Abstract
The main goal of this paper is to discuss the generalized Bosbach states on EQ-algebras. The notions of generalized Bosbach states of type I and generalized Bosbach states of type II (resp. III, IV, V, VI and VII) are proposed. The equivalent characterizations of these generalized Bosbach states on special EQ-algebras are given. The properties of these generalized Bosbach states and the relationship between them are investigated.
Keywords
Introduction
In [30], Nov
In order to measure the average truth-value of propositions in Łukasiewicz logic, Mundici [28] introduced the notion of states on MV-algebras, which generalizes the probability measures on Boolean algebras. It is possible that random experiments can follow the rules of other logical system. Based on this, the notion of states has been extended to many logical algebras such as BL-algebras [31], R0-algebras [23], MTL-algebras [22], Rℓ-monoids [11], semi-divisible residuated lattices [32], residuated lattices [3, 21], their non-commutative cases [8–10, 18] and EQ-algebras [1]. These generalization mainly concentrated on the study of Rie
Totally different from the research direction above, Flaminio and Montagna [17] used the same MV-algrbas as the domain and codomain of a state and presented the notion of internal states on MV-algebras, which preserved the usual properties of states in the original states of Mundici. For detailed investigations, see [6, 14]. Afterwards, the notions of internal states were introduced into other algebraic structures [2, 20].
Recently, Georgescu, Mureşan and Ciungu [5, 19] noted that Bosbach states could be defined by the canonical structure of the standard MV-algebras [0, 1], replaced the codomain [0, 1] of Rie
The paper is motivated by the following considerations: (1) Whether these generalized Bosbach states exist on EQ-algebras; (2) which properties do these generalized Bosbach states possess in the case of existence? (3) whether these generailzed Bosbach states preserve the corresponding properties on residuated lattices. Based on this, we put forward the notion of generalized Bosbach states of type I and generalized Bosbach states of type II (resp. III, IV, V, VI, VII) on EQ-algebras and dedicate to the algebraric study of these generalized Bosbach states.
Preliminaries
(E1) (E, ∧ , 1) is a ∧-semilattice with top element 1;
(E2) (E, ⊗ , 1) is a commutative monoid and ⊗ is isotone in both arguments w.r.t. a ≤ b;
(E3) a ∼ a = 1;
(E4) ((a ∧ b) ∼ c) ⊗ (d ∼ a) ≤ c ∼ (d ∧ b);
(E5) (a ∼ b) ⊗ (c ∼ d) ≤ (a ∼ c) ∼ (b ∼ d);
(E6) (a ∧ b ∧ c) ∼ a ≤ a ∧ b ∼ a;
(E7) a ⊗ b ≤ a ∼ b.
Clearly, (E, ≤) is a partially ordered set, where a ≤ b iff a ∧ b = a. For all a, b ∈ E, we put a → b = (a ∧ b) ∼ a,
In what follows, E is an EQ-algebra unless otherwise specified. We appoint that E has the bottom element 0 and denote ¬a = a ∼ 0.
For convenience, for all a ∈ E and n ∈ N, we define if n = 0, then a0 = 1; if n ≥ 1, then a n = an-1 ⊗ a.
seperated, if for all a, b ∈ E, a ∼ b = 1 implies a = b. good, if for all residuated, if for all a, b, c ∈ E, (a ⊗ b) ∧ c = a ⊗ b iff a ∧ ((b ∧ c) ∼ b) = a. involutive (or an IEQ-algebra), if for all a ∈ E, ¬¬ a = a. a lattice-order EQ-algebra, if it has a lattice reduct. a lattice EQ-algebra (or ℓEQ-algebra), if it is a lattice-order EQ-algebra and satisfies: for all a, b, c, d ∈ E, ((a ∨ b) ∼ c) ⊗ (d ∼ a) ≤ c ∼ (d ∨ b). prelinear, if for all a, b ∈ E, 1 is the unique upper bound of the set {a → b, b → a}.
EQ-algebras have many interesting properties, we list some which are used in the sequel.
