State maps on semihoops

Definition 2.1

([14]). An algebra (H, ⊙, →, ∧, 1) of type (2, 2, 2, 0) is called a semihoop if it satisfies the following conditions:

1. (H, ∧, 1) is a ∧-semilattice with upper bound 1,

2. (H, ⊙, 1) is a commutative monoid,

3. (x ⊙ y) → z = x → (y → z), for allx, y, z ∈ H.

Proposition 2.2

([14, 30]). In any semihoop (H, ⊙, →, ∧, 1), the following properties hold: for allx, y, z ∈ H,

1. x ⊙ y ≤ ziffx ≤ y → z,

2. x ⊙ y ≤ x ∧ y, x ≤ y → x,

3. 1 → x = x, x → 1 = 1,

4. x ⊙ (x → y) ≤ y,

5. Ifx ≤ y, theny → z ≤ x → z, z → x ≤ z → yandx ⊙ z ≤ y ⊙ z,

6. x ≤ (x → y) → y,

7. ((x → y) → y) → y = x → y,

8. x → (y → z) = y → (x → z),

9. x → y ≤ (z → x) → (z → y), x → y ≤ (y → z) → (x → z),

10. x → (x ∧ y) = x → y,

11. x ⊙ y = x ⊙ (x → x ⊙ y).

Proposition 2.3

([4, 30]). In a bounded semihoop (H, ⊙, →, ∧, 0, 1), the following properties hold: for allx, y, z ∈ H,

1. 1^∗ = 0, 0^∗ = 1,

2. x ≤ x^∗∗, wherex^∗∗ = (x^∗)^∗,

3. x ⊙ x^∗ = 0, x^∗∗∗ = x^∗,

4. x ≤ yimpliesy^∗ ≤ x^∗,

5. x → y ≤ y^∗ → x^∗,

6. (x → y^∗∗)^∗∗ = x → y^∗∗,

7. x^∗∗ ⊙ y^∗∗ ≤ (x ⊙ y)^∗∗,

8. (x^∗∗ ⊙ y)^∗ = (x ⊙ y)^∗.

Proposition 2.4

([14]). Let (H, ⊙, →, ∧, 1) be a semihoop and for allx, y ∈ H, we definex ⊔ y = ((x → y) → y) ∧ ((y → x) → x). Then the following conditions are equivalent:

1. ⊔ is an associative operation onH,

2. x ≤ yimpliesx ⊔ z ≤ y ⊔ zfor allx, y, z ∈ H,

3. x ⊔ (y ∧ z) ≤ (x ⊔ y) ∧ (x ⊔ z) for allx, y, z ∈ H,

4. ⊔ is the join operation onH.

Definition 2.5

([14]). A semihoop is called a ⊔-semihoop if it satisfies one of the equivalent conditions of Proposition 2.4.

Proposition 2.6

([30]). In a ⊔-semihoop, the following properties hold: for allx, y, z ∈ H,

1. x ⊙ (y ⊔ z) = (x ⊙ y) ⊔ (x ⊙ z),

2. x ⊔ (y ⊙ z) ≥ (x ⊔ y) ⊙ (x ⊔ z),

3. x ⊔ yⁿ ≥ (x ⊔ y)ⁿandx^m ⊔ yⁿ ≥ (x ⊔ y)^mn for any natural numbersm, n.

Proof

The proofs are easy, and we hence omit the details. □

Definition 2.7

([1, 2]). Let (H, →, ⊙, 1) be a hoop. His called:

1. a basic hoop if (x → y) → z ≤ ((y → x) → z) → zfor anyx, y, z ∈ H.

2. a Wajsberg hoop (x → y) → y = (y → x) → xfor anyx, y ∈ H.

3. a Gödel hoop ifx ⊙ x = xfor anyx ∈ H.

Proposition 2.8

([3]). Let (H, ⊙, →, 1) be a bounded hoop. Then

1. bounded basic hoops are definitionally equivalent to BL-algebras.

2. bounded Wajsberg hoops are definitionally equivalent to MV-algebras.

Definition 2.9

([4, 28]). Let (H, →, ⊙, ∧, 1) be a semihoop. His called:

1. a simple semihoop if it has exactly two filters: {1} andH.

2. a local semihoop if it has only one maximal filter.

Definition 3.1

Let (X, ⊙₁, →₁, ∧₁, 1₁) and (Y, ⊙₂, →₂, ∧₂, 1₂) be two semihoops. A mapσ : X → Yis called a state map fromX to Y, which is denoted simply by S-map, if it is satisfies the following conditions:

1. x ≤₁yimpliesσ (x) ≤₂σ (y);

2. σ(x →₁y) = σ((x →₁y) →₁y) →₂σ(y);

3. σ(x ⊙₁y) = σ(x) ⊙₂σ(x →₁ (x ⊙₁y));

4. σ(x) ⊙₂σ(y) ∈ σ(X);

5. σ (x) ∧₂σ (y) ∈ σ(X);

6. σ (x) →₂σ (y) ∈ σ(X).

for allx, y ∈ X.

The pair (X, Y, σ) is said to be a S-map semihoop. Moreover, ifX = Yandσ² = σ, thenσis called an internal state map onX, simply IS-map onX, in this case, (H,σ) is said to be an IS-map semihoop.

Example 3.2

LetH₁andH₂be two semihoops. Then the map 1_H1, defined by 1_H1(x) = 1₂for allx ∈ H₁, is a S-map fromH₁toH₂.

Example 3.3

LetHbe a semihoop. One can check thatid_His a S-map onH.

Example 3.4

LetH₁ = {0₁, a₁, b₁, c₁, 1₁} andH₂ = {0₂, a₂, b₂, c₂, 1₂}, where 0₁ ≤ a₁ ≤ b₁, c₁ ≤ 1₁and 0₂ ≤ a₂ ≤ b₂ ≤ c₂ ≤ 1₂. Define operations ⊙_iand →_ifori = 1, 2 as follows:

Then (H₁, →₁, ⊙₁, ∧₁, 1₁) and (H₂, →₂, ⊙₂, ∧₂, 1₂) are semihoops. Now, we define a mapσ : H₁ → H₂as follows:

One can check thatσis a S-map fromH₁toH₂.

Example 3.5

LetH = [0, 1] be the real interval. If forx, y ∈ H, we definex ⊙ y = max{0, x + y−1} andx → y = min{1, 1 − x + y}, then (H, ⊙, →, 0, 1) becomes a hoop, and hence it is a semihoop. Now we defineσ : H₁ → Has follows:

whereH₁is given in Example 3.4. One can easily check thatσis a S-map fromH₁toH.

