An important notion in fuzzy set theory is that of triangular norms: $t$-norms are used to define a generalized intersection. By a fuzzy subset $\mu$ in a given universe $X$ we understand a mapping $\mu$: $X\rightarrow [0, 1]$, the membership degrees $\mu(x)$ can in a natural way be understood as the truth value (in fuzzy logic) of the statement "$x$ belongs to $\mu$". In the same way, the intersection $\mu\cap \nu$ of fuzzy sets $\mu, \nu$ can be viewed as having the membership degree $(\mu\cap \nu) (x)$ corresponding to the truth degree of the statement"$x$ belongs to $\mu$" and"$x$ belongs to $\nu$". Here AND refers to a suitably defined conjunction connective, defined according to the different possibilities which one has to determine the membership degrees $(\mu\cap \nu) (x)$. For example, AND can be understood as taking the minimum or as taking the (usual, i.e., algebraic) product, or more generally, it also can be understood as a $t$-norm. Accordingly, the $t$-norms were considered as the candidates for generalized conjunction connectives of the background many-valued logic. The fuzzy logic based on $t$-norms, especially left continuous $t$-norms, was developed significantly by Hájek, Esteva et al. (see [1-4]). Also, there are many applications of triangular norms in several fields of mathematics and artificial intelligence [5].

In 1971, Rosenfeld [6] used the concept of fuzzy sets to formulate the notion of fuzzy groups. Since then, many other fuzzy algebraic concepts of fuzzy groups have been developed. Anthony and Sherwood [7] redefined fuzzy subgroups, which we call $T$-fuzzy subgroups in this note, in terms of $t$-norm $T$ which replaced the minimum operation and they [7, 8] characterized basic properties of $T$-fuzzy subgroups. In [9], Hu studied $T$-fuzzy groups with thresholds. Chon [10] characterized necessary and sufficient conditions whereby a fuzzy subgroup of a Cartesian product of groups is the product of fuzzy subgroups under minimum operation. He pointed out that finding necessary and sufficient conditions for $T$-fuzzy subgroups under a $t$-norm is still an open problem. In 2011, Yamak et al.[11] solved this open problem and identified necessary and sufficient conditions for $TL$-subgroups of a Cartesian product of groups, which can be represented as a $T$-product of $TL$-subgroups under $t$-norm operation. In the same paper, they also pointed out that the same problem could be studied in other algebraic structures such as rings and lattices.

In [6], the idea of a least fuzzy subgroupoid containing a given fuzzy set was also introduced. Consequently, Rosenfeld constructed the lattice of all fuzzy subgroupoids of a given group. In a recent paper [12], Jahan established that the lattice of all fuzzy ideals of a ring is modular. In fact, the proof of modularity is heavily based on the property of the unit interval that it is a dense chain. However, modularity of the lattices of $L$-normal subgroups of a group and $L$-ideals of a ring remains an open question. In 2011, Jahan [13] answered the question of modularity of the lattice of $L$-ideals of a ring.

With the development of theories of fuzzy algebra, Swamy [14] discussed the correspondence relation between fuzzy ideals and fuzzy congruences in a distributive lattice. In 2008, Koguep et al.[15] studied the notion of fuzzy prime ideal and highlighted the difference between fuzzy prime ideal and prime fuzzy ideal of a lattice. However, not much attention was paid to the studies of the lattices of fuzzy ideals of a lattice and the modularity of them.

The present work has been started as a continuation of these studies. In this paper, we will discuss modularity of the lattices of fuzzy ideals in a lattice. Moreover, we will explore necessary and sufficient conditions for a fuzzy ideal of a Cartesian product of lattices to be a $T$-product of fuzzy ideals of lattices under a left continuous $t$-norm $T$ on a complete lattice $L$.

In this section, we recall some notions and definitions that will be used in the sequel

Let $(L, \wedge, \vee, \leq, 0, 1)$ denote a complete lattice with the top and bottom elements $1$ and $0$, respectively.

Definition 2.1 [5] A binary operation $T$ on $L$ is called a $t$-norm if it satisfies the following conditions: for any $a, b, c\in L$,

(T1) $aT1=a$;

(T2) $aTb=bTa$;

(T3) $(aTb)Tc=aT(bTc)$;

(T4) if $b\leq c$, then $aTb\leq aT c$.

Because of associative and commutative properties, for any $a_{1}, a_{2}, \cdots, a_{n}\in L$ $(n\geq 1)$, $a_{1}Ta_{2}T\cdots T a_{n}$ is well defined and its value is irrelevant to the order of $a_{1}, a_{2}, \cdots, a_{n}$. We write $T^{n}_{i=1}a_{i}=a_{1}Ta_{2}T\cdots T a_{n}$. If $aT(\vee_{i\in I} b_{i})=\vee_{i\in I}(aTb_{i})$, for $a, b_{i}\in L$, where $I$ is the set of natural numbers, then $T$ is called a left continuous $t$-norm, see [2].

In what follows, let $L$=[0,1], there are some examples of the most popular $t$-norms: for any x, y$ \in $

(1) the Lukasiewicz $t$-norm: $xT_{L}y=\max\{x + y-1, 0\}$;

(2) the algebraic product: $xT_{P}y=xy$;

(3) the standard min operation: $xT_{M}y=\min\{x, y\}$.

Throughout this paper, unless otherwise stated, $(L, \wedge, \vee, \leq, 0, 1)$ always represents a given complete lattice with a left continuous $t$-norm $T$.

An $L$-fuzzy subset of $X$ is a mapping from $X$ to $L$. The family of all $L$-subsets of $X$ is denoted by $LF[X]$ (see [16]). When $L=^{[<xref ref-type="bibr" rid="b0">0</xref>, <xref ref-type="bibr" rid="b1">1</xref>]}$, the $L$-subsets of $X$ are known as fuzzy subsets of $X$ (see [17]). Let $\mu, \nu \in LF[X]$ be given, $\mu$ is said to be included in $\nu$ and written as $\mu\subseteq \nu$ if $\mu(x)\leq \nu(x)$ for all $x\in X$.