(P1)
(P2) a → b = a → a ∧ b.
(P3) If a ≤ b, then a → b = 1, a ∼ b = b → a, c → a ≤ c → b, b → c ≤ a → c.
(P4) a ⊗ b ≤ a ∧ b.
(P5) a → b ≤ (c → a) → (c → b), a → b ≤ (b → c) → (a → c), a → b ≤ ¬ b → ¬ a.
(P6) a → b ≤ a ∧ c → b ∧ c.
(P7)
(P8) b ≤ a → b.
If E is good, then
(P9) a → (b → c) = b → (a → c).
(P10) a ≤ (a → b) → b. Specially a ≤ ¬¬ a.
(P11) a ⊗ (a → b) ≤ a, b.
If E is a ℓEQ-algebra, then
(P12) a → b = a ∨ b → b = (a ∨ b) ∼ b.
(P13) a → b ≤ (a ∨ c) → (b ∨ c).
If E is prelinear and good, then
(P14) a ∼ b = d E (a, b) = (a ∨ b) → (a ∧ b).
If E is a good ℓEQ-algebra, then
(P15) a ∨ b → c = (a → c) ∧ (b → c). ¬ (a ∨ b) = ¬ a ∧ ¬ b.
If E is residuated, then
(P16) a ≤ b → a ⊗ b.
(P17) (a ⊗ b) → c = a → (b → c).
(1) E is separated.
(2) a ≤ b iff a → b = 1 for all a, b ∈ E.
(1) E is good.
(2) 1 → a = a for all a ∈ E.
(1) s (0) =0 and s (1) =1;
(2) s (a) + s (a → b) = s (b) + s (b → a).
(1) s is a Bosbach state.
(2) b ≤ a implies s (a → b) =1 - s (a) + s (b).
(3) s (a → b) =1 - s (a) + s (a ∧ b).
(1) a ≤ b implies s (a) ≤ s (b).
(2) s (¬ a) =1 - s (a).
(3) s (¬¬ a) = s (a).
(4)
Generalized Bosbach states on EQ-algebras
Throughout this section, let E, L be EQ-algebras with the bottom element 0 and s : E ⟶ L an arbitrary mapping.
Type I states
(1) s (0) =0;
(2) s (a → b) = s (a) → s (a ∧ b).
Then E is an EQ-algebra ([30]). The mapping id E : E ⟶ E is always an order-preserving type I state. Define s : E ⟶ E as s (0) =0, s (a) = s (b) = s (d) = s (1) =1, s (c) = d. It is routine to verify that s is a type I state but not order-preserving.
Conversely, by a ∧ b ≤ a and Proposition 2.3 (P2), s (a → b) = s (a → a ∧ b) = s (a) → s (a ∧ b).
Conversely, suppose b ≤ a, then s (a → b) = s (a ∨ b) → s (b) = s (a) → s (b).
Conversely, suppose b ≤ a, then s (a → b) = s ((a → b) ∧ (b → a)) = s (d E (a, b)) = s (a ∨ b) → s (a ∧ b) = s (a) → s (b).
(1) s (1) =1.
(2) s (¬ a) = ¬ s (a) , s (¬¬ a) = ¬¬ s (a).
(3) s ((a → b) → b) = s (a → b) → s (b).
(4) s (a → a2) = s (a) → s (a2).
(5) s (a → a ⊗ b) = s (a) → s (a ⊗ b).
If E is good, then
(6) s (((a → b) → b) → a) = s ((a → b) → b) → s (a). Specially, s (¬¬ a → a) = s (¬¬ a) → s (a).
If E, L is good, then
(7) s (a) ⊗ s (a → a ⊗ b) ≤ s (a ⊗ b).
If E is a ℓEQ-algebra, then
(8) s (a ∨ b → a) = s (b) → s (a ∧ b).
(9) s ((a → b) → b) = (s (a ∨ b) → s (b)) → s (b).
(2) s (¬ a) = s (a) → s (a ∧ 0) = s (a) → s (0) = ¬ s (a) , s (¬¬ a) = ¬ s (¬ a) = ¬¬ s (a).