Proposition 3.6

LetH_i, i = 1, 2 be semihoops andσbe a S-map fromH₁toH₂. Then we have: for anyx, y ∈ H₁,

1. σ(1₁) = 1₂;

2. σ(x ⊙₁y) ≥ σ(x) ⊙₂σ(y);

3. σ(x →₁y) ≤₂σ(x) →₂σ(y) and ifx ≤₁y, thenσ(x →₁y) = σ(x) →₂σ(y);

4. σ(H₁) is a subalgebra ofH₂.

Proof

1. Applying (SM2), we have σ(1₁) = σ(0₁ →₁ 0₁) = σ((0₁ →₁ 0₁) →₁ 0₁) →₂σ(0₁) = σ(1₁ →₁ 0₁) →₂σ(0₁) = σ(0₁) →₂σ(0₁) = 1₂.

2. From x ⊙₁y ≤ x ⊙₁y, we get y ≤₁x →₁ (x ⊙₁y) by Proposition 2.2(1). By (SM1), we have σ(y) ≤₂σ(x →₁( x ⊙₁y)). Applying (SM3), we get σ(x ⊙₁y) = σ(x) ⊙₂σ(x →₁ (x ⊙₁y)) ≥₂σ(x) ⊙₂σ(y).

3. By (SM2), we deduce σ(x →₁y) = σ((x →₁y) →₁y) →₂σ(y) ≤₂σ(x) →₂σ(y) by (5) and (6) of Proposition 2.2. If x ≤₁y, then σ(x) ≤₂σ(y). This means σ(x) →₂σ(y) = 1. Moreover, σ(x →₁y) = σ((x →₁y) →₁y) →₂σ(y) = σ(1₁ →₁y) →₂σ(y) = σ(y) →₂σ(y) = 1₂. Thus σ(x →₁y) = σ(x) →₂σ(y).

4. It follows from (SM4), (SM5), (SM6) and (1). □

Definition 3.7

LetH₁andH₂be two bounded semihoops. A S-mapσfromH₁toH₂is called a regular if it satisfiesσ(0₁) = 0₂.

Theorem 3.8

LetH_i, i = 1, 2 be two bounded Wajsberg semihoops andσbe a S-map fromH₁toH₂. Then the following are equivalent:

1. σis regular,

2. σ(x^∗₁) = (σ(x))^∗₂for anyx, y ∈ H₁,

3. x ⊥₁yimpliesσ(x) ⊥₂σ(y) for anyx, y ∈ H₁.

Proof

(1) ⇒ (2) By (1) and (SM2), we get σ(x^∗₁) = σ(x →₁ 0₁) = σ((x →₁ 0₁) →₁ 0₁) →₂σ(0₁) = σ(x) →₂ 0₂ = (σ(x))^∗₂.

(2) ⇒ (3) Suppose that x ⊥₁y. Then y^∗₁∗₁ ≤₁x^∗₁, it follows that σ(y^∗₁∗₁) ≤₁σ(x^∗₁). By (2) we have (σ(y))^∗₂∗₂ ≤₂ (σ(x))^∗. Hence we have σ(x) ⊥₂σ(y).

(3) ⇒ (1) Since 0₁^∗₁∗₁ = 1^∗₁, we get 1₁ ⊥₁ 0₁. By (3) we have σ(1₁) ⊥₂σ(0₁), and so σ(0₁)^∗₂∗₂ ≤₂σ(1₁)^∗₂. From Proposition 3.6(1), σ(0₁)^∗₂∗₂ ≤₂σ(1₁)^∗₂ = 1₂^∗₂ = 0₂, and hence σ(0₁)^∗₂∗₂ = 0₂. It follows that σ(0₁)^{∗₂∗₂∗₂} = 1₂. By Proposition 2.2(7), σ(0₁)^{∗₂∗₂∗₂} = σ(0₁)^∗₂ = 1₂, that is, σ(0₁) →₂ 0₂ = 1₂. This shows that σ(0₁) ≤₂ 0₂, so σ(0₁) = 0₂. □

Proposition 3.9

LetHbe a semihoop andσbe an IS-map onH. Then we have: for anyx, y ∈ H,

1. σ(1) = 1;

2. σ(x ⊙ y) ≥ σ(x) ⊙ σ(y);

3. σ(x → y) ≤ σ(x) → σ(y) and ifx ≤ y, thenσ(x → y) = σ(x) → σ(y);

4. σ(σ(x) ⊙ σ(y)) = σ(x) ⊙ σ(y);

5. σ(σ(x) ∧ σ(y)) = σ(x) ∧ σ(y);

6. σ(σ(x) → σ(y)) = σ(x) → σ(y);

7. σ(H) = Fix(σ), whereFix(σ) = {x ∈ H ∣ σ(x) = x};

8. σ(H) is a subalgebra ofH;

9. Ker(σ) is a filter ofH, whereKer(σ) = {x ∈ H∣σ(x) = 1}.

Proof

1. It follows from Proposition 3.6(1).

2. It follows from Proposition 3.6(2).

3. It follows from Proposition 3.6(3).

4. From (SM4), we have σ(x) ⊙ σ(y) = σ(z) for some z ∈ H. Hence σ(σ(x) ⊙ σ(y)) = σ²(z) = σ(z) = σ(x) ⊙ σ(y) by the definition of the IS-maps.

5. It is similar to (4).

6. It is similar to (1).

7. Let ∈ σ(H). Then x = σ(z) for some z ∈ H. Hence σ(x) = σ²(z) = σ(z) = x. So x ∈ Fix(σ). Conversely assume x ∈ Fix(σ). Then x = σ(x) ∈ σ(H). This shows that (7) is true.

8. It follows from (1), (2), (3) and (4).

9. It is straightforward. □

Theorem 3.10

LetHbe a semihoop. Then the following are equivalent:

1. His a hoop;

2. every IS-mapσonHsatisfiesσ(x) ⊙ σ(x → y) = σ(y) ⊙ σ(y → x) for allx, y ∈ H.

Proof

The proof is similar to that of He et al [30].(Theorem 4.7 ). □

Theorem 3.11

LetHbe a semihoop. Then the following are equivalent:

1. His idemopent;

2. every IS-mapσonHsatisfiesσ(x ∧ y) = σ(y) ⊙ σ(y) = σ(x) ⊙ σ(x → y) for allx, y ∈ H.

Proof

The proof is similar to that of He et al [30].(Theorem 4.8 ). □

Definition 3.12

([30]). LetHbe a bounded semihoop. A Riečan state onHis a founctions : H ⟶ [0, 1] such that the following conditions hold: for allx, y ∈ H,

1. s(1) = 1,

2. ifx ⊥ y, thens(x + y) = s(x) + s(y).

Theorem 3.13

LetHbe a semihoop andσbe an IS-map onH. Then there is a one-to-one correspondence betweenσ-compatible Riečan states onHand Riečan states onσ(H).

