In this section, we introduce the notions of Rie*č*an states and state-morphisms on pseudo equality algebras. We mainly study some of their properties and investigate the relations between Rie*č*an states, state-morphisms and Bosbach states.

#### Definition 4.1

([21]). *Let* (*X*; ∼, ∽, ∧, 0, 1) *be a bounded pseudo equality algebra*. *A function s* : *X* → [0, 1] *is said to be a Bosbach state on X*, *if the following hold*:

*for all x*, *y* ∈ *x*.

#### Example 4.2

*Let* (*Y*; ∼_{2}, ∽_{2}, ∧_{2}, 0_{2}, 1_{2}) *be a bounded pseudo equality algebra given by Example 3.5*. *Define a function s* : *y* → [0, 1] *by s*(0_{2}) = 0, s(*a*_{2}) = *s*(*b*_{2}) = 0.5, *s*(1_{2}) = 1. *Then one can check that s is a unique Bosbach state on Y*.

#### Example 4.3

*Let* (*X*; ∼_{1}, ∽_{1}, ∧_{1}, 0_{1}, 1_{1}) *be a pseudo equality algebra defined by Example 3.5*. *Then one can check that the function s*: *X* → [0, 1] *defined by s*(0_{1}) = 0, *s*(*a*_{1}) = *s*(*b*_{1}) = *s*(*c*_{1}) = *s*(1_{1}) = 1 *is a unique Bosbach state on X*.

The following example shows that not every pseudo equality algebra has a Bosbach state.

#### Example 4.4

*Let* (*X*; ∼_{1}, ∽_{1}, ∧_{1}, 0_{1}, 1_{1}) *be a bounded pseudo equality algebra given by Example 3.3*, *and the function s* : *X* → [0, 1] *defined by s*(0_{1}) = 0, *s*(*a*_{1}) = *α*, *s*(*b*_{1}) = *β*, *s*(*c*_{1}) = *γ*, *s*(1_{1}) = 1, *be a Bosbach state on X*. *In (BS2)*, *(BS3)*, *taking x* = 0_{1}, *y* = *b*_{1}, *we obtain* *β* = 1 *and* *β* + *γ* = 1, *irespectively*. *Hence* *γ* = 0. *On the other hand*, *taking x* = 0_{1}, *y* = *a*_{1}, *we get* *α* + *β* = 1 *and* *α* + *γ* = 1, respectively. *According to* *β* = 1, *it implies* *α* = 0 *and so* *γ* = 1, *which is a contraction*. *This shows that* *X admits no Bosbach state*.

#### Proposition 4.5

([21]). *Let* (*X*; ∼, ∽, ∧, 0, 1) *be a bounded pseudo equality algebra and s be a Bosbach state on X*. *Then for all x*, *y* ∈ *x*, *the following hold*:

*If x* ≤ *y*, *then s*(*x*) ≤ *s*(*y*);

*If x* ≤ *y*, *then s*(*y* → *x*) = 1 – *s*(*y*) + *s*(*x*) = *s*(*y* ⤳ *x*);

*s*(*x* → *y*) = 1 – *s*(*x*) + *s*(*x* ∧ *y*) = *s*(*x* ⤳ *y*);

*s*(*x*^{−}) = 1 – *s*(*x*) = *s*(*x*^{∼});

*s*(*x*^{−∼}) = *s*(*x*) = *s*(*x*^{∼−}).

#### Proposition 4.6

*Let* (*X*; ∼, ∽, ∧, 0, 1) *be a bounded pseudo equality algebra and s* : *X* → [0, 1] *be a function such that s*(0) = 0. *Then the following conditions are equivalent*:

*s is a Bosbach state*;

*If x* ≤ *y*, *then s*(*y* → *x*) = 1 + *s*(*x*) – *s*(*y*) = *s*(*y* ⤳ *x*);

*s*(*x* → *y*) = 1 – *s*(*x*) + *s*(*x* ∧ *y*) = *s*(*x* ⤳ *y*).