By b ≤ a → b, a2 ≤ a, a ⊗ b ≤ a and Theorem 3.3, we have (3),(4),(5).
(6) By Proposition 2.3 (P10) and Theorem 3.3, s (((a → b) → b) → a) = s ((a → b) → b) → s (a). Put b = 0, then s (¬¬ a → a) = s (¬¬ a) → s (a).
(7) By (5) and Proposition 2.3 (P11), s (a) ⊗ s (a → a ⊗ b) = s (a) ⊗ (s (a) → s (a ⊗ b)) ≤ s (a ⊗ b).
(8) By Proposition 2.3 (P12) and Theorem 3.3, s (a ∨ b) → s (a) = s (a ∨ b → a) = s (b → a) = s (b) → s (b ∧ a) = s (b) → s (a ∧ b).
(9) By Proposition 2.3 (P8), Theorem 3.3 and Theorem 3.4, s ((a → b) → b) = s (a → b) → s (b) = (s (a ∨ b) → s (b)) → s (b).
(1) s (a → b) ≤ s (a) → s (b).
(2) s (d E (a, b)) ≤ d E (s (a) , s (b)).
(3) s (a → b) ⊗ s (b → a) ≤ d E (s (a) , s (b)).
(2) s (d E (a, b)) = s ((a → b) ∧ (b → a)) ≤ s (a → b) ∧ s (b → a) ≤ (s (a) → s (b)) ∧ (s (b) → s (a)) = d E (s (a) , s (b)).
(3) s (a → b) ⊗ s (b → a) ≤ (s (a) → s (b)) ⊗ (s (b) → s (a)) ≤ d E (s (a), s (b)).
Type II states
(1) s (0) =0;
(2) s (a) = s (a → b) → s (a ∧ b).
Then E is an EQ-algebra ([30]). Define s : E ⟶ E as s (0) = s (a) =0, s (b) = b, s (c) = s (d) = s (1) =1, It is routine to verify that s is a type II state.
Conversely, by Proposition 2.3 (P2) and a ∧ b ≤ a, s (a) = s (a → a ∧ b) → s (a ∧ b) = s (a → b) → s (a ∧ b).
Conversely, by b ≤ a ∨ b and Theorem 3.11, s (a ∨ b) = s (a ∨ b → b) → s (b) = s (a → b) → s (b).
(1) s (a → b) → s (b) = s (b → a) → s (a).
(2) b ≤ a implies s (a) = s (a → b) → s (b).
(3) s (a ∨ b) = s (a → b) → s (b).
If b ≤ a, then s (a) =1 → s (a) = s (b → a) → s (a) = s (a → b) → s (b).
(2)⇒(3)
See the proof of Theorem 3.12.
(3)⇒(1)
s (a → b) → s (b) = s (a ∨ b) = s (b ∨ a) = s (b → a) → s (a).
(1) s (1) =1.
(2) s is order-preserving: b ≤ a implies s (b) ≤ s (a).
(3) s (a) = ¬ s (¬ a).
(4) s (a → b) = s ((a → b) → b) → s (b).
(5) If E is residuated, then s (a ⊗ b) = ¬ s (¬ (a ⊗ b)) = ¬ s (a → ¬ b).
(2) Suppose b ≤ a, then s (b) ≤ s (a → b) → s (b) = s (a).
(3) By 0 ≤ a and Theorem 3.11, s (a) = s (a → 0) → s (0) = s (¬ a) →0 = ¬ s (¬ a).
(4) Analogous to (3), by b ≤ a → b, s (a → b) = s ((a → b) → b) → s (b).
(5) s (a ⊗ b) = ¬ s (¬ (a ⊗ b)) = ¬ s (a ⊗ b → 0) = ¬ s (a → (b → 0)) = ¬ s (a → ¬ b).
