alpha_{1},\beta_{1}] \geq
[\alpha_{2},\beta_{2}] \Longrightarrow x \in
\overline{U}(\widetilde{\mu},[\alpha_{2},\beta_{2}])$.

\newpage

So we have the following result :
\begin{Proposition}
If $[\alpha_{1},\beta_{1}]$ and $[\alpha_{2},\beta_{2}]$ be two interval numbers such that $[\alpha_{1},\beta_{1}] > [\alpha_{2},\beta_{2}]$ then $\overline{U}(\widetilde{\mu},[\alpha_{1},\beta_{1}] \subseteq \overline{U}(\widetilde{\mu},[\alpha_{2},\beta_{2}])$.
\end{Proposition}

\begin{Definition}$[8]$
The interval Min-norm is a function $Min^{i} : D[0,1] \times
D[0,1] \longrightarrow D[0,1]$, defined by :\\
$Min^{i}(\widetilde{a},\widetilde{b}) = [min(a^{-},b^{-}), min(a^{+},b^{+})]$ for all $\widetilde{a}, \widetilde{b} \in
D[0,1]$, where $\widetilde{a} = [a^{-},a^{+}]$ and $\widetilde{b} =
[b^{-},b^{+}]$.
\end{Definition}

\begin{Definition}
The interval Max-norm is a function $Max^{i} : D[0,1] \times
D[0,1] \longrightarrow D[0,1]$, defined by :\\
$Max^{i}(\widetilde{a},\widetilde{b}) = [max(a^{-},b^{-}), max(a^{+},b^{+})]$ for all $\widetilde{a}, \widetilde{b} \in
D[0,1]$, where $\widetilde{a} = [a^{-},a^{+}]$ and $\widetilde{b} =
[b^{-},b^{+}]$.
\end{Definition}

\begin{Definition}
Let $\widetilde{\mu_{1}}$ and $\widetilde{\mu_{2}}$ be two $i.v.$ fuzzy subsets
of a non-empty set $X$ and $`\cdot$' be a binary composition on
$X$. Then the composition of these two $i.v.$ fuzzy subsets is an $i.v.$ fuzzy subset
of $X$, defined by :
  $$ (\widetilde{\mu_{1}} \circ  \widetilde{\mu_{2}})(x) = \begin{cases}
          \displaystyle\sup_{x = a \cdot b}\Big\{Min^{i}\Big(\widetilde{\mu_{1}}(a),\widetilde{\mu_{2}}(b)\Big)\Big\}, &\text{when} \quad x =
          a \cdot b ~\text{for some} \quad a, b \in X.\\
            \widetilde{0}, &\text{otherwise}.\\
         \end{cases} $$
				
\vspace{.2in}

When we consider the composition of two $i.v.$ fuzzy subsets of a semiring,
the second case will not arise as we consider a semiring with
identity element in this paper.
\end{Definition}

Thus the composition of two $i.v.$ fuzzy subsets $\widetilde{\mu}_{1}$ and
$\widetilde{\mu}_{2}$ of a semiring $S$ is given by
$(\widetilde{\mu}_{1} \circ \widetilde{\mu}_{2})(x) =
\displaystyle\sup_{x=ab}\Big\{Min^{i}\Big(\widetilde{\mu}_{1}(a),
\widetilde{\mu}_{2}(b)\Big)\Big\}$ for all $x \in S$ such that $x =
ab$ for some $a, b \in S$.

\vspace{.2in}

$\bullet$ {\bf Throughout this paper we assume that any two
interval numbers in $D[0,1]$ are comparable i.e. for any two
interval numbers $\widetilde{a}$ and $\widetilde{b}$ in $D[0,1]$, we
have either $\widetilde{a} \leq \widetilde{b}$ or $\widetilde{a} >
\widetilde{b}$}.



\section{Interval-valued fuzzy prime ideal of a semiring :}
\begin{Definition}
A non-empty $i.v.$ fuzzy subset $\widetilde{\mu}$ of a semiring $S$ (i.e.
$\widetilde{\mu}(x) \neq \widetilde{0}$ for some $x \in S$) is said
to be interval-valued fuzzy ideal of $S$ if
\begin{enumerate}[(i)]
\item $\widetilde{\mu}(x + y) \geq
Min^{i}(\widetilde{\mu}(x),\widetilde{\mu}(y))$ \item
$\widetilde{\mu}(xy) \geq
Max^{i}(\widetilde{\mu}(x),\widetilde{\mu}(y))$ for all $x,y \in S$.
\end{enumerate}
\end{Definition}

\begin{Example}
We consider the the semiring $\mathbb{N}_{0}$ of non-negative
integers with respect to usual addition and multiplication. Let
$\widetilde{\mu}$ be an $i.v.$ fuzzy subset of $\mathbb{N}_{0}$, defined by :
 $$ \widetilde{\mu}(x) = \begin{cases}
          \widetilde{1} &\text{if} \quad x = 0, \\
           \hfill [0.5,0.6] &\text{if} \quad x ~\text{is non-zero even}, \\
         \hfill [0.3,0.4] &\text{if} \quad x ~\text{is odd}. \\
         \end{cases} $$
Then $\widetilde{\mu}$ is an $i.v.$ fuzzy ideal of $\mathbb{N}_{0}$.
\end{Example}

\begin{Remark}
Let $\widetilde{\mu}$ be an $i.v.$ fuzzy ideal of a semiring $S$. Then
$\widetilde{\mu}(0_{S}) \geq \widetilde{\mu}(x)$ for all $x \in S$.
\end{Remark}

%\begin{proof}
%$\widetilde{\mu}(0_{S}) = \widetilde{\mu}(0_{S}x) \geq
%Max^{i}(\widetilde{\mu}(0_{S}),\widetilde{\mu}(x))$ for all $x \in
%S$ (since $\widetilde{\mu}$ is an interval-valued fuzzy ideal of
%$S$). Thus $\widetilde{\mu}(0_{S}) \geq \widetilde{\mu}(x)$ for all
%$x \in S$.
%\end{proof}

\begin{Lemma}
Let $S$ be a semiring and $A$ be a subset of $S$. Then $A$ is an
ideal of $S$ if and only if $\widetilde{\chi}_{A}$ is an
$i.v.$ fuzzy ideal of $S$.
\end{Lemma}

\begin{proof}
%$\underline{Case1}$: Let $x \in A$ and $y \in A$. Then $\widetilde{\chi}_{A}(x) = \widetilde{1} = \widetilde{\chi}_{A}(y)$. Since $A$ is an
%ideal of $S$, $x + y \in A$. This implies that $\widetilde{\chi}_{A}(x + y) =
%\widetilde{1} = Min^{i}(\widetilde{\chi}_{A}(x),
%\widetilde{\chi}_{A}(y))$. Also since $A$ is an ideal of $S$, $xy
%\in A$. Then $\widetilde{\chi}_{A}(xy) = \widetilde{1} =
%Max^{i}(\widetilde{\chi}_{A}(x), \widetilde{\chi}_{A}(y))$.\\
%$\underline{Case2}$: Let one of $x$ and $y$ be not in $A$. Then one of $\widetilde{\chi}_{A}(x)$ and $\widetilde{\chi}_{A}(y)$ is $\widetilde{0}$. So we have
%$\widetilde{\chi}_{A}(x + y) \geq \widetilde{0} =
%Min^{i}(\widetilde{\chi}_{A}(x), \widetilde{\chi}_{A}(y))$. Since
%$A$ is an ideal of $S$, $xy \in A$. Therefore
%$\widetilde{\chi}_{A}(xy) =
%\widetilde{1} = Max^{i}(\widetilde{\chi}_{A}(x), \widetilde{\chi}_{A}(y))$.\\
%$\underline{Case3}$: Let $x \notin A$ and $y \notin A$. Then $\widetilde{\chi}_{A}(x) = \widetilde{0} = \widetilde{\chi}_{A}(y)$. This shows that
%$\widetilde{\chi}_{A}(x + y) \geq \widetilde{0} =
%Min^{i}(\widetilde{\chi}_{A}(x), \widetilde{\chi}_{A}(y))$. Also
%$\widetilde{\chi}_{A}(xy) \geq \widetilde{0} =
%Max^{i}(\widetilde{\chi}_{A}(x),
%\widetilde{\chi}_{A}(y))$.\\
%Thus in all cases we have $\widetilde{\chi}_{A}(x + y) \geq
%Min^{i}(\widetilde{\chi}_{A}(x),\widetilde{\chi}_{A}(y))$ and
%$\widetilde{\chi}_{A}(xy) \geq
%Max^{i}(\widetilde{\chi}_{A}(x),\widetilde{\chi}_{A}(y))$.
Let $A$ be an ideal of $S$. Then $0_{S} \in A$. So
$\widetilde{\chi}_{A}(0_{S}) = \widetilde{1}$ and hence
$\widetilde{\chi}_{A}$ is non-empty.
Now suppose that $x, y \in S$.\\
$\underline{Case1}$: Let $Max^{i}(\widetilde{\chi}_{A}(x),
\widetilde{\chi}_{A}(y)) = \widetilde{0}$. Then
$\widetilde{\chi}_{A}(x) = \widetilde{0}$ and
$\widetilde{\chi}_{A}(y) = \widetilde{0}$. So
$\widetilde{\chi}_{A}(xy) \geq \widetilde{0} =
Max^{i}(\widetilde{\chi}_{A}(x), \widetilde{\chi}_{A}(y))$ and
$\widetilde{\chi}_{A}(x + y) \geq \widetilde{0} =
Min^{i}(\widetilde{\chi}_{A}(x), \widetilde{\chi}_{A}(y))$.\\
$\underline{Case2}$: Let $Max^{i}(\widetilde{\chi}_{A}(x),
\widetilde{\chi}_{A}(y)) = \widetilde{1}$. Then
$\widetilde{\chi}_{A}(x) = \widetilde{1}$ or $\widetilde{\chi}_{A}(y)
= \widetilde{1}$. This implies that $x \in A$ or $y \in A$. Then
$xy \in A$, since $A$ is an ideal of $S$. This shows that
$\widetilde{\chi}_{A}(xy) = \widetilde{1} =
Max^{i}(\widetilde{\chi}_{A}(x), \widetilde{\chi}_{A}(y))$. Now
$Max^{i}(\widetilde{\chi}_{A}(x), \widetilde{\chi}_{A}(y)) =
\widetilde{1} \Longrightarrow Min^{i}(\widetilde{\chi}_{A}(x),
\widetilde{\chi}_{A}(y)) = \widetilde{0}$ or $\widetilde{1}$.
$Min^{i}(\widetilde{\chi}_{A}(x), \widetilde{\chi}_{A}(y)) =
\widetilde{0} \Longrightarrow \widetilde{\chi}_{A}(x + y) \geq
Min^{i}(\widetilde{\chi}_{A}(x), \widetilde{\chi}_{A}(y))$.
$Min^{i}(\widetilde{\chi}_{A}(x), \widetilde{\chi}_{A}(y)) =
\widetilde{1} \Longrightarrow \widetilde{\chi}_{A}(x) =
\widetilde{1}$ and $\widetilde{\chi}_{A}(y) = \widetilde{1}
\Longrightarrow x \in A$ and $y \in A \Longrightarrow x + y \in A$
(since $A$ is an ideal of $S$) $\Longrightarrow
\widetilde{\chi}_{A}(x + y) = \widetilde{1} =
Min^{i}(\widetilde{\chi}_{A}(x), \widetilde{\chi}_{A}(y))$.
Consequently, $\widetilde{\chi}_{A}$ is an $i.v.$ fuzzy ideal of $S$.\\
Conversely, let $\widetilde{\chi}_{A}$ be an $i.v.$ fuzzy ideal of $S$. Then
$\widetilde{\chi}_{A}$ is non-empty. So $\widetilde{\chi}_{A}(s)
\neq \widetilde{0}$ for some $s \in S$. This implies that
$\widetilde{\chi}_{A}(s) = \widetilde{1}$ for some $s \in S$. So $s
\in A$. Hence $A$ is non-empty. Let $x, y \in A$. Then
$\widetilde{\chi}_{A}(x) = \widetilde{1} = \widetilde{\chi}_{A}(y)$.
Now since $\widetilde{\chi}_{A}$ is an $i.v.$ fuzzy ideal of $S$, we have
$\widetilde{\chi}_{A}(x + y) \geq
Min^{i}(\widetilde{\chi}_{A}(x),\widetilde{\chi}_{A}(y)) =
Min^{i}(\widetilde{1}, \widetilde{1}) = \widetilde{1}$. So
$\widetilde{\chi}_{A}(x + y) \geq \widetilde{1}$. Also
$\widetilde{\chi}_{A}(x + y) \leq \widetilde{1}$, since
$\widetilde{\chi}_{A}(z) \leq \widetilde{1}$ for all $z \in S$. Thus
$\widetilde{\chi}_{A}(x + y) = \widetilde{1}$. So we find that $x +
y \in A$. Now, let $a \in A$ and $s_{1} \in S$. Then
$\widetilde{\chi}_{A}(a) = \widetilde{1}$. Now since
$\widetilde{\chi}_{A}$ is an $i.v.$ fuzzy of $S$, we have
$\widetilde{\chi}_{A}(s_{1}a) \geq
Max^{i}(\widetilde{\chi}_{A}(a),\widetilde{\chi}_{A}(s_{1}) =
\widetilde{1}$. So $\widetilde{\chi}_{A}(s_{1}a) \geq \widetilde{1}$.
Again $\widetilde{\chi}_{A}(s_{1}a) \leq \widetilde{1}$. Thus we
find that $\widetilde{\chi}_{A}(s_{1}a) = \widetilde{1}$.
Consequently, $s_{1}a  \in A$. Similarly, we can show that $as_{1}
\in A$. Hence $A$ is an ideal of $S$.
\end{proof}