The following are the most popular operators on $L$-fuzzy sets: for all $\mu, \nu \in LF[X]$, $x\in X$, $(\mu\cup\nu)(x)=\mu(x)\vee\nu(x)$, $(\mu\cap\nu)(x)=\mu(x)\wedge\nu(x)$.

In this section, we shall introduce the notion of $TL$-fuzzy ideals in lattices and give some properties of them that will be used in the sequel.

Definition 3.1  Let $(X, \wedge, \vee, \leq)$ be a lattice and $\mu$ be an $L$-fuzzy subset of $X$. Then $\mu$ is called a $TL$-fuzzy ideal of $X$ if it satisfies the following conditions: for all $x, y\in X$,

(ⅰ) $\mu(x\vee y)\geq \mu(x)T \mu(y)$,

(ⅱ) $\mu(x\wedge y)\geq \mu(x)\vee \mu(y)$.

We shall denote the set of all $TL$-fuzzy ideals of the lattice $X$ as $TLFI[X]$.

In particular, a $TL$-fuzzy ideal is called an $L$-fuzzy ideal when $T=\wedge$. Moreover, when $L=^{[<xref ref-type="bibr" rid="b0">0</xref>, <xref ref-type="bibr" rid="b1">1</xref>]}$, a $TL$-fuzzy ideal and an $L$-fuzzy ideal of $X$ are, respectively, referred to as a $T$-fuzzy ideal and fuzzy ideal of the lattice $X$.

Now, the following result gives an equivalent version of the concept of $TL$-fuzzy ideals in lattices.

Theorem 3.2  Let $(X, \wedge, \vee, \leq)$ be a lattice and $\mu$ be an $L$-fuzzy subset of $X$. Then $\mu$ is a $TL$-fuzzy ideal of $X$ if and only if it satisfies the following conditions: for all $x, y\in X$,

(ⅰ) $\mu(x\vee y)\geq \mu(x)T \mu(y)$,

(ⅱ) if $y\leq x$, then $\mu(y)\geq \mu(x)$.

Proof  The proof is straightforward.

Example 3.3  $(1)$ Let $X=\{0, a, b, c, 1\}$ be a lattice, the partial order on $X$ is defined as shown in Fig. 1. Let $L=\{1, 2, 3, 4, 5, 6\}$, the partial order on $L$ is defined as shown in Fig. 2, $T=\wedge$. Define two $L$-fuzzy subsets $\mu$ and $\nu$ of $X$ as follows: $\mu(0)=6$, $\mu(a)=2$, $\mu(b)=5$, $\mu(c)=2$, $\mu(1)=2$. By routine calculations, it is easy to check that $\mu$ is a $TL$-fuzzy ideal of $X$.

$(2)$ Let $X=\{0, a, b, 1\}$ be a lattice, the partial order on $X$ is defined as shown in Fig. 3. Let $L=^{[<xref ref-type="bibr" rid="b0">0</xref>, <xref ref-type="bibr" rid="b1">1</xref>]}$ and $T=T_{L}$, that is $xT_{L}y =\max\{x + y-1, 0\}$, for any $x, y \in L=^{[<xref ref-type="bibr" rid="b0">0</xref>, <xref ref-type="bibr" rid="b1">1</xref>]}$. Define two $L$-fuzzy subsets $\mu$ and $\nu$ of $X$ as follows: $\mu(0)=\frac{4}{5}$, $\mu(a)=\frac{3}{5}$, $\mu(b)=\frac{3}{10}$, $\mu(1)=\frac{3}{10}$. By routine calculations, it is easy to check that $\mu$ is a $T_{L}$-fuzzy ideal of $X$.

In the following, we give some properties of $TL$-fuzzy ideals, which will be used in the sequel.

Proposition 3.4  Let $\mu_{i}$($i\in I$) be $TL$-fuzzy ideals of a lattice $X$. Then $\cap_{i\in I}\mu_{i}$ is a $TL$-fuzzy ideal of $X$.

Proof  The proof is straightforward.

By the following example we show that the union of two $TL$-fuzzy ideals is not a $TL$-fuzzy ideal.

Example 3.5 Let $X=\{0, a, b, 1\}$ be a lattice, the partial order on $X$ is defined as shown in Fig. 3 in Example 3.3 (2). Let $L=^{[<xref ref-type="bibr" rid="b0">0</xref>, <xref ref-type="bibr" rid="b1">1</xref>]}$ and $T=T_{M}$. Define two $L$-fuzzy subsets $\mu$ and $\nu$ of $X$ as follows: $\mu(0)=\frac{3}{5}$, $\mu(a)=\frac{1}{2}$, $\mu(b)=\frac{1}{5}$, $\mu(1)=\frac{1}{5}$; $\nu(0)=\frac{7}{10}$, $\nu(a)=\frac{1}{10}$, $\nu(b)=\frac{3}{10}$, $\nu(1)=\frac{1}{10}$. Then we can check that both $\mu$ and $\nu$ are $TL$-fuzzy ideals of $X$. Now $(\mu\cup \nu)(0)=\frac{7}{10}$, $(\mu\cup \nu)(a)=\frac{1}{2}$, $(\mu\cup \nu)(b)=\frac{3}{10}$, $(\mu\cup \nu)(1)=\frac{1}{5}$. Since $(\mu\cup \nu)(a\vee b)=(\mu\cup \nu)(1)=\frac{1}{5}< (\mu\cup \nu)(a)\wedge(\mu\cup \nu)(b)=\frac{3}{10}$, $\mu\cup\nu$ is not a $TL$-fuzzy ideal of $X$.

Now, we give a procedure to construct the $TL$-fuzzy ideal generated by an $L$-fuzzy subset. And we shall discuss the algebraic structure of the set of all $TL$-fuzzy ideals in lattices.