Conversely, replacing a with a ∧ b in the assumption, it follows that 1 = s (a ∧ b → b) = (s (b → a ∧ b) → s (a ∧ b)) → s (b) = (s (b → a) → s (a ∧ b)) → s (b). That is, s (b → a) → s (a ∧ b) ≤ s (b). Replacing b with a ∧ b, it follows that s (a → a ∧ b) = (s (a ∧ b → a) → s (a)) → s (a ∧ b). That is s (a → b) = s (a) → s (a ∧ b), Thus s (b → a) = s (b) → s (a ∧ b). Hence s (b → a) → s (a ∧ b) = (s (b) → s (a ∧ b)) → s (a ∧ b) ≥ s (b). This shows that s (b) = s (b → a) → s (a ∧ b).
(1) b ≤ a implies s (a → b) = s (a) → s (b).
(2) s ((a → b) → b) = s ((b → a) → a) = s (a ∨ b).
(3) s (¬¬ a) = s (a) , s (¬ a) = ¬ s (a). If L is good, then ¬¬ s (a) = s (a).
(4) b ≤ a and s (a) = s (b) implies s (a → b) =1. Specially, s (¬¬ a → a) =1.
If L is separated, then
(5) b ≤ a and s (a) = s (b) implies s (a ∘ c) = s (b ∘ c) for ∘ ∈ {∧ , ∨}. Specially s (¬¬ a ∘ c) = s (a ∘ c).
(6) b ≤ a and s (a) = s (b) implies s (c → a) = s (c → b). Specially, s (c → ¬¬ a) = s (c → a). s (c → a ∨ b) = s (c → ((a → b) → b)).
(7) b ≤ a and s (a) = s (b) implies s (a → c) = s (b → c). Specially s (¬¬ a → c) = s (a → c).
(8) s (¬ a → ¬ b) = s (b → a).
(2) s (a ∨ b) = s (a → b) → s (b) = s ((a → b) → b). Similarly, s (b ∨ a) = s ((b → a) → a). Again s (a ∨ b) = s (b ∨ a). Thus s ((a → b) → b) = s ((b → a) → a) = s (a ∨ b).
(3) Put b = 0 in (2), we have s (¬¬ a) = s (a). And s (¬ a) = s (a → 0) = s ((a → 0) →0) → s (0) = s (¬¬ a) →0 = ¬ s (¬¬ a) = ¬ s (a). If L is good, then ¬¬ s (a) = ¬¬ ¬ s (¬ a) = ¬ s (¬ a) = s (a).
(4) By (1) and (3).
(5) Let ∘ ∈ {∧ , ∨}. It is clear that s (b ∘ c) ≤ s (a ∘ c) by the monotonicity of ∘ and the order-preservation of s. By Proposition 2.3 (P6),(P13), we have 1 = s (a) → s (b) = s (a → b) ≤ s (a ∘ c → b ∘ c) = s (a ∘ c) → s (b ∘ c). Thus s (a ∘ c) ≤ s (b ∘ c). Hence s (a ∘ c) = s (b ∘ c). It is easy that s (¬¬ a ∘ c) = s (a ∘ c) from (3) and a ≤ ¬¬ a.
(6) Suppose b ≤ a, then c → b ≤ c → a. By the order-preservation of s, we have s (c → b) ≤ s (c → a). It follows from Proposition 2,3 (P5) that 1 = s (a) → s (b) = s (a → b) ≤ s ((c → a) → (c → b)) = s (c → a) → (c → b). Hence s (c → a) ≤ s (c → b). Thus s (c → a) = s (c → b). By (2) and (3), we have s (c → ¬¬ a) = s (c → a). s (c → a ∨ b) = s (c → ((a → b) → b)).
(7) The proof is similar to that of (6) by using the inequalities a → b ≤ (b → c) → (a → c).
(8) By Proposition 2.3 (P9), ¬a → ¬ b = ¬ a → (b → 0) = b → ¬¬ a . Thus (8) follows from s (b → ¬¬ a) = s (b → a).
(1) s is an order-preserving type I state.
(2) s is a type II state.
Type III states
(1) s (0) =0, s (1) =1;
(2) s (b) → s (a) = s (a → b) → s (b → a).