\begin{Lemma}
A non-empty $i.v.$ fuzzy subset $\widetilde{\mu}$ of a semiring $S$ is an
$i.v.$ fuzzy ideal of $S$ if and only if
$\overline{U}(\widetilde{\mu},[\alpha,\beta])$ are ideals of $S$
for all $[\alpha,\beta] \in Im\widetilde{\mu}$.
\end{Lemma}

\begin{proof}
First suppose that $\widetilde{\mu}$ is an $i.v.$ fuzzy ideal of $S$. Let
$[\alpha,\beta]$ be an arbitrary element in $Im\widetilde{\mu}$.
Now consider the level subset
$\overline{U}(\widetilde{\mu},[\alpha,\beta])$. Since
$[\alpha,\beta] \in Im\widetilde{\mu}$, we have
$\widetilde{\mu}(s_{0}) = [\alpha,\beta]$ for some $s_{0} \in S$.
This implies that $s_{0} \in
\overline{U}(\widetilde{\mu},[\alpha,\beta])$. So,
$\overline{U}(\widetilde{\mu},[\alpha,\beta])$ is non-empty. Now
take $x, y \in \overline{U}(\widetilde{\mu},[\alpha,\beta])$. Then
we have $\widetilde{\mu}(x) \geq [\alpha,\beta]$ and
$\widetilde{\mu}(y) \geq [\alpha,\beta]$. Since $\widetilde{\mu}$ is
an $i.v.$ fuzzy ideal of $S$, we have $\widetilde{\mu}(x + y) \geq
Min^{i}(\widetilde{\mu}(x), \widetilde{\mu}(y)) \geq [\alpha,
\beta]$. So we get $x + y \in
\overline{U}(\widetilde{\mu},[\alpha,\beta])$. Again let $a \in
\overline{U}(\widetilde{\mu},[\alpha,\beta])$ and $s_{1} \in S$.
Then $\widetilde{\mu}(a) \geq [\alpha,\beta]$. Since
$\widetilde{\mu}$ is an $i.v.$ fuzzy ideal of $S$, we have
$\widetilde{\mu}(s_{1}a) \geq
Max^{i}(\widetilde{\mu}(s_{1}),\widetilde{\mu}(a)) \geq
[\alpha,\beta]$. This implies that $s_{1}a \in
\overline{U}(\widetilde{\mu},[\alpha,\beta])$. Similarly, we can
show that $as_{1} \in
\overline{U}(\widetilde{\mu},[\alpha,\beta])$. Thus
$\overline{U}(\widetilde{\mu},[\alpha,\beta])$ is an ideal of $S$.
Since $[\alpha, \beta]$ is arbitrary, it follows that
$\overline{U}(\widetilde{\mu},[\alpha,\beta])$ are ideals of $S$
for all $[\alpha, \beta] \in Im\widetilde{\mu}$.\\
Conversely, suppose that
$\overline{U}(\widetilde{\mu},[\alpha,\beta])$ are ideals of $S$
for all $[\alpha,\beta] \in Im\widetilde{\mu}$. Let $x, y \in S$
and let $\widetilde{\mu}(x) = [\alpha_{1},\beta_{1}]$ and
$\widetilde{\mu}(y) = [\alpha_{2},\beta_{2}]$. This shows that $x
\in \overline{U}(\widetilde{\mu},[\alpha_{1},\beta_{1}])$ and $y
\in \overline{U}(\widetilde{\mu},[\alpha_{2},\beta_{2}])$. Without
loss of generality, we consider $[\alpha_{1},\beta_{1}]
> [\alpha_{2},\beta_{2}]$. Then by Proposition $2.17$, we have
$\overline{U}(\widetilde{\mu},[\alpha_{1},\beta_{1}]) \subseteq
\overline{U}(\widetilde{\mu},[\alpha_{2},\beta_{2}])$. So we find
that $x, y \in
\overline{U}(\widetilde{\mu},[\alpha_{2},\beta_{2}])$. Now since
$\overline{U}(\widetilde{\mu},[\alpha,\beta])$ are ideals of $S$
for all $[\alpha,\beta] \in Im\widetilde{\mu}$,
$\overline{U}(\widetilde{\mu},[\alpha_{2},\beta_{2}])$ is an ideal
of $S$. Thus $x, y \in
\overline{U}(\widetilde{\mu},[\alpha_{2},\beta_{2}])$ implies that
$x + y \in \overline{U}(\widetilde{\mu},[\alpha_{2},\beta_{2}])$.
Therefore $\widetilde{\mu}(x + y) \geq [\alpha_{2},\beta_{2}] =
Min^{i}([\alpha_{1},\beta_{1}],[\alpha_{2},\beta_{2}]) =
Min^{i}(\widetilde{\mu}(x),\widetilde{\mu}(y))$. Now let $s, t \in
S$ be such that $\widetilde{\mu}(t) = [\alpha_{3},\beta_{3}]$. Then
$t \in \overline{U}(\widetilde{\mu},[\alpha_{3},\beta_{3}])$.
Therefore $st \in
\overline{U}(\widetilde{\mu},[\alpha_{3},\beta_{3}])$, since
$\overline{U}(\widetilde{\mu},[\alpha_{3},\beta_{3}])$ is an ideal
of $S$. So $\widetilde{\mu}(st) \geq [\alpha_{3},\beta_{3}] =
\widetilde{\mu}(t)$. Similarly, if we take $\widetilde{\mu}(s) =
[\alpha_{3},\beta_{3}]$, we can prove that $\widetilde{\mu}(st)
\geq \widetilde{\mu}(s)$. Consequently, $\widetilde{\mu}(st) \geq
Max^{i}(\widetilde{\mu}(s),\widetilde{\mu}(t))$. Hence
$\widetilde{\mu}$ is an $i.v.$ fuzzy ideal of $S$.
\end{proof}

\begin{Definition}
If $\widetilde{\mu}$ be an $i.v.$ fuzzy ideal of a semiring $S$, then the
ideals $\overline{U}(\widetilde{\mu},[\alpha,\beta])$ of $S$, where
$[\alpha,\beta] \in Im\widetilde{\mu}$, are called the level ideals
of $\widetilde{\mu}$.
\end{Definition}

\begin{Lemma}
Let $S$ be a semiring and $\widetilde{\mu}$ be an $i.v.$ fuzzy ideal of $S$.
Then for any $x, y \in S$; $\widetilde{\mu}(x) \geq
\widetilde{\mu}(y)$ whenever $x \in$ $<y>$, the principal ideal
generated by $y$.
\end{Lemma}

\begin{proof}
Since $S$ is a semiring with identity, we find that $<y>$ $=
\Big\{\displaystyle\sum_{i = 1}^{n}r_{i}ys_{i} : r_{i}, s_{i} \in
S$ and $n \in \mathbb{N}\Big\}$. Now $x \in$ $<y>$ implies that $x
= \displaystyle\sum_{i = 1}^{n}r_{i}ys_{i}$ for some $r_{i}, s_{i}
\in S$ and $n \in \mathbb{N}$. Then\\
\begin{align*} \widetilde{\mu}(x) & =
\widetilde{\mu}\Big(\displaystyle\sum_{i =
1}^{n}r_{i}ys_{i}\Big)\\
 &= \widetilde{\mu}(r_{1}ys_{1} + r_{2}ys_{2} + \ldots + r_{n}ys_{n})\\
 &= \widetilde{\mu}\Big((r_{1}ys_{1} + r_{2}ys_{2} + \ldots + r_{n-1}ys_{n-1}) + r_{n}ys_{n}\Big)\\
 &\geq  Min^{i}\Big(\widetilde{\mu}(r_{1}ys_{1} + r_{2}ys_{2} + \ldots +
 r_{n-1}ys_{n-1}), \widetilde{\mu}(r_{n}ys_{n})\Big)  ~~ \text{(Since $\widetilde{\mu}$ is an $i.v.$ fuzzy ideal of $S$)}\\
 %&\hspace{11cm} \text{fuzzy ideal of S)}\\
 &\geq  Min^{i}\Big(\widetilde{\mu}(r_{1}ys_{1}+ \ldots +
 r_{n-1}ys_{n-1}),Max^{i}(\widetilde{\mu}(r_{n}y),\widetilde{\mu}(s_{n}))\Big) ~~ \text{(Since $\widetilde{\mu}$ is an $i.v.$ fuzzy ideal of $S$)}\\
 %&\hspace{11cm} \text{fuzzy ideal of S)}\\
 &\geq  Min^{i}\Big(\widetilde{\mu}(r_{1}ys_{1}+ \ldots +
 r_{n-1}ys_{n-1}),Max^{i}(Max^{i}(\widetilde{\mu}(r_{n}),\widetilde{\mu}(y)),\widetilde{\mu}(s_{n}))\Big)\\
 &\geq  Min^{i}\Big(\widetilde{\mu}(r_{1}ys_{1}+ \ldots +
 r_{n-1}ys_{n-1}), \widetilde{\mu}(y)\Big)\\
 &\vdots\\
 &\geq \widetilde{\mu}(y).
\end{align*}
Thus we get that $\widetilde{\mu}(x) \geq \widetilde{\mu}(y)$.
\end{proof}

\begin{Lemma}
Let $I$ be an ideal of a semiring $S$ and $[\alpha,\beta] \leq
[\gamma,\delta] \neq \widetilde{0}$ be any two interval numbers on
$[0,1]$. Then the $i.v.$ fuzzy subset $\widetilde{\mu}$ of $S$, defined by :

$$ \widetilde{\mu}(x) = \begin{cases}
         [\gamma,\delta] &\text{when} \quad x \in I,\\
          \hfill  [\alpha,\beta] &\text{otherwise},\\
         \end{cases} $$
is an $i.v.$ fuzzy ideal of $S$.
\end{Lemma}