Definition 4.1   Let $\mu$ be an $L$-fuzzy subset in a lattice $X$. A $TL$-fuzzy ideal $\nu$ of the lattice $X$ is said to be generated by $\mu$, if $\mu\subseteq \nu$ and for any $TL$-fuzzy ideal $\omega$ of $X$, $\mu\subseteq \omega$ implies $\nu\subseteq \omega$. The $TL$-fuzzy ideal generated by $\mu$ will be denoted by $(\mu]_{TL}$.

It follows from Definition $4.1$ that $(\mu]_{TL}$ is the smallest $TL$-fuzzy ideal of the lattice $X$ containing $\mu$. And we can easily get that $(\mu]_{TL}=\cap_{i\in I}\{\mu_{i}\in TLFI[X]|\mu_{i}\supseteq \mu, i\in I\}$.

It is easy to verify that for $\mu$ and $\nu$ be $L$-fuzzy subsets of $X$. Then

(1) if $\mu$ is a $TL$-fuzzy ideal of the lattice $X$, then $(\mu]_{TL}=\mu$,

(2) $\mu\subseteq \nu$ implies $(\mu]_{TL}\subseteq (\nu]_{TL}$.

In what follows, we give the formula for calculating the $TL$-fuzzy ideals generated by $L$-fuzzy subsets.

Theorem 4.2   Let $\mu$ be an $L$-fuzzy subset in the lattice $X$. Then for any $x\in X$,

$(\mu]_{TL}(x)=\vee\{\mu(a_{1})T \mu(a_{2})T\cdots T \mu(a_{n})|x\leq a_{1}\vee a_{2}\vee \cdots \vee a_{n}, a_{i}\in X, i\in I\}.$

Proof  Let

$\nu(x)=\vee\{\mu(a_{1})T \mu(a_{2})T\cdots T \mu(a_{n})|x\leq a_{1}\vee a_{2}\vee \cdots \vee a_{n}, a_{i}\in X, i\in I\}.$

First, we prove that $\nu$ is a $TL$-fuzzy ideal of $X$.

For any $x, y\in X$, we have that

$\begin{eqnarray*}\nu(x)T\nu(y)&=&(\vee\{\mu(a_{1})T \mu(a_{2})T\cdots T \mu(a_{n})|x\leq a_{1}\vee a_{2}\vee \cdots \vee a_{n}, a_{i}\in X, i\in I\})T\\ &&(\vee\{\mu(b_{1})T \mu(b_{2})T\cdots T \mu(b_{m})|y\leq b_{1}\vee b_{2}\vee \cdots \vee b_{m}, b_{j}\in X, j\in I\})\\ &=& \vee\{\mu(a_{1})T \mu(a_{2})T\cdots T \mu(a_{n})T \mu(b_{1})T \mu(b_{2})T\cdots T \mu(b_{m})|x\\ &\leq& a_{1}\vee a_{2}\vee \cdots \vee a_{n}, y \leq b_{1}\vee b_{2}\vee \cdots \vee b_{m}, a_{i}, b_{j}\in X, i, j\in I\}\\ &\leq&\vee\{\mu(c_{1})T \mu(c_{2})T\cdots T \mu(c_{l})|x\vee y\leq c_{1}\vee c_{2}\vee \cdots \vee c_{l}, c_{i}\in X, l\in I\}\\ &=&\nu(x\vee y).\end{eqnarray*}$

If $y\leq x$ and $x\leq a_{1}\vee a_{2}\vee \cdots \vee a_{n}$ for some $a_{1}, a_{2}, \cdots, a_{n}\in X$, then $y\leq a_{1}\vee a_{2}\vee \cdots \vee a_{n}$. It follows that $\nu(y)\geq \nu(x)$. By Theorem 3.2, we can get that $\nu$ is a $TL$-fuzzy ideal of $X$.

Next, since $x\leq x$, we have that $\nu(x)\geq \mu(x)$. So $\mu\subseteq \nu$.

Finally, suppose that $\omega$ is a $TL$-fuzzy ideal of $X$ with $\mu\subseteq \omega$. Then for any $x\in X$, $\mu(x)\leq \omega(x)$. Moreover, for any $a_{1}, a_{2}, \cdots a_{n}\in X$ with $x\leq a_{1}\vee a_{2}\vee \cdots \vee a_{n}$, we have $\mu(a_{1})T \mu(a_{2})T\cdots T \mu(a_{n})\leq \omega(a_{1})T\omega(a_{2})T\cdots T \omega(a_{n})\leq \omega(a_{1}\vee a_{2}\vee \cdots \vee a_{n})\leq \omega(x)$, since $\omega$ is a $TL$-fuzzy ideal of $X$. It follows that $\nu(x)\leq \omega(x)$. Thus, $\nu\subseteq \omega$.

Summarizing the above facts, we obtain that $\nu$ is the smallest $TL$-fuzzy ideal in the lattice $X$ with $\mu \subseteq \nu$, that is, $\nu$ is the $TL$-fuzzy ideal generated by $\mu$ in $X$. Therefore, $\nu=(\mu]_{TL}$.

Example 4.3   Let $X=\{0, a, b, 1\}$ be a lattice, the partial order on $X$ is defined as shown in Fig. 3 in Example 3.3 (2). Let $L=\{1, 2, 3, 4\}$, the partial order on $L$ is defined as shown in Fig. 4 and $T=\wedge$. Define an $L$-fuzzy subset $\mu$ of $X$ as follows: $\mu(0)=3$, $\mu(a)=2$, $\mu(b)=4$, $\mu(1)=1$. One can easily check that the $TL$-fuzzy ideal $(\mu]_{TL}$ generated by $\mu$ as follows: $(\mu]_{TL}(0)=4$, $(\mu]_{TL}(a)=2$, $(\mu]_{TL}(b)=4$, $(\mu]_{TL}(1)=2$.