(1) s (a → b) ≤ s (a) → s (b).
(2) If L is sparated, a ≤ b implies s (a) ≤ s (b).
(3) If L is good, then b ≤ a implies s (a → b) = s (a) → s (b).
(2) Suppose a ≤ b, by (1), 1 = s (1) = s (a → b) ≤ s (a) → s (b). Thus s (a) ≤ s (b).
(3) Suppose b ≤ a, then s (a) → s (b) = s (b → a) → s (a → b) =1 → s (a → b) = s (a → b).
(1) s (¬ a) = ¬ s (a).
(2) s ((a → b) → b) = s (a → b) → s (b).
(3) s (a → a2) = s (a) → s (a2).
(4) s (a → a ⊗ b) = s (a) → s (a ⊗ b).
(5) s (a → a ∧ b) = s (a) → s (a ∧ b).
Type IV states
(1) s (0) =0, s (1) =1;
(2) s (a ∧ b) = s (a) ⊗ s (a → b).
Then E is a good EQ-algebra ([27]). Define s : E ⟶ E as s (0) = s (a) =0, s (b) = s (c) = b, s (d) = d, s (1) =1. It is routine to verify that s is a type IV state.
(1) s is a type IV state.
(2) s (a) ⊗ s (a → b) = s (b) ⊗ s (b → a).
(3) b ≤ a implies s (a) ⊗ s (a → b) = s (b).
It is clear by exchanging a and b.
(2)⇒(3)
Suppose b ≤ a, then s (a) ⊗ s (a → b) = s (b) ⊗ s (b → a) = s (b) ⊗1 = s (b).
(3)⇒(1)
Since a ∧ b ≤ a, then s (a) ⊗ s (a → a ∧ b) = s (a ∧ b). That is, s (a) ⊗ s (a → b) = s (a ∧ b).
(1) s (a ∨ b) ⊗ s (a → b) = s (b).
(2) s is a type IV state.
(3) s (a ∨ b) ⊗ s (d E (a, b)) = s (a ∧ b).
Replacing b with a ∧ b, it follows that s (a ∧ b) = s (a ∨ (a ∧ b)) ⊗ s (a → a ∧ b) = s (a) ⊗ s (a → b).
(2)⇒(3)
s ((a ∨ b) ∧ (a ∧ b)) = s (a ∨ b) ⊗ s ((a ∨ b) → (a ∧ b)) = s (a ∨ b) ⊗ s (d E (a, b)).
(3)⇒(1)
Replacing a with a ∨ b in the assumption, then s (b) = s ((a ∨ b) ∧ b) = s ((a ∨ b) ∨ b) ⊗ s ((a ∨ b → b) ∧ (b → a ∨ b)) = s (a ∨ b) ⊗ s (a → b).
(1) s is order-preserving.
(2) s (a) ⊗ s (¬ a) =0.
(3) If E is residuated, then s (a) ⊗ s (b) ≤ s (a ⊗ b), s (a) ⊗ s (a → b) = s (a ⊗ (a → b)).
(2) s (a) ⊗ s (a → 0) = s (a ∧ 0) =0.
(3) s (a) ⊗ s (b) ≤ s (a) ⊗ s (a → a ⊗ b) = s (a ∧ (a ⊗ b)) = s (a ⊗ b). And s (a) ⊗ s (a → b) ≤ s (a ⊗ (a → b)) ≤ s (a ∧ b) ≤ s (a) ⊗ s (a → b). Thus s (a) ⊗ s (a → b) = s (a ⊗ (a → b)).
Type V states
(1) s (0) =0;
(2) s (a → b) = s (a) → (s (b) ⊗ s (b → a)).
Then E is an EQ-algebra ([33]). Define s : E ⟶ E as s (0) =0, s (a) = s (b) = a, s (1) =1. It is routine to verify that s is a type V state.
(1) s (1) =1.
(2) s (a → b) ≤ s (a) → s (b).
(3) b ≤ a implies s (a → b) = s (a) → s (b).
(4) If L is separated, then s is order-preserving.
