In [16], we defined the notion of derivation on quantales as follows:

Let *d* be a mapping on a quantale *Q*, then d is a *derivation* on *Q*, if it satisfies the following conditions

$$d\left({\vee}_{i}{b}_{i}\right)={\vee}_{i}d\left({b}_{i}\right)\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\mathrm{a}\mathrm{n}\mathrm{d}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}d\left(a\mathrm{\&}b\right)=\left(a\mathrm{\&}d\left(b\right)\right)\vee \left(d\left(a\right)\mathrm{\&}b\right).$$for all *a, b* ∈ *Q* and {*b*_{i}} ⊆ *Q*.

According to the above definition, we obtain the following basic properties:

*Let d be a derivation on a quantale Q and a, b, c* ∈ *Q, we have*:

*b* ≤ *d*(*a*) *→*_{r} d(*a&b*)*, a* ≤ *d*(*b*) → _{l}*d*(*a&b*).

*a →*_{r} b ≤ *d*(*a*) *→*_{r} d(*b*)*, d*(*a →*_{r} b) ≤ *a →*_{r} d(*b*).

*a* → _{l}*b* ≤ *d*(*a*) → _{l}*d*(*b*)*, d*(*a* → _{l}*b*) ≤ *a* → _{l}*d*(*b*).

*If a* ≤ *b, then d*(*a*) ≤ *d*(*b*).

*If a&b = b* (*or b&a* = *b*) *and d*(*a*) ≥ *a, then d*(*b*) ≥ *b*.

*If c* ≤ *a →*_{r} b, then c ≤ *d*(*a*) *→*_{r} d(*b*) *and a* ≤ *d*(*c*) → _{l} *d(b)*.

*If a&a = a, then a* ≤ *d*(*a*) *→*_{r} d(*a*) *and a* ≤ *d*(*a*) → _{l}*d*(*a*).

*If Q* is *right (left) unital, then a&d*(⊤) ≤ *d*(*a*) (*d*(⊤)&*a* ≤ *d*(*a*)).

*If Q is right (left) unital and d*(1) *=* 1, *then a* ≤ *d*(*a*) and *d*(⊤) *=* ⊤.

*If Q is right* (*left*) *unital and d*(1) *=* 1, *then d*(*a&*⊤) *= d*(*a*)& =(*d*(⊤*&a)* = ⊤*&d*(*a*)).

The following corollary follows from Proposition 3.2 (5).

*Let d be a derivation on a right* (*left*) *unital quantale Q. If d*(1) ≥ 1, *then d*(*a*) ≥ *a for all a* ∈ *Q*.

*Let d be a derivation on a quantale Q*.

*d*(0) = 0.

*d*(⊤) *=* ⊤ *if Q is right* (*left*) *unital and d*(1) ≥ 1.

*Proof*. (1) By the definition of derivation and 0&*x* = 0 and *x*&0 = 0 for all *x* ∈ *Q*, we have *d*(0) = *d*(0&0) = (*d*(0)&0) ∨ (0&*d*(0)) = 0 ∨ 0 = 0.

(2) Since *Q* is right (left) unital and *d*(1) ≥ 1, by Corollary 3.3, we have *d*(⊤) ≥ ⊤. Since T is the top element, we get *d(*⊤) *=* ⊤. □

*Let d be a derivation on a quantale Q. If a* ∈ *Q is right* (*left, two*) *sided, then d*(*a*) *is right* (*left, two*) *sided*.

*Proof*. Since *a* is right sided, then *a*&⊤ ≤ *a*. By the Proposition 3.2 (4), we have *d*(*a*) ≥ *d*(*a*&⊤) *=* (*d*(*a)&* ⊤) ∨ (*a&d*(⊤)) ≥ *d*(*a*)& ⊤, so *d*(*a*) is right sided. □

*Let d be a derivation on a right unital quantale Q and d*(1) *=* 1. *If a* ∈ *Q is right* (*left, two*) *sided, then d*(*a*) *is strictly right* (*left, two*) *sided*.