#### Proof

(1) ⇒ (2) By Proposition 4.5 (2).

(2) ⇒ (3) By this proof of Proposition 4.5 (3).

(3) ⇒ (1) Assume (3) holds and *x*, *y* ∈ *x*. Then *s*(1) = *s*(*x* → 1) = 1 – *s*(*x*) + *s*(*x* ∧ 1) = 1 – *s*(*x*) + *s*(*x*) = 1. Also, *s*(*x*) + *s*(*x* → *y*) = *s*(*x*) + 1 – *s*(*x*) + *s*(*x* ∧ *y*) = 1 + *s*(*x* ∧ *y*) = *s*(*y*) + 1 – *s*(*y*) + *s*(*x* ∧ *y*) = *s*(*y*) + *s*(*y* → *x*). Similarly, we can prove (BS3). Thus *s* is a Bosbach state on *X*.

Let (*X*; ∼, ∧, 0, 1) be a bounded pseudo equality algebra and *s* : *X* → [0, 1] be a Bosbach state on *X*, we define the kernel of *s* by *Ker*(*s*) := {*x* ∈ *X* : *s*(*x*) = 1}. □

#### Proposition 4.7

*Let* (*X*; ∼, ∽, ∧, 0, 1) *be a bounded pseudo equality algebra and s be a Bosbach state on X*. *Then Ker*(*s*) *is a deductive system of X*.

#### Proof

Assume that *s* is a Bosbach state on *X*. Then for any *x*, *y* ∈ *x*, it follows from *s*(1) = 1 that 1 ∈ *Ker*(*s*). Let *x*, *x* → *y* ∈ *Ker*(*s*). Then *s*(*x*) = *s*(*x* → *y*) = 1. Since *x* ≤ *y* → *x*, then 1 = *s*(*x*) ≤ *s*(*y* → *x*). This implies that *s*(*y* → *x*) = 1. Again applying (BS2), we obtain *s*(*y*) = 1 and thus *y* ∈ *Ker*(*s*). Therefore, *Ker*(*s*) is a deductive system of *X*.

Let (X; ∼, ∧, 0, 1) be a bounded equality algebra and *s* be a Bosbach state on *X*. Then by Proposition 4.7 and [20] Proposition 3.9, the relation *θ* : x*θ**y* iff *x* ∼ *y* ∈ *Ker*(*s*) is a congruence relation on *X*. In this case, we denote the quotient algebra *X*/*θ* by *X*/*Ker*(*s*) and the congruence class of *x* ∈ *x* by *x*/*Ker*(*s*), where *x*/*Ker*(*s*) ∧ *y*/*Ker*(*s*) = (*x* ∧ *y*)/*Ker*(*s*), *x*/*Ker*(*s*) ∼ *y*/*Ker*(*s*) = (*x* ∼ *y*)/*Ker*(*s*), *x*/*Ker*(*s*) ≤ *y*/*Ker*(*s*) iff *x*/*Ker*(*s*) ∧ *y*/*Ker*(*s*) = *x*/*Ker*(*s*). □

#### Definition 4.8

*A pseudo equality algebra* (*X*; ∼, ∽, ∧, 1) *is said to be*

#### Example 4.9

*In Example 3.5 it is evident that* (*X*; ∼_{1}, ∽_{1}, ∧_{1}, 1_{1}) *is an involutive pseudo equality algebra*.

*Let X* = {0, *a*, *b*, *c*, 1} *with* 0 < *a* < *b* < *c* < 1 *and the operations* ∼,∽ *be given as follows*:

$$\begin{array}{}\overline{)\begin{array}{cccccc}\sim & \mathit{0}& \mathit{a}& \mathit{b}& \mathit{c}& \mathit{1}\\ \mathit{0}& \mathit{1}& \mathit{b}& \mathit{b}& \mathit{0}& \mathit{0}\\ \mathit{a}& \mathit{1}& \mathit{1}& \mathit{c}& \mathit{a}& \mathit{a}\\ \mathit{b}& \mathit{1}& \mathit{1}& \mathit{1}& \mathit{b}& \mathit{b}\\ \mathit{c}& \mathit{1}& \mathit{1}& \mathit{1}& \mathit{1}& \mathit{c}\\ \mathit{1}& \mathit{1}& \mathit{1}& \mathit{1}& \mathit{1}& \mathit{1}\end{array}}& & & \overline{)\begin{array}{cccccc}\backsim & \mathit{0}& \mathit{a}& \mathit{b}& \mathit{c}& \mathit{1}\\ \mathit{0}& \mathit{1}& \mathit{1}& \mathit{1}& \mathit{1}& \mathit{1}\\ \mathit{a}& \mathit{b}& \mathit{1}& \mathit{1}& \mathit{1}& \mathit{1}\\ \mathit{b}& \mathit{b}& \mathit{b}& \mathit{1}& \mathit{1}& \mathit{1}\\ \mathit{c}& \mathit{0}& \mathit{b}& \mathit{b}& \mathit{1}& \mathit{1}\\ \mathit{1}& \mathit{0}& \mathit{a}& \mathit{b}& \mathit{c}& \mathit{1}\end{array}}\end{array}$$

*One can check that* (*X*; ∼, ∽, ∧, 1) *is a good pseudo equality algebra*, *but it is not involutive*, *since a*^{−∼} = *a*^{∼−} = *b* ≠ *a*.

#### Theorem 4.10

*Let* (*X*; ∼, ∧, 0, 1) *be a bounded equality algebra and* *s* *be a Bosbach state on* *X*. *Then* (*X**/Ker*(*s*); ∼,∧, 0*/Ker*(*s*), 1*/Ker*(*s*)) *is an involutive equality algebra*.

#### Proof

Assume *s* is a Bosbach state on *X*. First, it follows that (*X**/Ker*(*s*); ∼, ∧, 0*/Ker*(*s*), 1*/Ker*(*s*)) is a bounded equality algebra. In the following, we prove that *X**/Ker*(*s*) is involutive. By (*S2*), we have *s*(*X*)+*s*(*X* → *x*^{––}) = *s*(*X*^{––}) + *s*(*X*^{––} → *x*). Since *s*(*X*) = *s*(*X*^{––}) by Proposition 4.5 (3), then *s*(*X* → *x*^{––}) = *s*(*X*^{––} → *x*). Again, *x* ≤ *x*^{––} by Proposition 2.6 (3), we get *s*(*X* → *x*^{––}) = *s*(1) = 1. This implies *s*(*X*^{––} → *x*) = 1 and so *x*^{––} → *x* ∈ Ker(*s*). Notice that *x* ≤ *x*^{––}, we obtain *x*^{––} ∼ *x* = *x*^{––} → *x* ∈ Ker(*s*). Hence *x*^{––} *θ* *x*. This shows *x*^{––}*/Ker*(*s*) = *X**/Ker*(*s*). Therefore, (*X**/Ker*(*s*))^{––} = *x*^{––}*/Ker*(*s*) = *X**/Ker*(*s*) and this proof is complete. □

#### Definition 4.11

*Let* (*X*; ∼, ∽, ∧, 0, 1) *be a bounded pseudo equality algebra*. *A state-morphism on* *X* *is a function* *m* : *X* → [0,1] *such that*

#### Proposition 4.12

*A state-morphism m is a Bosbach state on a bounded pseudo equality algebra* *X*.