\begin{proof}
Since $I$ is an ideal of $S$, we have $0_{S} \in I$. Then
$\widetilde{\mu}(0_{S}) = [\gamma,\delta] \neq \widetilde{0}$. So
$\widetilde{\mu}$ is non-empty. Now let $x, y \in S$.\\
$\underline{Case1}$: Let
$Max^{i}(\widetilde{\mu}(x),\widetilde{\mu}(y)) = [\alpha, \beta]$.
Then $\widetilde{\mu}(x) = [\alpha, \beta]$ and $\widetilde{\mu}(y)
= [\alpha, \beta] \Longrightarrow \widetilde{\mu}(xy) \geq [\alpha,
\beta] = Max^{i}(\widetilde{\mu}(x),\widetilde{\mu}(y))$ and
$\widetilde{\mu}(x + y) \geq [\alpha, \beta] =
Min^{i}(\widetilde{\mu}(x),\widetilde{\mu}(y))$.\\
$\underline{Case2}$: Let
$Max^{i}(\widetilde{\mu}(x),\widetilde{\mu}(y)) = [\gamma, \delta]$.
Then $\widetilde{\mu}(x) = [\gamma, \delta]$ or $\widetilde{\mu}(y)
= [\gamma, \delta] \Longrightarrow x \in I$ or $y \in I
\Longrightarrow xy \in I$ (since $I$ is an ideal of $S$)
$\Longrightarrow \widetilde{\mu}(xy) = [\gamma, \delta] =
Max^{i}(\widetilde{\mu}(x),\widetilde{\mu}(y))$. Now
$Max^{i}(\widetilde{\mu}(x),\widetilde{\mu}(y)) = [\gamma, \delta]
\Longrightarrow Min^{i}(\widetilde{\mu}(x),\widetilde{\mu}(y)) =
[\alpha, \beta]$ or $[\gamma, \delta]$.
$Min^{i}(\widetilde{\mu}(x),\widetilde{\mu}(y)) = [\alpha, \beta]
\Longrightarrow \widetilde{\mu}(x + y) \geq [\alpha, \beta] =
Min^{i}(\widetilde{\mu}(x),\widetilde{\mu}(y))$. Again
$Min^{i}(\widetilde{\mu}(x),\widetilde{\mu}(y)) = [\gamma, \delta]
\Longrightarrow \widetilde{\mu}(x) = [\gamma, \delta]$ and
$\widetilde{\mu}(y) = [\gamma, \delta] \Longrightarrow x \in I$ and
$y \in I \Longrightarrow x + y \in I$ (since $I$ is an ideal of
$S$) $\Longrightarrow \widetilde{\mu}(x + y) = [\gamma, \delta] =
Min^{i}(\widetilde{\mu}(x),\widetilde{\mu}(y))$.
%$\underline{Case1}$: Let $x, y \in I$. Then $\widetilde{\mu}(x) =
%[\gamma,\delta] = \widetilde{\mu}(y)$. Since $I$ is an ideal of
%$S$, $x + y \in I$. This implies that $\widetilde{\mu}(x + y) =
%[\gamma,\delta] = Min^{i}(\widetilde{\mu}(x),\widetilde{\mu}(y))$.
%Also since $I$ is an ideal of $S$, $xy \in I$. This shows that
%$\widetilde{\mu}(xy) = [\gamma,\delta] =
%Max^{i}(\widetilde{\mu}(x),\widetilde{\mu}(y))$.\\
%$\underline{Case2}$: Let one of $x$ and $y$ be not in $I$. Then
%one of $\widetilde{\mu}(x)$ and $\widetilde{\mu}(y)$ is
%$[\alpha,\beta]$. Thus we get $\widetilde{\mu}(x + y) \geq
%[\alpha,\beta] = Min^{i}([\alpha,\beta],[\gamma,\delta]) =
%Min^{i}(\widetilde{\mu}(x),\widetilde{\mu}(y))$. Now since $I$ is an
%ideal of $S$, $xy \in I$. So, $\widetilde{\mu}(xy) =
%[\gamma,\delta] = Max^{i}(\widetilde{\mu}(x),\widetilde{\mu}(y))$.\\
%$\underline{Case3}$: Let $x \notin I$ and $y \notin I$. Then
%$\widetilde{\mu}(x) = [\alpha,\beta] = \widetilde{\mu}(y)$. Then
%$\widetilde{\mu}(x + y) \geq [\alpha,\beta] =
%Min^{i}(\widetilde{\mu}(x),\widetilde{\mu}(y))$. Also
%$\widetilde{\mu}(xy) \geq [\alpha,\beta] =
%Max^{i}(\widetilde{\mu}(x),\widetilde{\mu}(y))$.\\
Thus in all cases we find that $\widetilde{\mu}(x + y) \geq
Min^{i}(\widetilde{\mu}(x),\widetilde{\mu}(y))$ and
$\widetilde{\mu}(xy) \geq
Max^{i}(\widetilde{\mu}(x),\widetilde{\mu}(y))$. Consequently,
$\widetilde{\mu}$ is an $i.v.$ fuzzy ideal of $S$.
\end{proof}

\begin{Lemma}
Let $\widetilde{\mu}$ be an $i.v.$ fuzzy ideal of a semiring $S$. Then the
set $\widetilde{\mu}_{0} = \{x \in S: \widetilde{\mu}(x) =
\widetilde{\mu}(0_{S})\}$ is an ideal of $S$.
\end{Lemma}

\begin{proof}
Since $0_{S} \in \widetilde{\mu}_{0}$, $\widetilde{\mu}_{0}$ is
non-empty. Let $x, y \in \widetilde{\mu}_{0}$. Then
$\widetilde{\mu}(x) = \widetilde{\mu}(0_{S}) = \widetilde{\mu}(y)$.
Now, since $\widetilde{\mu}$ is an $i.v.$ fuzzy ideal of $S$, we have
$\widetilde{\mu}(x + y) \geq Min^{i}(\widetilde{\mu}(x),
\widetilde{\mu}(y)) = \widetilde{\mu}(0_{S})$. Also by Remark $3.3$,
we have $\widetilde{\mu}(0_{S}) \geq \widetilde{\mu}(x + y)$. Thus
$\widetilde{\mu}(x + y) = \widetilde{\mu}(0_{S})$. So $x + y \in
\widetilde{\mu}_{0}$. Let $s \in S$ and $t \in \widetilde{\mu}_{0}$.
Then $\widetilde{\mu}(t) = \widetilde{\mu}(0_{S})$. Now since
$\widetilde{\mu}$ is an $i.v.$ fuzzy ideal of $S$, we have
$\widetilde{\mu}(st) \geq
Max^{i}(\widetilde{\mu}(s),\widetilde{\mu}(t)) =
Max^{i}(\widetilde{\mu}(s),\widetilde{\mu}(0_{S})) =
\widetilde{\mu}(0_{S})$ (since $\widetilde{\mu}(0_{S}) \geq
\widetilde{\mu}(s)$, by Remark $3.3$). Also $\widetilde{\mu}(0_{S})
\geq \widetilde{\mu}(st)$, by Remark $3.3$. So $\widetilde{\mu}(st) =
\widetilde{\mu}(0_{S})$. Thus $st \in \widetilde{\mu}_{0}$.
Similarly, we can show that $ts \in \widetilde{\mu}_{0}$.
Consequently, $\widetilde{\mu}_{0}$ is an ideal of $S$.
\end{proof}

\begin{Definition}
An $i.v.$ fuzzy ideal $\widetilde{\mu}$ of a semiring $S$ is said to be an
interval-valued fuzzy prime ideal of $S$ if $\widetilde{\mu}$ is
not a constant function (i.e. $|Im\widetilde{\mu}| \geq 2$) and for
any two $i.v.$ fuzzy ideals $\widetilde{\mu_{1}}$ and $\widetilde{\mu_{2}}$ of
$S$, $\widetilde{\mu_{1}} \circ \widetilde{\mu_{2}} \subseteq
\widetilde{\mu}$ implies that either $\widetilde{\mu_{1}} \subseteq
\widetilde{\mu}$ or $\widetilde{\mu_{2}} \subseteq \widetilde{\mu}$.
\end{Definition}

\begin{Theorem}
Let $I$ be a prime ideal of a semiring $S$ and $[\alpha,\beta] \in
D[0,1] \setminus \{\widetilde{1}\}$. Let $\widetilde{\mu}$ be an
$i.v.$ fuzzy subset of $S$, defined by :

  $$ \widetilde{\mu}(x) = \begin{cases}
          \widetilde{1} &\text{when} \quad x \in I, \\
           \hfill [\alpha,\beta] &\text{otherwise}. \\
         \end{cases} $$

Then $\widetilde{\mu}$ is an $i.v.$ fuzzy prime ideal of $S$.
\end{Theorem}
\begin{proof}
By Lemma $3.8$, it follows that $\widetilde{\mu}$ is an $i.v.$ fuzzy ideal
of $S$. Clearly, $\widetilde{\mu}$ is non-constant. Let
$\widetilde{\mu_{1}}$ and $\widetilde{\mu_{2}}$ be two $i.v.$ fuzzy ideals of
$S$ such that $(\widetilde{\mu_{1}} \circ \widetilde{\mu_{2}})
\subseteq \widetilde{\mu}$. We have to prove that either
$\widetilde{\mu_{1}} \subseteq \widetilde{\mu}$ or
$\widetilde{\mu_{2}} \subseteq \widetilde{\mu}$. If possible, let
$\widetilde{\mu_{1}} \nsubseteq \widetilde{\mu}$ and
$\widetilde{\mu_{2}} \nsubseteq \widetilde{\mu}$. Then there exist
$x, y \in S$ such that $\widetilde{\mu_{1}}(x) \nleq
\widetilde{\mu}(x)$ and $\widetilde{\mu_{2}}(y) \nleq
\widetilde{\mu}(y)$. Now according to our assumption, any two
interval numbers in $D[0,1]$ are comparable. So we find that
$\widetilde{\mu_{1}}(x) > \widetilde{\mu}(x)$ and
$\widetilde{\mu_{2}}(y) > \widetilde{\mu}(y)$. This implies that
$\widetilde{\mu}(x) \neq \widetilde{1}$ and $\widetilde{\mu}(y) \neq
\widetilde{1}$. Hence $\widetilde{\mu}(x) = \widetilde{\mu}(y) =
[\alpha, \beta]$. So $x \notin I$ and $y \notin I$. Since $I$ is a
prime ideal of $S$, there exists $s \in S$ such that $xsy \notin
I$, by Proposition $2.5$. Therefore
$\widetilde{\mu}(xsy) = [\alpha, \beta]$. Now,\\
$\begin{array}{cccc}(\widetilde{\mu_{1}} \circ
\widetilde{\mu_{2}})(xsy) & = & \displaystyle\sup_{xsy=pq}\Big\{Min^{i}\Big(\widetilde{\mu_{1}}(p),\widetilde{\mu_{2}}(q)\Big)\Big\} &\\
& \geq & Min^{i}\Big(\widetilde{\mu_{1}}(x),\widetilde{\mu_{2}}(sy)\Big) &\\
 & \geq &
Min^{i}\Big(\widetilde{\mu_{1}}(x),\widetilde{\mu_{2}}(y)\Big) & \text{(Since $\widetilde{\mu}_{2}$ is an $i.v.$ fuzzy ideal of $S$)}\\
% & & & \text{of $S$)}\\
 & > & Min^{i}\Big(\widetilde{\mu}(x),\widetilde{\mu}(y)\Big) &\\
 & = &[\alpha,\beta] ~= ~\widetilde{\mu}(xsy) &.
\end{array}$
\\
This contradicts the fact that $(\widetilde{\mu_{1}} \circ
\widetilde{\mu_{2}}) \subseteq \widetilde{\mu}$. Consequently,
either $\widetilde{\mu_{1}} \subseteq \widetilde{\mu}$ or
$\widetilde{\mu_{2}} \subseteq \widetilde{\mu}$. Hence
$\widetilde{\mu}$ is an $i.v.$ fuzzy prime ideal of $S$.
\end{proof}

From the above theorem, we can easily produce an example of an
$i.v.$ fuzzy prime ideal of a semiring as follows :

\begin{Example}
Let $\widetilde{\mu}$ be an $i.v.$ fuzzy subset of the set of non-negative
integers $\mathbb{N}_{0}$ defined by:

  $$ \widetilde{\mu}(x) = \begin{cases}
          \widetilde{1} &\text{when} \quad x \in 3\mathbb{N}_{o}, \\
           \hfill [0.5,0.6] &\text{otherwise}. \\
         \end{cases} $$
				