Let $X$ be a lattice. For any $\mu_{1}, \mu_{2}\in TLFI[X]$, we define $\mu_{1}\oplus \mu_{2}$ and $\mu_{1}\otimes \mu_{2}$ as follows: $\mu_{1}\oplus \mu_{2}=\mu_{1}\cap \mu_{2}$, $\mu_{1}\otimes \mu_{2}=\cap\{\mu\in TLFI[X]|\mu\supseteq \mu_{1}\cup \mu_{2}\}=(\mu_{1}\cup \mu_{2}]_{TL}$. In general, for any $\mu_{i}\in TLFI[X]$, where $i\in I$, we define $\oplus\{\mu_{i}|i\in I\}=\cap\{\mu_{i}|i\in I\}$, $\otimes\{\mu_{i}|i\in I\}=\cap\{\mu\in TLFI[X]|\mu\supseteq \cup_{i\in I} \mu_{i}\}=(\cup_{i\in I} \mu_{i}]_{TL}$. Therefore, we can get the following result.

Theorem 4.4 $(TLFI[X], \oplus, \otimes)$ is a complete lattice, which is called the lattice of $TL$-fuzzy ideals.

In particular, when $L=^{[<xref ref-type="bibr" rid="b0">0</xref>, <xref ref-type="bibr" rid="b1">1</xref>]}$, we can give the simple formulas for calculating the $T$-fuzzy ideals generated by the union of $T$-fuzzy ideals.

Theorem 4.5 Let $\mu_{1}, \mu_{2}$ be $T$-fuzzy ideals of a lattice $X$. Then for any $x\in X$,

$(\mu_{1}\cup \mu_{2}]_{T}(x)=\sup\{\{\mu_{1}(a)| x\leq a\}\cup \{\mu_{2}(b)|x\leq b\}\cup \{\mu_{1}(a)T \mu_{2}(b)|x\leq a\vee b\} \}.$

Proof  By Theorem 4.2, we have

$\begin{eqnarray*}(\mu_{1}\cup \mu_{2}]_{T}(x)&=&\sup\{(\mu_{1}\cup \mu_{2})(a_1)T (\mu_{1}\cup \mu_{2})(a_2)T\cdots T(\mu_{1}\cup \mu_{2})(a_n)|x\\ &\leq&a_1\vee \cdots \vee a_n, a_{i}\in X, i\in I\}.\end{eqnarray*}$

Given an arbitrary small $\epsilon >0$, we have the following three cases:

Case 1   There exist $a_1, \cdots, a_n\in X$, satisfying

(1) $x\leq a_1\vee a_2\vee \cdots\vee a_n$,

(2) $(\mu_{1}\cup \mu_{2}]_{T}(x)<\epsilon + \max\{\mu_{1}(a_1), \mu_{2}(a_1)\}T \cdots T \max\{\mu_{1}(a_n), \mu_{2}(a_n)\}$,

(3) $\mu_{2}(a_i)\leq \mu_{1}(a_i), i=1, \cdots, n$.

Thus $(\mu_{1}\cup \mu_{2}]_{T}(x)<\epsilon +\mu_{1}(a_1)T\mu_{1}(a_2) \cdots T \mu_{1}(a_n)$.

Denote $a= a_1\vee \cdots\vee a_n$. Since $\mu_{1}$ is a $T$-fuzzy ideal of a lattice $X$, we have

$\mu_{1}(a)\geq \mu_{1}(a_1)T\mu_{1}(a_2) \cdots T \mu_{1}(a_n), $

it follows that $x\leq a$ and $(\mu_{1}\cup \mu_{2}]_{T}(x)<\epsilon +\mu_{1}(a)$.

Case 2   There exist $b_1, \cdots, b_m\in X$, satisfying

(1) $x\leq b_1\vee b_2\vee \cdots\vee b_m$,

(2) $(\mu_{1}\cup \mu_{2}]_{T}(x)<\epsilon + \max\{\mu_{1}(b_1), \mu_{2}(b_1)\}T \cdots T \max\{\mu_{1}(b_m), \mu_{2}(b_m)\}$,

(3) $\mu_{1}(b_i)\leq \mu_{2}(b_i), i=1, \cdots, m$.

Thus $(\mu_{1}\cup \mu_{2}]_{T}(x)<\epsilon +\mu_{2}(b_1)T\mu_{2}(b_2)T \cdots T \mu_{2}(b_m)$.

Denote $b=b_1\vee b_2\vee \cdots\vee b_m$. Since $\mu_{2}$ is a $T$-fuzzy ideal of a lattice $X$, we have $\mu_{2}(b)\geq \mu_{2}(b_1)T\mu_{2}(b_2) \cdots T \mu_{2}(b_m)$, it follows that $x\leq b$ and $(\mu_{1}\cup \mu_{2}]_{T}(x)<\epsilon +\mu_{2}(b)$.

Cases 3   There exist $a_1, \cdots, a_s, b_1, \cdots, b_t\in X$, satisfying

(1) $x\leq a_1\vee \cdots \vee a_s\vee b_1\vee \cdots \vee b_t$,

(2)

$\begin{eqnarray*} (\mu_{1}\cup \mu_{2}]_{T}(x)&<& \epsilon + \max\{\mu_{1}(a_1), \mu_{2}(a_1)\}T \cdots T \max\{\mu_{1}(a_s), \mu_{2}(a_s)\}\\ &&T \max\{\mu_{1}(b_1), \mu_{2}(b_1)\}T \cdots \max\{\mu_{1}(b_t), \mu_{2}(b_t)\}, \end{eqnarray*}$

(3) $\mu_{2}(a_i)\leq \mu_{1}(a_i), \mu_{1}(b_j)\leq\mu_{2}(b_j), i=1, 2, \cdots, s, j=1, 2, \cdots t $.