#### Proof

Let *m* be a state-morphism on *X*. For any *x*, *y* ∈ *X*, *m*(1) = *m*(*X* → *x*) = *min*{1, 1 – *m*(*X*) + *m*(*X*)} = 1, and *m*(*X*) + *m*(*X* → *y*) = *m*(*X*) + *min*{1, 1 – *m*(*X*) + *m*(*y*)} = *min*{1 + *m*(*X*), 1 + *m*(*y*)} = *m*(*y*) + *min*{1, 1 – *m*(*y*) + *m*(*X*)} = *m*(*y*) + *m*(*y* → *x*). Similarly, we can prove (*BS3*). This shows *m* is a Bosbach state on *X*. □

#### Proposition 4.13

*Let* (*X*; ∼, ∽, ∧, 0, 1) *be a bounded pseudo equality algebra and* *s* *be a Bosbach state on* *X*. *Then* *s* *is a state-morphism on* *X* *if and only if* *s*(*X* ∧ *y*) = *min*{*s*(*X*), *s*(*y*)} *for all* *x*, *y* ∈ *x*.

#### Proof

Let *s* be a state-morphism on *X*. Then by Proposition 4.6, *s*(*X* ∧ *y*) = *s*(*X*) + *s*(*X* → *y*) – 1 = *s*(*X*) + *min*{1, 1 – *s*(*X*) + *s*(*y*)} – 1 = *min*{*s*(*X*), *s*(*y*)} for all *x*, *y* ∈ *x*. Conversely, let *s*(*X* ∧ *y*) = *min*{*s*(*X*),*s*(*y*)} for all *x*, *y* ∈ *x*. Taking *x* = *y* = 0, then *s*(0) = 0. Again by Proposition 4.6, we obtain *s*(*X* → *y*) = *s*(*X* ⤳ *y*) = 1 – *s*(*X*) + *s*(*X* ∧ *y*) = 1 – *m*(*X*) + *min*{*s*(*X*), *s*(*y*)} = *min*{1, 1 – *m*(*X*) + *m*(*y*)} = *m*(*X*) →_{R} *m*(*y*). Thus *s* is a state-morphism on *X*. □

#### Example 4.14

*Let* (*Y*;∼_{2}, ∽_{2}, ∧_{2}, 0_{2}, 1_{2}) *be a bounded pseudo equality algebra given by Example 3.3*. *Define a function s* : *Y* → [0, 1] *by* *s*(0_{2}) = 0, *s*(*a*_{2}) = *s*(*b*_{2}) = 0.5, *s*(1_{2}) = 1. *Then one can check that* *s* *is a Bosbach state on* *Y*, *but it is not a state-morphism on* *Y since s*(*a*_{2} ∧ *b*_{2}) = *s*(0_{2}) = 0 ≠ 0.5 = *min*{*s*(*a*_{2}), *s*(*b*_{2})}.

#### Example 4.15

*Let* *X* = {0, *a*, *b*, 1} *in which the Hasse diagram and the operation* ∼ *on* *X* *is below*:

$$\begin{array}{}\overline{)\begin{array}{ccccc}\sim & 0& a& b& 1\\ 0& 1& b& a& 0\\ a& b& 1& 0& a\\ b& a& 0& 1& b\\ 1& 0& a& b& 1\end{array}}& & & \overline{)\begin{array}{ccccc}\to & 0& a& b& 1\\ 0& 1& 1& 1& 1\\ a& b& 1& b& 1\\ b& a& a& 1& 1\\ 1& 0& a& b& 1\end{array}}\end{array}$$

*Then* (*X*; ∼, ∧,1) *is an equality algebra [20]*, *where the derived operation* → *as the above*. *The function s* : *X* → [0, 1] *is given by* *s*(0) = *s*(*a*) = 0, *s*(*b*) = *s*(1) = 1. *Then* *s* *is a Bosbach state on* *X*. *Furthermore*, *s* *is a state-morphism on* *X*, *since* *s*(*a* ∧ *b*) = *s*(0) = 0 = *min*{*s*(*a*), *s*(*b*)}.

#### Corollary 4.16

*In any linearly ordered bounded pseudo equality algebra* (*X*; ∼, ∽, ∧, 0, 1), *the Bosbach states coincide with the state-morphisms*.