Then $\widetilde{\mu}$ is an $i.v.$ fuzzy prime ideal of $\mathbb{N}_{0}$.
\end{Example}

\begin{Theorem}
If $\widetilde{\mu}$ be an $i.v.$ fuzzy prime ideal of a semiring $S$, then
\begin{enumerate}[(i)]
\item $\widetilde{\mu}(0_{S}) = \widetilde{1}$. \item $|Im
\widetilde{\mu}| = 2$. \item $\widetilde{\mu}_{0} = \Big\{x \in S:
\widetilde{\mu}(x) = \widetilde{\mu}(0_{S})\Big\}$ is a prime ideal
of $S$.
\end{enumerate}
\end{Theorem}

\begin{proof}
Let $\widetilde{\mu}$ be an $i.v.$ fuzzy prime ideal of a semiring $S$.
\begin{enumerate}[$(i)$]
\item If possible, let $\widetilde{\mu}(0_{S}) \neq \widetilde{1}$.
Since any two interval numbers in $D[0,1]$ are comparable, it
follows that $\widetilde{\mu}(0_{S}) < \widetilde{1}$. Since
$\widetilde{\mu}$ is an $i.v.$ fuzzy prime ideal of $S$, $\widetilde{\mu}$ is
non-constant. Also $\widetilde{\mu}(0_{S}) \geq \widetilde{\mu}(x)$
for all $x \in S$, by Remark $3.3$. Since $\widetilde{\mu}$ is
non-constant, there exists $s \in S$ such that
$\widetilde{\mu}(0_{S})
> \widetilde{\mu}(s)$. Now we construct two $i.v.$ fuzzy subsets $\widetilde{\mu_{1}}$ and $\widetilde{\mu_{2}}$ of $S$ as
follows :

$ \widetilde{\mu_{1}}(x) = \begin{cases}
          \widetilde{1} &\text{when} \quad x \in \widetilde{\mu}_{0},\\
           \hfill \widetilde{0} &\text{otherwise}, \\
         \end{cases} $  $\hspace{1cm}$ and \; \; $\widetilde{\mu_{2}}(x) = \widetilde{\mu}(0_{S})$ for all
$x \in S$.\\
Since $\widetilde{\mu}$ is an $i.v.$ fuzzy ideal of $S$, we have
$\widetilde{\mu}_{0}$ is an ideal of $S$, by Lemma $3.9$. Then by
Lemma $3.8$, we have $\widetilde{\mu_{1}}$ is an $i.v.$ fuzzy ideal of $S$.
Also $\widetilde{\mu_{2}}$ is an $i.v.$ fuzzy ideal of $S$, since it is a
constant function. Let $x \in S$ be such that $\widetilde{\mu}(x) =
\widetilde{\mu}(0_{S})$. This implies that $\widetilde{\mu_{1}}(x) =
\widetilde{1}$. Then for any $y \in S$,
$Min^{i}\Big(\widetilde{\mu_{1}}(x),\widetilde{\mu_{2}}(y)\Big) =
\widetilde{\mu_{2}}(y) = \widetilde{\mu}(0_{S}) =
\widetilde{\mu}(xy)$. (Since $\widetilde{\mu}$ is an $i.v.$ fuzzy prime ideal of
$S$, it is an $i.v.$ fuzzy ideal of $S$. So $\widetilde{\mu}(0_{S}) \geq
\widetilde{\mu}(xy)$, by Remark $3.3$ and $\widetilde{\mu}(xy) \geq
\widetilde{\mu}(x) = \widetilde{\mu}(0_{S})$. So we get
$\widetilde{\mu}(xy) =
\widetilde{\mu}(0_{S})$). Now, let $x \in S$ be such that $\widetilde{\mu}(x) \neq
\widetilde{\mu}(0_{S})$. This implies that $\widetilde{\mu_{1}}(x) =
\widetilde{0}$. Then for any $y \in S$,
$Min^{i}\Big(\widetilde{\mu_{1}}(x),\widetilde{\mu_{2}}(y)\Big) =
\widetilde{0} \leq \widetilde{\mu}(xy)$. Thus for any $z \in S$,
$\displaystyle\sup_{z=xy}\Big\{Min^{i}\Big(\widetilde{\mu_{1}}(x),\widetilde{\mu_{2}}(y)\Big)\Big\}
\leq \widetilde{\mu}(z)$ i.e. $(\widetilde{\mu_{1}} \circ \widetilde{\mu_{2}})(z) \leq \widetilde{\mu}(z)$. This implies that $\widetilde{\mu_{1}}
\circ \widetilde{\mu_{2}} \subseteq \widetilde{\mu}$. Since
$\widetilde{\mu}$ is an $i.v.$ fuzzy prime ideal of $S$, it follows that either
$\widetilde{\mu_{1}} \subseteq \widetilde{\mu}$ or
$\widetilde{\mu_{2}} \subseteq \widetilde{\mu}$. But
$\widetilde{\mu_{1}}(0_{S}) = \widetilde{1} > \widetilde{\mu}(0_{S})$
and $\widetilde{\mu_{2}}(s) = \widetilde{\mu}(0_{S}) >
\widetilde{\mu}(s)$. Thus we arrive at a contradiction.
Consequently, $\widetilde{\mu}(0_{S}) = \widetilde{1}$. \item Since
$\widetilde{\mu}$ is an $i.v.$ fuzzy prime ideal of $S$, it follows that
$\widetilde{\mu}$ is non-constant. So $|Im \widetilde{\mu}| \geq 2$.
If possible, let $|Im \widetilde{\mu}| > 2$. Let $a, b \in S$ be
such that $\widetilde{1}
> \widetilde{\mu}(a) > \widetilde{\mu}(b)$. Now we construct two
$i.v.$ fuzzy subsets $\widetilde{\mu_{1}}$ and $\widetilde{\mu_{2}}$ of $S$ as
follows :\\

  $ \widetilde{\mu_{1}}(x)  = \begin{cases}
          \widetilde{1} &\text{when} \quad x \in \;<a>, \\
           \hfill \widetilde{0} &\text{otherwise}, \\
         \end{cases} $ $\hspace{1cm}$ and \; \; $\widetilde{\mu_{2}}(x) = \widetilde{\mu}(a)$ for all
$x \in S$.\\
By Lemma $3.8$, we have $\widetilde{\mu_{1}}$ is an $i.v.$ fuzzy ideal of
$S$. Also $\widetilde{\mu_{2}}$ is an $i.v.$ fuzzy ideal of $S$, since it is
a constant function. Now suppose $x \in$ $<a>$ for $x \in S$. Then
$\widetilde{\mu_{1}}(x) = \widetilde{1}$. So for any $y \in S$,
$Min^{i}\Big(\widetilde{\mu_{1}}(x),\widetilde{\mu_{2}}(y)\Big) =
\widetilde{\mu_{2}}(y) = \widetilde{\mu}(a) \leq
\widetilde{\mu}(xy)$. (Since $x \in$ $<a>$, we have
$\widetilde{\mu}(x) \geq \widetilde{\mu}(a)$, by Lemma $3.7$. Also
since $\widetilde{\mu}$ is an $i.v.$ fuzzy ideal of $S$, we have
$\widetilde{\mu}(xy) \geq \widetilde{\mu}(x)$ which implies that
$\widetilde{\mu}(xy) \geq \widetilde{\mu}(a)$). Now let $x \in S$ be
such that $x \notin$ $<a>$. This implies that
$\widetilde{\mu_{1}}(x) = \widetilde{0}$. Then for any $y \in S$,
$Min^{i}\Big(\widetilde{\mu_{1}}(x),\widetilde{\mu_{2}}(y)\Big) =
\widetilde{0} \leq \widetilde{\mu}(xy)$. Therefore for any $z \in
S$,
$\displaystyle\sup_{z=xy}\Big\{Min^{i}\Big(\widetilde{\mu_{1}}(x),\widetilde{\mu_{2}}(y)\Big)\Big\}
\leq \widetilde{\mu}(z)$ i.e. $(\widetilde{\mu_{1}} \circ \widetilde{\mu_{2}})(z) \leq \widetilde{\mu}(z)$. This implies that $\widetilde{\mu_{1}}
\circ \widetilde{\mu_{2}} \subseteq \widetilde{\mu}$. Now since
$\widetilde{\mu}$ is an $i.v.$ fuzzy prime ideal of $S$, we find that either
$\widetilde{\mu_{1}} \subseteq \widetilde{\mu}$ or
$\widetilde{\mu_{2}} \subseteq \widetilde{\mu}$. But
$\widetilde{\mu_{1}}(a) = \widetilde{1} > \widetilde{\mu}(a)$ and
$\widetilde{\mu_{2}}(b) = \widetilde{\mu}(a) > \widetilde{\mu}(b)$.
This contradicts the fact that either $\widetilde{\mu_{1}}
\subseteq \widetilde{\mu}$ or $\widetilde{\mu_{2}} \subseteq
\widetilde{\mu}$. Hence $|Im \widetilde{\mu}| = 2$. \item Since
$\widetilde{\mu}$ is an $i.v.$ fuzzy prime ideal of $S$, $\widetilde{\mu}_{0}$ is
an ideal of $S$, by Lemma $3.9$. Also $\widetilde{\mu}$ is
non-constant. So $\widetilde{\mu}_{0}$ is a proper ideal of $S$.
Now let $A, B$ be any two ideals of $S$ such that $AB \subseteq
\widetilde{\mu}_{0}$. Since $A, B$ are ideals of $S$,
$\widetilde{\chi}_{A}$ and $\widetilde{\chi}_{B}$ are $i.v.$ fuzzy ideals of
$S$, by Lemma $3.4$. Let $x \in S$. If $(\widetilde{\chi}_{A} \circ
\widetilde{\chi}_{B})(x) = \widetilde{0}$, we find that
$(\widetilde{\chi}_{A} \circ \widetilde{\chi}_{B})(x) = \widetilde{0}
\leq \widetilde{\chi}_{\widetilde{\mu}_{0}}(x)$. If
$(\widetilde{\chi}_{A} \circ \widetilde{\chi}_{B})(x) \neq
\widetilde{0}$, $(\widetilde{\chi}_{A} \circ \widetilde{\chi}_{B})(x)
=
\displaystyle\sup_{x=pq}\Big\{Min^{i}\Big(\widetilde{\chi}_{A}(p),
\widetilde{\chi}_{B}(q)\Big)\Big\} \neq \widetilde{0}$. This shows
that
$\displaystyle\sup_{x=pq}\Big\{Min^{i}\Big(\widetilde{\chi}_{A}(p),
\widetilde{\chi}_{B}(q)\Big)\Big\} = \widetilde{1}$. Then
 $Min^{i}\Big(\widetilde{\chi}_{A}(p),\widetilde{\chi}_{B}(q)\Big)
= \widetilde{1}$ for some $p, q \in S$ such that $x = pq$, where
$\widetilde{\chi}_{A}(p) = \widetilde{1}$ and
$\widetilde{\chi}_{B}(q) = \widetilde{1}$. This implies that $p \in
A$ and $q \in B$ i.e. $pq \in AB$. Since $AB \subseteq
\widetilde{\mu}_{0}$ we obtain that $x = pq \in
\widetilde{\mu}_{0}$. Thus $\widetilde{\chi}_{\widetilde{\mu}_{0}}(x)
= \widetilde{1}$. Then $(\widetilde{\chi}_{A} \circ
\widetilde{\chi}_{B})(x) =
\widetilde{\chi}_{\widetilde{\mu}_{0}}(x)$. Consequently,
$(\widetilde{\chi}_{A} \circ \widetilde{\chi}_{B})(x) \leq
\widetilde{\chi}_{\widetilde{\mu}_{0}}(x)$ for all $x \in S$. So we
get that $\widetilde{\chi}_{A} \circ \widetilde{\chi}_{B} \subseteq
\widetilde{\chi}_{\widetilde{\mu}_{0}}$. Let  $y \in S$. If
$\widetilde{\chi}_{\widetilde{\mu}_{0}}(y) = \widetilde{0}$ we find
that $\widetilde{\chi}_{\widetilde{\mu}_{0}}(y) \leq
\widetilde{\mu}(y)$. If $\widetilde{\chi}_{\widetilde{\mu}_{0}}(y) =
\widetilde{1}$, it follows that $y \in \widetilde{\mu}_{0}$. Then we
get that $\widetilde{\mu}(y) = \widetilde{\mu}(0)$. Again since
$\widetilde{\mu}$ is an $i.v.$ fuzzy prime ideal of $S$, we find that
$\widetilde{\mu}(0) = \widetilde{1}$, by Theorem $3.13(i)$. Thus
$\widetilde{\chi}_{\widetilde{\mu}_{0}}(y) = \widetilde{\mu}(y)$. So
we see that $\widetilde{\chi}_{\widetilde{\mu}_{0}}(y) \leq
\widetilde{\mu}(y)$ for all $y \in S$. Consequently,
$\widetilde{\chi}_{\widetilde{\mu}_{0}} \subseteq \widetilde{\mu}$.
Thus we get $\widetilde{\chi}_{A} \circ \widetilde{\chi}_{B}
\subseteq \widetilde{\chi}_{\widetilde{\mu}_{0}} \subseteq
\widetilde{\mu}$ i.e. $\widetilde{\chi}_{A} \circ
\widetilde{\chi}_{B} \subseteq \widetilde{\mu}$. Since
$\widetilde{\mu}$ is an $i.v.$ fuzzy prime ideal of $S$, we find that either
$\widetilde{\chi}_{A} \subseteq \widetilde{\mu}$ or
$\widetilde{\chi}_{B} \subseteq \widetilde{\mu}$. Suppose
$\widetilde{\chi}_{A} \subseteq \widetilde{\mu}$. Let $z \in A$.
Then $\widetilde{\chi}_{A}(z) = \widetilde{1}$. Also since
$\widetilde{\chi}_{A} \subseteq \widetilde{\mu}$, we obtain that
$\widetilde{\chi}_{A}(z) \leq \widetilde{\mu}(z)$. This implies that
$\widetilde{\mu}(z) \geq \widetilde{1}$ i.e. $\widetilde{\mu}(z) =
\widetilde{1}$. Since $\widetilde{\mu}$ is an $i.v.$ fuzzy prime ideal of $S$, we
have $\widetilde{\mu}(0) = \widetilde{1}$, by Theorem $3.13(i)$.
Thus $\widetilde{\mu}(z) = \widetilde{\mu}(0)$. Consequently, $z \in
\widetilde{\mu}_{0}$. So we find that $A \subseteq
\widetilde{\mu}_{0}$. Similarly, we can show that $B \subseteq
\widetilde{\mu}_{0}$ whenever $\widetilde{\chi}_{B} \subseteq
\widetilde{\mu}$. Thus we see that for any two ideals $A, B$ of
$S$, $AB \subseteq \widetilde{\mu}_{0} \Longrightarrow$ either $A
\subseteq \widetilde{\mu}_{0}$ or $B \subseteq \widetilde{\mu}_{0}$.
Hence $\widetilde{\mu}_{0}$ is a prime ideal of $S$.\end{enumerate}
\end{proof}