Proof

1. Suppose that s is a Riečan state on σ(H). Define a mapping φ : RS[σ(H)] → RS_σ[H] as follows: φ(s)(x) := s(σ(x)) for all x ∈ H. We will prove that φ(s) is a Riečan state on H. Clearly, φ(s)(1) = s(σ(1)) = s(1) = 1. Next, we will show that φ(s)(x + y) = φ(s)(x)+φ(s)(y) when x ⊥ y. In order to do this, we prove that σ(x + y) = σ(x) + σ(y) for x ⊥ y. Now, suppose that x ⊥ y. From Theorem 3.8(3), we have σ(x) ⊥ σ(y). Then σ(x) + σ(y) = (σ(x))^∗ → (σ(y))^∗∗. Moreover, σ(x + y) = σ(x^∗y^∗∗) = σ(x^∗) → σ(x^∗ ∧ y^∗∗). Since x ⊥ y, then y^∗∗ ≤ x^∗. It follows that σ(x + y) = σ(x^∗) → σ(y^∗∗) = (σ(x))^∗ → (σ(y))^∗∗ = σ(x) + σ(y). Now, we prove that φ(s)(x + y) = φ(s)(x)+φ(s)(y) when x ⊥ y. Since σ(x + y) = σ(x) + σ(y) for x ⊥ y, we have that φ(s)(x + y) = s(σ(x + y)) = s(σ(x) + σ(y)) = s(σ(x)) + s(σ(y)) = φ(s)(x)+φ(s)(y). Therefore, φ(s) is a Riečan state on H. Moreover, let σ(x) = σ(y) for all x, y ∈ H, then φ(s)(x) = s(σ(x)) = s(σ(y)) = φ(s)(y). Thus, φ(s) is a σ-compatible state on H. Therefore, the mapping φ is well defined.

2. Assume that s is a σ-compatible Riečan state on H. The mapping ψ : RS_σ[H] → RS[σ(H)] is defined by ψ(s)(σ(x)) := s(x) for all x ∈ H. Let σ(x) = σ(y), then s(x) = s(y) for all x, y ∈ H. Now, we show that ψ(s) is a Riečan state on σ(H). Let σ(x) ⊥ σ(y). Then σ(σ(x) + σ(y)) = σ((σ(x))^∗ → (σ(y))^∗∗) = σ(σ(x^∗) → σ(y^∗∗)) = σ(x)^∗ → σ(y)^∗∗ = σ(x) + σ(y). Based on this, we have that ψ(s)(σ(x) + σ(y)) = ψ(s)(σ(σ(x) + σ(y))) = s(σ(x) + σ(y)) = s(σ(x)) + s(σ(y)) = ψ(s)(σ(σ(x))) + ψ(s)(σ(σ(y))) = ψ(s)(σ(x)) + ψ(s)(σ(y)). Moreover, ψ(s)(σ(1)) = s(1) = 1. That means that ψ(s) is a Riečan state on σ(H). Therefore, ψ is a mapping of RS_σ[H] into RS[σ(H)].

3. Let s₁, s₂ be σ-compatible states on H and ψ(s₁) = ψ(s₂). Then we have ψ(s₁)(σ(x)) = ψ(s₂)(σ(x)), which implies s₁(x) = s₂(x) for all x ∈ H. Thus, s₁ = s₂. Now, suppose that s is a Riečan state on σ(H), then we have that (ψ(φ(s))(σ(x)) = φ(s)(x) = s(σ(x)). Therefore, ψ is a bijective mapping from RS_σ[H] onto RS[σ(H)] and ψ⁻¹ = φ. □

Definition 4.1

([10]). A Bosbach state on a bounded pseudo-hoop (A, ⊙, →, ⇝, 0, 1) is a functions : A → [0, 1] such that the following conditions hold: for anyx, y ∈ A:

1. s(x) + s(x → y) = s(y) + s(y → x);

2. s(x) + s(x ⇝ y) = s(y) + s(y ⇝ x);

3. s(0) = 0 ands(1) = 1.

Proposition 4.2

([10]). LetAbe a bounded pseudo-hoop andsbe a Bosbach state onA. Then for allx, y ∈ Athe following properties hold:

1. y ≤ ximpliess(y) ≤ s(x) ands(x → y) = s(x ⇝ y) = 1 − s(x) + s(y);

2. s(x⁻) = s(x^∼) = 1 − s(x), wherex⁻ = x → 0 andx^∼ = x ⇝ 0.

Definition 4.3

A state-morphism map on a bounded hoopAis a functions : A → [0, 1] such that:

1. m(0) = 0;

2. m(x → y) = min{1, 1 − m(x) + m(y)}.

Proposition 4.4

Every state-morphism map on a bounded hoopAis a Bosbach state onA.

Proof

Let m be a state-morphism map on A. Then m(1) = m(0⁻) = m(0 → 0) = min{1, 1 − m(0) + m(0) = 1}. (B3) holds. Consider m(x) + m(x → y) = m(x)+min{1, 1 − m(x) + m(y) = min{1 + m(x), 1 + m(y)} = m(y) + m(y → x). Hence (B1) is true. Since A is a hoop, (B2) is true, too. Combining the above arguments we get that m is a Bosbach state on A. □

Proposition 4.5

Every state-morphism on a bounded hoopAis a state map fromAto the hoopH = ([0, 1], ⊙, →, 0, 1) given in Example 3.5.

Proof

Assume m is a state-morphism map on a bounded hoop A. By Propositions 4.2 and 4.4, (SM1) holds.

Now we check (SM2). Let x, y ∈ A. By definition of 4.3, we have

For (SM3), we have

For (SM4), we have m(x) ⊙ m(y) = max{0, m(x) + m(y) − 1} = 1 − m(y) + m(x) = min{1, 1 − m(y) + m(x)} = m(y → x) and hence m(x) ⊙ m(y) ∈ m(A). This shows that (SM4) holds.

Note that m(x) → m(y) = min{1, 1 − m(x) + m(y)} = m(x → y). It follows that m(x) → m(y) ∈ m(A), that is (SM6).

For (SM5), we have m(x) ∧ m(y) = m(x) ⊙ (m(x) → m(y)). From (SM4) and (SM6), we get that (SM5) holds. □

Definition 4.6

([30]). A state semihoop is a pair (H,σ) whereHis a bounded semihoop andσ : H → His a mapping, called state operator, such that for anyx, y ∈ Hthe following conditions are satisfied:

1. σ(0) = 0;

2. x ≤ yimpliesσ(x) ≤ σ(y);

3. σ(x → y) = σ(x) → σ(x ∧ y);

4. σ(x ⊙ y) = σ(x) ⊙ σ(x → x ⊙ y);

5. σ(σ(x) ⊙ σ(y)) = σ(x) ⊙ σ(y);

6. σ(σ(x) ∧ σ(y)) = σ(x) ∧ σ(y).