Thus

$(\mu_{1}\cup \mu_{2}]_{T}(x)<\epsilon +\mu_{1}(a_1)T\mu_{1}(a_2) \cdots T \mu_{1}(a_s)T\mu_{2}(b_1)T\mu_{2}(b_2)T \cdots T \mu_{2}(b_t).$

Denote $a=a_1\vee \cdots \vee a_s$ and $b=b_1\vee \cdots \vee b_t$, then $x\leq a\vee b$. Since $\mu_{1}$ and $\mu_{2}$ are $T$-fuzzy ideals of a lattice $X$, we have

$\mu_{1}(a)\geq \mu_{1}(a_1)T\mu_{1}(a_2) \cdots T \mu_{1}(a_s), \mu_{2}(b)\geq \mu_{2}(b_1)T\mu_{2}(b_2) \cdots T \mu_{2}(b_t).$

It follows that $x\leq a\vee b$ and $(\mu_{1}\cup \mu_{2}]_{T}(x)<\epsilon +\mu_{1}(a)T \mu_{2}(b)$. Summarizing the above results we obtain

$(\mu_{1}\cup \mu_{2}]_{T}(x)\leq \sup\{\{\mu_{1}(a)| x\leq a\}\cup \{\mu_{2}(b)|x\leq b\}\cup \{\mu_{1}(a)T \mu_{2}(b)|x\leq a\vee b\} \}.$

Conversely,

$\begin{eqnarray*}&& \sup\{\mu_{1}(a)|x\leq a\}\\ &\leq &\sup\{\max\{\mu_{1}(a), \mu_{2}(a)\}|x\leq a\}\\ &\leq& \sup\{\max\{\mu_{1}(a_1), \mu_{2}(a_1)\}T \cdots T \max\{\mu_{1}(a_n), \mu_{2}(a_n)\}|x\leq a_1\vee \cdots \vee a_n\}.\end{eqnarray*}$

Similarly, we have

$\begin{eqnarray*}&&\sup\{\mu_{2}(b)|x\leq b\}\\ &\leq& \sup\{\max\{\mu_{1}(a_1), \mu_{2}(a_1)\}T \cdots T \max\{\mu_{1}(a_n), \mu_{2}(a_n)\}|x\leq a_1\vee \cdots \vee a_n\}.\end{eqnarray*}$

Since $\mu_{1}(a)T \mu_{2}(b)\leq \max\{\mu_{1}(a), \mu_{2}(a)\}T \max\{\mu_{1}(b), \mu_{2}(b)\}$, we have

$\begin{eqnarray*}&& \sup\{\mu_{1}(a)T \mu_{2}(b)|x\leq a\vee b\}\\ &\leq& \sup\{\max\{\mu_{1}(a), \mu_{2}(a)\}T \max\{\mu_{1}(b), \mu_{2}(b)\}|x\leq a\vee b\}\\ &\leq &\sup\{\max\{\mu_{1}(a_1), \mu_{2}(a_1)\}T \cdots T \max\{\mu_{1}(a_n), \mu_{2}(a_n)\}|x\leq a_1\vee \cdots \vee a_n\}.\end{eqnarray*}$

Therefore, $\sup\{\{\mu_{1}(a)| x\leq a\}\cup \{\mu_{2}(b)|x\leq b\}\cup \{\mu_{1}(a)T \mu_{2}(b)|x\leq a\vee b\} \}\leq (\mu_{1}\cup \mu_{2}]_{T}(x)$. This completes the proof.

When $L=^{[<xref ref-type="bibr" rid="b0">0</xref>, <xref ref-type="bibr" rid="b1">1</xref>]}$ and $T=T_{M}$, that is, $xT_{M}y=\min\{x, y\}$ for all $x, y\in ^{[<xref ref-type="bibr" rid="b0">0</xref>, <xref ref-type="bibr" rid="b1">1</xref>]}$, we can obtain the following main result.

Theorem 4.6   $(TLFI[X], \oplus, \otimes)$ is a complete modular lattice if $X$ is a modular lattice.

Proof   From Theorem 4.4, we have that $(TLFI[X], \oplus, \otimes)$ is a complete lattice. To verify that $(TLFI[X], \oplus, \otimes)$ is a modular lattice, we should show that it satisfies modular law. Now, assume $\mu_{1}, \mu_{2}, \mu_{3}\in TLFI[X]$, where $\mu_{1}\supseteq\mu_{2}$ and $x\in X$, the inequality $\mu_{1}\oplus (\mu_{2}\otimes \mu_{3})\supseteq \mu_{2}\otimes (\mu_{1}\oplus \mu_{3})$ is trivial. We only need to prove that $\mu_{1}\oplus (\mu_{2}\otimes \mu_{3})\subseteq \mu_{2}\otimes (\mu_{1}\oplus \mu_{3})$.

Given an arbitrarily small $\epsilon >0$, by Theorem 4.5, we have the following three cases:

Case 1   There exists $a\in X$ such that $x\leq a$ and $(\mu_{2}\otimes \mu_{3})(x)<\epsilon +\mu_{2}(a)$. And so

$(\mu_{1}\oplus (\mu_{2}\otimes \mu_{3}))(x)\leq\epsilon+\min\{\mu_{1}(x), \mu_{2}(a)\}.$

Since $\mu_{2}(a)\leq\mu_{2}(x)\leq\mu_{1}(x)$, it follows that $\min\{\mu_{1}(x), \mu_{2}(a)\}=\mu_{2}(a)$, hence

$(\mu_{1}\oplus (\mu_{2}\otimes \mu_{3}))(x)\leq\epsilon+\mu_{2}(a).$

Combining $x\leq a$, we obtain

$(\mu_{1}\oplus (\mu_{2}\otimes \mu_{3}))(x)\leq\epsilon+\mu_{2}(x)\leq\epsilon +(\mu_{2}\otimes (\mu_{1}\oplus \mu_{3}))(x).$

Case 2   There exists $b\in X$ such that $x\leq b$ and $(\mu_{1}\otimes \mu_{3})(x)<\epsilon +\mu_{3}(b)$. From $x\leq b$ it follows that $\mu_{3}(b)\leq\mu_{3}(x)$, hence

$(\mu_{1}\oplus (\mu_{2}\otimes \mu_{3}))(x)\leq\epsilon+\min\{\mu_{1}(x), \mu_{3}(x)\}.$

Combining $x\leq x$ and the definition of $\mu_{2}\otimes (\mu_{1}\oplus \mu_{3})$, we have

