On generalized Ehresmann semigroups

Lemma 2.1

A semigroup (S, ⋅) is Ehresmann if and only if there are two unary operations “+” and “*” on S such that the following identities hold:

In this case, S is an Ehresmann semigroup with distinguished semilattice {x⁺|x ∈ S} (={x^*|x ∈ S}), and we shall call (S⋅, +, *) an Ehresmann semigroup.

Now we consider regular semigroups with an inverse transversal. Let S be a regular semigroup and S^∘ an inverse transversal of S. For any x ∈ S, we use x^∘ to denote the unique inverse of x in S^∘ and let x^∘∘ = (x^∘)^∘. We can consider the induced tri-unary semigroup (S, ⋅, +, *, −), where

Observe that x^∘∘∘ = x^∘ and x^∘∘x^∘, x^∘x^∘∘ ∈ E(S^∘). Here we are only interested in normal orthodox semigroups with an inverse transversal. Recall that a normal orthodox semigroup means a regular semigroup whose idempotents form a normal band, and a band B is called left normal (resp. right normal, normal) if efg = egf (resp. efg = feg, efge = egfe) for all e, f, g ∈ B. Observe that a band B is normal if and only if efgh = egfh for all e, f, g, h ∈ B. The following proposition can be deduced from the text [20]. However, we give its direct proof for the sake of completeness.

Proposition 2.2

Let S be a normal orthodox semigroup with an inverse transversal S^∘. The tri-unary semigroup (S, ⋅, +, *, −) satisfies the following identities:

Proof

By symmetry, we only need to show (1)–(8).

1. This is equivalent to the equality (xx^∘)x = x.

2. As E(S) is a normal band, we have

   x+y+=(xx∘)[(yy∘)(y∘∘y∘)](y∘∘y∘)=(xx∘)[(y∘∘y∘)(yy∘)](y∘∘y∘)=xx∘y∘∘y∘. (1)

   This implies that

   x+y+x+=x+(y+x+)=xx∘(yyox∘∘x∘)=(xxoyy∘)(x∘∘x∘)=xx∘(y∘∘y∘)(x∘∘x∘)=xx∘(x∘∘x∘)(y∘∘y∘)=xx∘y∘∘y∘=x+y+.

3. We first observe that

   (xy)∘=y∘x∘ for all x,y∈S. (2)

   In fact, since S is orthodox, y^∘x^∘ is an inverse of xy. But y^∘x^∘ ∈ S^∘, so (xy)^∘ = y^∘x^∘. By (1) and (2), we have

   (x+y+)+=(xx∘y∘∘y∘)(xx∘y∘∘y∘)∘=xx∘y∘∘y∘y∘∘y∘∘∘x∘∘x∘=xx∘y∘∘y∘x∘∘x∘=xx∘x∘∘x∘y∘∘y∘=xx∘y∘∘y∘=x+y+.

4. By (2), we have

   (xy+)+=(xyy∘)(xyy∘)∘=xyy∘y∘∘y∘x∘=xyy∘x∘=xy(xy)∘=(xy)+.

5. From (2),

   (x+)∗=(xx∘)∘(xx∘)=x∘∘x∘xx∘=x∘∘x∘=x∘∘x∘∘∘=x¯+.

6. It follows that x+¯ = (xx^∘)^∘∘ = x^∘∘x^∘ = x^∘∘(x^∘∘)^∘ = x⁺ from (2).

7. This is equivalent to the statement x = (xx^∘)x^∘∘(x^∘x).

8. Since E(S) is a normal band, using (1) and its dual, we have

   x∗y+=x∘xyy∘=x∘x∘∘x∘xyy∘y∘∘y∘=(x∘x∘∘yy∘)(x∘xy∘∘y∘)=(x∘x∘∘y∘∘y∘)(x∘x∘∘y∘∘y∘)=x∘x∘∘y∘∘y∘=((x∘∘)∘x∘∘)(y∘∘(y∘∘)∘)=x¯∗y¯+,

   as required. □

   We shall term any tri-unary semigroup (S, ⋅, +, *, −) that satisfies the identities in Table 1 a generalized Ehresmann semigroup. By Proposition 2.2, any normal orthodox semigroup S with an inverse transversal S^∘ induces the generalized Ehresmann semigroup (S, ⋅, +, *, −) by setting x⁺ = xx^∘, x^* = x^∘x and x = x^∘∘. The following example gives a very special case of this kind of generalized Ehresmann semigroups.