Theorem 4.7

LetHbe a bounded semihoop andσ : H → Hbe a mapping onHpreserving → . Then the following conditions are equivalent:

1. (H,σ) is an IS-map semihoop;

2. (H,σ) is a state semihoop.

Proof

(1) ⇒ (2) If H is a bounded semihoop and σ : H → H is a mapping on H preserving →. Then σ(x → y) = σ(x → x ∧ y) = σ(x) → σ(x ∧ y). From proposition 3.9 and definition 4.6, we can obtain that (H,σ) a state semihoop.

(2) ⇒ (1) Let (H,σ) be a state semihoop and σ preserving →. We only need to prove that (SM2) holds. Since ((x → y) → y) → y = x → y, so we have σ((x → y) → y) → σ(y) = σ(((x → y) → y) → y) = σ(x → y). Thus σ is an IS-map on H and hence (H,σ) is an IS-map semihoop. □

Definition 4.8

([18]). A state residuated lattice is a pair (A, σ) whereAis a residuated lattice andσ: A → Ais a mapping, called state operator, such that for anyx, y ∈ Athe following conditions are satisfied:

1. σ(0) = 0;

2. x → y = 1 impliesσ(x) → σ(y) = 1;

3. σ(x → y) = σ(x) → σ(x ∧ y);

4. σ(x ⊙ y) = σ(x) ⊙ σ(x → x ⊙ y);

5. σ(σ(x) ⊙ σ(y)) = σ(x) ⊙ σ(y);

6. σ(σ(x) → σ(y)) = σ(x) → σ(y);

7. σ(σ(x) ∨ σ(y)) = σ(x) ∨ σ(y);

8. σ(σ(x) ∧ σ(y)) = σ(x) ∧ σ(y).

Theorem 4.9

LetHbe a bounded ⊔-semihoop andσ : H → Hbe a mapping onHpreserving → . Then the following conditions are equivalent:

1. σis an IS-map onH;

2. (H,σ) is a state residuated lattice.

Proof

(1) ⇒ (2) If H is a bounded ⊔-semihoop and σ : H → H is a mapping on H preserving →. Then σ(x → y) = σ(x → x ∧ y) = σ(x) → σ(x ∧ y). Moreover, by Proposition 3.9(5),(6), we have σ(σ(x) ⊔ σ(y)) = σ(((σ(x) → σ(y)) → σ(y)) ∧ ((σ(y) → σ(x)) → σ(x))) = σ(x) ⊔ σ(y). Therefore, (H,σ) is a state residuated lattice.

(2) ⇒ (1) Let (H,σ) be a state residuated lattice and σ preserving →. We only need to prove that (SM2) holds. Since ((x → y) → y) → y = x → y, so we have σ((x → y) → y) → σ(y) = σ(((x → y) → y) → y) = σ(x → y). Thus σ is an IS-map on H. □

Corollary 4.10

LetHbe a bounded basic hoop andσ : H → Hbe a mapping onHpreserving → . Then the following conditions are equivalent:

1. σis an IS-map onH;

2. (H,σ) is a state BL-algebra.

Proof

It follows from Proposition 2.8(1) and Theorem 4.9. □

Corollary 4.11

LetHbe a bounded Wajsberg hoop andσ : H → Hbe a mapping onH. Then the following conditions are equivalent:

1. σis an IS-map onHpreserving → ;

2. (H,σ) is a state MV-algebra.

Proof

It follows from Proposition 2.8(2) and Theorem 4.9. □

Definition 4.12

([5]). A state BCK-meet semilattice is a pair (A, σ) whereAis a BCK-meet semilattices andσ : A → Ais a mapping, called state operator, such that for anyx, y ∈ Athe following conditions are satisfied:

1. x → y = 1 impliesσ(x) → σ(y) = 1;

2. σ(x → y) = σ(x → y) → y) → σ(y);

3. σ(σ(x) → σ(y)) = σ(x) → σ(y);

4. σ(σ(x) ∧ σ(y)) = σ(x) ∧ σ(y).

Proposition 4.13

LetHbe a hoop andσ : H → Hbe an IS-map onH. Then the {→, ∧} subreduct of (H,σ) is a state BCK-meet semilattice.

Proof

It follows from Definition 3.1 and Definition 4.12. □

Definition 4.14

([27]). A state equality algebra is a pair (A, σ) whereAis an equality algebra andσ : A → Ais a mapping, called state operator, such that for anyx, y ∈ Athe following conditions are satisfied:

1. x ≤ yimpliesσ(x) ≤ σ(y);

2. σ(x∼ x ∧ y) = σ(x∼ x ∧ y)∼ y)∼σ(y);

3. σ(σ(x) → σ(y)) = σ(x) → σ(y);

4. σ(σ(x) ∧ σ(y)) = σ(x) ∧ σ(y).

Proposition 4.15

LetHbe a hoop andσ : H → Hbe an IS-map onH. Then the {∼, ∧} subreduction of (H,σ) is a state equality algebra, wherex ∼ y = x → (x ∧ y).

Proof

It follows from Definition 4.14. □

Definition 5.1

LetH₁andH₂be semihoops, σ : H₁ → H₂be a S-map fromH₁toH₂, Fbe a filter ofH₁. Ifσ⁻¹(σ(F)) ⊆ F, we callFto be a SM-filter of (H₁, H₂, σ).

Example 5.2

Consider the Example 3.4, one can easily check that the SM-filter of (H₁, H₂, σ) are {a₁, b₁, c₁, 1}, {1₁} andH₁.

Example 5.3

LetH₁andH₂be semihoops andσbe a S-map fromH₁toH₂. ThenKer(σ) = {x ∈ H₁∣σ(x) = 1₂} is a SM-filter of (H₁, H₂, σ).

Proof

Let K = Ker(σ) and x, y ∈ K. Then σ(x) = 1₂ and σ(y) = 1₂. By Proposition 3.6(2) we have σ(x ⊙₁y) ≥₂σ(x) ⊙₂σ(y) = 1₂ ⊙₂₁₂ = 1₂. This means x ⊙₁y ∈ K. Let x ∈ K and x ≤ y. Then 1₂ = σ(x) ≤ σ(y) and hence σ(y) = 1₂. This shows that y ∈ K. It follows that K is a filter of H₁. Moreover let x ∈ σ⁻¹σ(K). Then σ(x) ∈ σ(K) = {1₂} and hence σ(x) = 1₂. Therefore x ∈ K. This shows that σ⁻¹σ(K) ⊆ K, or K is a SM-filter of (H₁, σ). □

Definition 5.4

LetHbe a semihoop andσbe an IS-map onH.

1. A filterFofHis called state filter of (H,σ) ifx ∈ Fimpliesσ(x) ∈ Ffor allx ∈ H[31],

2. A filterFofHis called dual state filter of (H, σ) ifσ(x) ∈ Fimpliesx ∈ Ffor allx ∈ H,

3. A filterFofHis called strong state filter of (H, σ) if it is both a state filter and a dual state filter of (H, σ).

Proposition 5.5

LetHbe a semihoop andσbe an IS-map onH. Then each SM-filter ofHis a state filter onH.