\vspace{.2in}

From the above two theorems we have the following result :
\begin{Theorem}
If $\widetilde{\mu}$ be an $i.v.$ fuzzy subset of a semiring $S$, then
$\widetilde{\mu}$ is an $i.v.$ fuzzy prime ideal of $S$ if and only if
$Im\widetilde{\mu} = \{\widetilde{1}, [\alpha,\beta]\}$ where
$[\alpha,\beta] \in D[0,1] \setminus \{\widetilde{1}\}$ and
$\widetilde{\mu}_{0}$ is a prime ideal of $S$.
\end{Theorem}

\begin{Theorem}
Let $I (\neq S)$ be a subset of a semiring $S$. Then $I$ is a
prime ideal of $S$ if and only if $\widetilde{\chi}_{I}$ is an
$i.v.$ fuzzy prime ideal of $S$.
\end{Theorem}

\begin{proof}
Let $I$ be a prime ideal of $S$. Now in Theorem $3.11$, if we
replace $[\alpha,\beta]$ by $\widetilde{0}$, we find that
$\widetilde{\chi}_{I}$ is an $i.v.$ fuzzy prime ideal of
$S$.\\
Conversely, let $\widetilde{\chi}_{I}$ be an $i.v.$ fuzzy prime ideal of $S$.
Then $(\widetilde{\chi}_{I})_{0} = \Big\{x \in S:
\widetilde{\chi}_{I}(x) = \widetilde{\chi}_{I}(0_{S})\Big\} =
\Big\{x \in S: \widetilde{\chi}_{I}(x) = \widetilde{1}\Big\} = I$.
Hence $I$ is a prime ideal of $S$, by Theorem $3.13(iii)$.
\end{proof}

\begin{Lemma}
Let $\widetilde{\mu}$ be an $i.v.$ fuzzy prime ideal of a semiring $S$. Then for
any $a, b \in S$; $Inf\Big\{\widetilde{\mu}(asb): s \in S\Big\} =
Max^{i}\Big(\widetilde{\mu}(a),\widetilde{\mu}(b)\Big)$.
\end{Lemma}
\begin{proof}
Since $\widetilde{\mu}$ is an $i.v.$ fuzzy prime ideal of $S$, we have
$Im\widetilde{\mu} = \{\widetilde{1}, [\alpha,\beta]\}$, where
$[\alpha,\beta] \in D[0,1] \setminus \{\widetilde{1}\}$ and
$\widetilde{\mu}_{0}$ is a prime ideal of
$S$, by Theorem $3.14$.\\
$\underline{Case1}$: Let
$Max^{i}\Big(\widetilde{\mu}(a),\widetilde{\mu}(b)\Big) =
\widetilde{1}$. This implies that either $\widetilde{\mu}(a) =
\widetilde{1}$ or $\widetilde{\mu}(b) = \widetilde{1}$ i.e.
$\widetilde{\mu}(a) = \widetilde{\mu}(0_{S})$ or $\widetilde{\mu}(b)
= \widetilde{\mu}(0_{S})$, by Theorem $3.13(i)$. Therefore either $a
\in \widetilde{\mu}_{0}$ or $b \in \widetilde{\mu}_{0}$. This
implies that $asb \in \widetilde{\mu}_{0}$ for all $s \in S$, by
Proposition $2.5$. Therefore $\widetilde{\mu}(asb) =
\widetilde{\mu}(0_{S}) = \widetilde{1}$ for all $s \in S$, by
Theorem $3.13(i)$. This shows that $Inf\Big\{\widetilde{\mu}(asb):
s \in S\Big\} =\widetilde{1} =
Max^{i}\Big(\widetilde{\mu}(a),\widetilde{\mu}(b)\Big)$.\\
$\underline{Case2}$: Let
$Max^{i}\Big(\widetilde{\mu}(a),\widetilde{\mu}(b)\Big) =
[\alpha,\beta]$. Then $\widetilde{\mu}(a) = [\alpha,\beta]$ and
$\widetilde{\mu}(b) = [\alpha,\beta]$, since $Im\widetilde{\mu} =
\{\widetilde{1}, [\alpha, \beta]\}$. This implies that $a \notin
\widetilde{\mu}_{0}$ and $b \notin \widetilde{\mu}_{0}$. Since
$\widetilde{\mu}_{0}$ is a prime ideal of $S$, there exists $s_{0}
\in S$ such that $as_{0}b \notin \widetilde{\mu}_{0}$, by
Proposition $2.5$. Then $\widetilde{\mu}(as_{0}b) =
[\alpha,\beta]$. Consequently, $Inf\Big\{\widetilde{\mu}(asb): s
\in S\Big\} =
Max^{i}\Big(\widetilde{\mu}(a),\widetilde{\mu}(b)\Big)$.
\end{proof}

\begin{Theorem}
Let $\widetilde{\mu}$ be an $i.v.$ fuzzy prime ideal of a semiring $S$ and $a,
b$ be any two elements of $S$. Then the following are equivalent :
\begin{enumerate}[(i)]
\item $\widetilde{\mu}(ab) =
Max^{i}\Big(\widetilde{\mu}(a),\widetilde{\mu}(b)\Big)$. \item
$\widetilde{\mu}(ab) = \widetilde{\mu}(ba)$.
\end{enumerate}
\end{Theorem}

\begin{proof}
$\underline{(i) \Longrightarrow (ii)}$ : Let $\widetilde{\mu}(ab) =
Max^{i}\Big(\widetilde{\mu}(a),\widetilde{\mu}(b)\Big)$. Then
$\widetilde{\mu}(ab) =
Max^{i}\Big(\widetilde{\mu}(a),\widetilde{\mu}(b)\Big) =
Max^{i}\Big(\widetilde{\mu}(b),\widetilde{\mu}(a)\Big) =
\widetilde{\mu}(ba)$.\\
$\underline{(ii) \Longrightarrow (i)}$ : Let $\widetilde{\mu}(ab) =
\widetilde{\mu}(ba)$ for any two elements $a, b$ of $S$. Then
$\widetilde{\mu}(asb) = \widetilde{\mu}(bas)$ for all $s \in S$.
Also since $\widetilde{\mu}$ is an $i.v.$ fuzzy prime ideal of $S$,
$\widetilde{\mu}(asb) = \widetilde{\mu}(bas) \geq
Max^{i}(\widetilde{\mu}(ba), \widetilde{\mu}(s)) \geq
\widetilde{\mu}(ba) = \widetilde{\mu}(ab)$ for all $s \in S$. Hence
$Inf\Big\{\widetilde{\mu}(asb) : s \in S\Big\} \geq
\widetilde{\mu}(ab)$. Now by Lemma $3.16$, we have
$Max^{i}\Big(\widetilde{\mu}(a),\widetilde{\mu}(b)\Big) =
Inf\Big\{\widetilde{\mu}(asb) : s \in S\Big\}$. So
$Max^{i}\Big(\widetilde{\mu}(a),\widetilde{\mu}(b)\Big) \geq
\widetilde{\mu}(ab)$. Again $\widetilde{\mu}(ab) \geq
Max^{i}\Big(\widetilde{\mu}(a),\widetilde{\mu}(b)\Big)$.
Consequently, $\widetilde{\mu}(ab) =
Max^{i}\Big(\widetilde{\mu}(a),\widetilde{\mu}(b)\Big)$.
%Since
%$\widetilde{\mu}$ is an interval-valued fuzzy ideal of $S$, we have
%$\widetilde{\mu}(ab) \geq
%Max^{i}\Big(\widetilde{\mu}(a),\widetilde{\mu}(b)\Big)$. Thus we
%find that $Max^{i}\Big(\widetilde{\mu}(a),\widetilde{\mu}(b)\Big) =
%Inf\Big\{\widetilde{\mu}(asb) : s \in S\Big\} \geq
%\widetilde{\mu}(ab) \geq
%Max^{i}\Big(\widetilde{\mu}(a),\widetilde{\mu}(b)\Big)$.
%Consequently, $\widetilde{\mu}(ab) =
%Max^{i}\Big(\widetilde{\mu}(a),\widetilde{\mu}(b)\Big)$.
\end{proof}

\section{Interval-valued fuzzy completely prime ideal of a \\ semiring :}

\begin{Definition}
Let $S$ be a semiring and $x \in S$. Let $\widetilde{a} \in D[0,1]
\setminus \{\widetilde{0}\}$. Then an $i.v.$ fuzzy subset $x_{\widetilde{a}}$
of $S$ is called an interval-valued
fuzzy point of $S$ if\\
$$x_{\widetilde{a}}(y) =
\left\{
\begin{array}{cc}
\widetilde{a} & if~ x~ = ~ y,\\
\widetilde{0} & otherwise.
\end{array}
\right.$$ An interval-valued fuzzy point $x_{\widetilde{a}}$ is
said to be contained in an $i.v.$ fuzzy subset $\widetilde{\mu}$ of $S$,
denoted by $x_{\widetilde{a}} \in \widetilde{\mu}$, if $\widetilde{a}
\leq \widetilde{\mu}(x)$.
\end{Definition}

\begin{Remark}
Let $x_{\widetilde{a}}$ be an $i.v.$ fuzzy point of a semiring $S$ and
$\widetilde{\mu} $ be an $i.v.$ fuzzy subset of $S$. Then $x_{\widetilde{a}}
\in \widetilde{\mu}$ if and only if $x_{\widetilde{a}} \subseteq
\widetilde{\mu}$.
\end{Remark}