Example 2.3

Let S be a rectangular band. Fix an element u in S. Consider the tri-unary semigroup (S, ⋅, +, *, −) where x⁺ = xu, x^* = ux, x = u. Then it is routine to check that the identities in Table 1 are satisfied and so (S, ⋅, +, *, −) is a generalized Ehresmann semigroup. In fact, S is indeed a normal band and {u} is an inverse transversal of S.

Example 2.4

Any Ehresmann semigroup (S, ⋅, +, *) also induces a generalized Ehresmann semigroup, which justifies our term “generalized Ehresmann semigroups” In fact, for an Ehresmann semigroup (S, ⋅, +, *), we define the third unary operation “−” on S by x = x. Then we have the tri-unary semigroup (S, ⋅, +, *, −) and it is easy to see that the identities in Table 1 are all satisfied by Lemma 2.1.

Since a rectangular band having more than one element must not be an Ehresmann semigroup, the class of generalized Ehresmann semigroups contains the class of Ehresmann semigroups and the class of rectangular bands as proper subclasses by the above two examples. We also observe that a generalized Ehresmann semigroup which is also regular may not contain any inverse transversal. In fact, any monoid S with the identity 1 is always a (generalized) Ehresmann semigroup by setting x⁺ = x^* = 1 and x = x for all x ∈ S. Obviously, a regular monoid may not contain any inverse transversal. Here is an example.

Example 2.5

Let M = {1, b, c, x} (taken from Exercise 10 in Chapter VI of [24]) with the multiplication

Then M is a monoid and

It is easy to check that M contains no inverse transversal.

In the remainder of this section, we consider some properties associated to generalized Ehresmann semigroups which will be used in the next sections. Let (S, ⋅, +, *, −) be a generalized Ehresmann semigroup. Denote

Lemma 2.6

Let (S, ⋅, +, *, −) be a generalized Ehresmann semigroup.

1. i⁺ = i, i^* = i and λ^* = λ, λ⁺ = λ for all i ∈ I_S and λ ∈Λ_S.

2. ES∘={x¯∗|x∈S}=IS∩ΛS,

3. x⁺𝓛x⁺ and x^*𝓡x^*.

4. I_S is a left normal band, Λ_S is a right normal band and ES∘ is a subsemilattice of S, respectively.

Proof

1. Using identities (1),(4) and (3) in Table 1, we have

   x+=(x+x)+=(x+x+)+=x+x+

   and so x⁺ ∈ E(S). Take i ∈ I_S. Then i = x⁺ for some x ∈ S. This implies that

   i+=(x+)+=(x+x+)+=x+x+=x+

   by x⁺ ∈ E(S) and the identity (3). Moreover,

   i∗=(x+)∗=x¯+=x+¯=i¯

   by the identities (5) and (6). By symmetry, λ^* = λ and λ⁺ = λ for all λ ∈ Λ_S.

2. By identity (5), x⁺ = (x⁺)^* ∈ I_S ∩ Λ_S for all x ∈ S. Now let u = x⁺ ∈ I_S ∩ Λ_S for some x ∈ S. Using item (a) and the identity (5), we have

   u=u∗=(x+)∗=x¯+∈ES∘.

   Thus ES∘ = I_S ∩ Λ_S. Dually, {x^*|x ∈ S} = I_S ∩ Λ_S.

3. Using identities (7), (4), (3), (4), (5)′ and (1)′ in Table 1 in that order, we have

   x+=(x+x¯x∗)+=(x+(x¯x∗)+)+=x+(x¯x∗)+=x+(x¯(x∗)+)+=x+(x¯x¯∗)+=x+x¯+.