Proof

Let x ∈ F. Then σ(x) ∈ σ (F). Therefore, σ(σ(x)) ∈ σ(F), that is σ(x) ∈ σ⁻¹(σ(F)) ⊆ F. So σ(x) ∈ F. □

Example 5.6

LetH = {0, a, b, 1} with 0 ≤ a, b ≤ 1. Consider the operation → and ⊙ as follows:

ThenHis semihoop. Now, we defineσfollows: σ0 = a, σa = a, σb = 1, σ 1 = 1. One can easily check thatσis an IS-map onH. It is clear that {a, 1} is a state filter of (H, σ), but it is not a SM-filter of (H, σ).

Proposition 5.7

LetHbe a semihoop, σbe an IS-map onHandF ⊆ H. Then the following are equivalent:

1. Fis a SM-filter ofH,

2. Fis a strong state filter onH.

Proof

(1) ⇒ (2) Let F be a SM-filter of (H, σ). By Proposition 5.5 we only need to prove that σ(x) ∈ F implies x ∈ F. Let σ(x) ∈ F. Then σ(x) = σ(σ(x)) ∈ σ(F). Hence there is t ∈ F such that σ(x) = σ(t). It follows from (1) that x ∈ σ⁻¹(σ(t)) ⊆ σ⁻¹(σ(F)) ⊆ F. That is x ∈ F.

(2)⇒ (1) Assume that F is a strong state filter on H. For x ∈ σ⁻¹(σ(F)), we have σ(x) ∈ σ(F). Since F is strong filter of H, we get x ∈ F and hence σ⁻¹(σ(F)) ⊆ F. □

Theorem 5.8

LetHbe a semihoop, σbe an IS-map onHandX ⊆ H. Then

1. 〈X〉_S = {x ∈ H ∣ x ≥ x₁ ⊙ σ(x₁) ⊙ ⋯ ⊙ x_n ⊙ σ(x_n), x_i ∈ X, m ∈ N},

2. (X)_DSis a dual state filter of (H, σ) containingX, and hence 〈X〉_DS ⊆ (X)_DS,

3. 〈X〉_σ = 〈X〉_S ∪ (X)_DS.

Proof

1. The proof is similar to that of He et al [30].(Theorem 4.13).

2. Let x, y ∈ (X)_DS. Then σ(x) ≥ x₁ ⊙ σ(x₁) ⊙ ⋯ ⊙ x_n ⊙ σ(x_n) for some x_i ∈ X, n ∈ N and σ(y) ≥ y₁ ⊙ σ(y₁) ⊙ ⋯ ⊙ y_m ⊙ σ(y_m) for some y_j ∈ X, m ∈ N. Hence σ(x ⊙ y) ≥ σ(x) ⊙ σ(y) ≥ x₁ ⊙ σ(x₁) ⊙ ⋯ ⊙ x_n ⊙ σ(x_n) ⊙ y₁ ⊙ σ(y₁) ⊙ ⋯ ⊙ y_m ⊙ σ(y_m). So x ⊙ y ∈ (X)_DS. Assume x ≤ y and x ∈ (X)_DS. Then σ(y) ≥ σ(x) ≥ x₁ ⊙ σ(x₁) ⊙ ⋯ ⊙ x_n ⊙ σ(x_n) for some x_i ∈ X. It follows that y ∈ (X)_DS. This shows that (X)_DS is a filter of H. Moreover, let σ(x) ∈ (X)_DS. Then σ(σ(x)) ≥ x₁ ⊙ σ(x₁) ⊙ ⋯ ⊙ x_n ⊙ σ(x_n) for some x_i ∈ X and hence σ(x) ≥ x₁ ⊙ σ(x₁) ⊙ ⋯ ⊙ x_n ⊙ σ(x_n) for some x_i ∈ X. This shows that x ∈ (X)_DS and hence (X)_DS is a dual state filter of (H, σ). Clearly X ⊆ (X)_DS.

3. Denote B = 〈X〉_S ∪ (X)_DS. Let x, y ∈ B. If x, y ∈ 〈X〉_S, then x ⊙ y ∈ 〈X〉_S ⊆ B by (1). If x, y ∈ (X)_DS, then x ⊙ y ∈ (X)_DS ⊆ B by (2). Let x ∈ 〈X〉_S and y ∈ (X)_DS. Then σ(x) ∈ 〈X〉_S since 〈X〉_S is a state filter of (H, σ) by (1). Hence σ(x) ≥ x₁ ⊙ σ(x₁) ⊙ ⋯ ⊙ x_n ⊙ σ(x_n) for some x_i ∈ X and σ(y) ≥ y₁ ⊙ σ(y₁) ⊙ ⋯ ⊙ y_m ⊙ σ(y_m) for some y_j ∈ X and hence σ(x ⊙ y) ≥ σ(x) ⊙ σ(y) ≥ x₁ ⊙ σ(x₁) ⊙ ⋯ ⊙ x_n ⊙ σ(x_n) ⊙ y₁ ⊙ σ(y₁) ⊙ ⋯ ⊙ y_m ⊙ σ(y_m) for some x_i,y_j ∈ X. It follows that x ⊙ y ∈ (X)_DS ⊆ B. Combining the above arguments we get that B is closed on ⊙. It is easy to check that if x ∈ B and x ≤ y then y ∈ B. Clearly X ⊆ B. Now we prove that B is a state filter. Let x ∈ B. If x ∈ 〈X〉_S, then σ(x) ∈ 〈X〉_S since 〈X〉_S is a state filter. If x ∈ (X)_DS, then σ(x) ∈ 〈X〉_S ⊆ B. So B is a state filter. Moreover we prove that B is a dual state filter. Let σ(x) ∈ B. If σ(x) ∈ 〈X〉_S, then x ∈ (X)_DS ⊆ B. Let σ(x) ∈ (X)_DS. Then x ∈ (X)_DS since (X)_DS is a dual state filter by (2). This shows that B a dual state filter. By Proposition 5.7, B a SM-filter. Let F be a SM-filter of (H, σ) containing X and x ∈ B. If x ∈ 〈X〉_S, then x ≥ x₁ ⊙ σ(x₁) ⊙ ⋯ ⊙ x_n ⊙ σ(x_n) for x_i ∈ X. Since X ⊆ F and F is a SM-filter of (H, σ), we have x₁ ⊙ σ(x₁) ⊙ ⋯ ⊙ x_n ⊙ σ(x_n) ∈ F. So x ∈ F. If x ∈ (X)_DS, then σ(x) ∈ 〈X〉_S by (1). If σ(x) ≥ x₁ ⊙ σ(x₁) ⊙ ⋯ ⊙ x_n ⊙ σ(x_n) for x_i ∈ X. Since F is a SM-filter of (H, σ) containing X, then x₁ ⊙ σ(x₁) ⊙ ⋯ ⊙ x_n ⊙ σ(x_n) ∈ F and hence σ(x) ∈ F. Note that F is also a dual state filter, we have x ∈ F. Combining the above arguments we get B ⊆ F. It follows that B = 〈X〉_σ. □

