**2. Preliminaries**

In this section, we provide some notions and results that will be used in this paper.

**Definition 1** ([1])**.** *An effect algebra is a system* (*<sup>E</sup>*, ⊕, 0, 1) *consisting of a set E with two special elements* 0, 1 ∈ *E, called the zero and the unit, and with a partially defined binary operation* ⊕ *satisfying the following conditions: for any x*, *y*, *z* ∈ *E,*


In the following, an effect algebra (*<sup>E</sup>*, ⊕, 0, 1) is denoted by *E* unless otherwise specified.

**Definition 2** ( [33])**.** *A nonempty subset S of an effect algebra E is called a subalgebra if it satisfies the following conditions:*


Let *E* be an effect algebra. If *x* ⊕ *y* is defined, then we say *x* ⊥ *y* for all *x*, *y* ∈ *E*. Define a binary relation on *E* by *x* ≤ *y* if for some *z* ∈ *E*, *x* ⊕ *z* = *y*, which turns out to be a partial ordering on *E* such that 0 and 1 is the smallest element and the greatest element of *E*, respectively. If the poset (*<sup>E</sup>*, ≤) is a lattice, then *E* is called a lattice-ordered effect algebra.

Let *L* be a complete lattice. We denote the minimal element and the maximal element of *L* by ⊥*L* and *L*, respectively. An element *λ* ∈ *L* is called co-prime if *λ* ≤ *δ* ∨ *ξ* implies *λ* ≤ *δ* or *λ* ≤ *ξ*. The set of nonzero co-prime elements in *L* is denoted by *J*(*L*). An element *λ* ∈ *L* is called prime if *λ* ≥ *δ* ∧ *ξ* implies *λ* ≥ *δ* or *λ* ≥ *ξ*. The set of nonunit prime elements in *L* is denoted by *<sup>P</sup>*(*L*). From [34], we know that each element of *L* is the sup of co-prime elements and the inf of prime elements.

Let *δ*, *ξ* ∈ *L*, the symbol *δ* ≺ *ξ* (*δ* is wedge below *ξ*) means that for every *H* ⊆ *L*, *ξ* ≤ *H* implies the existence of *η* ∈ *H* such that *δ* ≤ *η*. A complete lattice *L* is completely distributive [34] if and only if *ξ* = {*δ* | *δ* ≺ *ξ*} for each *ξ* ∈ *L*. The set {*δ* | *δ* ≺ *ξ*}, denoted by *β*(*ξ*), is called the greatest minimal family of *ξ* in the sense of [34]. Let *β*∗(*ξ*) = *β*(*ξ*) ∩ *J*(*L*). Moreover, define a binary relation ≺*op* as follows: for *ξ*, *δ* ∈ *L*, *ξ* ≺*op δ* if and only if for every subset *H* ⊆ *L*, *H* ≤ *ξ* implies *λ* ≤ *δ* for some *λ* ∈ *H*. The set {*δ* ∈ *L* | *ξ* ≺*op <sup>δ</sup>*}, denoted by *<sup>α</sup>*(*ξ*), is the greatest maximal family of *ξ* in the sense of [34]. Let *α*<sup>∗</sup>(*ξ*) = *α*(*ξ*) ∩ *<sup>P</sup>*(*L*). We know that *α* is an − mapping and *β* is a union-preserving mapping, it holds that *ξ* = *β*(*ξ*) = *β*∗(*ξ*) = *α*(*ξ*) = *α*<sup>∗</sup>(*ξ*) for each *ξ* ∈ *L* (see [34]). From [35], we have *<sup>α</sup>*(*L*) = ∅ and *β*(<sup>⊥</sup>*L*) = ∅.

In the following, a completely distributive lattice (*<sup>L</sup>*, ∧, ∨) is denoted by *L* unless otherwise specified.

For an effect algebra *E*, each mapping *A* : *E* −→ *L* is called an *L*-subset of *E*, and we denote the collection of all *L*-subsets of *E* by *LE*. *L<sup>E</sup>* is also a complete lattice by defining "≤" point-wisely. Furthermore, the smallest element and the largest element in *L<sup>E</sup>* are denoted by ⊥ and , respectively. The mapping *<sup>g</sup>*<sup>→</sup>*L* : *LE*1 −→ *LE*2 is induced by *g* : *E*1 −→ *E*2 as follows:

$$\forall A \in L^{E\_1}, \forall y \in E\_2, \ g\_L^{\rightarrow}(A)(y) = \bigvee\_{\emptyset(x) = y} A(x).$$

*<sup>g</sup>*<sup>←</sup>*L*: *LE*2 −→ *LE*1 is induced by the mapping *g* as follows:

$$\forall B \in L^{E\_2}, \forall \mathfrak{x} \in E\_1, \ g\_L^{\leftarrow}(B)(\mathfrak{x}) = B(\mathfrak{g}(\mathfrak{x})).$$

**Definition 3** ([36])**.** *Let E be an effect algebra and A* ∈ *LE. For any λ* ∈ *L, define <sup>A</sup>*[*λ*] = {*x* ∈ *E* | *<sup>A</sup>*(*x*) ≥ *<sup>λ</sup>*}, *A*(*λ*) = {*x* ∈ *E* | *<sup>A</sup>*(*x*) *<sup>λ</sup>*}*,*

*<sup>A</sup>*(*λ*) = {*x* ∈ *E* | *λ* ∈ *β*(*A*(*x*))}, *A*[*λ*] = {*x* ∈ *E* | *λ* ∈/ *<sup>α</sup>*(*A*(*x*))}*.*

The right adjoint → of the meet operation ∧ is a mapping from *L* × *L* to *L* defined as

$$
\lambda \to \mu = \bigvee \{ \delta \in L | \lambda \land \delta \le \mu \}.
$$

Some basic properties of the operation → are listed in the following [37,38].

(1) *λ* → *μ* = *L* ⇔ *λ* ≤ *μ*;


Let ⊥, ∈ *L<sup>E</sup>* represent <sup>⊥</sup>(*x*) = ⊥*L* and (*x*) = *L* for all *x* ∈ *E*. Next, we recall (*<sup>L</sup>*, *M*)-fuzzy convexities in [27], which are more general fuzzy convexities. Let *L* and *M* be two completely distributive lattices. An (*<sup>L</sup>*, *M*)-fuzzy convexity on an effect algebra *E* is defined as follows:

**Definition 4** ([27])**.** *A mapping* C : *L<sup>E</sup>* −→ *M is called an* (*<sup>L</sup>*, *M*)*-fuzzy convexity if it satisfies the following conditions:*

(LMC1) C(⊥) = C() = *M;*

(LMC2) *if* {*Ai* | *i* ∈ *I*} ⊆ *L<sup>E</sup> is nonempty, then* C*<sup>i</sup>*∈*<sup>I</sup>Ai*≥ *<sup>i</sup>*∈*<sup>I</sup>*C(*Ai*)*;*

(LMC3) *if* {*Ai* | *i* ∈ *I*} ⊆ *L<sup>E</sup> is nonempty and directed, then* C*<sup>i</sup>*∈*<sup>I</sup> Ai*≥ *<sup>i</sup>*∈*<sup>I</sup>* C(*Ai*)*.*

*In this case, the pair* (*<sup>E</sup>*, C) *is called an* (*<sup>L</sup>*, *M*)*-fuzzy convex space. An* (*<sup>L</sup>*, *L*)*-fuzzy convexity is briefly called an L-fuzzy convexity.*

An (*<sup>L</sup>*, **2**)-fuzzy convexity is an *L*-convexity in [15]. An (*I*, **2**)-fuzzy convexity is a fuzzy convexity in [14], where *I* = [0, 1]. A (**<sup>2</sup>**, *M*)-fuzzy convexity is an *M*-fuzzifying convexity in [22]. A (**<sup>2</sup>**, **2**)-fuzzy convexity is a convexity in [7].