*Proof*. Since *a* is right sided, then *a*&⊤ ≤ *a*. By the Proposition 3.2 (4), we have *d*(*a*) ≥ *d*(*a*&⊤) = (*d*(*a*)& ⊤) ∨ (*a*&*d*(⊤)) ≥ *d*(*a*)& ⊤ and *d*(*a*) = *d*(*a&*1) ≤ *d*(*a&*⊤). That *d*(*a*&⊤) = *d*(*a*)& ⊤ follows from Proposition 3.2 (10). From the above, we get *d*(*a*) *= d*(*a*)& ⊤ which implies that *d*(*a*) is strictly right sided. □

In the following, we give some definitions about derivation which are similar to those on quantic nucleus.

*Let d be a derivation on a quantale Q, a* ∈ *Q*.

*d is right sided iff d*(⊤*&a*) ≤ *d*(*a*).

*d is left sided iff d*(*a*&⊤) ≤ *d*(*a*).

*d is two sided iff d is both right and left sided*.

*d is idempotent iff d*(*a&a*) *= d*(*a*).

*Let Q be the complete lattice shown in Fig. 1 and the operations* & *on Q is shown in . It is straightforward to verify that* (*Q, &*) *is a quantale. We define d*_{1} *: Q* → *Q by d*_{1}(0) = 0, *d*_{1}(*a*) = *a, d*_{1}(⊤) = ⊤ *and d*_{2} : *Q → Q by* *d*_{2}(0) = 0, *d*_{2}(*a*) *= d*_{2}(⊤) = ⊤. *It is easy to prove that d1 and* d_{2} *are derivations on* (*Q*, &). *It is clear that d*_{1} *is left sided and d*_{2} *is two sided. Obviously, any derivation on* (*Q, &*) *is idempotent*.

*Let d be a derivation on a quantale Q.If d is idempotent, right* (*left*) *sided and d*(*a*) ≥ *a for all a* ∈ *Q, then d*(*a*) *is an idempotent element of Q*.

*Proof*. Since *d* is a right-sided derivation, we have *d*(*a*&⊤) = (*d*(*a*)& ⊤) ∨ (*a*&*d*) ∨ ≤ *d*(*a*) for all *a* ∈ *Q*. So *d*(*a*)& ⊤ ≤ *d*(*a*), and then *d*(*a*)*&d*(*a*) ≤ *d*(*a*). Making use of the idempotence of *d, d*(*a*) *= d*(*a&a*) ∨ (*d*(*a*)*&a*) ∨ (*a&d*(*a*)) ≤ *d*(*a*)*&d*(*a*). Therefore, *d*(*a*)*&d*(*a*) *= d*(*a*) for all *a* ∈ *Q. □*

*Let d be a derivation on a quantale Q. If d*(*a*) ≥ *a for all a* ∈ *Q, then d is right* (*left*) *sided iff d*(*a*) *is right* (*left*) *sided*.

*Proof*. If *d*(*a*) is right sided for all *a* ∈ *Q*, then *d*(*a*)& ⊤ ≤ *d*(*a*). So *a&d*(⊤) ≤ *a*&⊤ ≤ *d*(*a*)& ⊤ ≤ *da*. Thus *d*(*a*&⊤) = (*d*(*a*)& ⊤)∨ (*a*&*d*(⊤)) ≤ *d*(*a*).

Conversely, if *d* is right sided, we have *d*(*a*&⊤) = (*d*(*a*)& ⊤) ∨ (*a*&*d*(⊤)) ≤ *d*(*a*), then *d*(*a*)&⊤ ≤ *d*(*a*). □

Let .(*Q*, &) be a quantale, we denote the collection of all derivations on *Q* by *D*(*Q*). Define the operation "⋁" and "o" on *D*(*Q*) by (*d*_{1} ⋁ *d*_{2}).() = *d*_{1}.⋁*d*_{2}.() and *.d*_{1} ◦ *d*_{2} () *= d*_{1}()&*d*_{2}() for *d*_{1}*;d*_{2} ∈ *D*(*Q*).