Proposition 5.9

LetHbe a semihoop, σbe an IS-map andFbe state filters of (H, σ) anda ∉ F. Then

1. 〈a〉_σ = {x ∈ H ∣ x ≥ (a ⊙ σ(a))ⁿ, n ≥ 1} ∪ {x ∈ H ∣ σ(x) ≥ (a ⊙ σ(a))ⁿ, n ≥ 1},

2. 〈F, {a}〉_σ = {x ∈ H ∣ x ≥ f ⊙ (a ⊙ σ(a))ⁿ, f ∈ F, n ≥ 1} ∪ {x ∈ H ∣ σ(x) ≥ f ⊙ (a ⊙ σ(a))ⁿ, f ∈ F, n ≥ 1},

3. ifa ≤ b, then 〈b〉_σ ⊆ 〈a〉_σ,

4. 〈a ⊙ a〉_σ = 〈a〉_σ,

5. 〈σ(a)〉_σ = 〈a〉_σ,

6. 〈a ⊙ σ(a)〉_σ = 〈a〉_σ,

7. ifHis a ⊔-semihoop, then 〈a〉_σ ∩ 〈b〉_σ = 〈(a ⊙ σ(a)) ⊔ (b ⊙ σ(b))〉_σ.

Proof

The proofs of (1)–(4) are obvious.

(5) Let x ∈ 〈σ(a)〉_σ. Then x ≥ (σ(a) ⊙ σ²(a))ⁿ = (σ(a))²ⁿ ≥ (a ⊙ σ(a))²ⁿ or σ(x) ≥ (a ⊙ σ(a))²ⁿ and hence x ∈ 〈a〉_σ. Conversely, let x ∈ 〈a〉_σ. Then x ≥ (a ⊙ σ(a))ⁿ or σ(x) ≥ (a ⊙ σ(a))ⁿ. Hence σ(x) ≥ (σ(a) ⊙ σ²(a))ⁿ. It follows that σ(x) ∈ 〈σ(a)〉_σ. Since 〈σ(a)〉_σ is a dual state filter we have x ∈ 〈σ(a)〉_σ.

(6) Since a ⊙ σ(a) ≤ a we have 〈a〉_σ ⊆ 〈a ⊙ σ(a)〉_σ by (3). Conversely, by use of (3), (4) and (5) we have 〈a ⊙ σ(a)〉_σ = 〈σ(a ⊙ σ(a))〉_σ ⊆ 〈σ(a) ⊙ σ²(a)〉_σ = 〈σ(a) ⊙ σ(a)〉_σ = 〈 σ(a)〉_σ = 〈a〉_σ.

(7) Suppose that H is a ⊔-semihoop. From a ⊙ σ(a) ≤ (a ⊙ σ(a)) ⊔ (b ⊙ σ(b)), we have that 〈(a ⊙ σ(a)) ⊔ (b ⊙ σ(b))〉_σ ⊆ 〈a ⊙ σ(a)〉_σ = 〈a〉_σ. Similarly, we can prove 〈(a ⊙ σ(a)) ⊔ (b ⊙ σ(b))〉_σ ⊆ 〈b〉_σ. Thus, 〈(a ⊙ σ(a)) ⊔ (b ⊙ σ(b))〉_σ ⊆ 〈a〉_σ ∩ 〈b〉_σ. Conversely, let x ∈ 〈a〉_σ ∩ 〈b〉_σ. Then there exist n, m, s, t ≥ 1, such that x ≥ (a ⊙ σ(a))ⁿ or σ(x) ≥ (a ⊙ σ(a))ⁿ, and x ≥ (b ⊙ σ(b))ⁿ or σ(x) ≥ (b ⊙ σ(b))ⁿ. To complete the proof, we divide four cases as following:

1. Let x ≥ (a ⊙ σ(a))ⁿ and x ≥ (b ⊙ σ(b))ⁿ. Then x ≥ (a ⊙ σ(a))ⁿ ⊔ (b ⊙ σ(b))ⁿ ≥ ((a ⊙ σ(a)) ⊔ (b ⊙ σ(b)))^ns ≥ (((a ⊙ σ(a)) ⊔ (b ⊙ σ(b))) ⊙ σ((a ⊙ σ(a)) ⊔ (b ⊙ σ(b))))^ns by Proposition 2.6(3). We deduce that x ∈ 〈(a ⊙ σ(a)) ⊔ (b ⊙ σ(b))〉_σ.

2. Let σ(x) ≥ (a ⊙ σ(a))ⁿ and σ(x) ≥ (b ⊙ σ(b))ⁿ. Similarly to (a) we can get σ(x) ∈ 〈(a ⊙ σ(a)) ⊔ (b ⊙ σ(b))〉_σ. Since 〈(a ⊙ σ(a)) ⊔ (b ⊙ σ(b))〉_σ is a dual state filter we have x ∈ 〈(a ⊙ σ(a)) ⊔ (b ⊙ σ(b))〉_σ.

3. Let x ≥ (a ⊙ σ(a))ⁿ and σ(x) ≥ (b ⊙ σ(b))ⁿ. Then σ(x) ≥ (σ(a) ⊙ σ(a))ⁿ ≥ (a ⊙ σ(a))²ⁿ and σ(x) ≥ (b ⊙ σ(b))ⁿ. Thus σ(x) ≥ (a ⊙ σ(a))²ⁿ ⊔ (b ⊙ σ(b))ⁿ ≥ ((a ⊙ σ(a)) ⊔ (b ⊙ σ(b)))^2nt ≥ (((a ⊙ σ(a)) ⊔ (b ⊙ σ(b))) ⊙ σ((a ⊙ σ(a)) ⊔ (b ⊙ σ(b))))^2nt. It follows that σ(x) ∈ 〈(a ⊙ σ(a)) ⊔ (b ⊙ σ(b))〉_σ. Since 〈(a ⊙ σ(a)) ⊔ (b ⊙ σ(b))〉_σ is a dual state filter we have x ∈ 〈(a ⊙ σ(a)) ⊔ (b ⊙ σ(b))〉_σ.