**Definition 5** ([27])**.** *Let* (*<sup>E</sup>*, C) *and* (*<sup>F</sup>*, D) *be two* (*<sup>L</sup>*, *M*)*-fuzzy convex spaces. If g* : *E* −→ *F is a mapping between E and F, then*

(i) *g* : (*<sup>E</sup>*, C) −→ (*<sup>F</sup>*, D) *is called an* (*<sup>L</sup>*, *M*)*-fuzzy convexity preserving mapping provided that* <sup>C</sup>(*g*<sup>←</sup>*L* (*A*)) ≥ D(*A*) *for all A* ∈ *LF.*

(ii) *g* : (*<sup>E</sup>*, C) −→ (*<sup>F</sup>*, D) *is called an* (*<sup>L</sup>*, *M*)*-fuzzy convex-to-convex mapping provided that* <sup>D</sup>(*g*<sup>→</sup>*L*(*B*)) ≥ C(*B*) *for all B* ∈ *LE.*

*An* (*<sup>L</sup>*, *L*)*-fuzzy convexity preserving mapping and an* (*<sup>L</sup>*, *L*)*-fuzzy convex-to-convex mapping are briefly called an L-fuzzy convexity preserving mapping and an L-fuzzy convex-to-convex mapping, respectively.*

An (*<sup>L</sup>*, **2**)-fuzzy convexity preserving mapping is an *L*-convexity preserving mapping in [15]. An (*<sup>L</sup>*, **2**)-fuzzy convex-to-convex mapping is an *L*-convex-to-convex mapping in [15].

**Definition 6** ([1])**.** *Let* {*Ei*}*<sup>i</sup>*∈<sup>Λ</sup> *be a family of effect algebras, define* ∏ *Ei as:*

$$\prod\_{i \in \Lambda} E\_i = \{ \mathbf{x} \mid \mathbf{x} : \Lambda \longrightarrow \bigcup\_{i \in \Lambda} E\_i \text{ s.t.}\\ \forall i \in \Lambda, \mathbf{x}\_i = \mathbf{x}(i) \in E\_i \}.$$

*i*∈Λ

*Define the operation* ⊕ *on* ∏ *i*∈Λ *Ei as: for x*, *y* ∈ ∏ *i*∈Λ *Ei, x* ⊥ *y iff xi* ⊥ *yi for all i* ∈ Λ*. In this case,* (*x* ⊕ *y*)(*i*) = *xi* ⊕ *yi and x*(*i*) = *xi. Further,* **0***i* = 0*i*, **1***i* = 1*i where* 0*i and* 1*i are the minimal element and the maximal element of Ei. Then* ( ∏ *i*∈Λ *Ei*, ⊕, **0**, **1**) *is called the direct product ofeffectalgebras.*

It is easy to check that ( ∏ *i*∈Λ *Ei*, ⊕, **0**, **1**) is an effect algebra.

**Definition 7** ([39])**.** *Let* {*Ei*}*ni*=<sup>1</sup> *be a family of effect algebras and Ai be an L-subset of Ei for all i* ∈ Λ*, then the L-subset* ∏ *i*∈Λ *Ai of* ∏ *i*∈Λ *Ei is defined by* ( ∏ *i*∈Λ *Ai*)(*x*) = *i*∈Λ *Ai*(*xi*)*.*

#### **3.** *L***-Fuzzy Subalgebra Degree on Effect Algebras**

In this section, we introduce the concept of *L*-fuzzy subalgebra degree on effect algebras by means of the implication operator of *L*. We define an *L*-fuzzy subalgebra provided that its *L*-fuzzy subalgebra degree is equal to *L*. Moreover, we give some characterizations of *L*-subalgebra degree in terms of four kinds of cut sets of *L*-subsets.

**Definition 8.** *Let E be an effect algebra and A* ∈ *LE. Then the L-fuzzy subalgebra degree* E(*A*) *of A is defined as:*

$$\mathcal{E}(A) = \bigwedge\_{\mathbf{x}, \mathbf{y} \in \mathbf{E}, \mathbf{x} \perp \mathbf{y}} \{ [A(\mathbf{x}) \to A(\mathbf{0})] \land [A(\mathbf{x}) \to A(\mathbf{x'})] \land [(A(\mathbf{x}) \land A(\mathbf{y})) \to A(\mathbf{x} \oplus \mathbf{y})] \}.$$

**Example 1.** *Let E* = {0, *x*, *y*, 1} *and* ⊕ *be given by:*


*Then* (*<sup>E</sup>*, ⊕, 0, 1) *is an effect algebra. Let L* = [0, 1] *and define the L-subsets of E as follows:*

(i) *<sup>A</sup>*1(0) = 0.8, *<sup>A</sup>*1(*x*) = 0.5, *<sup>A</sup>*1(*y*) = 0.5, *<sup>A</sup>*1(1) = 0.8*. By Definition 8, we can obtain* E(*<sup>A</sup>*1) = 1*.*

*Indeed, since for any z* ∈ *E, <sup>A</sup>*1(*z*) ≤ *<sup>A</sup>*1(0)*, we have <sup>A</sup>*1(*z*) → *<sup>A</sup>*1(0) = 1 *for all z* ∈ *E. We can easily know x* = *x*, *y* = *y*, 0 = 1 *and* 1 = 0*, it follows from <sup>A</sup>*1(0) = *<sup>A</sup>*1(1) *that <sup>A</sup>*1(*z*) = *<sup>A</sup>*1(*z*)*, so <sup>A</sup>*1(*z*) → *<sup>A</sup>*1(*z*) = 1 *for all z* ∈ *E. We can routinely prove* (*A*(*z*) ∧ *<sup>A</sup>*(*w*)) → *<sup>A</sup>*(*z* ⊕ *w*) = 1 *for all z*, *w* ∈ *E with z* ⊥ *w. Therefore,* E(*<sup>A</sup>*1) = 1*.*


From the properties (3) of the implication operator, the following lemma is obvious.

**Lemma 1.** *Let E be an effect algebra and A* ∈ *LE. Then* E(*A*) ≥ *λ (λ* ∈ *L) if and only if for any x*, *y* ∈ *E satisfying x* ⊥ *y,*

$$A(\mathbf{x}) \land A(\mathbf{y}) \land \lambda \le A(\mathbf{x} \oplus \mathbf{y}),\\ A(\mathbf{x}) \land \lambda \le A(\mathbf{x'}) \text{ and } A(\mathbf{x}) \land \lambda \le A(\mathbf{0}).$$

By Lemma 1, we can obtain the following theorem.

**Theorem 1.** *Let E be an effect algebra and A* ∈ *LE. Then*

=

E(*A*) = {*λ* ∈ *L* | ∀*<sup>x</sup>*, *y* ∈ *E s*.*t*. *x* ⊥ *y*, *<sup>A</sup>*(*x*) ∧ *<sup>A</sup>*(*y*) ∧ *λ* ≤ *<sup>A</sup>*(*x* ⊕ *y*), *<sup>A</sup>*(*x*) ∧ *λ* ≤ *<sup>A</sup>*(*x*) ∧ *<sup>A</sup>*(0)}.

> We use four kinds of cut sets of *L*-subsets to characterize the *L*-fuzzy subalgebra degree in the following Theorem.

**Theorem 2.** *Let E be an effect algebra and A* ∈ *LE. Then*


**Proof.** (i) For any *x*, *y* ∈ *E* satisfying *x* ⊥ *y*, assume that *λ* ∈ *L* with the property of *<sup>A</sup>*(*x*) ∧ *<sup>A</sup>*(*y*) ∧ *λ* ≤ *<sup>A</sup>*(*x* ⊕ *y*) and *<sup>A</sup>*(*x*) ∧ *λ* ≤ *<sup>A</sup>*(*x*) ∧ *<sup>A</sup>*(0). For any *μ* ≤ *λ*, suppose *<sup>A</sup>*[*μ*] = ∅ and let *x*, *y* ∈ *<sup>A</sup>*[*μ*] with *x* ⊥ *y*, then

$$
\mu = \mu \land \lambda \le A(\mathbf{x}) \land A(\mathbf{y}) \land \lambda \le A(\mathbf{x} \oplus \mathbf{y}) \text{ and } \mu = \mu \land \lambda \le A(\mathbf{x}) \land \lambda \le A(\mathbf{x'}) \land A(\mathbf{0}),
$$

which implies *x* ⊕ *y* ∈ *<sup>A</sup>*[*μ*] and *<sup>x</sup>*, 0 ∈ *<sup>A</sup>*[*μ*]. Hence *<sup>A</sup>*[*μ*] is a subalgebra of *E*. This gives that

$$\mathcal{E}(A) \le \bigvee \{ \lambda \in L \mid \forall \mu \le \lambda, A\_{[\mu]} = \mathcal{Q} \text{ or it is a subalgebra of } E \}.$$