   Since x⁺ ∈ Λ_S by (b), we have (x⁺)^* = x⁺ by (a). Using the identities (5), (8) and (6), we have

   x¯+x+=(x+)∗x+=(x+¯)∗x¯+=(x¯+)∗x¯+=x¯+x¯+=x¯+.

   This shows that x⁺𝓛x⁺. Dually, x^*𝓡x^*

4. In view of the proof of item (a), every element in I_S is idempotent. By the identity (3), x⁺y⁺ = (x⁺y⁺)⁺ ∈ I_S for all x⁺, y⁺ ∈ I_S. So I_S is a subband of S. By the identity (2) and (2)' and item (b), ES∘ is a subsemilattice of S. Moreover, for x, y, z ∈ S, by identities (5), (8) and (6) and items (b) and (a), we have

   x¯+y+=(x+)∗y+=x+¯∗y¯+=(x¯+)∗y¯+=x¯+y¯+.

   Similarly, y⁺z⁺ = y⁺z⁺. In view of item (c) and the fact that ES∘ is a subsemilattice,

   x+y+z+=x+x¯+y+z+=x+x¯+y¯+z+=x+x¯+y¯+z¯+=x+x¯+y¯+z¯+=x+y¯+z¯+=x+z¯+y¯+.

   Similarly, we have x⁺z⁺y⁺ = x⁺z⁺y⁺. This yields that I_S is a left normal band. Dually, Λ_S is a right normal band. □

   Let S be a semigroup. Recall that the natural partial order “≤” on E(S) is defined as follows:

   e≤f if and only if ef=fe=e for all e,f∈E(S). (3)

Lemma 2.7

Let (S, ⋅, +, *, −) be a generalized Ehresmann semigroup and x, y ∈ S.

1. (xy)⁺ = x⁺(xy)⁺, (xy)* = (xy)^*y^*.

2. (xy)⁺ ≤ x⁺, (xy)* ≤ y^*.

3. xy = xy.

Proof

1. Using the identities (4), (7), (8), (1)′, (4) and (3) in that order, we have

   (xy)+=(xy+)+=(x+x¯x∗y+)+=(x+x¯x¯∗y¯+)+=(x+x¯y¯+)+=(x+x¯y¯)+=(x+(x¯y¯)+)+=x+(x¯y¯)+.

   Dually, (xy)* = (xy)^*y^*.

2. Since (xy)⁺ = x⁺(xy)⁺ by item (a), we have x⁺(xy)⁺ = (xy)⁺. Moreover, by the identity (2),

   (xy)+x+=x+(x¯y¯)+x+=x+(x¯y¯)+=(xy)+.

   So (xy)⁺ ≤ x⁺. Dually, (xy)* ≤ y^*.

3. In view of the identity (8) and items (b), (d) in Lemma 2.6, we have (x+)∗(x¯y¯)+=x+¯∗x¯y¯¯+ ∈ ES∘ so that ((x⁺)^*(xy)⁺)^* = (x⁺)^*(xy)⁺ by Lemma 2.6 (a). Using the identity (5), item (a) of this lemma, the identities (4)′, (5), (3), (4), (1) in that order, we obtain that

   xy¯+=((xy)+)∗=(x+(x¯y¯)+)∗=((x+)∗(x¯y¯)+)∗=(x+)∗(x¯y¯)+=x¯+(x¯y¯)+=(x¯+(x¯y¯)+)+=(x¯+x¯y¯)+=(x¯y¯)+.

   Dually, xy^* = (xy)^*. Using the identities (1) and (1)′, Lemma 2.6 (c), the identities (7), (8), (1), (1)′, item (b) of this lemma, Lemma 2.6 (c), the identities (1) and (1)′ in that order, we get

   xy¯=xy¯+xy¯xy¯∗=xy¯+(xy)+xy¯(xy)∗xy¯∗=xy¯+xyxy¯∗=xy¯+x+x¯x∗y+y¯y∗xy¯∗=xy¯+x+x¯x¯∗y¯+y¯y∗xy¯∗=xy¯+x+x¯y¯y∗xy¯∗=(x¯y¯)+x+x¯y¯y∗(x¯y¯)∗=(x¯y¯)+x¯+x+x¯y¯y∗y¯∗(x¯y¯)∗=(x¯y¯)+x¯+x¯y¯y¯∗(x¯y¯)∗=(x¯y¯)+x¯y¯(x¯y¯)∗=x¯y¯,