%\begin{proof}
%Suppose $x_{\widetilde{a}} \in \widetilde{\mu}$. Then we have
%$\widetilde{a} \leq \widetilde{\mu}(x)$. Let $y \in S$.
%$x_{\widetilde{a}}(y) = \widetilde{a}$ if $x = y$. Then
%$x_{\widetilde{a}}(y) = x_{\widetilde{a}}(x) = \widetilde{a} \leq
%\widetilde{\mu}(x)$. If $y \neq x$, we find that
%$x_{\widetilde{a}}(y) = \widetilde{0} \leq \widetilde{\mu}(y)$. Thus
%$x_{\widetilde{a}}(y) \leq \widetilde{\mu}(y)$ for all $y \in S$. So
%$x_{\widetilde{a}} \subseteq \widetilde{\mu}$.\\
%Conversely, let $x_{\widetilde{a}} \subseteq \widetilde{\mu}$. Then
%$x_{\widetilde{a}}(y) \leq \widetilde{\mu}(y)$ for all $y \in S$. So
%$x_{\widetilde{a}}(x) \leq \widetilde{\mu}(x)$ i.e. $\widetilde{a}
%\leq \widetilde{\mu}(x)$. Consequently, $x_{\widetilde{a}} \in
%\widetilde{\mu}$.
%\end{proof}

\begin{Remark}
Let $x_{\widetilde{a}}$ and $y_{\widetilde{b}}$ be two
$i.v.$ fuzzy points of a semiring $S$. Then\\
$(x_{\widetilde{a}} \circ y_{\widetilde{b}}) =
\Big(xy\Big)_{Min^{i}(\widetilde{a},\; \widetilde{b})}$.
\end{Remark}

%\begin{proof}
%Let $p \in S$. $(x_{\widetilde{a}} \circ y_{\widetilde{b}})(p) =
%\displaystyle\sup_{p=qr}\Big\{Min^{i}\Big(x_{\widetilde{a}}(q),y_{\widetilde{b}}(r)\Big)\Big\}$\\
%$$ =
%\left\{
%\begin{array}{cc}
%Min^{i}(\widetilde{a},\widetilde{b}) & \text{if} ~p = xy\\
%\widetilde{0} & \text{otherwise.}
%\end{array}
%\right.$$ Also,
%$$\Big(xy\Big)_{Min^{i}(\widetilde{a},\widetilde{b})}(p) =
%\left\{
%\begin{array}{cc}
%Min^{i}(\widetilde{a},\widetilde{b}) & \text{if} ~p = xy\\
%\widetilde{0} & \text{otherwise.}
%\end{array}
%\right.$$ Consequently, $(x_{\widetilde{a}} \circ y_{\widetilde{b}})
%= \Big(xy\Big)_{Min^{i}(\widetilde{a},\widetilde{b})}$.
%\end{proof}

\begin{Definition}
A non-constant $i.v.$ fuzzy ideal $\widetilde{\mu}$ of a semiring $S$ is
said to be an interval-valued fuzzy completely prime ideal of $S$
if for any two $i.v.$ fuzzy points $x_{\widetilde{a}}$ and $y_{\widetilde{b}}$
of $S$, $x_{\widetilde{a}} \circ y_{\widetilde{b}} \in
\widetilde{\mu} \Longrightarrow$ either $x_{\widetilde{a}} \in
\widetilde{\mu}$ or $y_{\widetilde{b}} \in \widetilde{\mu}$.
\end{Definition}

\begin{Theorem}
Every $i.v.$ fuzzy completely prime ideal of a semiring $S$ is an $i.v.$ fuzzy prime ideal of $S$.
\end{Theorem}

\begin{proof}
Let $\widetilde{\mu}$ be an $i.v.$ fuzzy completely prime ideal of $S$. Then
$\widetilde{\mu}$ is non-constant. Let $\widetilde{\mu_{1}}$ and
$\widetilde{\mu_{2}}$ be any two $i.v.$ fuzzy ideals of $S$ such that
$\widetilde{\mu_{1}} \circ \widetilde{\mu_{2}} \subseteq
\widetilde{\mu}$. Let $\widetilde{\mu_{1}} \nsubseteq
\widetilde{\mu}$. This implies that there exists $x \in S$ such
that $\widetilde{\mu_{1}}(x) \nleq \widetilde{\mu}(x)$. Now since
according to our assumption, any two interval numbers of $D[0,1]$
are comparable, we have $\widetilde{\mu_{1}}(x) >
\widetilde{\mu}(x)$. This shows that $x_{\widetilde{\mu_{1}}(x)}
\notin \widetilde{\mu}$. Now for any $y
\in S$ and $w \in S$\\
$(x_{\widetilde{\mu_{1}}(x)} \circ y_{\widetilde{\mu_{2}}(y)})(w) =
\Big(xy\Big)_{Min^{i}(\widetilde{\mu_{1}}(x),\;
\widetilde{\mu_{2}}(y))}(w)$ (by Remark 4.3)

$\hspace{2.6cm}$ $= \begin{cases}
          Min^{i}(\widetilde{\mu_{1}}(x),\widetilde{\mu_{2}}(y)) &\text{when} \quad w = xy, \\
            \widetilde{0} &\text{otherwise.} \\
         \end{cases} $


Now $\widetilde{\mu}(w) \geq (\widetilde{\mu_{1}} \circ
\widetilde{\mu_{2}})(w) =
\displaystyle\sup_{w=xy}\Big\{Min^{i}(\widetilde{\mu_{1}}(x),\widetilde{\mu_{2}}(y))\Big\}
\geq Min^{i}(\widetilde{\mu_{1}}(x),\widetilde{\mu_{2}}(y))$ (when
$w = xy$) $= (x_{\widetilde{\mu_{1}}(x)} \circ
y_{\widetilde{\mu_{2}}(y)})(w)$. Also when $w \neq xy$, we have
$\widetilde{\mu}(w) \geq (x_{\widetilde{\mu_{1}}(x)} \circ
y_{\widetilde{\mu_{2}}(y)})(w)$. This shows that
$(x_{\widetilde{\mu_{1}}(x)} \circ y_{\widetilde{\mu_{2}}(y)})
\subseteq \widetilde{\mu}$ i.e. $(x_{\widetilde{\mu_{1}}(x)} \circ
y_{\widetilde{\mu_{2}}(y)}) \in \widetilde{\mu}$, by Remark $4.2$.
Now since $\widetilde{\mu}$ is an $i.v.$ fuzzy completely prime ideal of $S$ and
$x_{\widetilde{\mu_{1}}(x)} \notin \widetilde{\mu}$, we have
$y_{\widetilde{\mu_{2}}(y)} \in \widetilde{\mu}$. This implies that
$\widetilde{\mu_{2}}(y) \leq \widetilde{\mu}(y) \Longrightarrow
\widetilde{\mu_{2}} \subseteq \widetilde{\mu}$. Consequently,
$\widetilde{\mu}$ is an $i.v.$ fuzzy prime ideal of $S$.
\end{proof}

In general, the converse of the Theorem $4.5$ for an arbitrary
semiring is not true.

\begin{Example}
Consider the semiring $M_{2}(\mathbb{R}^{+}_{0})$, where
$M_{2}(\mathbb{R}^{+}_{0})$ denotes the set of all $2 \times 2$
matrices with non-negative real entries.\\
Let $\widetilde{\mu}$ be an $i.v.$ fuzzy subset of
$M_{2}(\mathbb{R}^{+}_{0})$, defined by :
$$\widetilde{\mu}(A) = \begin{cases}
          \widetilde{1}  &\text{when  $A$ is the null matrix},\\
            \widetilde{0} &\text{otherwise}.\\
         \end{cases} $$
Then it can be shown that $\widetilde{\mu}$ is an $i.v.$ fuzzy prime ideal of
$M_{2}(\mathbb{R}^{+}_{0})$ but $\widetilde{\mu}$ is not an
$i.v.$ fuzzy completely prime ideal of $M_{2}(\mathbb{R}^{+}_{0})$.

%Now $\widetilde{\mu}_{0} = \Big\{A \in M_{2}(\mathbb{R}^{+}_{0}) :
%\widetilde{\mu}(A) =
%\widetilde{\mu}(0_{M_{2}(\mathbb{R}^{+}_{0})})\Big\} = \Big\{A \in
%M_{2}(\mathbb{R}^{+}_{0}) : \widetilde{\mu}(A) = \widetilde{1}\Big\}
%= \Big\{0_{M_{2}(\mathbb{R}^{+}_{0})}\Big\}$. So we find that
%$\widetilde{\mu}_{0}$ is a proper ideal of
%$M_{2}(\mathbb{R}^{+}_{0})$. Since $\mathbb{R}^{+}_{0}$ is a
%semifield, it has only two ideals, namely
%$M_{2}(\mathbb{R}^{+}_{0})$ and
%$\Big\{0_{M_{2}(\mathbb{R}^{+}_{0})}\Big\}$. If we consider $AB
%\subseteq \widetilde{\mu}_{0}$ for any two ideals $A, B$ of
%$M_{2}(\mathbb{R}^{+}_{0})$, we find that  $AB \subseteq
%\Big\{0_{M_{2}(\mathbb{R}^{+}_{0})}\Big\}$ i.e. $AB =
%0_{M_{2}(\mathbb{R}^{+}_{0})}$. Then we get either $A =
%0_{M_{2}(\mathbb{R}^{+}_{0})}$ or $B =
%0_{M_{2}(\mathbb{R}^{+}_{0})}$, since only possibilities for $A$
%and $B$ are $0_{M_{2}(\mathbb{R}^{+}_{0})}$ and
%$M_{2}(\mathbb{R}^{+}_{0})$. This implies that
%$\widetilde{\mu}_{0}$ is a prime ideal of
%$M_{2}(\mathbb{R}^{+}_{0})$. So we get $Im\widetilde{\mu} =
%\{\widetilde{0}, \widetilde{1}\}$ and $\widetilde{\mu}_{0}$ is a
%prime ideal of $M_{2}(\mathbb{R}^{+}_{0})$. Hence by Theorem
%$3.14$ we find that $\widetilde{\mu}$ is an interval-valued fuzzy
%prime ideal of $M_{2}(\mathbb{R}^{+}_{0})$.\\
%Let us take  $A, B \in M_{2}(\mathbb{R}^{+}_{0})$, where $A =
%\left(
%\begin{array}{cc}
%1 & 0\\
%0 & 0 \end{array} \right)$ and  $B = \left(
%\begin{array}{cc}
%0 & 0 \\
%0 & 1 \end{array} \right)$. Then we find that $AB = \left(
%\begin{array}{cc}
%1 & 0\\
%0 & 0 \end{array} \right) \left(
%\begin{array}{cc}
%0 & 0\\
%0 & 1 \end{array} \right) = \left(
%\begin{array}{cc}
%0 & 0\\
%0 & 0 \end{array} \right) = 0_{M_{2}(\mathbb{R}^{+}_{0})}$. So
%$\widetilde{\mu}(AB) = \widetilde{1}$. This implies that
%$(AB)_{\widetilde{1}} \in \widetilde{\mu}$. Then by Remark $4.3$, we
%get $A_{\widetilde{1}} \circ B_{\widetilde{1}} \in \widetilde{\mu}$.
%But $A_{\widetilde{1}} \notin \widetilde{\mu}$ and $B_{\widetilde{1}}
%\notin \widetilde{\mu}$, since $\widetilde{\mu}(A) = \widetilde{0} <
%\widetilde{1}$ and $\widetilde{\mu}(B) = \widetilde{0} <
%\widetilde{1}$. Consequently, $\widetilde{\mu}$ is not an
%interval-valued fuzzy completely prime ideal of
%$M_{2}(\mathbb{R}^{+}_{0})$.
\end{Example}

But in particular, for commutative semiring we have the following
result :

\begin{Theorem}
Every $i.v.$ fuzzy prime ideal of a commutative semiring $S$ is an
$i.v.$ fuzzy completely prime ideal of $S$.
\end{Theorem}

\begin{proof}
Let $S$ be a commutative semiring and $\widetilde{\mu}$ be an
$i.v.$ fuzzy prime ideal of $S$. Then $\widetilde{\mu}$ is non-constant. Let
$x_{\widetilde{a}}$ and $y_{\widetilde{b}}$ be any two $i.v.$ fuzzy points of
$S$ such that $x_{\widetilde{a}} \circ y_{\widetilde{b}} \in
\widetilde{\mu}$. Then by Remark $4.3$, we get
$(xy)_{Min^{i}(\widetilde{a},\; \widetilde{b})} \in \widetilde{\mu}$.
So we have $Min^{i}(\widetilde{a},\widetilde{b}) \leq
\widetilde{\mu}(xy)$.\\
Now we construct two $i.v.$ fuzzy subsets $\widetilde{\mu_{3}}$ and
$\widetilde{\mu_{4}}$ of $S$ as follows :