Conversely, assume *λ* ∈ *L* and for each *μ* ≤ *λ*, *<sup>A</sup>*[*μ*] = ∅ or it is a subalgebra of *E*. For any *x*, *y* ∈ *E* satisfying *x* ⊥ *y*, let *μ* = *<sup>A</sup>*(*x*) ∧ *<sup>A</sup>*(*y*) ∧ *λ* and *γ* = *<sup>A</sup>*(*x*) ∧ *λ*, then *μ* ≤ *λ* and *γ* ≤ *λ*. It follows that *x*, *y* ∈ *<sup>A</sup>*[*μ*] and *x* ∈ *<sup>A</sup>*[*γ*]. Since *<sup>A</sup>*[*μ*] and *<sup>A</sup>*[*γ*] are subalgebras of *E*, we have *x* ⊕ *y* ∈ *<sup>A</sup>*[*μ*] and *<sup>x</sup>*, 0 ∈ *<sup>A</sup>*[*γ*], i.e., *<sup>A</sup>*(*x*) ∧ *<sup>A</sup>*(*y*) ∧ *λ* ≤ *<sup>A</sup>*(*x* ⊕ *y*) and *<sup>A</sup>*(*x*) ∧ *λ* ≤ *<sup>A</sup>*(*x*) ∧ *<sup>A</sup>*(0). So we have

$$\mathcal{E}(A) \ge \bigvee \{ \lambda \in L \mid \forall \mu \le \lambda, A\_{[\mu]} = \mathcal{Q} \text{ or it is a subalgebra of } E \}.$$

(ii) For any *x*, *y* ∈ *E* satisfying *x* ⊥ *y*, assume that *λ* ∈ *L* with the property of *<sup>A</sup>*(*x*) ∧ *<sup>A</sup>*(*y*) ∧ *λ* ≤ *<sup>A</sup>*(*x* ⊕ *y*) and *<sup>A</sup>*(*x*) ∧ *λ* ≤ *<sup>A</sup>*(*x*) ∧ *<sup>A</sup>*(0). For any *μ* ∈ *P*(*L*) with *μ λ*, suppose *A*(*μ*) = ∅ and let *x*, *y* ∈ *A*(*μ*) with *x* ⊥ *y*, i.e., *<sup>A</sup>*(*x*) *μ* and *<sup>A</sup>*(*y*) *μ*, then *<sup>A</sup>*(*x*) ∧ *<sup>A</sup>*(*y*) ∧ *λ μ* since *μ* is prime. By *<sup>A</sup>*(*x*) ∧ *<sup>A</sup>*(*y*) ∧ *λ* ≤ *<sup>A</sup>*(*x* ⊕ *y*) and *<sup>A</sup>*(*x*) ∧ *λ* ≤

*<sup>A</sup>*(*x*) ∧ *<sup>A</sup>*(0), it follows that *<sup>A</sup>*(*x* ⊕ *y*) *μ* and *<sup>A</sup>*(*x*) ∧ *A*(0) *μ*, i.e., *x* ⊕ *y*, *<sup>x</sup>*, 0 ∈ *A*(*μ*). Hence *A*(*μ*) is a subalgebra of *E*. This shows that

$$\mathcal{E}(A) \le \bigvee \{ \lambda \in L \mid \forall \mu \in P(L), \mu \not\supseteq \lambda, A^{(\mu)} = \bigotimes \text{ or it is a subalgebra of } E \}.$$

Conversely, assume that *λ* ∈ *L* and *A*(*μ*) = ∅ or it is a subalgebra of *E* for *μ* ∈ *P*(*L*) with *μ λ*. For any *x*, *y* ∈ *E* satisfying *x* ⊥ *y*, let *μ* ∈ *P*(*L*) and *<sup>A</sup>*(*x*) ∧ *<sup>A</sup>*(*y*) ∧ *λ μ*, then *x*, *y* ∈ *A*(*μ*) and *λ μ*. Since *A*(*μ*) is a subalgebra of *E*, it holds that *x* ⊕ *y* ∈ *<sup>A</sup>*(*μ*), i.e., *<sup>A</sup>*(*x* ⊕ *y*) *μ*. Hence *<sup>A</sup>*(*x*) ∧ *<sup>A</sup>*(*y*) ∧ *λ* ≤ *<sup>A</sup>*(*x* ⊕ *y*). Similarly, we can prove *<sup>A</sup>*(*x*) ∧ *λ* ≤ *<sup>A</sup>*(*x*) ∧ *<sup>A</sup>*(0). This shows

$$\mathcal{E}(A) \ge \bigvee \{ \lambda \in L \mid \forall \mu \in P(L), \mu \not\supseteq \lambda, A^{(\mu)} = \bigotimes \text{ or it is a subalgebra of } E \}.$$

(iii) For any *x*, *y* ∈ *E* satisfying *x* ⊥ *y*, assume that *λ* ∈ *L* with the property of *<sup>A</sup>*(*x*) ∧ *<sup>A</sup>*(*y*) ∧ *λ* ≤ *<sup>A</sup>*(*x* ⊕ *y*) and *<sup>A</sup>*(*x*) ∧ *λ* ≤ *<sup>A</sup>*(*x*) ∧ *<sup>A</sup>*(0). For any *μ* ∈/ *<sup>α</sup>*(*λ*), suppose *A*[*μ*] = ∅ and let *x*, *y* ∈ *A*[*μ*] with *x* ⊥ *y*, i.e., *μ* ∈/ *α*(*A*(*x*)) and *μ* ∈/ *<sup>α</sup>*(*A*(*y*)), then by *μ* ∈/ *<sup>α</sup>*(*λ*), we have *μ* ∈/ *α*(*A*(*x*)) ∪ *α*(*A*(*y*)) ∪ *<sup>α</sup>*(*λ*). From

$$a(A(\mathbf{x})) \cup a(A(\mathbf{y})) \cup a(\lambda) = a(A(\mathbf{x}) \wedge A(\mathbf{y}) \wedge \lambda) \text{ and } a(A(\mathbf{x})) \cup a(\lambda) = a(A(\mathbf{x}) \wedge \lambda),$$

we know *μ* ∈/ *α*(*A*(*x*) ∧ *<sup>A</sup>*(*y*) ∧ *λ*) and *μ* ∈/ *α*(*A*(*x*) ∧ *<sup>λ</sup>*). By *<sup>A</sup>*(*x*) ∧ *<sup>A</sup>*(*y*) ∧ *λ* ≤ *<sup>A</sup>*(*x* ⊕ *y*) and *<sup>A</sup>*(*x*) ∧ *λ* ≤ *<sup>A</sup>*(*x*) ∧ *<sup>A</sup>*(0), we have *μ* ∈/ *α*(*A*(*x* ⊕ *y*)), *μ* ∈/ *α*(*A*(*x*)) and *μ* ∈/ *<sup>α</sup>*(*A*(0)), i.e., *x* ⊕ *y*, *<sup>x</sup>*, 0 ∈ *A*[*μ*]. Hence *A*[*μ*] is a subalgebra of *E*. This shows

> E(*A*) ≤ {*λ* ∈ *L* | ∀*μ* ∈/ *<sup>α</sup>*(*λ*), *A*[*μ*] = ∅ *or it is a subalgebra o f <sup>E</sup>*}.

Conversely, assume that *λ* ∈ *L* and *A*[*μ*] = ∅ or it is a subalgebra of *E* for *μ* ∈/ *<sup>α</sup>*(*λ*). Now we prove *<sup>A</sup>*(*x*) ∧ *<sup>A</sup>*(*y*) ∧ *λ* ≤ *<sup>A</sup>*(*x* ⊕ *y*) and *<sup>A</sup>*(*x*) ∧ *λ* ≤ *<sup>A</sup>*(*x*) ∧ *A*(0) for all *x*, *y* ∈ *E* satisfying *x* ⊥ *y*. Let *μ* ∈ *L* and *μ* ∈/ *α*(*A*(*x*) ∧ *<sup>A</sup>*(*y*) ∧ *<sup>λ</sup>*), then *μ* ∈/ *α*(*A*(*x*)) ∪ *α*(*A*(*y*)) ∪ *<sup>α</sup>*(*λ*). Thus *x*, *y* ∈ *A*[*μ*] and *μ* ∈/ *<sup>α</sup>*(*λ*). Since *A*[*μ*] is a subalgebra of *E*, it holds that *x* ⊕ *y* ∈ *<sup>A</sup>*[*μ*], i.e., *μ* ∈/ *α*(*A*(*x* ⊕ *y*)). This means *α*(*A*(*x*) ∧ *<sup>A</sup>*(*y*) ∧ *λ*) ⊇ *α*(*A*(*x* ⊕ *y*)). Hence

$$A(\mathbf{x}) \wedge A(\mathbf{y}) \wedge \lambda = \bigwedge \mathfrak{a}(A(\mathbf{x}) \wedge A(\mathbf{y}) \wedge \lambda) \le \bigwedge \mathfrak{a}(A(\mathbf{x} \oplus \mathbf{y})) = A(\mathbf{x} \oplus \mathbf{y}).$$