#### Proposition 4.17

*Let* (*X*; ∼, ∽, ∧, 0, 1) *be a bounded involutive pseudo equality algebra and* *s* *be a Bosbach state on X*. *Then the following are equivalent*:

*s* *is a state-morphism on* *X*;

*s*(*X*^{∼} → *y*^{–∼}) = *min*{1,*s*(*X*) + *s*(*y*)};

*s*(*y*^{–} ⤳ *x*^{∼–}) = *min*{1, *s*(*X*) + *s*(*y*)},

*for all* *x*, *y* ∈ *x*.

#### Proof

(1) ⇒ (2) Let *s* be a state-morphism on *X*. Then by (M2) and Proposition 4.5, we get *s*(*X*^{∼} → *y*^{–∼}) = *min*{1, 1 – *s*(*X*^{∼}) + *s*(*y*^{–∼})} = *min*{1, 1 – 1 + *s*(*X*) + *s*(*y*)} = *min*{1, *s*(*X*) + *s*(*y*)}, for all *x*, *y* ∈ *x*.

(2) ⇒ (3) Assume that (2) holds. By Proposition 2.6 (7), *x*^{∼} → *y*^{–∼} = *y*^{–} ⤳ *x*^{∼–}. Hence *s*(*y*^{–} ⤳ *x*^{∼–}) = *s*(*X*^{∼} → *y*^{–∼}) = *min*{1, *s*(*X*) + *s*(*y*)}.

(3) ⇒ (1) Assume that (3) holds. Since *X* is involutive, then *s*(*X* ⤳ *y*) = *s*(*X*^{∼–} ⤳ *y*^{∼–}) = *min*{1, *s*(*y*) + *s*(*X*^{∼})} = *min*{1, 1 – *s*(*X*) + *s*(*y*)}. Again since *s* is a Bosbach state, we have *s*(0) = 0. Hence *s* is a state-morphism on *X*. □

#### Definition 4.18

*Let* (*X*; ∼, ∽, ∧, 0, 1) *be a bounded pseudo equality algebra*. *Two elements* *x*, *y* ∈ *X* *are said to be orthogonal*, *if* *x*^{–∼} ≤ *y*^{∼}, *we write by* *x* ⊥ *y*. *If* *x*, *y* ∈ *X* *are orthogonal*, *we define a binary operation* + *on* *X* *by* *x* + *y* := *y*^{∼} → *x*^{–∼}.

#### Proposition 4.19

*In any bounded pseudo equality algebra* (*X*; ∼, ∽, ∧, 0, 1), *the following properties hold for all* *x*, *y* ∈ *X*:

*x* ⊥ *y* *iff* *y*^{∼–}≤ *x*^{–};

*x* ⊥ *y iff x* ≤ *y*^{∼} *and* *x* ⊥ *y* *iff* *y* ≤ *x*^{–};

*x* ⊥ *y* *implies* *x* + *y* = *x*^{–} ⤳ *y*^{∼–};

*x* ⊥ *x*^{–} *and* *x* + *x*^{–} = 1;

*x*^{∼} ⊥ *x* *and* *x*^{∼} + *x* = 1;

0 ⊥ *x* *and* 0 + *x* = *x*^{∼–};

*x* ⊥ 0 *and* *x* + 0 = *x*^{–∼};

*x* ≤ *y* *implies* *x* ⊥ *y*^{–} *and* *x* + *y*^{–} = *y*^{–∼} → *x*^{–∼};

*x* ≤ *y* *implies* *y*^{∼} ⊥ *x* *and* *y*^{∼} + *x* = *y* ⤳ *x*^{∼–}.

#### Proof

Let *x* ⊥ *y*, *x*, *y* ∈ *x*. Then *x*^{–∼} ≤ *y*^{∼}. By Proposition 2.6 (3) and (4), *y*^{∼–} ≤ *x*^{∼–} = *x*^{–}. Conversely, let *y*^{∼–} ≤ *x*^{–}. Using Proposition 2.6 (3) and (4) again, we get *x*^{–∼} ≤ *y*^{∼–∼} = *y*^{∼}.

Let *x* ⊥ *y*. Then by (2) and Proposition 2.6 (3), *x* ≤ *x*^{–∼} ≤ *y*^{∼} and *y* ≤ *y*^{∼–} ≤ *x*^{–}.