4. Let σ(x) ≥ (a ⊙ σ(a))ⁿ and x ≥ (b ⊙ σ(b))ⁿ. Similarly to the case (c) we can get x ∈ 〈(a ⊙ σ(a)) ⊔ (b ⊙ σ(b))〉_σ.

Combining the above arguments we can prove 〈a〉_σ ∩ 〈b〉_σ ⊆ 〈(a ⊙ σ(a)) ⊔ (b ⊙ σ(b))〉_σ. Therefore 〈a〉_σ ∩ 〈b〉_σ = 〈(a ⊙ σ(a)) ⊔ (b ⊙ σ(b))〉_σ. □

Definition 5.10

LetH₁andH₂be two semihoops andσbe a S-map fromH₁toH₂. A proper SM-filterFof (H₁, H₂, σ) is called a prime SM-filter of (H₁, H₂, σ), if for all SM-filtersF₁, F₂of (H₁, σ) such thatF₁ ∩ F₂ ⊆ F, thenF₁ ⊆ ForF₂ ⊆ F.

Example 5.11

Consider the Example 3.4, one can check thatF = {a₁, b₁, c₁, 1₁} is a prime SM-filter of (H₁, σ).

Theorem 5.12

LetHbe a ⊔-semihoop, σbe an IS-map andFbe a proper SM-filter of (H, σ). Then the following are equivalent:

1. Fis a prime SM-filter of (H, σ),

2. if ((x ⊙ σ(x)) ⊔ (y ⊙ σ(y))) ⊙ σ((x ⊙ σ(x)) ⊔ (y ⊙ σ(y))) ∈ Ffor somex, y ∈ H, thenx ∈ Fory ∈ F.

Proof

(1) ⇒ (2) Let ((x ⊙ σ(x)) ⊔ (y ⊙ σ(y))) ⊙ σ((x ⊙ σ(x)) ⊔ (y ⊙ σ(y))) ∈ F for some x, y ∈ H. Then 〈x〉_σ ∩ 〈y〉_σ = 〈(x ⊙ σ(x)) ⊔ (y ⊙ σ(y))〉_σ ⊆ F. Since F is a prime SM-filter of (H, σ), then 〈x〉_σ ⊆ F or 〈y〉_σ ⊆ F. Therefore, x ∈ F or y ∈ F.

(2) ⇒ (1) Suppose that F₁, F₂ ∈ SMF[L] such that F₁ ∩ F₂ ⊆ F and F₁ ⊈ F and F₂ ⊈ F. Then there exist x ∈ F₁ and y ∈ F₂ such that x, y ∉ F. Since F₁, F₂ are SM-filter of (H, σ), then x ⊙ σ(x) ∈ F₁ and y ⊙ σ(y) ∈ F₂. From x ⊙ σ(x), y ⊙ σ(y) ≤ (x ⊙ σ(x)) ⊔ (y ⊙ σ(y)), we obtain (x ⊙ σ(x)) ⊔ (y ⊙ σ(y)) ∈ F₁ ∩ F₂ ⊆ F and hence ((x ⊙ σ(x)) ⊔ (y ⊙ σ(y))) ⊙ σ((x ⊙ σ(x)) ⊔ (y ⊙ σ(y))) ∈ F₁ ∩ F₂ ⊆ F. By (2), we get that x ∈ F or y ∈ F, which is a contradiction. Therefore, F is a prime SM-filter of (H, σ). □

Definition 5.13

LetH₁andH₂be two semihoops andσbe a S-map fromH₁toH₂. A proper SM-filter of (H₁, H₂, σ) is called a maximal SM-filter if it not strictly contained in any proper SM-filter of (H₁, H₂, σ).

Example 5.14

LetH₁andH₂be two semihoops andσbe a S-map fromH₁toH₂in Example 3.4. One can easily check thatF = {a₁, b₁, c₁, 1₁} is a maximal SM-filter of (H₁, H₂, σ).

Proposition 5.15

LetHbe a bounded ⊔-semihoop, σbe an IS-map andFbe a proper SM-filter of (H, σ). Then the following are equivalent:

1. Fis a maximal SM-filter of (H, σ),

2. for anya ∉ F, there is an integern ≥ 1 such that (σ(a)ⁿ)^∗ ∈ F.

Proof

(1) ⇒ (2) Suppose that F is a maximal SM-filter of (H, σ), and let a ∉ F. Then 〈F,a〉_σ = H, which implies 0 ∈ 〈F, a〉_σ. Then there is f ∈ F and an integer n ≥ 1 such that 0 = f ⊙ (a ⊙ σ(a))ⁿ. So we have 0 = σ(0) ≥ σ(f) ⊙ σ(a)²ⁿ. Therefore, σ(f) ≥ (σ(a)²ⁿ)^∗. Thus, (σ(a)²ⁿ)^∗ ∈ F.

(2) ⇒ (1) Let a satisfy the condition. Since (σ(a)ⁿ)^∗ ⊙ (a ⊙ σ(a))ⁿ ≤ (σ(a)ⁿ)^∗ ⊙ (σ(a))ⁿ = 0 and (σ(a)ⁿ)^∗ ∈ F, we obtain 0 ∈ 〈F,a〉_σ, that is, 〈F, a〉_σ = H. Therefore, F is a maximal SM-filter of (H, σ). □

Proposition 5.16

LetH₁andH₂be two bounded semihoops andσbe a S-map fromH₁toH₂.

1. IfF₂is filter ofσ(H₁), thenσ⁻¹(F₂) is a SM-filter of (H₁, H₂, σ).

2. Ifσis an IS-map onHandFis a maximal filter ofσH, thenσ⁻¹(F) is a maximal SM-filter ofH.

Proof

1. Suppose that F₂ is a filter of σ(H₁). If x, y ∈ σ⁻¹(F₂), then σ(x),σ(y) ∈ F₂. It follows that σ(x) ⊙ σ(y) ∈ F₂. Since σ(x ⊙ y) ≥ σ(x) ⊙ σ(y) and σ(x ⊙ y) ∈ σ(H₁), we have σ(x ⊙ y) ∈ F₂, that is, x ⊙ y ∈ σ⁻¹(F₂). Let x, y ∈ H₁ such that x ∈ σ⁻¹(F₂) and x ≤ y. Then σ(x) ≤ σ(y). Since σ(x) ∈ F and σ(y) ∈ σ(H₁), we can obtain that σ(y) ∈ F₂, that is, y ∈ σ⁻¹(F₂). Thus, σ⁻¹(F₂) is a filter of H₁. Note that σ(σ⁻¹(x) = x for any x ∈ H₁. Hence σ⁻¹(σ(σ⁻¹(F)) = σ⁻¹(F). Thus σ⁻¹(F₂) is a SM-filter of H₁.

2. Now, suppose that F is a maximal filter of σ(H). Let a ∉ σ⁻¹(F), thus σ(a) ∉ F. By the maximality of F, there is an integer n ≥ 1 such that (σ(a)ⁿ)^∗ ∈ F ⊆ σ(H). Since σ ((σ(a)ⁿ)^∗) = (σ(a)ⁿ)^∗ ∈ F, we have (σ(a)ⁿ)^∗ ∈ σ⁻¹(F). Therefore, σ⁻¹(F) is a maximal SM-filter of H. □

Proposition 5.17

LetHbe a bounded semihoop andσbe an IS-map onHpreserving ⊙.