Similarly, we can prove *<sup>A</sup>*(*x*) ∧ *λ* ≤ *<sup>A</sup>*(*x*) ∧ *<sup>A</sup>*(0). This shows

$$\mathcal{E}(A) \ge \bigvee \{ \lambda \in L \mid \forall \mu \notin a(\lambda), A^{\left[\mu\right]} = \bigotimes \text{ or it is a subalgebra of } E \}.$$

(iv) By (iii) we first obtain

$$\mathcal{E}(A) \le \bigvee \{ \lambda \in L \mid \forall \mu \in P(L), \mu \notin \mathfrak{a}(\lambda), A^{|\mu|} = \bigotimes \text{ or it is a subalgebra of } E \}.$$

Conversely, assume that *λ* ∈ *L* and *A*[*μ*] = ∅ or it is a subalgebra of *E* for all *μ* ∈ *P*(*L*) satisfying *μ* ∈/ *<sup>α</sup>*(*λ*). Now we prove *<sup>A</sup>*(*x*) ∧ *<sup>A</sup>*(*y*) ∧ *λ* ≤ *<sup>A</sup>*(*x* ⊕ *y*) and *<sup>A</sup>*(*x*) ∧ *λ* ≤ *<sup>A</sup>*(*x*) ∧ *A*(0) for all *x*, *y* ∈ *E* satisfying *x* ⊥ *y*. Let *μ* ∈ *P*(*L*) and *μ* ∈/ *α*<sup>∗</sup>(*A*(*x*) ∧ *<sup>A</sup>*(*y*) ∧ *<sup>λ</sup>*). It follows that

$$
\mu \notin \mathfrak{a}(A(\mathfrak{x}) \land A(\mathfrak{y}) \land \lambda) = \mathfrak{a}(A(\mathfrak{x})) \cup \mathfrak{a}(A(\mathfrak{y})) \cup \mathfrak{a}(\lambda).
$$

Thus *x*, *y* ∈ *A*[*μ*] and *μ* ∈/ *<sup>α</sup>*(*λ*). Since *A*[*μ*] is a subalgebra of *E*, it holds that *x* ⊕ *y* ∈ *<sup>A</sup>*[*μ*], i.e., *μ* ∈/ *α*(*A*(*x* ⊕ *y*)). So *μ* ∈/ *α*<sup>∗</sup>(*A*(*x* ⊕ *y*)), this means *α*<sup>∗</sup>(*A*(*x*) ∧ *<sup>A</sup>*(*y*) ∧ *λ*) ⊇ *α*<sup>∗</sup>(*A*(*x* ⊕ *y*)). Hence

$$A(\mathbf{x}) \wedge A(\mathbf{y}) \wedge \lambda = \bigwedge \mathfrak{a}^\*(A(\mathbf{x}) \wedge A(\mathbf{y}) \wedge \lambda) \le \bigwedge \mathfrak{a}^\*(A(\mathbf{x} \oplus \mathbf{y})) = A(\mathbf{x} \oplus \mathbf{y}).$$

Similarly, we can prove *<sup>A</sup>*(*x*) ∧ *λ* ≤ *<sup>A</sup>*(*x*) ∧ *<sup>A</sup>*(0). This shows

$$\mathcal{E}(A) \ge \bigvee \{ \lambda \in L \mid \forall \mu \in P(L), \mu \notin a(\lambda), A^{\left[\mu\right]} = \bigotimes \text{ or it is a subalgebra of } E \}.$$

(v) For any *x*, *y* ∈ *E* satisfying *x* ⊥ *y*, assume that ⊥*L* = *λ* ∈ *L* with the property of *<sup>A</sup>*(*x*) ∧ *<sup>A</sup>*(*y*) ∧ *λ* ≤ *<sup>A</sup>*(*x* ⊕ *y*) and *<sup>A</sup>*(*x*) ∧ *λ* ≤ *<sup>A</sup>*(*x*) ∧ *<sup>A</sup>*(0). For any *μ* ∈ *β*(*λ*), suppose *<sup>A</sup>*(*μ*) = ∅ and let *x*, *y* ∈ *<sup>A</sup>*(*μ*) with *x* ⊥ *y*, i.e., *μ* ∈ *β*(*A*(*x*)) and *μ* ∈ *β*(*A*(*y*)). Then by *μ* ∈ *β*(*λ*), we have *μ* ∈ *β*(*A*(*x*)) ∩ *β*(*A*(*y*)) ∩ *β*(*λ*). From

$$
\beta(A(\mathbf{x})) \cap \beta(A(\mathbf{y})) \cap \beta(\lambda) = \beta(A(\mathbf{x}) \land A(\mathbf{y}) \land \lambda) \text{ and } \beta(A(\mathbf{x})) \cap \beta(\lambda) = \beta(A(\mathbf{x}) \land \lambda),
$$

we know *μ* ∈ *β*(*A*(*x*) ∧ *<sup>A</sup>*(*y*) ∧ *λ*) and *μ* ∈ *β*(*A*(*x*) ∧ *<sup>λ</sup>*). By *<sup>A</sup>*(*x*) ∧ *<sup>A</sup>*(*y*) ∧ *λ* ≤ *<sup>A</sup>*(*x* ⊕ *y*) and *<sup>A</sup>*(*x*) ∧ *λ* ≤ *<sup>A</sup>*(*x*) ∧ *<sup>A</sup>*(0), we have *μ* ∈ *β*(*A*(*x* ⊕ *y*)), *μ* ∈ *β*(*A*(*x*)) and *μ* ∈ *β*(*A*(0)), i.e., *x* ⊕ *y*, *<sup>x</sup>*, 0 ∈ *<sup>A</sup>*(*μ*). Hence *<sup>A</sup>*(*μ*) is a subalgebra of *E*. This shows

$$\mathcal{E}(A) \le \bigvee \{ \lambda \in L \mid \forall \mu \in \mathcal{B}(\lambda), A\_{(\mu)} = \bigotimes \text{ or } \text{it is a subalgebra of } E \}.$$

Conversely, assume that *λ* ∈ *L* and *<sup>A</sup>*(*μ*) = ∅ or it is a subalgebra of *E* for *μ* ∈ *β*(*λ*). Now we prove *<sup>A</sup>*(*x*) ∧ *<sup>A</sup>*(*y*) ∧ *λ* ≤ *<sup>A</sup>*(*x* ⊕ *y*) and *<sup>A</sup>*(*x*) ∧ *λ* ≤ *<sup>A</sup>*(*x*) ∧ *A*(0) for all *x*, *y* ∈ *E* satisfying *x* ⊥ *y*. The statement holds obviously when *<sup>A</sup>*(*x*) ∧ *<sup>A</sup>*(*y*) ∧ *λ* = ⊥*L*. Assume *<sup>A</sup>*(*x*) ∧ *<sup>A</sup>*(*y*) ∧ *λ* = ⊥*L*, it implies *<sup>A</sup>*(*x*) = ⊥*L*, *<sup>A</sup>*(*y*) = ⊥*L* and *λ* = ⊥*L*. Take *μ* ∈ *L* with *μ* ∈ *β*(*A*(*x*) ∧ *<sup>A</sup>*(*y*) ∧ *<sup>λ</sup>*), then *μ* ∈ *β*(*A*(*x*)) ∩ *β*(*A*(*y*)) ∩ *β*(*λ*). Thus *x*, *y* ∈ *<sup>A</sup>*(*μ*) and *μ* ∈ *β*(*λ*). Since *<sup>A</sup>*(*μ*) is a subalgebra of *E*, it holds that *x* ⊕ *y* ∈ *<sup>A</sup>*(*μ*), i.e., *μ* ∈ *β*(*A*(*x* ⊕ *y*)). This means *β*(*A*(*x*) ∧ *<sup>A</sup>*(*y*) ∧ *λ*) ⊆ *β*(*A*(*x* ⊕ *y*)). Hence

$$A(\mathbf{x}) \land A(\mathbf{y}) \land \lambda = \bigvee \beta(A(\mathbf{x}) \land A(\mathbf{y}) \land \lambda) \le \bigvee \beta(A(\mathbf{x} \oplus \mathbf{y})) = A(\mathbf{x} \oplus \mathbf{y}) \dots$$

Similarly, we can prove *<sup>A</sup>*(*x*) ∧ *λ* ≤ *<sup>A</sup>*(*x*) ∧ *<sup>A</sup>*(0). This shows

$$\mathcal{E}(A) \ge \bigvee \{ \lambda \in L \mid \forall \mu \in \mathcal{B}(\lambda), A\_{(\mu)} = \bigotimes \text{ or it is a subalgebra of } E \}.$$

This completes the proof.