$(\mu_{1}\oplus (\mu_{2}\otimes \mu_{3}))(x)\leq \epsilon+ (\mu_{2}\otimes (\mu_{1}\oplus \mu_{3}))(x).$

Case 3   There are $a, b\in X$ such that $x\leq a\vee b$ and $(\mu_{2}\otimes \mu_{3})(x)<\epsilon +\min\{\mu_{2}(a), \mu_{3}(b)\}$. Hence $(\mu_{1}\oplus (\mu_{2}\otimes \mu_{3}))(x)\leq \epsilon+ \min\{\mu_{1}(x), \mu_{2}(a), \mu_{3}(b)\}$. Denote $b_1=(x\vee a)\wedge b$, then $b_1\leq x\vee a$ and $b_1\leq b$. Notice that $X$ is a modular lattice, $a\vee b_1=a\vee ((x\vee a)\wedge b)=(x\vee a)\wedge (a\vee b)\geq x$. Since $\mu_{1}, \mu_{3}$ are fuzzy ideals, we have $\mu_{3} (b)\leq\mu_{3}(b_1)$ and $\mu_{1}(x\vee a)\leq \mu_{1}(b_1)$. It follows that

$\begin{eqnarray*}&& (\mu_{1}\oplus_1(\mu_{2}\otimes \mu_{3}))(x)\\ &\leq&\epsilon+ \min\{\mu_{2}(a), \mu_{1}(a), \mu_{1}(x), \mu_{3}(b)\}\\ &=&\epsilon+ \min\{\mu_{2}(a), \min\{\mu_{1}(a), \mu_{1}(x)\}, \mu_{3}(b)\}\\ &\leq&\epsilon+ \min\{\mu_{2}(a), \mu_{1}(x\vee a), \mu_{3}(b)\}\\ &\leq& \epsilon+ \min\{\mu_{2}(a), \mu_{1}(b_1), \mu_{3}(b_1)\}\\ &=&\epsilon+ \min\{\mu_{2}(a), \min\{\mu_{1}(b_1), \mu_{3}(b_1)\}\}\\ &=&\epsilon+ \min\{\mu_{2}(a), (\mu_{1}\oplus \mu_{3})(b_1)\}\\ &\leq&\epsilon+ (\mu_{2}\otimes (\mu_{1}\oplus \mu_{3}))(x).\end{eqnarray*}$

Since $\epsilon$ is arbitrary, we have $(\mu_{1}\oplus (\mu_{2}\otimes \mu_{3}))(x)$ $\leq (\mu_{2}\otimes (\mu_{1}\oplus \mu_{3}))(x)$. Therefore,

$\mu_{1}\oplus (\mu_{2}\otimes \mu_{3})\subseteq \mu_{2}\otimes (\mu_{1}\oplus \mu_{3}).$

Summarizing the above facts, we get that for any $x\in X$ and given an arbitrarily small $\epsilon >0$,

$(\mu_{1}\oplus (\mu_{2}\otimes \mu_{3}))(x)\leq \epsilon+ (\mu_{2}\otimes (\mu_{1}\oplus \mu_{3}))(x).$

Therefore, $(\mu_{1}\oplus (\mu_{2}\otimes \mu_{3}))(x)\leq (\mu_{2}\otimes (\mu_{1}\oplus \mu_{3}))(x)$. So $\mu_{1}\oplus (\mu_{2}\otimes \mu_{3})\subseteq\mu_{2}\otimes (\mu_{1}\oplus \mu_{3})$, that is, $(TLFI[X], \oplus, \otimes)$ is a complete modular lattice.

In this section, as a continuation of the work [11], we will explore necessary and sufficient conditions for a fuzzy ideal of a Cartesian product of lattices to be a $T$-product of fuzzy ideals of lattices under a left continuous $t$-norm $T$ on a complete lattice $L$.

First, let us recall the Cartesian product of lattices for the sake of completeness.

Let $(X_{1}, \wedge_{1}, \vee_{1}, \leq_{1})$ and $(X_{2}, \wedge_{2}, \vee_{2}, \leq_{2})$ be two lattices. Define two binary operations $\wedge$ and $\vee$ on $X_{1}\times X_{2}$ as follows: for any $(x_{1}, x_{2})$, $(y_{1}, y_{2})\in X_{1}\times X_{2}$,

$(x_{1}, x_{2})\wedge(y_{1}, y_{2})=(x_{1}\wedge_{1}y_{1}, x_{2}\wedge_{2}y_{2}), (x_{1}, x_{2})\vee(y_{1}, y_{2})=(x_{1}\vee_{1}y_{1}, x_{2}\vee_{2}y_{2}). $

Then $X_{1}\times X_{2}$ is a lattice, which is called the Cartesian product lattice of $X_{1}$ and $X_{2}$. The corresponding partial order $\leq$ on $X_{1}\times X_{2}$ as follows:

$(x_{1}, x_{2})\leq(y_{1}, y_{2}) \Longleftrightarrow x_{1}\leq_{1}y_{1}, x_{2}\leq_{2} y_{2}.$

Definition 5.1 Let $\mu_{i}\in LF[X_{i}]$, $i=1, 2$. Then the $T$-product of $\mu_{i}$ ($i=1, 2$) denoted by $\mu_{1}\times_{T}\mu_{2}$ is defined as the $L$-fuzzy subset of $X_{1}\times X_{2}$ that satisfies: for any $(x_{1}, x_{2})\in X_{1}\times X_{2}$, $\mu_{1}\times_{T}\mu_{2}(x_{1}, x_{2})=\mu_{1}(x_{1})T\mu_{2}(x_{2})$.

Theorem 5.2   Let $\mu_{i}$ be a $TL$-fuzzy ideal of a lattice $X_{i}$, $i=1, 2$. Then $\mu_{1}\times_{T}\mu_{2}$ is a $TL$-fuzzy ideal of $X_{1}\times X_{2}$.

