**Proof.**

(a) Let *T* = {*<sup>t</sup>*1, *<sup>t</sup>*2} , *R* = {(*<sup>t</sup>*1, *<sup>t</sup>*2)} . Let *I* be a set of indexes. For any *i*: *Ti* = *T*, *Ri* = *R*. Let *k*, *k* > 1, be a certain index of state of knowledge. Let F = {*Vi*}*<sup>i</sup>*∈*<sup>I</sup>* be a class of functions satisfied the following conditions:

$$\text{for any } i \text{ such that } i \le k \text{ holds } p \notin V\_i(t\_1), \tag{39}$$

> and

$$\text{for any } i \text{ such that } k < i \text{ holds } p \in V\_i(t\_1). \tag{40}$$

The *Vi* valuations are therefore selected so that the sentence *p* is true at the time of *t*1 in the states of knowledge with index not greater than *k* and at the same time it was not true at the time of *t*1 in the states of knowledge with index greater than *k*.

Let T = *<sup>T</sup>*, *<sup>R</sup>*. Let M = {*mi* : *i* ∈ *I*} . From the construction of the M model, we ge<sup>t</sup> that there are states of knowledge in the M which level of knowledge is not lesser than the level of knowledge of *m*1 in which at the moment *t*1 *p* is true and there are states of knowledge with a level of knowledge not lesser than the level of knowledge of *m*1, in which at the moment *t*1 is not true that *p*. Therefore, it is not true that in any state of knowledge *<sup>m</sup>*<sup>∗</sup>1 (∈ M) holds M, *<sup>m</sup>*<sup>∗</sup>1, *t*1 *p*. Therefore, by Definition **??** we ge<sup>t</sup> M, *m*1, *t*1 <sup>¬</sup>*p*. From the construction of the M model we ge<sup>t</sup> that in any state of knowledge *<sup>m</sup>*<sup>∗</sup>1 (∈ M) holds M, *<sup>m</sup>*<sup>∗</sup>1, *t*1 <sup>¬</sup>*p*. Because *<sup>t</sup>*1*R*<sup>∗</sup>1 *t*2, therefore, by the Definition **??** we have M, *<sup>m</sup>*<sup>∗</sup>1, *t*2 *P*¬*p*. By the Definition **??** we ge<sup>t</sup>

$$
\partial \mathfrak{M}, m\_1, t\_2 \models \neg P \neg p. \tag{41}
$$

Because the moment *t*1 is such that *t*1*R*1*t*2 and M, *m*1, *t*1 *p* so by the Definition **??**

$$2\mathfrak{N}, m\_1, t\_2 \nmid H \, p. \tag{42}$$

From (41), (42) and the Definition **??**: we have M, *m*1, *t*2 ¬*P*¬*p* → *H p*. Therefore **IKt** ¬*P*¬*p* → *H p*.

(b) The M model proposed in the proof of a) will be used to prove that ¬*H*¬*p* → *Pp* is not a tautology of **IKt**. Please note that from the construction of the model and by Definition **??** we have M, *<sup>m</sup>*<sup>∗</sup>1, *t*1 <sup>¬</sup>*p*. Because in any state of knowledge *<sup>m</sup>*<sup>∗</sup>1 (∈ M) the only time before *t*2 is the time *t*1, so by the Definition **??** for any *<sup>m</sup>*<sup>∗</sup>1 holds M, *<sup>m</sup>*<sup>∗</sup>1, *t*2 *H*¬*p*. Hence, by the Definition **??**

$$2\mathfrak{R}, m\_1, t\_2 \models \neg H \neg p. \tag{43}$$

From the construction of the model M we have M, *m*1, *t*1 *p*. Because *t*1*R*1*t*2, therefore by the Definition **??**

$$2\mathfrak{R}, m\_1, t\_2 \nleq p.c.\tag{44}$$

From (43), (44) and the Definition of **??** we obtain: M, *m*1, *t*2 ¬*H*¬*p* → *Pp*. Therefore **IKt**¬*H*¬*p* → *Pp*.