*Let Q be a quantale. Then*,

(*D*(*Q*), ◦) *is a complete lattice;*

*if Q is two sided and right commutative, then* (*D*(*Q*), ◦) *is a quantale*.

*Proof*. (1) We prove that *D*(*Q*) is a complete lattice under the pointwise order.

Let {*d*_{i}} ⊆ *D*(*Q*) and {*a*_{j}} ⊆ *Q*, we have ⋁_{i}*d*_{i}(⋁_{j}*a*_{j}) *=* ⋁*i*(*d*_{i}(⋁_{i}*a*_{j})) = ⋁_{i}(⋁_{j} (*d*_{i}(*a*_{j}))) = ⋁_{j}(⋁_{i}*d*_{i}(*a*_{j})). Let *a*, *b* ∈ Q, we have ⋁_{i} *d*_{i}(*a&b*) *=* \/_{i}((*d*_{i}(*a*)&b) ∨ (*a*&*d*_{i}(b))) *=* (⋁_{i}(*d*_{i}(*a*)&b)) ∨ (⋁_{i}(*a&d*_{i}(*b*))) *=* ((⋁_{i}*d*_{i}(*a*))*&b*) ∨ *(a&*(⋁_{i}*d*_{i}(*b*))). So ⋁_{i}*d*_{i} ∈ *D*(*Q*).

Since *d* : *Q* → *Q* defined by *d*(*a*) = 0 for all *a* ∈ *Q* is a derivation, then *D*(*Q*) has a bottom element. Therefore, *D*(*Q*) is a complete lattice.

(2) We need to prove "◦" is a binary operation on *D*(*Q*).

Let *d*_{1}*, d*_{2} ∈ *D*(*Q*)*, a, b* ∈ *Q*, then (*d*_{1} ◦ *d*_{2})(*a&b*) *= d*_{1}(*a&b*)*&d*_{2}(*a&b*) *=* ((*d*_{1}(*a*)*& b*) ∨ (*a&d*_{1}(*b*)))&((*d*_{2}(*a*)*&b*) ∨ (*a&d*_{2}(*b*)))*=*((*d*_{1}(*a*)*&b*)&(*d*_{2}(*a*)*&b*)) ∨ ((*a*&*d*_{1}(*b*)) & (*d*_{2}(*a*)*&b*)) ∨ ((*d*_{1}(*a*)*&b*)&(*a&d*_{2}(*b*))) ∨ ((*a*&*d*_{1}(*b*))&(*a*&*d*_{2}(*b*))).

Since *Q* is right sided, we have ((*d*_{1}(*a*)*&b*)&(*d*_{2}(*a*)*&b*)) ∨ ((*a&d*_{1}(*b*))&(*d*_{2}(*a*)*&b*)) *=* ((*d*_{1}(*a*)&*b*) ∨ (*a&d*_{1}(*b*)))&(*d*_{2}(*a*)*&b*) *= d*_{1}(*a&b*)&(*d*_{2}(*a*)*&b*) ≤ *d*_{1}(*a*&⊤)&(*d*_{2}*(a)&b*) ≤ (*d*_{1}(*a*)& *d*_{2}(*a*))*&b =* (*d*_{1} ◦ *d*_{2})(*a*)&*b*.

Since *Q* is left sided and right commutative, then ((*d*_{1}(*a*)*&b*)&(*a&d*_{2}(*b*))) ∨ ((*a&d*_{1}(*b*))&(*a&d*_{2}(*b*))) *=* ((*d*_{1}(*a*)*&b*) ∨ (*a&d*_{1}(*b*)))&(*a&d*_{2}(*b*)) *= d*_{1}(*a&b*)&(*a&d*_{2}(*b*)) ≤ *d*_{1}(⊤*&b*)&(*a&d*_{2}(*b*)) ≤ *d*_{1}(*b*)&*a*&*d*_{2}(*b*) *= a&d*_{1}(*b*)*&d*_{2}(*b*) *= a&*(*d*_{1} ◦ *d*_{2})(*b*).