Proof  Assume that $\mu_{i}$ be a $TL$-fuzzy ideal of a lattice $X_{i}$, $i=1, 2$. For any $(x_{1}, x_{2})$, $(y_{1}, y_{2})\in X_{1}\times X_{2}$, then we have that

$\begin{eqnarray*}&& \mu_{1}\times_{T}\mu_{2}((x_{1}, x_{2})\vee(y_{1}, y_{2})) =\mu_{1}\times_{T}\mu_{2}((x_{1}\vee_{1} y_{1}, x_{2}\vee_{2} y_{2}))\\ &=&\mu_{1}(x_{1}\vee_{1} y_{1})T\mu_{2}(x_{2}\vee_{2} y_{2}) \geq(\mu_{1}(x_{1})T\mu_{1}(y_{1}))T(\mu_{2}(x_{2})T\mu_{2}(y_{2}))\\ &=&(\mu_{1}(x_{1})T\mu_{2}(x_{2}))T(\mu_{1}(y_{1})T\mu_{2}(y_{2}))\\ &=&(\mu_{1}\times_{T}\mu_{2})(x_{1}, x_{2})T(\mu_{1}\times_{T}\mu_{2})(y_{1}, y_{2}).\end{eqnarray*}$

On the other hand, if $(x_{1}, x_{2})\leq(y_{1}, y_{2})$, that is $x_{1}\leq_{1}y_{1}, x_{2}\leq_{2} y_{2}$. Then we have

$\mu_{1}\times_{T}\mu_{2}(x_{1}, x_{2})=\mu_{1}(x_{1})T\mu_{2}(x_{2}) \geq\mu_{1}(y_{1})T\mu_{2}(y_{2}) =\mu_{1}\times_{T}\mu_{2}(y_{1}, y_{2}).$

Therefore, $\mu_{1}\times_{T}\mu_{2}$ is a $TL$-fuzzy ideal of $X_{1}\times X_{2}$.

In what follows, we introduce the concepts of the projection and the cut shadow of an $L$-fuzzy set that are instrumental to determine necessary and sufficient conditions under $t$-norm operation.

Definition 5.3 Let $\mu\in LF[X_{1}\times X_{2}]$. Then the projection of $\mu$ on $X_{i}$ ($i=1, 2$) denoted by $\mu_{X_{i}}$ is defined as the $L$-fuzzy subset of $X_{i}$ ($i=1, 2$) that satisfies, respectively, $\mu_{X_{1}}(x)=\bigvee_{b\in X_{2}}\mu(x, b)$ for any $x\in X_{1}$ and $\mu_{X_{2}}(y)=\bigvee_{a\in X_{1}}\mu(a, y)$ for any $y\in X_{2}$.

Theorem 5.4 Let $X_{1}$ and $X_{2}$ be two lattices and $\mu$ be a $TL$-fuzzy ideal of $X_{1}\times X_{2}$. Then $\mu_{X_{i}}$ is a $TL$-fuzzy ideal of $X_{i}$, $i=1, 2$.

Proof  Assume that $\mu$ be a $TL$-fuzzy ideal of a lattice $X_{1}\times X_{2}$. For any $x, y \in X_{1}$, then we have that

$\begin{eqnarray*}&& \mu_{X_{1}}(x\vee_{1}y)=\bigvee_{b\in X_{2}}\mu(x\vee_{1}y, b) \geq \bigvee_{b_{1}, b_{2}\in X_{2}}\mu(x\vee_{1}y, b_{1}\vee_{2} b_{2})\\ &=&\bigvee_{b_{1}, b_{2}\in X_{2}}\mu((x, b_{1})\vee (y, b_{2})) \geq\bigvee_{b_{1}, b_{2}\in X_{2}}[\mu(x, b_{1})T \mu(y, b_{2})]\\ &=&(\bigvee_{b_{1}\in X_{2}}\mu(x, b_{1}))T (\bigvee_{b_{2}\in X_{2}}\mu(y, b_{2})) =\mu_{X_{1}}(x)T \mu_{X_{1}}(y).\end{eqnarray*}$

On the other hand, we obtain

$\begin{eqnarray*}&& \mu_{X_{1}}(x\wedge_{1}y)=\bigvee_{b\in X_{2}}\mu(x\wedge_{1}y, b) \geq\bigvee_{b_{1}, b_{2}\in X_{2}}\mu(x\wedge_{1}y, b_{1}\wedge_{2} b_{2})\\ &=&\bigvee_{b_{1}, b_{2}\in X_{2}}\mu((x, b_{1})\wedge (y, b_{2})) \geq\bigvee_{b_{1}, b_{2}\in X_{2}}[\mu(x, b_{1})\vee\mu(y, b_{2})]\\ &=&(\bigvee_{b_{1}\in X_{2}}\mu(x, b_{1}))\vee (\bigvee_{b_{2}\in X_{2}}\mu(y, b_{2})) =\mu_{X_{1}}(x)\vee \mu_{X_{1}}(y).\end{eqnarray*}$

Therefore, $\mu_{X_{1}}$ is a $TL$-fuzzy ideal of $X_{1}$. Dually, we have that $\mu_{X_{2}}$ is a $TL$-fuzzy ideal of $X_{2}$.

Definition 5.5 Let $\mu\in LF[X_{1}\times X_{2}]$ and $a\in X_{1}$, $b\in X_{2}$. Then the cut shadow of $\mu$ with respect to $b$ denoted by $\mu_{1}|_{b}$ is defined as the $L$-fuzzy subset of $X_{1}$ that satisfies: for any $x\in X_{1}$, $\mu_{1}|_{b}(x)=\mu(x, b)$. Similarly, the cut shadow of $\mu$ with respect to $a$ denoted by $\mu_{2}|_{a}$ is defined as the $L$-fuzzy subset of $X_{2}$ that satisfies: for any $y\in X_{2}$, $\mu_{2}|_{a}(y)=\mu(a, y)$.

Theorem 5.6  Let $X_{1}$ and $X_{2}$ be two lattices and $\mu$ be a $TL$-fuzzy ideal of $X_{1}\times X_{2}$, let $a\in X_{1}$, $b\in X_{2}$. Then $\mu_{1}|_{b}$ is a $TL$-fuzzy ideal of $X_{1}$ and $\mu_{2}|_{a}$ is a $TL$-fuzzy ideal of $X_{2}$.