(c) We will now show that ¬*H p* → *P*¬*p* is not a tautology of **IKt**. Let *T*1 = {*<sup>t</sup>*1, *<sup>t</sup>*2} , *R*1 = {(*<sup>t</sup>*2, *<sup>t</sup>*1)} . Let the function *V*1 be such that *p* ∈ *V*1 (*<sup>t</sup>*2). States of knowledge in which the level of knowledge is not lower than the level of *m*1 we construct as follows:

$$T\_{i+1} = T\_i \cup \{t\_{i+2}\} \,. \tag{45}$$

$$R\_{i+1} = R\_i \cup \{ (t\_{i+2}, t\_1) \} \,. \tag{46}$$

$$V\_{i+1} \text{ is such that for } t \neq t\_{i+2}:\ p \in V\_{i+1}\left(t\right), \text{ and for } t = t\_{i+2}:\ p \notin V\_{i+1}\left(t\right). \tag{47}$$

State of knowledge *mi*+1 is an ordered triple *Ti*+1, *Ri*+1, *Vi*+<sup>1</sup>. Let F = {*Vi*}*<sup>i</sup>*∈*<sup>I</sup>* will be a class of functions satisfying the condition (47), T = *i*∈*I Ti*, *i*∈*I Ri*, M = {*mi* : *i* ∈ *I*} , the states of knowledge *mi* are constructed in accordance with conditions (45), (46) and (47). From the construction of the M model we ge<sup>t</sup> that in every state of knowledge *m*<sup>∗</sup> 1 (∈ <sup>M</sup>), there is a moment *t*, earlier than *t*1 such that *tR*∗ 1 *t*1 in which such that M, *m*<sup>∗</sup> 1, *t p*. So by the Definition **??** for any state of knowledge *m*<sup>∗</sup> 1 we have M, *m*<sup>∗</sup> 1, *t*1 *H p*. From the definition of **??** we have that:

$$
\exists \mathfrak{R}, m\_1, t\_2 \models \neg H \, p. \tag{48}
$$

From the construction of the model we have that if at some moment of time *t*, in any state of knowledge *mi* (∈ M) is that M, *mi*, *t p*, then in every state of knowledge *m*<sup>∗</sup> *i* (∈ M) holds M, *m*<sup>∗</sup> *i* , *t p*. In the state of knowledge *m*1 (∈ M) the only time before *t*1 is the moment *t*2. The moment *t*2 is such that *p* ∈/ *V*1 (*<sup>t</sup>*2). In the classical model, this would suffice to say that M, *m*1, *t*2 <sup>¬</sup>*p*. This is not the case in the temporal logic model built upon intuitionistic logic. From the way of constructing states of knowledge with no lower level of knowledge than the level of knowledge in the state *m*1 we have *p* ∈ *V*2 (*<sup>t</sup>*2). Therefore, by the Definition **??**

$$2\mathfrak{N}, m\_1, t\_2 \nleq\neg p.\tag{49}$$

By the Definition **??** we have M, *m*1, *t*1 ¬*H p* → *P* <sup>¬</sup>*p*.

We construct counter-examples for d), e) and f) in an analogous way.

In the **IKt** system, between the *G* and *F* and *H* and *P* operators there are no relationships usually found in temporal logic systems that are based on classical logic. However, the above conclusion is not sufficient to state that the operators *G* and *F* as well as *H* and *P* are not mutually definable in **IKt**. The conclusion is only that they do not occur between these operators definition relationships the same as those in *classical* tense logics. We will show that in intuitionistic temporal logic, temporal operators are not definable as they are in temporal logics based on classical propositional logic. We will show that intuitionistic temporal operators are not definable in any other way using sentence connectives and other intuitionistic temporal operators.

To show that a temporal operator is not definable in the **IKt**, two structures should be indicated such that the sentence with the considered operator at a moment *t* in one structure is true, and it is false in the other. On the other hand, all sentences in which the operator does not appear have the same logical value in both structures at the moment *t*.

**Theorem 17** ([**?** ])**.** *The intuitionistic temporal operators F and G as well as P and H are not each other definable in the* **IKt***.*

**Proof.** We will show first that the operator *F* is not definable if we use of intuitionistic sentence connectives and other temporal operators. We will show that *Fp* is not equivalent to any temporal formula in which the *F* operator does not occur.