From the above, we have (*d*_{1} ◦ *d*_{2})(*a&b*) *=* ((*d*_{1} ◦ *d*_{2})(*a*)*&b*) ∨ (*a&*(*d*_{1} ◦ *d*_{2})(*b*)). So *d*_{1} ◦ *d*_{2} ∈ *D*(*Q*).

It is clear that the operation o is an associative binary operation satisfying the distribution over arbitrary joins, so (*D*(*Q*), ◦) is a quantale. □

Given a derivation d on a quantale (*Q*, &), we denote the fixed set {*a* ∈ *Q |d*(*a*) *= a*} by *Q*_{d}.

Let *d be a derivation on a quantale* (*Q*, &), *then* (*Q*_{d}, &) *is a quantale*.

*Proof*. We first prove that *Q*_{d} is a complete lattice under the pointwise order of *Q*. Let {*b*_{i}} ⊆ *Q*_{d}, we have *d*(⋁_{i}*b*_{i}) = ⋁_{i}*d*(*b*_{i}) *=* ⋁_{i}*b*_{i}, then ⋁_{i} *b*_{i} ∈ *Q*_{d}. By Proposition 3.4, we have *d*(0) = 0, then 0 ∈ *Q*_{d}.

Then we prove that & is a binary operation on *Q*_{d}. Let *a*, *b* ∈ *Q*_{d}, we have *d*(*a*&*b*) = (*d*(*a*)&*b*) ∨ (*a&d*(*b*)) *= a&b*, so *a*&*b* ∈ *Q*_{d}.

From the above, we have (*Q*_{d}, &) is a quantale. □

Let *Q* = {0, ⊤} *and define a binary operation & on Q by* 0&0 = 0, 0&⊤ = 0, ⊤&0 = 0, ⊤&⊤ = ⊤. *The quantale* (*Q*, &) *is two sided, idempotent, unital and commutative. There are only zero derivation and identity derivation on Q*, *denoted by d*_{1} *and d*_{2}*.It is clear that D*(*Q*) *=* {*d*_{1}*, d*_{2}}*, Q*_{d1} = {0}*and Q*_{d2} *=* {0, ⊤} *are quantales*.

*Let d be a derivation on a quantale Q*. *If d is two-sided, idempotent and d*(*a*) ≥ *a for all a* ∈ *Q*, *then d is a quantale homomorphism from Q onto Q*_{d}.

*Proof*. By Proposition 3.9 and Proposition 3.10, we have *d*(*a*) is idempotent and two-sided for all *a* ∈ *Q*. Therefore, *d(d*(*a*)) *= d*(*d*(*a*)*&d*(*a*)) *=* (*d*(*d*(*a*))*&d*(*a*)) ∨ (*d*(*a*)*&d*(*d*(*a*))) ≤ (⊤&*d*(*a*)) ∨ (*d*(*a*)&⊤) ≤ *d*(*a*). On the other hand, *d*(*d(a*)) ≥ *d*(*a*). Thus *d*(*d*(*a*)) *= d*(*a*) which implies that *d*(*a*) ∈ *Q*_{d} for all *a* ∈ *Q*. So *d* is a mapping from *Q* onto *Q*_{d}.

We need to show that *d*(*a&b*) *= d*(*a*)*&d*(*b*) for all *a, b* ∈ *Q*. Since *d* is idempotent, we have *d*(*a&b*) *= d*(*d*(*a&b*)) *= d*((*d*(*a*)*&b*) ∨ (*a&d*(*b*))) *= d*(*d*(*a*)*&b*) ∨ *d*(*a&d*(*b*)) ≥ *d*(*d*(*a*)*&b*) *=* (*d*(*d*(*a*))*&b*) ∨ (*d*(*a*)*&d*(*b*)) ≥ *d*(*a*)*&d*(*b*). On the other hand, *d*(*a&b*) *=* (*d*(*a*)*&b*)∨(*a&d*(*b*)) ≤ (*d*(*a*)*&d*(*b*))∨(*d*(*a*)*&d*(*b*)) *= d*(*a*)*&d*(*b*). So *d*(*a&b*) *= d*(*a*)*&d*(*b*). Since *d* preserves arbitrary sups, we know that d is a quantale homomor phism from *Q* onto *Q*_{d}. □

By the proof of Theorem 3.14, we have the following corollary.