   as required. □

Definition 3.1

Let I (resp. Λ) be a left normal band (resp. a right normal band), E^∘ = I ∩ Λ a subsemilattice of I and Λ. The triple (I, Λ, E^∘) is called admissible if for all g ∈ I and f ∈ Λ, there exist g^∘, f^∘ ∈ E^∘ such that g𝓛g^∘ and f𝓡f^∘.

Remark 3.2

Let (I, Λ,E^∘) be an admissible triple. Since E^∘ is a subsemilattice, the elements g^∘ and f^∘ in Definition 3.1 are uniquely determined by g and f, respectively. In particular, i ∈ E^∘ if and only if i^∘ = i.

Remark 3.3

Let (S, .,+,*,−) be a generalized Ehresmann semigroup. By Lemma 2.6 (b), (c) and (d), (I_S, Λ_S, ES∘ ) is an admissible triple which will be called the admissible triple of S. In this case, for all i ∈ I_S and λ ∈ Λ_S, we have i^∘ = i^* and λ^∘ = λ⁺. In fact, if x⁺ ∈ I_S, then by the identity (5) in Table 1 and Lemma 2.6 (c), we have x⁺𝓛x⁺ = (x⁺)^*. The case for λ ∈ Λ_S can be showed dually.

To construct the semigroup C_(I,Λ,E^∘) of an admissible triple (I, Λ, E^∘), we need some preliminaries. First, we have the following basic facts on admissible triples.

Lemma 3.4

Let (I, Λ,E^∘) be an admissible triple and e,g ∈ I,f,h ∈ Λ. Then

Moreover, we have eE^∘e = eE^∘ and fE^∘f = E^∘f, which are subsemilattices of I and Λ, respectively.

Proof

Since g𝓛g^∘ and I is a left normal band, we have eg = egg^∘ = eg^∘g = eg^∘. This implies that e^∘g = e^∘g^∘, and so eg𝓛e^∘g = e^∘g^∘ ∈ E^∘ by the fact that e𝓛e^∘. This yields that (eg)^∘ = e^∘g^∘. Finally, it follows that eE^∘e = eE^∘ by the fact that I is a left normal band. Moreover, for i, j ∈ E^∘, we have

whence eE^∘ is a subsemilattice of I. The remaining facts of this lemma can be proved by symmetry. □

Let (I, Λ, E^∘) be an admissible triple. We use I¹ (resp. Λ¹) to denote I (resp. Λ) with the identity adjoined. Furthermore, we always assume that 1^∘ = 1 for the adjoined identity 1 on both I and Λ. A function η from I¹ to Λ is called order-preserving if xη ≤ yη in Λ for all x, y ∈ I¹ with x ≤ y in I¹, where ≤ is defined in (3). Denote the set of order-preserving functions from I¹ to Λ by 𝓞(I¹ → Λ). Dually, we also have 𝓞(Λ¹ → I). Similarly, we have 𝓞(I¹ → I) and 𝓞(Λ¹ →Λ). Moreover, we denote the set of the morphisms from I¹ to Λ by End (I¹ → Λ). Dually, we have End (Λ¹ → I). Obviously,

For every e ∈ I (resp. f ∈ Λ), define

Then it is easy to see that ρ_e ∈ 𝓞(I¹ → I) and σ_f ∈ 𝓞(Λ¹ → Λ) for all e ∈ I and f ∈ Λ as I is a left normal band and Λ is a right normal band. Moreover, we use ρ₁ and σ₁ to denote the identity maps on I and Λ, respectively.

Lemma 3.5

Let (I, Λ,E^∘) be an admissible triple. Define a multiplication “⋄” on 𝓞(I¹ → Λ)as follows: for all α,β ∈ 𝓞(I¹ → Λ),

Then 𝓞(I¹ → Λ) is a semigroup with respect to “⋄”. Dually, for α, β ∈ 𝓞(Λ¹ → I), define

then 𝓞(Λ¹ → I) forms a semigroup with respect to “⋆”.