**Definition 9.** *Let E be an effect algebra and A* ∈ *LE. Then A is called an L-subalgebra provided that* E(*A*) = *L. In particular, we say an L-subalgebra is a fuzzy subalgebra when L* = [0, 1]*.*

**Example 2.** *In Example 1, it follows from Definition 9 that A*1 *is a fuzzy subalgebra, but A*2 *and A*3 *are not fuzzy subalgebras; A*2 *is a fuzzy subalgebra in the degree of 0.3.*

From Definition 9 and Lemma 1, the following proposition is obvious.

**Proposition 1.** *Let E be an effect algebra and A* ∈ *LE. Then A is an L-subalgebra if and only if for any x*, *y* ∈ *E with x* ⊥ *y,*

$$A(0) \ge A(\mathbf{x}),\\ A(\mathbf{x'}) \ge A(\mathbf{x}) \text{ and } A(\mathbf{x} \oplus \mathbf{y}) \ge A(\mathbf{x}) \land A(\mathbf{y}).$$

**Corollary 1.** *Let A be an L-subalgebra of an effect algebra E. Then*


**Proof.** (i) Since *x* ⊥ *x* and *A* is an *L*-subalgebra of *E*, we have

$$A(1) = A(\mathfrak{x} \oplus \mathfrak{x}') \supseteq A(\mathfrak{x}) \land A(\mathfrak{x}') = A(\mathfrak{x}).$$

(ii) By Proposition 1 we first obtain *<sup>A</sup>*(*x*) ≥ *<sup>A</sup>*(*x*). On the other hand, by Definition 1 (E3), we know *x* = *x* and thus *<sup>A</sup>*(*x*) = *<sup>A</sup>*(*x*) ≥ *<sup>A</sup>*(*x*).

(iii) By (i) we know *A*(1) ≥ *<sup>A</sup>*(0). By Proposition 1, we have *A*(1) ≤ *<sup>A</sup>*(0), so *A*(1) = *<sup>A</sup>*(0).

By Definition 9 and Theorem 2, we can obtain the following two results

**Corollary 2.** *Let A be an L-subset of an effect algebra E. Then the following conditions are equivalent:*


**Corollary 3.** *Let A be an L-subset of an effect algebra E. If β*(*λ* ∧ *μ*) = *β*(*λ*) ∩ *β*(*μ*) *for all λ*, *μ* ∈ *L, then the following conditions are equivalent:*


Let *E* be an effect algebras and (2*<sup>E</sup>*)*<sup>L</sup>* represents the set of all mapping *H* : *L* −→ 2*E*. Then we have the following definitions.

**Definition 10** ([36])**.** *Let H* ∈ (2*<sup>E</sup>*)*L.*


By Theorem 2.3 in [36], we can prove the following two theorems.

**Theorem 3.** *Let L be completely distributive and let* {*A*(*λ*) | *λ* ∈ *L*}) *(A*(*λ*) = ∅*) be an Lα-nest of subalgebras on an effect algebra E. Then there exists an L-subalgebra B such that*


**Theorem 4.** *Let L be completely distributive and let* {*A*(*λ*) | *λ* ∈ *L*} *(A*(*λ*) = ∅*) be an <sup>L</sup>β-nest of subalgebras on an effect algebra E. Then there exists an L-subalgebra B such that*


In particular, when *L* = [0, 1], we have the following result.

**Corollary 4.** *Let* {*A*(*λ*) | *λ* ∈ (0, 1]} *(A*(*λ*) = ∅*) be a family of subalgebras on an effect algebra E. If λ* < *μ* ⇒ *<sup>A</sup>*(*μ*) ⊆ *<sup>A</sup>*(*λ*)*, then there exists a fuzzy subalgebra B satisfying*


#### **4.** *L***-Fuzzy Convexity on Effect Algebras**

In this section, we study the relationship between the *L*-fuzzy subalgebra degree and *L*-fuzzy convexity on an effect algebra. Further, we prove that a morphism between two effect algebras is an *L*-fuzzy convexity preserving mapping and a monomorphism is an *L*-fuzzy convex-to-convex mapping. Finally, we prove that the set of all *L*-subalgebras on an effect algebra is an *L*-convexity. For instance, we give its *L*-convex hull formula.

For each *A* ∈ *<sup>L</sup>E*, E can be naturally considered as a mapping E : *L<sup>E</sup>* −→ *L* defined by *A* \$−→ E(*A*). The following theorem will show that E is an *L*-fuzzy convexity on an effect algebra *E*.

**Theorem 5.** *Let E be an effect algebra. Then* E : *L<sup>E</sup>* −→ *L is an L-fuzzy convexity on E.*

**Proof. (LMC1)**. It is clear that E(⊥) = E() = *L* by Definition 8.

**(LMC2)**. Let {*Ai*}*<sup>i</sup>*∈*<sup>I</sup>* ⊆ *L<sup>E</sup>* be nonempty. Now we show E(*<sup>i</sup>*∈*<sup>I</sup> Ai*) ≥ *<sup>i</sup>*∈*<sup>I</sup>* E(*Ai*). Let *λ* ∈ *L* with *λ* ≤ *<sup>i</sup>*∈*<sup>I</sup>* E(*Ai*). Then for any *i* ∈ *I*, we have *λ* ≤ E(*Ai*), which implies *Ai*(*x*) ∧ *Ai*(*y*) ∧ *λ* ≤ *Ai*(*x* ⊕ *y*) and *Ai*(*x*) ∧ *λ* ≤ *Ai*(*x*) ∧ *Ai*(0) for all *x*, *y* ∈ *E* satisfying *x* ⊥ *y*. It follows that

$$
\bigwedge\_{i \in I} A\_i(\mathbf{x}) \land \bigwedge\_{i \in I} A\_i(y) \land \lambda \le \bigwedge\_{i \in I} A\_i(\mathbf{x} \oplus y),
$$

$$
\bigwedge\_{i \in I} A\_i(\mathbf{x}) \land \lambda \le \bigwedge\_{i \in I} A\_i(\mathbf{x'}) \land \bigwedge\_{i \in I} A\_i(\mathbf{0}).
$$

This gives that *λ* ≤ E(*<sup>i</sup>*∈*<sup>I</sup> Ai*). Hence E(*<sup>i</sup>*∈*<sup>I</sup> Ai*) ≥ *<sup>i</sup>*∈*<sup>I</sup>* E(*Ai*).

**(LMC3)** Let {*Ai*}*<sup>i</sup>*∈*<sup>I</sup>* ⊆ *L<sup>E</sup>* be nonempty and directed. Now, we show E(*<sup>i</sup>*∈*<sup>I</sup> Ai*) ≥ *<sup>i</sup>*∈*<sup>I</sup>* E(*Ai*). Let *λ* ∈ *L* with *λ* ≤ *<sup>i</sup>*∈*<sup>I</sup>* E(*Ai*). Then for any *i* ∈ *I*, *λ* ≤ E(*Ai*) implies *Ai*(*x*) ∧ *Ai*(*y*) ∧ *λ* ≤ *Ai*(*x* ⊕ *y*) and *Ai*(*x*) ∧ *λ* ≤ *Ai*(*x*) ∧ *Ai*(0) for all *x*, *y* ∈ *E* satisfying *x* ⊥ *y*. Now, we prove

$$\bigvee\_{i \in I} A\_i(x) \land \bigvee\_{i \in I} A\_i(y) \land \lambda \le \bigvee\_{i \in I} A\_i(x \oplus y).$$

Take *μ* ∈ *L* with

$$\mu \prec \bigvee\_{i \in I} A\_i(\mathfrak{x}) \land \bigvee\_{i \in I} A\_i(y) \land \lambda\_{\star}$$

then there exists *i*, *j* ∈ *I* such that *μ* ≤ *Ai*(*x*), *μ* ≤ *Aj*(*y*) and *μ* ≤ *λ*. Since {*Ai*}*<sup>i</sup>*∈*<sup>I</sup>* is directed, there exists *i*0 ∈ *I* such that *Ai* ≤ *Ai*0 and *Aj* ≤ *Ai*0 , we thus have *μ* ≤ *Ai*0 (*x*) ∧ *Ai*0 (*y*) ∧ *λ*. By the fact that *Ai*0 (*x*) ∧ *Ai*0 (*y*) ∧ *λ* ≤ *Ai*0 (*x* ⊕ *y*), so *μ* ≤ *Ai*0 (*x* ⊕ *y*) ≤ *<sup>i</sup>*∈*<sup>I</sup> Ai*(*x* ⊕ *y*). Hence

$$\bigvee\_{i \in I} A\_i(\mathfrak{x}) \land \bigvee\_{i \in I} A\_i(y) \land \lambda \le \bigvee\_{i \in I} A\_i(\mathfrak{x} \oplus y).$$

Similarly, we can prove

$$\bigvee\_{i \in I} A\_i(\mathbf{x}) \land \lambda \le \bigvee\_{i \in I} A\_i(\mathbf{x'}) \land \bigvee\_{i \in I} A\_i(\mathbf{0}).$$

Hence *λ* ≤ E(*<sup>i</sup>*∈*<sup>I</sup> Ai*), it implies E(*<sup>i</sup>*∈*<sup>I</sup> Ai*) ≥ *<sup>i</sup>*∈*<sup>I</sup>* E(*Ai*). Therefore, E : *L<sup>E</sup>* −→ *L* is an *L*-fuzzy convexity.