Let *x* ⊥ *y*. Since *x*^{–} ⤳ *y*^{∼–} = ∼ → *x*^{–∼} by Proposition 2.6 (7), we have *x* + *y* = *x*^{–} ⤳ *y*^{∼–}.

Since *x*^{–∼} ≤ *x*^{–∼}, then *x* ⊥ *x*^{–} and *x* + *x*^{–} = *x*^{–∼} → *x*^{–∼} = 1.

Since *x*^{∼–} ≤ *x*^{∼–}, then *x*^{∼} ⊥ *x* and *x*^{∼} + *x* = *x*^{∼–} → *x*^{∼–} = 1.

By Proposition 2.6 (2), 0^{–∼} = 0 ≤ *x*^{∼}. Hence 0 ⊥ *x*, and 0 + *x* = *x*^{∼} → 0^{–∼} = *x*^{∼} → 0 = *x*^{∼–}.

By Proposition 2.6 (1), *x*^{–∼} ≤ 1 = 0^{∼}. Hence *x* ⊥ 0. Again by Proposition 2.6 (10), we get that *x* + 0 = 0^{∼} → *x*^{–∼} = 1 → *x*^{–∼} = *x*^{–∼}.

Let *x* ≤ *y*. Then by Proposition 2.6 (5), *x*^{–∼} ≤ *y*^{–∼}. Hence *x* ⊥ *y*^{–} and *x* + *y*^{–} = *y*^{–∼} → *x*^{–∼}.

Let *x* ≤ *y*. Then by Proposition 2.6 (4) and (5), we have *y*^{∼} ≤ *x*^{∼} and so *y*^{∼–∼} = *y*^{∼} ≤ *x*^{∼}. Hence *y*^{∼} ⊥ *x* and *y*^{∼} + *x* = *x*^{∼} → *y*^{∼–∼} = *x*^{∼} → *y*^{∼} = *y* ⤳ *x*^{∼–} by Proposition 2.6 (8). □

#### Definition 4.20

*Let* (*X*; ∼, ∽, ∧, 0, 1) *be a good bounded pseudo equality algebra*. *A Riečan state on* *X* *is a function* *s* : *X* → [0, 1] *such that*

#### Example 4.21

*Consider the good bounded pseudo equality algebra* (*X*; ∼, ∽, ∧, 0, 1) *given by Example 3.7*. *Define the function* *s* : *X* → [0, 1] *by* *s*(0) = 0,*s*(*a*) = *s*(*b*) = 0.5,*s*(1) = 1, *then one can check that* *s* *is a Riečan state on* *X*.

#### Proposition 4.22

*Let* *s* *be a Riečan state on a good bounded pseudo equality algebra* (*X*; ∼, ∽, ∧, 0, 1). *Then for all* *x*, *y* ∈ *X*, *the following hold*:

*s*(*X*^{–}) = 1 – *s*(*X*) = *s*(*X*^{∼});

*s*(0) = 0;

*s*(*X*^{–∼}) = *s*(*X*) = *s*(*X*^{∼–});

*x* ≤ *y* *implies* *s*(*X*) ≤ *s*(*y*) *and* *s*(*y*^{–∼} → *x*^{–∼}) = 1 + *s*(*X*) – *s*(*y*) = *s*(*y* ⤳ *x*^{∼–});

*s*(*X*^{–∼} → (*X* ∧ *y*)^{–∼}) = 1 – *s*(*X*)+*s*(*X* ∧ *y*) = *s*(*X* ⤳(*X* ∧ *y*)^{∼–}).

#### Proof

By Proposition 4.19 (4), we have *s*(*X* + *x*^{–}) = *s*(*X*) + *s*(*X*^{–}) = *s*(1) = 1. Hence *s*(*X*^{–}) = 1 – *s*(*X*). Similarly, by Proposition 4.19 (5), *s*(*X*^{∼}) = 1 – *s*(*X*).