Proof

Since ∅ ≠ End (I¹ → Λ) ⊆ 𝓞(I¹ → Λ), it follows that 𝓞(I¹ → Λ) ≠ ∅. Observe that

by Lemma 3.4. Let x, y ∈ I¹ and x ≤ y. Then xα, yα ∈ Λ and xα ≤ yα as α is order-preserving. This implies that (xα)^∘ ≤ (yα)^∘. Observe that β is also order-preserving, it follows that ((xα)^∘)β ≤ ((yα)^∘)β. Thus α ⋄ β ∈ 𝓞(I¹ → Λ). Now let α, β, γ ∈ 𝓞(I¹ → Λ) and x, y ∈ I¹. Then

This implies that 𝓞(I¹ → Λ) is a semigroup with respect to “⋄”. Dually, 𝓞(Λ¹ → I) forms a semigroup with respect to “⋆”. □

Corollary 3.6

Let (I, Λ,E^∘) be an admissible triple, and “⋄” and “⋆” be defined as in Lemma 3.5. Then 𝓞(I¹ → Λ) × 𝓞(Λ¹ → I) forms a semigroup by defining

for all (α,β),(γ, δ) ∈ 𝓞(I¹ → Λ) × 𝓞(Λ¹ → I).

The semigroup 𝓞(I¹ → Λ) is partially ordered by “≤” where for all α, β ∈ 𝓞(I¹ → Λ),

Similarly, the semigroups 𝓞(Λ¹ → I), 𝓞(I¹ → I) and 𝓞(Λ¹ → Λ) can be partially ordered, respectively.

Consider the subset

of the product semigroup 𝓞(I¹ → Λ) × 𝓞(Λ¹ → I).

Lemma 3.7

Let (I, Λ, E^∘) be an admissible triple. For e ∈ I, define θ_e and τ_e as follows:

Then (θ_e, τ_e) ∈ C_(I,Λ,E^∘). Dually, for f ∈ Λ, define η f and ξ_f as follows:

Then (η_f, ξ_f) ∈ C_(I,Λ,E^∘). Moreover, if e ∈ I ∩ Λ = E^∘, then (θ_e, τ_e) = (η_e, ξ_e).

Proof

Firstly, by Lemma 3.4, we have e^∘x = e^∘x^∘ ∈ E^∘ ⊆ Λ for all x ∈ I, and x^∘ ∈ E^∘, ex^∘∈ I for all x ∈ Λ. Therefore θ_e and τ_e are well-defined. Secondly, since I is a left normal band, we have

for all x, y ∈ I¹. On the other hand, by Lemma 3.4, we have

for all x, y ∈ Λ¹. This shows that θ_e and τ_e are morphisms and so order-preserving. Thirdly, if x ∈ I, then xθ_e = e^∘x = e^∘x^∘ = x^∘θ_e. Similarly, yτ_e = ey^∘ = e(y^∘)^∘ = y^∘τ_e for all y ∈ Λ. Finally, let x ∈ I¹ and u ∈ Λ¹. Then by Lemma 3.4,

and

This implies that σ_xθe = τ_eρ_xθ_e for all x ∈ I¹. Similarly, ρ_yτe = θ_eσ_yτ_e for all y ∈ Λ¹. Thus (θ_e, τ_e) ∈ C_(I,Λ,E^∘). Dually, (η_f, ξ_f) ∈ C_(I,Λ,E^∘). If e ∈ E^∘, then e^∘ = e and so

for all x ∈ I¹. This shows that θ_e = η_e. Dually, τ_e = ξ_e. □

Lemma 3.8

C_(I,Λ,E^∘) is a subsemigroup of 𝓞(I¹ → Λ)× 𝓞(Λ¹ → I).