1. IfFis a SM-filter of (H, σ), thenσ(F) is a SM-filter of (σ(H),σ).

2. IfFis a maximal SM-filter of (H, σ), thenσ(F) is a maximal SM-filter of (σ(H),σ).

Proof

1. Let σ(x),σ(y) ∈ σ(F), then x, y ∈ σ⁻¹σ(F) ⊆ F. Since F is a filter, thus x ⊙ y ∈ F and hence σ(x) ⊙ σ(y) = σ(x ⊙ y) ∈ σ(F). Let σ(x), σ(y) ∈ σ(H) such that σ(x) ∈ σ(F) and σ(x) ≤ σ(y). Since σ(x) ∈ σ(F) we have x ∈ σ⁻¹σ(F) ⊆ F. So x ∈ F. By Proposition 5.7 we have σ(x) ∈ F. Since σ(x) ≤ σ(y) we get σ(y) ∈ F. Using Proposition 5.7 again we obtain y ∈ F, and so σ(y) ∈ σ(F). Thus, σ(F) is a filter of σ(H). Now let x ∈ σ(F). Then x = σ(t) for some t ∈ F and hence σ(x) = σ²(t) = σ(t) = x ∈ σ(F). It follows that σ(F) is a state filter of (H, σ). Let x ∈ σ(H) and σ(x) ∈ σ(F). Then x = σ(t) for some t ∈ H. Hence x = σ(t) = σ²(t) = σ(σ(t)) = σ(x) ∈ σ(F). This means that σ(F) is a dual state filter of (σ(H),σ). Therefore σ(H) is a strong state filter of (σ(H),σ). By Proposition 5.7 we have that σ(F) is a SM-filter of (σ(H),σ).

2. Now, let F be maximal and σ(a) ∉ σ(F). Then a ∉ F, and there is an integer n ≥ 1 such that (σ(a)ⁿ)^∗ ∈ F and hence σ ((σ(a)ⁿ)^∗) = (σ(a)ⁿ)^∗ ∈ σ(F). Since σ(σ(a)ⁿ) ≥ (σσ(a))ⁿ = (σ(a))ⁿ, we have (σ(a)ⁿ)^∗ ≥ σ ((σ(a)ⁿ)^∗). Hence (σ(a)ⁿ)^∗ ∈ σ(F). Therefore, σ(F) is a maximal SM-filter of (σ(H),σ). □

Corollary 5.18

LetHbe a bounded semihoop andσbe an IS-map onH.

1. IfFis a (maximal) filter ofσ(H), thenσ⁻¹(F) is a strong state (maximal)filter of (H, σ).

2. Ifσis preserving ⊙ andFis a strong state (maximal) filter of (H, σ), thenσ(F) is a strong state (maximal) filter of (σ(H),σ).

Proof

1. It follows from Proposition 5.7 and 5.16.

2. It follows from Proposition 5.7 and 5.17. □

Definition 5.19

LetHbe a semihoop andσ : H → Hbe an IS-map onH. If (H, σ) has exactly one maximal SM-filter, we call (H, σ) to be state local.

Theorem 5.20

LetHbe a semihoop andσbe an IS-map onH. Then the following are equivalent:

1. (H, σ) is state local;

2. σ(H) is local.

Proof

(1) ⇒ (2) Let F be the only maximal SM-filter of (H, σ). We prove that σ(F) is the only maximal filter of σ(H). First, σ(F) is a proper filter of σ(H). In fact, if σ(F) = σ(H), then 0 ∈ σ(F), which implies 0 ∈ F, a contradiction. Now, let G be a filter of σ(H), G ≠ σ(H) and let x ∈ G. It follows from Corollary 5.18(1) that σ⁻¹(G) is a SM-filter of (H, σ). Thus σ⁻¹(G) is a proper SM-filter of (H, σ). Moreover, if σ⁻¹(G) = H, then 0 ∈ σ⁻¹(G), so 0 ∈ G, a contradiction. It follows that σ⁻¹(G) ⊆ F. if x = σ(x) ∈ G, then x ∈ σ⁻¹(G), it follows that x ∈ F. But x = σ(x), so x ∈ σ(H). Thus G ⊆ σ(F). Hence σ(G) is the only maximal filter of σ(H). Therefore, σ(H) is local.

(2) ⇒ (1) Suppose that G is the only maximal filter of σ(H). By Corollary 5.18(1), we have that σ⁻¹(G) is a maximal SM-filter of (H, σ). We will prove that σ⁻¹(G) is the only maximal SM-filter of (H, σ). Let G be a SM-filter of (H, σ), F ≠ L. Then σ(F) is a proper filter of σ(H), so σ(F) ⊆ G. Let x ∈ F then σ(x) ∈ σ(F) ⊆ G. Thus, x ∈ σ⁻¹(G). It follows that F ⊆ σ⁻¹(G). Therefore, (H, σ) is state local. □

Definition 5.21

LetHa be semihoop andσ : H → Hbe an IS-map onH. If (H, σ) has two SM-filters {1} andH, we call (H, σ) to be simple.

Theorem 5.22

LetHa be semihoop andσ : H → Hbe an IS-map onHsuch thatσpreserving ⊙ . Then the following are equivalent:

1. (H, σ) is simple;

2. σ(H) is simple andKer(σ) = {1}.

Proof

(1) ⇒ (2) Let F be a filter of σ(H) and F ≠ {1}. It follows from Corollary 5.18(1) that σ⁻¹F is a SM-fiter of (H, σ). Since (H, σ) is state simple, we have that σ⁻¹(F) = {1} or σ⁻¹(F) = H. Notice that F ⊆ σ⁻¹F (if x ∈ F, then σx = x, that is, x ∈ σ⁻¹F, we obtain that σ⁻¹F ≠ {1}. Thus, σ⁻¹F = H. Then 0 ∈ σ⁻¹F, that is, 0 = σ 0 ∈ F. So we obtain that F = σH. Therefore, σH is simple.

By Example 5.3 we have Ker(σ) is a SM-filter of (H, σ) and Ker(σ) ≠ H. It follows that Ker(σ) = {1}.

(2) ⇒ (1) Let F be a SM-filter of (H, σ) and F ≠ {1}. By Corollary 5.18(2), we obtain that σF is a filter of σH. Since σH is simple, we obtain that σF = {1} or σF = σx. Since Ker(σ) = {1}, we have F ≠ {1}. Thus, σF = σx. Then 0 ∈ σF, that is, 0 ∈ F. It follows that F = H. Therefore (H, σ) is state simple. □