*F*: Let *T*1 = {*<sup>t</sup>*1, *<sup>t</sup>*2}, *T*2 = {*<sup>t</sup>*1, *t*2, *<sup>t</sup>*3}, *R*1 = {(*<sup>t</sup>*1, *<sup>t</sup>*2)}, *R*2 = {(*<sup>t</sup>*1, *<sup>t</sup>*2),(*<sup>t</sup>*1, *<sup>t</sup>*3)}, *T* = *T*1 ∪ *T*2. Let *V*1 : *T*1 → 2AP be such that *p* ∈ *V*1 (*<sup>t</sup>*2) while *V*2 : *T*2 → 2AP will be such a function that: *V*2 (*<sup>t</sup>*1) = *V*1 (*<sup>t</sup>*1), *V*2 (*<sup>t</sup>*2) = *V*1 (*<sup>t</sup>*2) ∪ {*p*}, *V*2 (*<sup>t</sup>*3) = *V*1 (*<sup>t</sup>*2). Let F = {*<sup>V</sup>*1, *<sup>V</sup>*2} , *m*1 = *<sup>T</sup>*1, *R*1, *<sup>V</sup>*1, *m*2 = *<sup>T</sup>*2, *R*2, *<sup>V</sup>*2, M = {*<sup>m</sup>*1, *<sup>m</sup>*2} .

By means of structural induction, it can be shown that for any *ϕ* without the *F* operator we have

*G*:

$$\mathfrak{P}\mathfrak{M}, m\_1, t\_1 \models \!\!\!\!/ \mathfrak{p} \quad \text{iff} \quad \mathfrak{M}, m\_2, t\_1 \models \!\!\!\!/ \mathfrak{p} \,. \tag{50}$$

At the same time, M, *m*2, *t*1 *Fp* and M, *m*1, *t*1 *Fp*. Therefore, the *F* operator is not definable in **IKt**. We will now show that the operator *G* is not definable if we use of intuitionistic sentence connectives and other temporal operators. We will show that *Gp* is not equivalent to any temporal formula in which the *G* operator is not present.

Let *T*1 = {*<sup>t</sup>*1, *t*2, *<sup>t</sup>*3}, *T*2 = {*<sup>t</sup>*1, *t*2, *<sup>t</sup>*3} , *R*1 = {(*<sup>t</sup>*1, *<sup>t</sup>*2)}, *R*2 = {(*<sup>t</sup>*1, *<sup>t</sup>*2),(*<sup>t</sup>*1, *<sup>t</sup>*3)} , *T* = *T*1 ∪ *T*2. Let *V*1 : *T*1 → 2AP will be such a function that *p* ∈ *V*1 (*<sup>t</sup>*2). Let *V*2 : *T*2 → 2AP will be such a function that: *V*2 (*<sup>t</sup>*1) = *V*1 (*<sup>t</sup>*1), *V*2 (*<sup>t</sup>*2) = *V*1 (*<sup>t</sup>*2) ∪ {*p*} , *V*2 (*<sup>t</sup>*3) = *V*1 (*<sup>t</sup>*3), *p* ∈ *V*1 (*<sup>t</sup>*3). Let F = {*<sup>V</sup>*1, *<sup>V</sup>*2} . Let *m*1 = *<sup>T</sup>*1, *R*1, *<sup>V</sup>*1, *m*2 = *<sup>T</sup>*2, *R*2, *<sup>V</sup>*2. Let M = {*<sup>m</sup>*1, *<sup>m</sup>*2} . By means of structural induction, it can be shown that for any *ϕ* sentence without the *G* operator we have:

$$\mathfrak{M}, m\_1, t\_1 \models q \quad \text{iff} \quad \mathfrak{M}, m\_2, t\_1 \models q. \tag{51}$$

At the same time, M, *m*2, *t*1 *Gp* and M, *m*1, *t*1 *Gp*. So the *G* operator is not definable in **IKt**.

Similarly, we can to show that *P* and *H* are not each other definable in **IKt**.

It is not, however, that the operators *G*, *F*, *H*, *P* are completely independent of each other. Certain relationships between the operators *H* and *P* and *G* and *F* occur in **IKt**. We will prove some of them:

## **Theorem 18.**