Proof

Denote C = C_(I,Λ,E^∘). We have seen that C is non-empty by Lemma 3.7. Let (α, β), (γ, δ) ∈ C. Then

We first show that σ_x(α⋄γ) ≤ (δ ⋆ β)ρ_x(α ⋄ γ) for all x ∈ I¹. Let u ∈ Λ¹ and x ∈ I¹. Then uδ ∈ I and (uδ)^∘ ∈ I ∩ Λ. Since σ_xα ≤ βρ_xα by the fact that (α, β) ∈ C, we have (uδ)^∘σ_xα≤(uδ)^∘βρ_xα whence (uδ)^∘(xα)≤(x((uδ)^∘β))α. In view of (4) and Lemma 3.4, we can obtain that

Since γ is order-preserving, it follows that

This shows that δρ(xα)^∘γ≤ (δ ⋆ β)ρ_x(α ⋄ γ). Since σ_vγ ≤ δρ_vγ for all v ∈ I¹ by the fact (γ, δ) ∈ C and (xα)^∘ ∈ I, it follows that

Finally, let x ∈ I. Since x^∘α = xα by the fact that (α, β) ∈ C, we get

By symmetry, we can obtain ρ_y(δ⋆β) ≤ (α ⋄ γ)σ_y (δ ⋆ β) and y^∘(δ ⋆ β) = y(δ ⋆ β) for all y ∈ Λ¹. Thus (α, β)(γ, δ) ∈ C. So C is a subsemigroup. □

Theorem 3.9

Define three unary operations on the semigroup C = C_(I,Λ,E^∘) as follows:

where

Then (C, ., ⁺, ^*, ⁻) is a generalized Ehresmann semigroup.

Proof

Let (α, β) ∈ C. By Lemma 3.7 and the fact that 1β ∈ I, 1α ∈ Λ, it follows that

This shows that “+″ and “*″ are well-defined.

Now, let (α, β) ∈ C. Then

for all x ∈ I¹ and y ∈ Λ¹. Let x, y ∈ I¹ and x ≤ y. Since α is order-preserving, xα ≤ yα. It follows that xα = (xα)^∘ ≤ (yα)^∘ = yα by (4). This shows that α is order-preserving. Dually, β is also order-preserving. Moreover, for all x ∈ I¹, we have

as x^∘α = xα for all x ∈ I¹. Dually, we have yβ = y^∘β for all y ∈ Λ¹. Now let x ∈ I¹. For u ∈ Λ¹, since σ_xα≤βρ_xα, we have uσ_{x α} ≤ uβρ_xα. That is, u(xα)≤(x(uβ))α. By (4), (u(x α))^∘≤((x(uβ))α)^∘. This implies that

by Lemma 3.4. This shows that σ_xα ≤ βρ_xα for all x ∈ I¹. Dually, ρ_yβ ≤ α σ_y β for all y ∈ Λ¹. Thus (α, β) ∈ C. This implies that “−″ is also well-defined.

We next show that the identities (1-8) in Table 1 are satisfied. By symmetry, the identities (1)′-(6)′ also hold. Let (α, β), (γ, δ) ∈ C.

1. For x ∈ I¹, we have σ_xα≤βρ_xα by the fact that (α, β) ∈ C. So

   xα=1σxα≤1(βρxα)=(x(1β))α.

   Since I is a left normal band, we have x(1β) ≤ x. This shows that (x(1β))α≤xα as α is order-preserving. So xα = (x(1β))α. Observe that v^∘α = vα for all v ∈ I¹ by the fact that (α, β) ∈ C, it follows that (x(1β))α = (x(1β))^∘α. Thus

   xα=(x(1β))α=(x(1β))∘α=(x∘(1β)∘)α=((1β)∘x)∘α=x(α+⋄α) (6)

   by lemma 3.4. We have shown that α⁺ ⋄ α = α. Dually, β ⋆ β⁺ = β. Thus

   (α,β)+(α,β)=(α+,β+)(α,β)=(α+⋄α,β⋆β+)=(α,β).