*Let d be a derivation on a quantale Q. If d is two-sided, idempotent and d*(*a*) ≥ *a for all a* ∈ *Q*, *then Q*_{p} = {*a* ∈ *Q|d*(*a*) ≥ *p*, *p*&*p* = *p*} *is a subquantale of Q*.

In the following, we shall give a description for a derivation on general quantales.

(Paseka and Kruml [8]). *Let* (*Q, &*) *be a quantale and Q*[*e*] *=*{*a* ∨ *k* : *a* ∈ *Q, k* ∈ {0, *e*}}, *where ∈ is an arbitrary element such that ∈* ∉ *Q. We define the supremum on Q* [*e*]*: ∈* ∨ 0 = ∈ *and*
$$\underset{i}{\overset{Q[e]}{\bigvee}}({a}_{i}\vee {k}_{i})=\left\{\begin{array}{ll}({\vee}_{i},{a}_{i})\vee e,& if\phantom{\rule{thinmathspace}{0ex}}\mathrm{\exists}{k}_{i}=e,\\ {\vee}_{i}{a}_{i},& otherwise.\end{array}\right.$$

*The multiplication &′ on Q*[*e*] *is as follows*:
$$(a\vee {k}^{\prime}){\mathrm{\&}}^{\prime}(b\vee {k}^{\u2033})=\left\{\begin{array}{ll}a\mathrm{\&}b,& if\phantom{\rule{thinmathspace}{0ex}}{k}^{\prime}={k}^{\u2033}=0,\\ (a\mathrm{\&}b)\vee b,& if{k}^{\prime}=e,\phantom{\rule{thinmathspace}{0ex}}{k}^{\u2033}=0,\\ (a\mathrm{\&}b)\vee a,& if{k}^{\prime}=0,\phantom{\rule{thinmathspace}{0ex}}{k}^{\u2033}=e,\\ ((a\mathrm{\&}b)\vee a\vee b)\vee e,& if\phantom{\rule{thinmathspace}{0ex}}{k}^{\prime}={k}^{\u2033}=e.\end{array}\right.$$

*Then* (*Q*[*e*]*; &′*) *is a unital quantale with the unit e*.

Let *Q* be a quantale and *d* : *Q → Q* a map on *Q*. We define the map *d* *: Q*[*e*] *→ Q* [*e*] such that
$$\overline{d}\left(a\right)=\left\{\begin{array}{ll}d\left(a\right),& \text{if}\phantom{\rule{thickmathspace}{0ex}}a\in Q,\\ d\left({a}^{\mathrm{\prime}}\right),& \text{if}\phantom{\rule{thickmathspace}{0ex}}a={a}^{\mathrm{\prime}}\vee e,{a}^{\mathrm{\prime}}\in Q.\end{array}\right.$$

It is clear that *d =* *d*|_{Q}.

*Let Q be a quantale and d : Q* → *Q a map on Q. Then d is a derivation on Q if and only if* *d* *is a derivation on Q* [*e*].

*Proof*. Clearly, if *d* is a derivation on *Q*[*e*], then *d is a* derivation on *Q*.