$\widetilde{\mu_{3}}(z) = \begin{cases}
          \widetilde{a}  &\text{when} \quad z \in \;<x>, \\
            \widetilde{0} &\text{otherwise}; \\
         \end{cases} $ \;\;\;\;and\;\;\;\; $\widetilde{\mu_{4}}(z) = \begin{cases}
          \widetilde{b}  &\text{when} \quad z \in \;<y>, \\
            \widetilde{0} &\text{otherwise.} \\
         \end{cases}$

Then from Lemma $3.8$, it follows that $\widetilde{\mu_{3}}$ and
$\widetilde{\mu_{4}}$ are $i.v.$ fuzzy ideals of $S$. Now
$(\widetilde{\mu_{3}} \circ \widetilde{\mu_{4}})(z) =
\displaystyle\sup_{z=uv}\Big\{Min^{i}(\widetilde{\mu_{3}}(u),\widetilde{\mu_{4}}(v))\Big\}$. When $z$ can not be expressed as $z = uv$, where $u \in \; <x>$ and $v \in \; <y>$, then $(\widetilde{\mu_{3}} \circ \widetilde{\mu_{4}})(z) = \widetilde{0} \leq \widetilde{\mu}(z)$. Let $z = uv$ for some $u, v \in S$, where $u \in$ $<x>$ and $v \in $
$<y>$. Then $(\widetilde{\mu_{3}} \circ \widetilde{\mu_{4}})(z) =
\displaystyle\sup_{z=uv}\Big\{Min^{i}(\widetilde{a},\widetilde{b})\Big\}
= Min^{i}(\widetilde{a},\widetilde{b})$. Since $S$ is a commutative
semiring with identity, $u \in$ $<x>$ and $v \in$ $<y>$
$\Longrightarrow u = s_{1}x$ and $v = s_{2}y$ for some $s_{1},
s_{2} \in S$. Then we have $\widetilde{\mu}(z) = \widetilde{\mu}(uv)
= \widetilde{\mu}(s_{1}xs_{2}y) = \widetilde{\mu}(s_{1}s_{2}xy) \geq
\widetilde{\mu}(xy) \geq Min^{i}(\widetilde{a}, \widetilde{b}) =
(\widetilde{\mu_{3}} \circ \widetilde{\mu_{4}})(z)$. This implies
that $(\widetilde{\mu_{3}} \circ \widetilde{\mu_{4}}) \subseteq
\widetilde{\mu}$. Since $\widetilde{\mu}$ is an $i.v.$ fuzzy prime ideal of $S$,
we have either $\widetilde{\mu_{3}} \subseteq \widetilde{\mu}$ or
$\widetilde{\mu_{4}} \subseteq \widetilde{\mu}$. This implies that
$\widetilde{\mu_{3}}(x) \leq \widetilde{\mu}(x)$ or
$\widetilde{\mu_{4}}(y) \leq \widetilde{\mu}(y) \Longrightarrow
\widetilde{a} \leq \widetilde{\mu}(x)$ or $\widetilde{b} \leq
\widetilde{\mu}(y) \Longrightarrow x_{\widetilde{a}} \in
\widetilde{\mu}$ or $y_{\widetilde{b}} \in \widetilde{\mu}$.
Consequently, $\widetilde{\mu}$ is an $i.v.$ fuzzy completely prime ideal of $S$.
\end{proof}

%\begin{Theorem}
%Every interval-valued fuzzy prime ideal of a commutative semiring
%$S$ is an interval-valued fuzzy completely prime ideal of $S$.
%\end{Theorem}
%
%\begin{proof}
%Let $S$ be a commutative semiring and $\widetilde{\mu}$ be an
%interval-valued fuzzy prime ideal of $S$. Then $\widetilde{\mu}$ is
%non-constant. Let $x_{\widetilde{a}}$ and $y_{\widetilde{b}}$ be any
%two interval-valued fuzzy points of $S$ such that
%$x_{\widetilde{a}} \circ y_{\widetilde{b}} \in \widetilde{\mu}$. Then
%by Remark $4.3$, we get $(xy)_{Min^{i}(\widetilde{a},\;
%\widetilde{b})} \in \widetilde{\mu}$. So we have
%$Min^{i}(\widetilde{a},\widetilde{b}) \leq
%\widetilde{\mu}(xy)$.\\
%Now we construct two interval-valued fuzzy subsets
%$\widetilde{\mu_{3}}$ and $\widetilde{\mu_{4}}$ of $S$ as follows :
%
%$\widetilde{\mu_{3}}(z) = \begin{cases}
%          \widetilde{a}  &\text{when} \quad z \in \;<x>, \\
%            \widetilde{0} &\text{otherwise}; \\
%         \end{cases} $ \;\;\;\;and\;\;\;\; $\widetilde{\mu_{4}}(z) = \begin{cases}
%          \widetilde{b}  &\text{when} \quad z \in \;<y>, \\
%            \widetilde{0} &\text{otherwise.} \\
%         \end{cases}$
%
%Then from Lemma $3.8$, it follows that $\widetilde{\mu_{3}}$ and
%$\widetilde{\mu_{4}}$ are interval-valued fuzzy ideals of $S$. Now
%$(\widetilde{\mu_{3}} \circ \widetilde{\mu_{4}})(z) =
%\displaystyle\sup_{z=uv}\Big\{Min^{i}(\widetilde{\mu_{3}}(u),\widetilde{\mu_{4}}(v))\Big\}$.
%Let $z = uv$ for some $u, v \in S$ and $u \in$ $<x>$, $v \in $
%$<y>$. Then $(\widetilde{\mu_{3}} \circ \widetilde{\mu_{4}})(z) =
%\displaystyle\sup_{z=uv}\Big\{Min^{i}(\widetilde{a},\widetilde{b})\Big\}
%= Min^{i}(\widetilde{a},\widetilde{b})$. Since $S$ is a commutative
%semiring with identity, $u \in$ $<x>$ and $v \in$ $<y>$
%$\Longrightarrow u = s_{1}x$ and $v = s_{2}y$ for some $s_{1},
%s_{2} \in S$. Then we have $\widetilde{\mu}(z) = \widetilde{\mu}(uv)
%= \widetilde{\mu}(s_{1}xs_{2}y) = \widetilde{\mu}(s_{1}s_{2}xy) \geq
%\widetilde{\mu}(xy) \geq Min^{i}(\widetilde{a}, \widetilde{b}) =
%(\widetilde{\mu_{3}} \circ \widetilde{\mu_{4}})(z)$. This implies
%that $(\widetilde{\mu_{3}} \circ \widetilde{\mu_{4}}) \subseteq
%\widetilde{\mu}$. Since $\widetilde{\mu}$ is an interval-valued
%fuzzy prime ideal of $S$, we have either $\widetilde{\mu_{3}}
%\subseteq \widetilde{\mu}$ or $\widetilde{\mu_{4}} \subseteq
%\widetilde{\mu}$. This implies that $\widetilde{\mu_{3}}(x) \leq
%\widetilde{\mu}(x)$ or $\widetilde{\mu_{4}}(y) \leq
%\widetilde{\mu}(y) \Longrightarrow \widetilde{a} \leq
%\widetilde{\mu}(x)$ or $\widetilde{b} \leq \widetilde{\mu}(y)
%\Longrightarrow x_{\widetilde{a}} \in \widetilde{\mu}$ or
%$y_{\widetilde{b}} \in \widetilde{\mu}$. Consequently,
%$\widetilde{\mu}$ is an interval-valued fuzzy completely prime
%ideal of $S$.
%\end{proof}

\begin{Proposition}
Let $\widetilde{\mu}$ be an $i.v.$ fuzzy comletely prime ideal of a semiring $S$. Then
$\widetilde{\mu}_{0} = \{x \in S: \widetilde{\mu}(x) =
\widetilde{\mu}(0_{S})\}$ is a completely prime ideal of $S$.
\end{Proposition}

\begin{proof}
Let $\widetilde{\mu}$ be an $i.v.$ fuzzy completely prime ideal of a semiring $S$. Then
$\widetilde{\mu}$ is an $i.v.$ fuzzy prime ideal of $S$, by Theorem $4.5$. So
$\widetilde{\mu}_{0}$ is a prime ideal of $S$, by Theorem
$3.13(iii)$ and hence a proper ideal of $S$. Let $x, y \in S$ be
such that $xy \in \widetilde{\mu}_{0}$. Then $\widetilde{\mu}(xy) =
\widetilde{\mu}(0_{S}) = \widetilde{1}$, by Theorem $3.13(i)$. This
implies that $(xy)_{\widetilde{1}} \in \widetilde{\mu}$. Thus by
Remark $4.3$, we get $(x_{\widetilde{1}} \circ y_{\widetilde{1}}) \in
\widetilde{\mu}$. This implies that either $x_{\widetilde{1}} \in
\widetilde{\mu}$ or $y_{\widetilde{1}} \in \widetilde{\mu}$, since
$\widetilde{\mu}$ is an $i.v.$ fuzzy completely prime ideal of $S$. Thus it follows that
either $\widetilde{1} \leq \widetilde{\mu}(x)$ or
$\widetilde{1} \leq \widetilde{\mu}(y)$.\\
$\Longrightarrow \widetilde{\mu}(x) = \widetilde{1}$ or
$\widetilde{\mu}(y) = \widetilde{1}$ i.e. $\widetilde{\mu}(x) = \widetilde{\mu}(0_{S})$ or $\widetilde{\mu}(y) = \widetilde{\mu}(0_{S})$.\\
$\Longrightarrow x \in \widetilde{\mu}_{0}$ or $y \in
\widetilde{\mu}_{0}$.\\ Hence $\widetilde{\mu}_{0}$ is a completely
prime ideal of $S$.
\end{proof}

\begin{Theorem}
Let $\widetilde{\mu}$ be an $i.v.$ fuzzy subset of a semiring $S$ such that
$Im\widetilde{\mu} = \{\widetilde{1},[\alpha,\beta]\}$; where
$[\alpha,\beta] \in D[0,1] \setminus \{\widetilde{1}\}$ and
$\widetilde{\mu}_{0} = \{x \in S: \widetilde{\mu}(x) =
\widetilde{\mu}(0_{S})\}$ is a completely prime ideal of $S$. Then
$\widetilde{\mu}$ is an $i.v.$ fuzzy completely prime ideal of $S$.
\end{Theorem}

\begin{proof}
Let $\widetilde{\mu}$ be an $i.v.$ fuzzy subset of a semiring $S$ such that
$Im\widetilde{\mu} = \{\widetilde{1},[\alpha,\beta]\}$; where
$[\alpha,\beta] \in D[0,1] \setminus \{\widetilde{1}\}$ and
$\widetilde{\mu}_{0} = \{x \in S: \widetilde{\mu}(x) =
\widetilde{\mu}(0_{S})\}$ is a completely prime ideal of $S$. Since $\widetilde{\mu}_{0}$ is a completely prime ideal of $S$, it is a prime ideal of $S$. Then by Thoeorem $3.14$, we get that $\widetilde{\mu}$ is an $i.v.$ fuzzy prime ideal and hence a non-constant $i.v.$ fuzzy ideal of $S$. Let
$x_{\widetilde{a}}$ and $y_{\widetilde{b}}$ be any two arbitrary
$i.v.$ fuzzy points of $S$ such that $x_{\widetilde{a}} \circ
y_{\widetilde{b}} \in \widetilde{\mu}$. Then by Remark $4.3$, we
find that $(xy)_{Min^{i}(\widetilde{a},\; \widetilde{b})} \in
\widetilde{\mu}$ i.e. $Min^{i}(\widetilde{a},\widetilde{b}) \leq
\widetilde{\mu}(xy)$. We have to prove that either
$x_{\widetilde{a}} \in \widetilde{\mu}$ or $y_{\widetilde{b}} \in
\widetilde{\mu}$. If possible, let $x_{\widetilde{a}} \notin
\widetilde{\mu}$ and $y_{\widetilde{b}} \notin \widetilde{\mu}$.
Since according to our assumption any two interval numbers are
comparable, it follows that $\widetilde{a}
> \widetilde{\mu}(x)$ and $\widetilde{b} > \widetilde{\mu}(y)$.
Therefore $\widetilde{\mu}(x) = [\alpha,\beta] =
\widetilde{\mu}(y)$, since $Im\widetilde{\mu} = \{\widetilde{1},
[\alpha, \beta]\}$. This implies that $x \notin
\widetilde{\mu}_{0}$ and $y \notin \widetilde{\mu}_{0}$. Now since
$\widetilde{\mu}_{0}$ is a completely prime ideal of $S$, we have
$xy \notin \widetilde{\mu}_{0}$. Since $\widetilde{\mu}$ is an $i.v.$ fuzzy prime ideal of $S$, $\widetilde{\mu}(0_{S}) =
\widetilde{1}$, by Theorem $3.13(i)$. So $xy \notin \widetilde{\mu}_{0} \Longrightarrow
\widetilde{\mu}(xy) \neq \widetilde{\mu}(0_{S}) = \widetilde{1}$.
Consequently, $\widetilde{\mu}(xy) = [\alpha,\beta]$, since
$Im\widetilde{\mu} = \{\widetilde{1},[\alpha, \beta]\}$. Now
$[\alpha,\beta] = \widetilde{\mu}(xy) \geq
Min^{i}(\widetilde{a},\widetilde{b}) >
Min^{i}(\widetilde{\mu}(x),\widetilde{\mu}(y)) = [\alpha,\beta]$.
Thus we arrive at a contradiction. Hence either $x_{\widetilde{a}}
\in \widetilde{\mu}$ or $y_{\widetilde{b}} \in \widetilde{\mu}$. Thus
$\widetilde{\mu}$ is an $i.v.$ fuzzy completely prime ideal of $S$.
\end{proof}