By (1) and Proposition 2.6 (1), *s*(0) = *s*(1^{–}) = 1 – *s*(1) = 1 – 1 = 0.

By (2) and Proposition 4.19 (6), *s*(*X*^{∼–}) = *s*(0 + *x*) = *s*(0) + *s*(*X*) = 0 + *s*(*X*) = *s*(*X*). Similarly, by Proposition 4.19 (7), *s*(*X*^{–∼}) = *s*(*X*).

Let *x* ≤ *y*. Then by Proposition 4.19 (9), *y*^{∼} ⊥ *x* and *y*^{∼} + *x* = *y* ⤳ *x*^{∼–}. Hence *s*(*y* ⤳ *x*^{∼–}) = *s*(*y*^{∼} + *x*) = 1 + *s*(*X*) – *s*(*y*), and so *s*(*X*) – *s*(*y*) = *s*(*y*^{∼} + *x*) – 1 ≤ 0. Hence *s*(*X*) ≤ *s*(*y*). It is similar that *s*(*y*^{–∼} → *x*^{–∼}) = 1 + *s*(*X*) – *s*(*y*) by Proposition 4.19 (8).

It follows from *x* ∧ *y* ≤ *x* and (4) □.

#### Theorem 4.23

*In any good bounded pseudo equality algebra* (*X*; ∼, ∽, ∧, 0, 1), *each Bosbach state on* *X* *is a Riečan state*.

#### Proof

Assume that *s* is a Bosbach state on *X*. Then *s*(1) = 1. Let *x* ⊥ *y* for *x*, *y* ∈ *x*. Then *x*^{–∼} ≤ *y*^{∼}. By (2),(4) and (5) of Proposition 4.5, we have *s*(*X*+*y*) = *s*(*y*^{∼} → *x*^{–∼}) = 1 – *s*(*y*^{∼}) + *s*(*X*^{–∼}) = 1 – (1 – *s*(*y*)) + *s*(*X*) = *s*(*X*) + *s*(*y*). Therefore *s* is a Rie*č*an state on *X* □.

Note that the converse of the above theorem is not true in general. Let us see the following example.

#### Example 4.24

*Let* (*X*; ∼, ∽, ∧, 0, 1) *be the good pseudo equality algebra given by 4.9 (2)*. *Define a map* *s* : *X* → [0, 1] *by* *s*(0) = 0, *s*(*a*) = *s*(*b*) = *s*(*c*) = 0.5,*s*(1) = 1. *Then* *s* *is a Riečan state on* *X*, *but* *s* *is not a Bosbach state on* *X*. *Taking* *x* = *a*, *y* = *b* *in (BS2)*, *we can obtain that* 0.5 + 1 = 0.5 + *s*(*c*) *and so* *s*(*c*) = 1, *which is a contradiction*.

#### Theorem 4.25

*In any bounded involutive pseudo equality algebra* (*X*; ∼, ∽, ∧, 0, 1), *the Bosbach states and the Riečan states coincide on* *X*.

#### Proof

Assume that *s* is a Rie*č*an state on *X*. Then *s*(1) = 1 and *s*(0) = 0 by Proposition 4.22 (2). Let *x* ≤ *y*, then by Proposition 4.22 (4), *s*(*y*^{–∼} → *x*^{–∼}) = 1 – *s*(*X*) + *s*(*y*) = *s*(*y* ⤳ *x*^{∼–}). Since *X* is involutive, we obtain *s*(*y* → *x*) = *s*(*y*^{–∼} → *x*^{–∼}) = 1 – *s*(*X*) + *s*(*y*) = *s*(*y* ⤳ *x*^{∼–}) = *s*(*y* ⤳ *x*). Hence by Proposition 4.6, it follows that *s* is a Bosbach state on *X* □.

## Comments (0)

General note:By using the comment function on degruyter.com you agree to our Privacy Statement. A respectful treatment of one another is important to us. Therefore we would like to draw your attention to our House Rules.