In order to investigate *L*-fuzzy convexity preserving mappings and *L*-fuzzy convexto-convex mappings, we first give following definition and lemma.

**Definition 11** ([40])**.** *Let E and F be effect algebras and g: E* −→ *F is called*


**Lemma 2.** *Let E and F be effect algebras and g: E* −→ *F is a morphism. Then g*(<sup>0</sup>*E*) = 0*F and g*(*x*) = *g*(*x*) *for all x* ∈ *E*

**Proof.** Since *g*(<sup>0</sup>*E* ⊕ <sup>1</sup>*E*) = *g*(<sup>0</sup>*E*) ⊕ *g*(<sup>1</sup>*E*) = *g*(<sup>0</sup>*E*) ⊕ 1*F* by Definition 11, it follows from Definition 1 (E3) that *g*(<sup>0</sup>*E*) = 0*F*. Since *x* ⊥ *x* for any *x* ∈ *E*, we have *g*(<sup>1</sup>*E*) = *g*(*x* ⊕ *x*) = *g*(*x*) ⊕ *g*(*x*), thus *g*(*x*) = *g*(*x*).

**Theorem 6.** *Let g* : *E* −→ *F be a morphism between two effect algebras E and F. Then*


**Proof.** (i) In order to prove *g*: (*<sup>E</sup>*, E*E*) −→ (*<sup>M</sup>*, E*F*) is *L*-fuzzy convexity preserving, we just need to prove for any *A* ∈ *<sup>L</sup>F*, <sup>E</sup>*E*(*g*<sup>←</sup>*L* (*A*)) ≥ E*F*(*A*).

Let *λ* ∈ *L* with *λ* ≤ E*F*(*A*). Then by Lemma 1 we obtain *<sup>A</sup>*(*x*) ∧ *<sup>A</sup>*(*y*) ∧ *λ* ≤ *<sup>A</sup>*(*x* ⊕ *y*) and *<sup>A</sup>*(*x*) ∧ *λ* ≤ *<sup>A</sup>*(*x*) ∧ *A*(0) for all *x*, *y* ∈ *F* satisfying *x* ⊥ *y*. So for any *a*, *b* ∈ *E* with *a* ⊥ *b*,

$$\begin{aligned} \lambda \wedge \operatorname{g}\_L^{\leftarrow}(A)(a) \wedge \operatorname{g}\_L^{\leftarrow}(A)(b) &= \lambda \wedge A(\operatorname{g}(a)) \wedge A(\operatorname{g}(b)) \\ &\leq \operatorname{A}(\operatorname{g}(a) \oplus \operatorname{g}(b)) \\ &= \operatorname{A}(\operatorname{g}(a \oplus b)) \\ &= \operatorname{g}\_L^{\leftarrow}(A)(a \oplus b). \end{aligned}$$

Similarly, we can obtain *λ* ∧ *<sup>g</sup>*<sup>←</sup>*L* (*A*)(*a*) ≤ *<sup>g</sup>*<sup>←</sup>*L* (*A*)(*a*) and *λ* ∧ *<sup>g</sup>*<sup>←</sup>*L* (*A*)(*a*) ≤ *<sup>g</sup>*<sup>←</sup>*L* (*A*)(<sup>0</sup>*E*), which implies *λ* ≤ <sup>E</sup>*E*(*g*<sup>←</sup>*L* (*A*)). Hence <sup>E</sup>*E*(*g*<sup>←</sup>*L* (*A*)) ≥ E*F*(*A*).

(ii) In order to prove *g*: (*<sup>E</sup>*, E*E*) −→ (*<sup>F</sup>*, E*F*) is *L*-fuzzy convex-to-convex, we just need to prove for any *A* ∈ *<sup>L</sup>E*, E*E*(*A*) ≤ <sup>E</sup>*F*(*g*<sup>→</sup>*L* (*A*)).

Let *λ* ∈ *L* with *λ* ≤ E*E*(*A*), then *<sup>A</sup>*(*a*) ∧ *A*(*b*) ∧ *λ* ≤ *<sup>A</sup>*(*a* ⊕ *b*) and *<sup>A</sup>*(*a*) ∧ *λ* ≤ *<sup>A</sup>*(*a*) ∧ *A*(0) for all *a*, *b* ∈ *E* satisfying *a* ⊥ *b*. It follows that for *x*, *y* ∈ *F* satisfying *x* ⊥ *y*,

$$\begin{aligned} (\mathcal{g}\_L^{\rightarrow}(A))(\mathbf{x}) \wedge (\mathcal{g}\_L^{\rightarrow}(A))(\mathbf{y}) \wedge \lambda &= \lambda \wedge \bigvee\_{\mathcal{g}(a)=\mathbf{x}} A(a) \wedge \bigvee\_{\mathcal{g}(b)=\mathbf{y}} A(b) \\ &= \bigvee \{\lambda \wedge A(a) \wedge A(b) \mid \mathbf{g}(a) = \mathbf{x}, \mathbf{g}(b) = \mathbf{y}\} \\ &\leq \bigvee \{A(a \oplus b) \mid \mathbf{g}(a) = \mathbf{x}, \mathbf{g}(b) = \mathbf{y}\} \\ &\leq \bigvee \{A(a \oplus b) \mid \mathbf{g}(a \oplus b) = \mathbf{x} \oplus \mathbf{y}\} \\ &\leq \bigvee \{A(c) \mid \mathbf{g}(c) = \mathbf{x} \oplus \mathbf{y}\} \\ &= (\mathcal{g}\_L^{\rightarrow}(A))(\mathbf{x} \oplus \mathbf{y}). \end{aligned}$$

Similarly, we can prove (*g*<sup>→</sup>*L* (*A*))(*x*) ∧ *λ* ≤ (*g*<sup>→</sup>*L* (*A*))(*x*) ∧ (*g*<sup>→</sup>*L* (*A*))(0). This gives that *λ* ≤ <sup>E</sup>*F*(*g*<sup>→</sup>*L* (*A*)). Hence E*E*(*A*) ≤ <sup>E</sup>*F*(*g*<sup>→</sup>*L* (*A*)). Therefore, *g* : (*<sup>E</sup>*, E*E*) −→ (*<sup>F</sup>*, E*F*) is *L*-fuzzy convex-to-convex.

**Corollary 5.** *Let* {*Ei*}*<sup>i</sup>*∈<sup>Λ</sup> *be a family of effect algebras and* ∏ *i*∈Λ *Ai be the direct product of* {*Ai*}*<sup>i</sup>*∈Λ*, where Ai* ∈ *LEi . If pi* : ∏ *i*∈Λ *Ei* −→ *Ei is the projection, then pi:* ( ∏ *i*∈Λ *Ei*, E ∏ *i*∈Λ *Ei*) −→ (*Ei*, <sup>E</sup>*Ei*) *is L-fuzzy convexity preserving for all i* ∈ Λ*.*

**Proof.** It is easy to check that *pi* : ∏ *i*∈Λ *Ei* −→ *Ei* is a morphism between effects algebras, so by Theorem 6, we know *pi* is *L*-fuzzy convexity preserving for all *i* ∈ Λ.