Conversely, we assume that *d* is a derivation on *Q*. Let {*b*_{i} ∨ *k*_{i}}_{i} ⊆ *Q*[*e*] *and b*_{i} ∈ *Q, k*_{i} ∈ {0, *e*}. We have
$$\begin{array}{c}\overline{d}(\underset{i}{\overset{Q[e]}{\bigvee}}\left({b}_{i}\vee {k}_{i}\right))=\left\{\begin{array}{ll}\overline{d}\left(\left({\vee}_{i}{b}_{i}\right)\vee e\right),& \text{if}\phantom{\rule{thinmathspace}{0ex}}\mathrm{\exists}{k}_{i}=e,\\ \overline{d}\left({\vee}_{i}{b}_{i}\right),& \text{otherwise}\text{.}\end{array}\right.\\ =d(\underset{i}{\bigvee}{b}_{i})=\underset{i}{\bigvee}d\left({b}_{i}\right)=\underset{i}{\bigvee}\overline{d}\left({b}_{i}\vee {k}_{i}\right)\end{array}$$

Let *a* ∨ *k′, b* ∨ *k″* ∈ *Q*[*e*] and *a*, *b* ∈ *Q, k′*_{1}*, k″* ∈ {0, *e*}. Then
$$\overline{d}\left(\left(a\vee {k}^{\prime}\right){\mathrm{\&}}^{\prime}\left(b\vee {k}^{\u2033}\right)\right)=\left\{\begin{array}{ll}d\left(a\mathrm{\&}b\right),& \text{if}\phantom{\rule{thickmathspace}{0ex}}{k}^{\prime}={k}^{\u2033}=0,\\ d\left(\left(a\mathrm{\&}b\right)\vee b\right),& \text{if}\phantom{\rule{thickmathspace}{0ex}}{k}^{\prime}=e,\phantom{\rule{thickmathspace}{0ex}}{k}^{\u2033}=0,\\ d\left(\left(a\mathrm{\&}b\right)\vee a\right),& \text{if}\phantom{\rule{thickmathspace}{0ex}}{k}^{\prime}=0,\phantom{\rule{thickmathspace}{0ex}}{k}^{\u2033}=e,\\ d\left(\left(a\mathrm{\&}b\right)\vee a\vee b\right),& \text{if}\phantom{\rule{thickmathspace}{0ex}}{k}^{\prime}=\phantom{\rule{thickmathspace}{0ex}}{k}^{\u2033}=e.\end{array}\right.$$

Case1:*k*′ = *k″* = 0,[(*d*(*a*∨*k′*))&*′*(*b*∨*k″*)]∨[(*a*∨*k′*)&′(*d*(*b*∨*k″*))] *=* [*d*(*a*)&′(*b*∨*k″*)] ∨ [(*a*∨*k′*)&*'d*(*b*)] *=* (*d*(*a*)*&b*) ∨ (*a&d*(*b*)) = *d*(*a&b)i*

Case *2: k′* = *e, k″* = 0, [(*d*(*a* ∨ *k′*))&′(*b* ∨ *k*″)] ∨ [(*a* ∨ *k′*)&′( *d*(*b* ∨ *k*"))] = [*d*(*a*)&′(*b* ∨ *k*″)] ∨ [(*a* ∨ *k*)&′*d*(*b*)] = (*d*(*a*)*&b*) ∨ (*a&d*(*b*)) ∨ *d*(b) = *d*((*a&b*) ∨ b);

Case 3: *k*′ = 0, *k*″ = ∈, [(*d*(*a* ∨ *k′*))&′(*b* ∨ *k*″)] ∨ [(*a* ∨ *k′*)&′(*d*(*b* ∨ *k*″))] = [*d*(*a*)&′(*b* ∨ *k″*)] ∨ [(*a* ∨ *k*′)&′*d*(b)] = (*d*(*a*)*&b*) ∨ *da* ∨ (*a&d*(*b*)) = *d*((*a&b*) ∨ *b*);

Case 4: *k′* = *e, k″* = *e*, [(*d*(*a* ∨ *k′*))*&′*(*b* ∨ *k*″)] ∨ [(*a* ∨ *k′*)&′(*d*(*b* ∨ *k*″))] = [*d*(*a*)&′(*b* ∨ *k*″)] ∨ [(*a* ∨ *k*)&′*d*(b)] = (*d*(*a*)*&b*) ∨ *da* ∨ (*a&d*(*b*)) ∨ *d*(*b*) = *d*((*a&b*) ∨ *a* ∨ *b*).

Therefore, *d* is a derivation on *Q* [*e*]. □

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