(5) If L is good, then s (a ∧ b) = s (a) ⊗ s (a → b).
(2) s (a → b) = s (a) → (s (b) ⊗ s (b → a)) ≤ s (a) → s (b).
(3) Suppose b ≤ a, then s (a → b) = s (a) → (s (b) ⊗ s (b → a)) = s (a) → (s (b) ⊗1) = s (a) → s (b).
(4) Suppose a ≤ b, then 1 = s (1) = s (a → b) = s (a) → (s (b) ⊗ s (b → a)), thus s (a) ≤ s (b) ⊗ s (b → a) ≤ s (b).
(5) Replacing a with a ∧ b, then 1 = s (a ∧ b → b) = s (a ∧ b) → (s (b) ⊗ s (b → a ∧ b)) = s (a ∧ b) → (s (b) ⊗ s (b → a)). Thus s (a ∧ b) ≤ s (b) ⊗ s (b → a). Then s (b ∧ a) ≤ s (a) ⊗ s (a → b). By (3), s (a → a ∧ b) = s (a) → s (a ∧ b). Thus s (a) ⊗ s (a → b) = s (a) ⊗ s (a → a ∧ b) = s (a) ⊗ (s (a) → s (a ∧ b)) ≤ s (a ∧ b). Hence s (a ∧ b) = s (a) ⊗ s (a → b).
Conversely, it is clear from Proposition 3.39 (4), (5).
Type VI states
(1) s (0) =0, s (1) =1;
(2) s (b) = s (a → b) ⊗ (s (b → a) → s (a)).
(1) a ≤ b implies s (a) ≤ s (b).
(2) s (b) = s (b → a) → s (b ∧ a).
(3) If E is a ℓEQ-algebra, then s (a ∨ b) = s (b → a) → s (a).
(4) If L is good, then s (a ∧ b) = s (a) ⊗ s (a → b).
(2) Replacing a with a ∧ b, then s (b) = s (a ∧ b → b) ⊗ s ((b → a ∧ b) → s (a ∧ b)) = 1⊗ (s (b → a) → s (a ∧ b)) = s (b → a) → s (b ∧ a).
(3) Replacing b with a ∨ b, then s (a ∨ b) = s (a → a ∨ b) ⊗ (s (a ∨ b → a) → s (a)) = 1⊗ (s (b → a) → s (a)) = s (b → a) → s (a).
(4) Replacing b with a ∧ b, then s (a ∧ b) = s (a → a∧ b) ⊗ (s (a∧ b → a) → s (a)) = s (a → b) ⊗ (s (1) → s (a)) = s (a) ⊗ s (a → b).
Conversely, suppose s is both a type II state and a type IV state, then s (a → b) ⊗ (s (b → a) → s (a)) = s (a → b) ⊗ s (b ∨ a) = s (b).
Conversely, replacing b with a ∧ b, then s (a ∧ b) = s (a ∧ b → a) → (s (a) ⊗ s (a → a ∧ b)) = 1→ (s (a) ⊗ s (a → b)) = s (a) ⊗ s (a → b). Thus s (b) = s (b → a) → s (b ∧ a). This show that s is both a type II state and a type IV state. By Theorem 3.49, we know that s is a type VI state.
Type VII states
(1) s (0) =0, s (1) =1;
(2) s (b) → s (a → b) = s (a) → s (b → a).
Conclusions
In this paper, we propose the notions of generalized Bosbach states of type I and generalized Bosbach states of type II (resp. III, IV, V, VI, VII) on EQ-algebras and give some specific examples of these generalized Bosbach states. We obtain their equivalent characterizations and discuss the relationship between them.
In summary, we devote to the algebraic studies of generalized Bosbach states on EQ-algebras. We mainly concentrate on the academic departments. Inspried by [24, 35], in the future work, we will consider combining EQ-algebras with relative knowledge of soft sets and use them in decision-making.
Footnotes
Acknowledgment
This work is supported by the National Natural Science Foundation of China, Grant No. 11701540, 61773019, 61379018.