**Theorem 7.** *Let* {*Ei*}*<sup>i</sup>*∈<sup>Λ</sup> *be a family of effect algebras and* ∏ *i*∈Λ *Ai be the direct product of* {*Ai*}*<sup>i</sup>*∈Λ*, where Ai* ∈ *LEi . Then* E ∏ *i*∈Λ *Ei*- ∏*i*∈Λ *Ai* ≥ *i*∈Λ <sup>E</sup>*Ei*(*Ai*)*.*

**Proof.** Let *pi* : ∏ *i*∈Λ *Ei* −→ *Ei* be the projection. It can be easily proved that ∏ *i*∈Λ *Ai* = *i*∈Λ *<sup>p</sup>*<sup>←</sup>*i* (*Ai*). So by the proof of **(LMC2)** in Theorem 8, we have

$$\mathcal{E}\_{\prod\_{i \in \Lambda} E\_i} \left( \prod\_{i \in \Lambda} A\_i \right) = \mathcal{E}\_{\prod\_{i \in \Lambda} E\_i} \left( \bigwedge\_{i \in \Lambda} p\_i^{\leftarrow} (A\_i) \right) \geq \bigwedge\_{i \in \Lambda} \mathcal{E}\_{\prod\_{i \in \Lambda} E\_i} (p\_i^{\leftarrow} (A\_i)) \geq \bigwedge\_{i \in \Lambda} \mathcal{E}\_{\mathcal{E}\_i} (A\_i).$$

The last inequality holds because *pi* is *L*-fuzzy convexity preserving by Corollary 4.

From paper [27], we know (*<sup>E</sup>*, E*λ*) is an *L*-convex space for all *λ* ∈ *<sup>L</sup>*\⊥*<sup>L</sup>*. In particular, (*<sup>E</sup>*, E*L* ) is an *L*-convex space. We should note that E*L* = {*A* ∈ *L<sup>E</sup>* | E(*A*) = *L*}, by Definition 9, we obtain that E*L*is the set of all *L*-subalgebras on an effect algebra *E*.

The *L*-convex hull operator with respect to the *L*-convex space (*<sup>E</sup>*, E*L* ) is as follows: for *A* ∈ *<sup>L</sup>E*, *co*(*A*) = {*B* ∈ *L<sup>E</sup>* | *A* ≤ *B* ∈ E*L* } [17], i.e., *co*(*A*) is the least *L*-subalgebra containing *A*. In the following, we will give the corresponding *L*-convex hull formula.

**Theorem 8.** *Let A be an L-subset of an effect algebra E. Define for any x* ∈ *E,*

$$
\sigma\_0 = A\_\prime \sigma\_1(\mathbf{x}) = \left(\bigvee\_{\mathbf{x} = \mathbf{x}\_1 \oplus \mathbf{x}\_2} [A(\mathbf{x}\_1) \lor A(\mathbf{x}\_1')] \land [A(\mathbf{x}\_2) \lor A(\mathbf{x}\_2')]\right) \lor A(\mathbf{x}),
$$

$$
\text{when } n \ge 2,\ \sigma\_n(\mathbf{x}) = \bigvee\_{\mathbf{x} = \mathbf{x}\_1 \oplus \mathbf{x}\_2} [\sigma\_{n-1}(\mathbf{x}\_1) \lor \sigma\_{n-1}(\mathbf{x}\_1')] \land [\sigma\_{n-1}(\mathbf{x}\_2) \lor \sigma\_{n-1}(\mathbf{x}\_2')].
$$

$$
(A) \quad \text{and}
$$

*Then co*(*A*) = <sup>∞</sup>*n*=<sup>0</sup> *σn.*

**Proof.** We first show <sup>∞</sup>*n*=<sup>0</sup> *σn* is an *L*-subalgebra of *E*. To achieve this, we give the following statements.

(i). For any *x* ∈ *E* and *n* ≥ 1, *<sup>σ</sup>n*(1) ≥ *<sup>σ</sup>n*(*x*). Let *n* = 1. For any *x* ∈ *E*, it follows from *x* ⊕ *x* = 1 that

$$
\sigma\_1(1) \ge A(\mathbf{x}) \lor A(\mathbf{x'}) \ge A(\mathbf{x}) = \sigma\_0(\mathbf{x}).
$$

Let *n* ≥ 2. For any *x* ∈ *E*, since *x* ⊕ *x* = 1, we have that for each *x* ∈ *E*.

$$\begin{aligned} \sigma\_n(1) &= \bigvee\_{1-y \in z} [\sigma\_{n-1}(y) \vee \sigma\_{n-1}(y')] \wedge [\sigma\_{n-1}(z) \vee \sigma\_{n-1}(z')] \\ &\ge \ [\sigma\_{n-1}(x) \vee \sigma\_{n-1}(x')] \wedge [\sigma\_{n-1}(x) \vee \sigma\_{n-1}(x')] \\ &= \ \sigma\_{n-1}(x) \vee \sigma\_{n-1}(x'). \end{aligned}$$

Since for any *x* ∈ *E*,

$$\sigma\_n(\mathbf{x}) = \bigvee\_{\mathbf{x} = \mathbf{x}\_1 \oplus \mathbf{x}\_2} [\sigma\_{n-1}(\mathbf{x}\_1) \vee \sigma\_{n-1}(\mathbf{x}\_1')] \wedge [\sigma\_{n-1}(\mathbf{x}\_2) \vee \sigma\_{n-1}(\mathbf{x}\_2')],$$

then for every pair (*<sup>x</sup>*1, *<sup>x</sup>*2) satisfying *x* = *x*1 ⊕ *x*2, it holds that *<sup>σ</sup>n*−<sup>1</sup>(*<sup>x</sup>*1) ∨ *<sup>σ</sup>n*−<sup>1</sup>(*x*1) ≤ *<sup>σ</sup>n*(1) and *<sup>σ</sup>n*−<sup>1</sup>(*<sup>x</sup>*2) ∨ *<sup>σ</sup>n*−<sup>1</sup>(*x*2) ≤ *<sup>σ</sup>n*(1), we thus obtain *<sup>σ</sup>n*(*x*) ≤ *<sup>σ</sup>n*(1). (ii). <sup>∞</sup>*n*=<sup>0</sup> *<sup>σ</sup>n*(*x*) ≤ <sup>∞</sup>*n*=<sup>0</sup> *<sup>σ</sup>n*(1).

Since *σ*0 = *A* and 1 = 1 ⊕ 0, we have *<sup>σ</sup>*1(1) ≥ *A*(1) ∨ *A*(0) ≥ *<sup>A</sup>*(1). By the above proof, we obtain *<sup>σ</sup>*1(1) ≥ *<sup>A</sup>*(*x*) ∨ *<sup>A</sup>*(*x*) ≥ *<sup>A</sup>*(*x*). This implies

$$\bigvee\_{n=0}^{\infty} \sigma\_{\mathfrak{n}}(\mathbf{x}) = A(\mathbf{x}) \vee \bigvee\_{n=1}^{\infty} \sigma\_{\mathfrak{n}}(\mathbf{x}) \le \sigma\_1(\mathbf{1}) \vee \bigvee\_{n=1}^{\infty} \sigma\_{\mathfrak{n}}(\mathbf{1}) \le \bigvee\_{n=0}^{\infty} \sigma\_{\mathfrak{n}}(\mathbf{1}).$$

(iii). *<sup>σ</sup>*0(*x*) ≤ *<sup>σ</sup>*1(*x*) ≤ *<sup>σ</sup>*2(*x*) ≤···

For any *x* ∈ *E*, from the construction of *<sup>σ</sup>*1(*x*), it is clear that *<sup>σ</sup>*1(*x*) ≥ *<sup>A</sup>*(*x*) = *<sup>σ</sup>*0(*x*). Let *n* ≥ 2. For any *x* ∈ *E*, by the fact that *x* = *x* ⊕ 0 and (i), we have

$$\begin{split} \sigma\_{\mathfrak{n}}(\mathbf{x}) &= \bigvee\_{\mathbf{x} = \mathbf{x}\_{1} \oplus \mathbf{x}\_{2}} [\sigma\_{n-1}(\mathbf{x}\_{1}) \vee \sigma\_{n-1}(\mathbf{x}\_{1}')] \wedge [\sigma\_{n-1}(\mathbf{x}\_{2}) \vee \sigma\_{n-1}(\mathbf{x}\_{2}')] \\ &\geq \ [\sigma\_{n-1}(\mathbf{x}) \vee \sigma\_{n-1}(\mathbf{x}')] \wedge [\sigma\_{n-1}(\mathbf{0}) \vee \sigma\_{n-1}(\mathbf{1})] \\ &\geq \ \sigma\_{n-1}(\mathbf{x}) .\end{split}$$