Proof  The proof is straightforward.

In order to obtain necessary and sufficient conditions for a $TL$-fuzzy ideal of a Cartesian product lattice to be a $T$-product of fuzzy ideals of lattices, we give the following lemma.

Lemma 5.7 Let $X_{1}$ and $X_{2}$ be two lattices and $\mu$ be a $TL$-fuzzy ideal of $X_{1}\times X_{2}$ such that $Im\mu \subseteq D_{T}$, let $a\in X_{1}$, $b\in X_{2}$. Then $\mu_{1}|_{b}\times_{T}\mu_{2}|_{a}\subseteq \mu \subseteq \mu_{X_{1}}\times_{T} \mu_{X_{2}}$.

Proof  Assume that $\mu$ is a $TL$-fuzzy ideal of a lattice $X_{1}\times X_{2}$.

First, we prove that $\mu \subseteq \mu_{X_{1}}\times_{T} \mu_{X_{2}}$. For any $(x, y)\in X_{1}\times X_{2}$, we have that

$\mu (x,y) \le \mathop \vee \limits_{b \in {X_2}} \mu (x,b) = {\mu _{{X_1}}}(x)$

and $\mu(x, y)\leq \bigvee_{a\in X_{1}}\mu(a, y)=\mu_{X_{2}}(y).$

Thus, we obtain $\mu(x, y)T \mu(x, y)\leq \mu_{X_{1}}(x)T \mu_{X_{2}}(y)$, then $\mu(x, y)\leq \mu_{X_{1}}\times_{T}\mu_{X_{2}}(x, y)$. Hence, $\mu \subseteq \mu_{X_{1}}\times_{T} \mu_{X_{2}}$.

Next, let us check $\mu_{1}|_{b}\times_{T}\mu_{2}|_{a}\subseteq \mu$. For any $(x, y)\in X_{1}\times X_{2}$, we can have

$\begin{eqnarray*}&& \mu_{1}|_{b}\times_{T}\mu_{2}|_{a}(x, y)=\mu_{1}|_{b}(x)T \mu_{2}|_{a}(y) =\mu(x, b)T \mu(a, y) \leq\mu((x, b)\vee(a, y))\\ &=&\mu((x\vee_{1} a, b\vee_{2} y)) =\mu((x, y)\vee (a, b)) \leq\mu(x, y).\end{eqnarray*}$

Thus $\mu_{1}|_{b}\times_{T}\mu_{2}|_{a}\subseteq \mu$. Combining the above arguments, we can obtain

$\mu_{1}|_{b}\times_{T}\mu_{2}|_{a}\subseteq \mu \subseteq \mu_{X_{1}}\times_{T} \mu_{X_{2}}$

for all $a\in X_{1}$, $b\in X_{2}$.

The following theorem gives one of the main results in this paper.

Theorem 5.8 Let $X_{1}$ and $X_{2}$ be two lattices with the bottom element $0$ and $\mu$ be a $TL$-fuzzy ideal of $X_{1}\times X_{2}$ such that $Im\mu \subseteq D_{T}$. Then $\mu$ is the $T$-product of a $TL$-fuzzy ideal of $X_{1}$ and a $TL$-fuzzy ideal of $X_{2}$ if and only if $\mu_{1}|_{0}\times_{T}\mu_{2}|_{0} = \mu_{X_{1}}\times_{T} \mu_{X_{2}}$.

Proof  Assume that $\mu=\mu_{1}\times_{T}\mu_{2}$, where $\mu_{1}$ and $\mu_{2}$ are $TL$-fuzzy ideals of $X_{1}$ and $X_{2}$, respectively. Then $\mu_{1}(x)\leq \mu_{1}(0)$ for any $x\in X_{1}$ and $\mu_{2}(y)\leq \mu_{2}(0)$ for any $y\in X_{2}$. Thus we have $\bigvee_{x\in X_{1}}\mu_{1}(x)=\mu_{1}(0)$ and $\bigvee_{y\in X_{2}}\mu_{2}(y)=\mu_{2}(0)$. Notice this, we can obtain that

$\begin{eqnarray*}&& \mu_{1}|_{0}(x)=\mu(x, 0)=\mu_{1}\times_{T}\mu_{2}(x, 0)=\mu_{1}(x)T \mu_{2}(0)=\mu_{1}(x)T (\bigvee_{y\in X_{2}}\mu_{2}(y))\\ &=& \bigvee_{y\in X_{2}}[\mu_{1}(x)T \mu_{2}(y)]=\bigvee_{y\in X_{2}} \mu(x, y)=\mu_{X_{1}}(x).\end{eqnarray*}$

Hence, $\mu_{1}|_{0}=\mu_{X_{1}}$. Similarly, we can get that $\mu_{2}|_{0}=\mu_{X_{2}}$. Therefore,

$\mu_{1}|_{0}\times_{T}\mu_{2}|_{0} = \mu_{X_{1}}\times_{T} \mu_{X_{2}}.$

Conversely, assume that $\mu_{1}|_{0}\times_{T}\mu_{2}|_{0} = \mu_{X_{1}}\times_{T} \mu_{X_{2}}$. By Lemma $5.7$, we can get $\mu=\mu_{X_{1}}\times_{T} \mu_{X_{2}}$. Since $\mu$ is a $TL$-fuzzy ideal of $X_{1}\times X_{2}$, it follows from Theorem $5.4$, we have $\mu_{X_{i}}$ is a $TL$-fuzzy ideal of $X_{i}$, $i=1, 2$. That is, $\mu$ is the $T$-product of a $TL$-fuzzy ideal of $X_{1}$ and a $TL$-fuzzy ideal of $X_{2}$. }

Open Problem  Whether the lattice of all $TL$-fuzzy ideals of a lattice $X$ forms a distributive lattice or even a modular lattice.