2. For x ∈ I¹, by Lemma 3.4 we have

   x(α+⋄γ+)=(xα+)∘γ+=((1β)∘x)∘γ+=((1β)∘x∘)γ+=(1δ)∘((1β)∘x∘)=(1δ)∘(1β)∘x∘ (7)

   and

   x((α+⋄γ+)∘α+)=(x(α+⋄γ+))∘α+=((1δ)∘(1β)∘x∘)∘α+=((1δ)∘(1β)∘x∘)α+=(1β)∘(1δ)∘(1β)∘x∘=(1δ)∘(1β)∘x∘.

   This implies that α⁺ ⋄ γ⁺ = α⁺ ⋄ γ⁺ ⋄ α⁺. Dually, β⁺ ⋆ δ⁺ ⋆ β⁺ = δ⁺ ⋆ β⁺. Thus

   (α,β)+(γ,δ)+(α,β)+=(α,β)+(γ,δ)+.

3. For all x ∈ I¹, we have x(α⁺0γ⁺) = (1δ)^∘(1β)^∘x^∘ by (7) and

   x(α+⋄γ+)+=(1(δ+⋆β+))∘x=((1β)(1δ)∘)∘x=((1δ)∘(1β)∘)x=(1δ)0(1β)∘x∘

   by Lemma 3.4. This yields that α⁺⋄γ⁺ = (α⁺⋄γ⁺)⁺. Dually, (δ⁺⋆β⁺)⁺ = δ⁺ ⋆ β⁺. Thus ((α, β)⁺(γ, δ)⁺)⁺ = (α, β)⁺(γ, δ)⁺.

4. For all x ∈ I¹, we have x(α ⋄ γ)⁺ = (1 (δ ⋆ β))^∘x = ((1δ)^∘β)^∘x and

   x(α⋄γ+)+=(1(δ+⋆β))∘x=((1δ+)∘β)∘x=((1δ)∘β)∘x.

   This implies that (α ⋄ γ)⁺ = (α ⋄ γ⁺)⁺. Dually, (δ ⋆ β)⁺ = (δ⁺⋆β)⁺. Thus

   ((α,β)(γ,δ))+=((α,β)(γ,δ)+)+.

5. For all x ∈ I¹, by Lemma 3.4 we have

   x(α+)∗=x∘(1α+)=x∘(1β)∘=(1β)∘x∘=(1β)∘∘x∘=(1β¯)∘x∘=(1β¯)∘x=xα¯+. (8)

   This shows that (α⁺)^* = α⁺. Dually, (β⁺)^* = β⁺. Thus ((α, β)⁺)^* = (α, β)⁺

6. For all x ∈ I¹, by Lemma 3.4 and (8) we have

   xα¯+=(1β)∘x∘=((1β)∘x)∘=(xα+)∘=xα+¯.

   So α¯+=α+¯.Dually,β¯+=β+¯.Thus α,β¯+=α,β+¯.

7. For all x ∈ I¹, by Lemma 3.4 and (6), we have

   x(α+⋄α¯⋄α∗)=(((1β)∘x∘)α)∘(1α)=(((1β)∘x)∘α)∘(1α)=(xα)∘(1α)=(xα)(1α).

   Since x ≤ 1 and α is order-preserving, we have (x α)(1α) = xα. This shows that x(α⁺⋄α⋄α^*) = xα. So α⁺⋄α⋄α^* = α. Dually, β^* ⋆ β ⋆ β⁺ = β. Thus

   (α,β)=(α,β)+(α,β)¯(α,β)∗.

8. For all x ∈ I¹, by Lemma 3.4 we have

   x(α∗⋄γ+)=(xα∗)∘γ+=(x∘(1α))∘γ+=(1δ)∘x∘(1α)∘

   and

   x(α¯∗⋄γ¯+)=(xα¯∗)∘γ¯+=(x∘(1α¯))∘γ¯+=(x∘(1α)∘)∘γ¯+=(x∘(1α)∘)γ¯+=(1δ¯)∘x∘(1α)∘=(1δ)∘x∘(1α)∘.

   This shows that α^* ⋄ γ⁺ = α^* ⋄ γ⁺. Dually, δ⁺ ⋆ β^* = δ⁺ ⋆ β^*. Thus

   (α,β)∗(γ,δ)+=(α,β)¯∗(γ,δ)¯+.