\begin{Theorem}
Let $\widetilde{\mu}$ be an $i.v.$ fuzzy ideal of a semiring $S$ with
$Im\widetilde{\mu} = \{\widetilde{1},[\alpha,\beta]\}$; where
$[\alpha,\beta] \in D[0,1] \setminus \{\widetilde{1}\}$. Then
$\widetilde{\mu}$ is an $i.v.$ fuzzy completely prime ideal of $S$ if and only if its
only proper level ideal
$\overline{U}(\widetilde{\mu},\widetilde{1})$ is a completely prime
ideal of $S$.
\end{Theorem}

\begin{proof}
Since $Im\widetilde{\mu} = \{\widetilde{1},[\alpha,\beta]\}$, where
$[\alpha,\beta] \in D[0,1] \setminus \{\widetilde{1}\}$, we have
$\overline{U}(\widetilde{\mu},[\alpha,\beta]) = S$. So only proper
level ideal of $\widetilde{\mu}$ is
$\overline{U}(\widetilde{\mu},\widetilde{1})$, by Lemma $3.5$. Let
$\widetilde{\mu}$ be an $i.v.$ fuzzy completely prime ideal of $S$. Then
$\widetilde{\mu}_{0}$ is a completely prime ideal of $S$, by
Proposition $4.8$. Now since $\widetilde{\mu}$ is an $i.v.$ fuzzy completely prime ideal
of $S$, we find that $\widetilde{\mu}$ is an $i.v.$ fuzzy prime ideal of $S$, by
Theorem $4.5$. Then $\widetilde{\mu}(0_{S}) = \widetilde{1}$, by
Theorem $3.13(i)$. Also we find that
$\overline{U}(\widetilde{\mu},\widetilde{1}) = \{x \in S:
\widetilde{\mu}(x) \geq \widetilde{1}\} = \{x \in S:
\widetilde{\mu}(x) = \widetilde{1}\} = \{x \in S : \widetilde{\mu}(x)
= \widetilde{\mu}(0_{S})\} = \widetilde{\mu}_{0}$. Hence
$\overline{U}(\widetilde{\mu},\widetilde{1})$ is a completely prime
ideal of $S$.\\
Conversely, let $\overline{U}(\widetilde{\mu},\widetilde{1})$ be a
completely prime ideal of $S$. Since $\widetilde{\mu}$ is an
$i.v.$ fuzzy ideal of $S$, it follows that that $\widetilde{\mu}(0_{S}) \geq
\widetilde{\mu}(x)$ for all $x \in S$, by Remark $3.3$. Again since
$Im\widetilde{\mu} = \{\widetilde{1}, [\alpha, \beta]\}$, where
$[\alpha,\beta] \in D[0,1] \setminus \{\widetilde{1}\}$, we find
that $\widetilde{\mu}(0_{S}) = \widetilde{1}$. Thus
$\overline{U}(\widetilde{\mu},\widetilde{1}) = \{x \in S:
\widetilde{\mu}(x) \geq \widetilde{1}\} = \{x \in S:
\widetilde{\mu}(x) = \widetilde{1}\} = \{x \in S : \widetilde{\mu}(x)
= \widetilde{\mu}(0_{S})\} = \widetilde{\mu}_{0}$. So
$\widetilde{\mu}_{0}$ is a completely prime ideal of $S$. Then by
Theorem $4.9$, we have $\widetilde{\mu}$ is an $i.v.$ fuzzy completely prime ideal of $S$.
\end{proof}

\begin{Theorem}
Let $\widetilde{\mu}$ be an $i.v.$ fuzzy prime ideal of a semiring $S$. Then
$\widetilde{\mu}$ is an $i.v.$ fuzzy completely prime ideal of a semiring $S$ if and only
if for any two $i.v.$ fuzzy points $x_{\widetilde{a}}$ and $y_{\widetilde{b}}$
of $S$, $x_{\widetilde{a}} \circ y_{\widetilde{b}} \in
\widetilde{\mu} \Longrightarrow y_{\widetilde{b}} \circ
x_{\widetilde{a}} \in \widetilde{\mu}$.
\end{Theorem}

\begin{proof}
Let $\widetilde{\mu}$ be an $i.v.$ fuzzy completely prime ideal of $S$. Since
$\widetilde{\mu}$ is an $i.v.$ fuzzy prime ideal of $S$, we find that
$Im\widetilde{\mu} = \{\widetilde{1},[\alpha,\beta]\}$, where
$[\alpha,\beta] \in D[0,1] \setminus \{\widetilde{1}\}$ and
$\widetilde{\mu}_{0}$ is a prime ideal of $S$, by Theorem $3.14$.
In fact we find that $\widetilde{\mu}_{0}$ is a completely prime
ideal of $S$, since $\widetilde{\mu}$ is an $i.v.$ fuzzy completely prime ideal  of $S$,
by Proposition $4.8$. Let $x_{\widetilde{a}}$ and
$y_{\widetilde{b}}$ be any two $i.v.$ fuzzy points of $S$ such that
$x_{\widetilde{a}} \circ y_{\widetilde{b}} \in \widetilde{\mu}$. Then
we have $\Big(xy\Big)_{Min^{i}(\widetilde{a},\; \widetilde{b})} \in
\widetilde{\mu}$, by Remark $4.3$. This implies that
$Min^{i}(\widetilde{a},\;\widetilde{b}) \leq \widetilde{\mu}(xy)$.
Now\\
$\underline{Case1}$: Let $xy \in \widetilde{\mu}_{0}$. Since
$\widetilde{\mu}$ is an $i.v.$ fuzzy prime ideal of $S$, we find that
$\widetilde{\mu}(0_{S}) = \widetilde{1}$, by Theorem $3.13(i)$. Then
$xy \in \widetilde{\mu}_{0}$ implies that $\widetilde{\mu}(xy) =
\widetilde{\mu}(0_{S}) = \widetilde{1}$. Now since the prime ideal
$\widetilde{\mu}_{0}$ of $S$ is also a completely prime ideal of
$S$ and $xy \in \widetilde{\mu}_{0}$, we find that $yx \in
\widetilde{\mu}_{0}$, by Proposition $2.7$. This shows that
$\widetilde{\mu}(yx) = \widetilde{\mu}(0_{S}) = \widetilde{1}$, by
Theorem $3.13(i)$ . Therefore, $Min^{i}(\widetilde{a},\;
\widetilde{b}) \leq \widetilde{\mu}(xy) = \widetilde{\mu}(yx)$. This
implies that $(yx)_{Min^{i}(\widetilde{a},\; \widetilde{b})} \in
\widetilde{\mu}$ i.e. $(yx)_{Min^{i}(\widetilde{b},\; \widetilde{a})}
\in \widetilde{\mu}$. Then by Remark $4.3$, it follows that
$y_{\widetilde{b}} \circ x_{\widetilde{a}} \in
\widetilde{\mu}$.\\
$\underline{Case2}$: Let $xy \notin \widetilde{\mu}_{0}$. This
implies that $\widetilde{\mu}(xy) \neq \widetilde{\mu}(0_{S})$ i.e.
$\widetilde{\mu}(xy) \neq \widetilde{1}$, by Theorem $3.13(i)$. Then
$\widetilde{\mu}(xy) = [\alpha,\beta]$, since $Im\widetilde{\mu} =
\{\widetilde{1},[\alpha, \beta]\}$. So we have $\widetilde{\mu}(yx)
\geq [\alpha,\beta] = \widetilde{\mu}(xy) \geq
Min^{i}(\widetilde{a},\widetilde{b}) \Longrightarrow
(yx)_{Min^{i}(\widetilde{a},\; \widetilde{b})} \in \widetilde{\mu}
\Longrightarrow (yx)_{Min^{i}(\widetilde{b},\; \widetilde{a})} \in
\widetilde{\mu}$. Then by Remark $4.3$, we get $y_{\widetilde{b}}
\circ
x_{\widetilde{a}} \in \widetilde{\mu}$.\\
Conversely, let $\widetilde{\mu}$ be an $i.v.$ fuzzy prime ideal of $S$ such
that for any two $i.v.$ fuzzy points $x_{\widetilde{a}}$ and
$y_{\widetilde{b}}$ of $S$, $x_{\widetilde{a}} \circ
y_{\widetilde{b}} \in \widetilde{\mu} \Longrightarrow
y_{\widetilde{b}} \circ x_{\widetilde{a}} \in \widetilde{\mu}$. Since
$\widetilde{\mu}$ is an $i.v.$ fuzzy prime ideal of $S$, $Im\widetilde{\mu} =
\{\widetilde{1},[\alpha,\beta]\}$, where $[\alpha,\beta] \in D[0,1]
\setminus \{\widetilde{1}\}$ and $\widetilde{\mu}_{0}$ is a prime
ideal of $S$, by Theorem $3.14$. Let $xy \in \widetilde{\mu}_{0}$.
Then $\widetilde{\mu}(xy) = \widetilde{\mu}(0_{S}) = \widetilde{1}$,
by Theorem $3.13(i)$. So $(xy)_{\widetilde{1}} \in \widetilde{\mu}
\Longrightarrow x_{\widetilde{1}} \circ y_{\widetilde{1}} \in
\widetilde{\mu}$ (by Remark $4.3$) $\Longrightarrow
y_{\widetilde{1}} \circ x_{\widetilde{1}} \in \widetilde{\mu}
\Longrightarrow (yx)_{\widetilde{1}} \in \widetilde{\mu}$ (by Remark
$4.3$) $\Longrightarrow \widetilde{1} \leq \widetilde{\mu}(yx)
\Longrightarrow \widetilde{\mu}(yx) = \widetilde{1} =
\widetilde{\mu}(0_{S})$ (by Theorem $3.13(i))$ $\Longrightarrow yx
\in \widetilde{\mu}_{0}$. Hence $\widetilde{\mu}_{0}$ is a
completely prime ideal of $S$, by Proposition $2.7$. Thus
$\widetilde{\mu}_{0}$ is a completely prime ideal of $S$ and
$Im\widetilde{\mu} = \{\widetilde{1},[\alpha,\beta]\}$, where
$[\alpha,\beta] \in D[0,1] \setminus \{\widetilde{1}\}$.
Consequently, by Theorem $4.9$ we find that $\widetilde{\mu}$ is an
$i.v.$ fuzzy completely prime ideal of $S$.
\end{proof}

\vspace{.2in}

%{\bf Acknowledgement :} The authors are grateful to Professor T.
%K. Dutta for his valuable suggestions and continuous help through
%out the preparation of this paper.

\begin{center}

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\end{document}