(iv). <sup>∞</sup>*n*=<sup>0</sup> *<sup>σ</sup>n*(*x*) ≤ <sup>∞</sup>*n*=<sup>0</sup> *<sup>σ</sup>n*(*x*). For any *x* ∈ *E* and *n* ≥ 2, since *x* = *x* ⊕ 0, we have

$$
\sigma\_n(\mathbf{x'}) \ge \left[ \sigma\_{n-1}(\mathbf{x}) \lor \sigma\_{n-1}(\mathbf{x'}) \right] \land \left[ \sigma\_{n-1}(\mathbf{0}) \lor \sigma\_{n-1}(\mathbf{1}) \right].
$$

By (i) we know that *<sup>σ</sup>n*−<sup>1</sup>(*x*), *<sup>σ</sup>n*−<sup>1</sup>(*x*) ≤ *<sup>σ</sup>n*−<sup>1</sup>(1), it follows that *<sup>σ</sup>n*(*x*) ≥ *<sup>σ</sup>n*−<sup>1</sup>(*x*). Thus <sup>∞</sup>*n*=<sup>1</sup> *<sup>σ</sup>n*(*x*) ≤ <sup>∞</sup>*n*=<sup>2</sup> *<sup>σ</sup>n*(*x*). By (iii) we obtain

$$\bigvee\_{n=0}^{\infty} \sigma\_{\mathfrak{n}}(\mathfrak{x}) = \bigvee\_{n=1}^{\infty} \sigma\_{\mathfrak{n}}(\mathfrak{x}) \le \bigvee\_{n=2}^{\infty} \sigma\_{\mathfrak{n}}(\mathfrak{x'}) = \bigvee\_{n=0}^{\infty} \sigma\_{\mathfrak{n}}(\mathfrak{x'}).$$

(v). For *x*, *y* ∈ *E* and *x* ⊥ *y*,

$$\bigvee\_{n=0}^{\infty} \sigma\_{\mathfrak{n}}(\mathfrak{x} \oplus y) \ge \left(\bigvee\_{n=0}^{\infty} \sigma\_{\mathfrak{n}}(\mathfrak{x})\right) \wedge \left(\bigvee\_{n=0}^{\infty} \sigma\_{\mathfrak{n}}(y)\right).$$

Let *t* ∈ *L* such that *t* ≺ -<sup>∞</sup>*n*=<sup>0</sup> *<sup>σ</sup>n*(*x*)∧ - <sup>∞</sup>*n*=<sup>0</sup> *<sup>σ</sup>n*(*y*), then we have *t* ≺ <sup>∞</sup>*n*=<sup>0</sup> *<sup>σ</sup>n*(*x*) and *t* ≺ <sup>∞</sup>*n*=<sup>0</sup> *<sup>σ</sup>n*(*y*). So there exist *i*, *k* ≥ 0 such that *t* ≤ *<sup>σ</sup>i*(*x*) and *t* ≤ *<sup>σ</sup>j*(*y*). Put *j* = max{*<sup>i</sup>*, *k*}, then by (iii) we obtain *t* ≤ *<sup>σ</sup>j*(*x*) and *t* ≤ *<sup>σ</sup>j*(*y*). It results that

$$\forall t \le \sigma\_j(\mathbf{x}) \land \sigma\_j(y) \le \left[\sigma\_j(\mathbf{x}) \lor \sigma\_j(\mathbf{x'})\right] \land \left[\sigma\_j(y) \lor \sigma\_j(y')\right] \le \sigma\_{j+1}(\mathbf{x} \oplus y) \le \bigvee\_{n=0}^{\infty} \sigma\_n(\mathbf{x} \oplus y).$$

$$\dots \qquad \dots \qquad \dots \qquad \dots$$

Hence <sup>∞</sup>*n*=<sup>0</sup> *<sup>σ</sup>n*(*<sup>x</sup>* ⊕ *y*) ≥ %<sup>∞</sup>*n*=<sup>0</sup> *σn*(*x*)(∧ %<sup>∞</sup>*n*=<sup>0</sup> *<sup>σ</sup>n*(*y*)(.(vi). <sup>∞</sup>*n*=<sup>0</sup> *<sup>σ</sup>n*(*x*) ≤ <sup>∞</sup>*n*=<sup>0</sup> *<sup>σ</sup>n*(0). By (ii) and (iv), we have,

$$\bigvee\_{n=0}^{\infty} \sigma\_n(0) = \bigvee\_{n=0}^{\infty} \sigma\_n(1') \ge \bigvee\_{n=0}^{\infty} \sigma\_n(1) \ge \bigvee\_{n=0}^{\infty} \sigma\_n(x).$$

Therefore, by (iii), (iv), (v), (vi) and Proposition 1, we can obtain that <sup>∞</sup>*n*=<sup>0</sup> *σn* is an *L*-subalgebra containing *A*. Next, we prove <sup>∞</sup>*n*=<sup>0</sup>*σn* is the least *L*-subalgebra containing *A*.

 Assume *B* ∈ *L<sup>E</sup>* is an *L*-subalgebra containing *A*, so *<sup>A</sup>*(*x*) ≤ *<sup>B</sup>*(*x*) for all *x* ∈ *E*. It is clear that *σ*0 ≤ *B*. Since

$$\begin{split} \sigma\_{1}(\mathbf{x}) &= \left(\bigvee\_{\mathbf{x} = \mathbf{x}\_{1} \oplus \mathbf{x}\_{2}} \left[A(\mathbf{x}\_{1}) \vee A(\mathbf{x}\_{1}')\right] \wedge \left[A(\mathbf{x}\_{2}) \vee A(\mathbf{x}\_{2}')\right]\right) \vee A(\mathbf{x}) \\ &\leq \left(\bigvee\_{\mathbf{x} = \mathbf{x}\_{1} \oplus \mathbf{x}\_{2}} \left[B(\mathbf{x}\_{1}) \vee B(\mathbf{x}\_{1}')\right] \wedge \left[B(\mathbf{x}\_{2}) \vee B(\mathbf{x}\_{2}')\right]\right) \vee B(\mathbf{x}) \\ &= \left(\bigvee\_{\mathbf{x} = \mathbf{x}\_{1} \oplus \mathbf{x}\_{2}} B(\mathbf{x}\_{1}) \wedge B(\mathbf{x}\_{2})\right) \vee B(\mathbf{x}) \qquad \text{(by Corollary 1 and } B \text{ is an } L\text{-subalgebra)} \\ &\leq \left(\bigvee\_{\mathbf{x} = \mathbf{x}\_{1} \oplus \mathbf{x}\_{2}} B(\mathbf{x}\_{1} \oplus \mathbf{x}\_{2})\right) \vee B(\mathbf{x}) = B(\mathbf{x}) \vee B(\mathbf{x}) = B(\mathbf{x}) .\end{split}$$

Assume *σk* ≤ *B* (*k* ≥ 2) holds. Now,

$$\begin{split} \sigma\_{k+1}(\mathbf{x}) &= \bigvee\_{\mathbf{x} = \mathbf{x}\_1 \oplus \mathbf{x}\_2} [\sigma\_k(\mathbf{x}\_1) \vee \sigma\_k(\mathbf{x}\_1')] \wedge [\sigma\_k(\mathbf{x}\_2) \vee \sigma\_k(\mathbf{x}\_2')] \\ &\leq \bigvee\_{\mathbf{x} = \mathbf{x}\_1 \oplus \mathbf{x}\_2} [B(\mathbf{x}\_1) \vee B(\mathbf{x}\_1')] \wedge [B(\mathbf{x}\_2) \vee B(\mathbf{x}\_2')] \\ &= \bigvee\_{\mathbf{x} = \mathbf{x}\_1 \oplus \mathbf{x}\_2} B(\mathbf{x}\_1) \wedge B(\mathbf{x}\_2) \\ &\leq \bigvee\_{\mathbf{x} = \mathbf{x}\_1 \oplus \mathbf{x}\_2} B(\mathbf{x}\_1 \oplus \mathbf{x}\_2) = B(\mathbf{x}). \end{split}$$

Thus for any *n* ≥ 0, it holds that *σn* ≤ *B*, which implies <sup>∞</sup>*n*=<sup>0</sup> *σn* ≤ *B*. This implies <sup>∞</sup>*n*=<sup>0</sup> *σn* is the least *L*-subalgebra containing *A*. Therefore, *co*(*A*) = <sup>∞</sup>*n*=<sup>0</sup> *σn*.
