# Solutions to the Exercises from Session 3

## Exercises

The following are a bit harder puzzles for the courageous.

1. We have just shown that $$\mathsf{O} (\mathsf{O}p \rightarrow p)$$ holds in Anderson’s system with reflexive accessibility relation. Does it hold in SDL? If yes, give an argument. If no, find a counter-model.
2. What about $$\mathsf{O}\mathsf{O}p \rightarrow \mathsf{O}p$$?
3. In view of the distribution principle

• $$\mathsf{O}(p \rightarrow q) \rightarrow (\mathsf{O}p \rightarrow \mathsf{O} q)$$

$$\mathsf{O}(\mathsf{O}p \rightarrow p)$$ implies $$\mathsf{O}\mathsf{O}p \rightarrow \mathsf{O}p$$. Do you see how this is so?

4. Is there an SDL models where $$\mathsf{O}\mathsf{O}p \rightarrow \mathsf{O}p$$ holds but $$\mathsf{O}(\mathsf{O}p \rightarrow p)$$ not?

## Solutions

No it doesn’t. We give a counter-model $$M = \langle \{w_{1}, w_{2}\}, R, v, w_{1} \rangle$$. The accessibility relation is given in the following graph:

Our assignment is as follows:

world $$p$$
$$w_1$$ 1
$$w_2$$ 0

Then: $$M,w_1 \models \neg \mathsf{O}(\mathsf{O}p \rightarrow p)$$. The reason is that $$M,w_2 \models \mathsf{O}p \wedge \neg p$$.

Additional information: the principle $$\mathsf{O}(\mathsf{O}p \rightarrow p)$$ can be warranted if we impose secondary or step reflexivity on the accessibility relation. It says: for all $$w$$ and $$w’$$, if $$w R w^{\prime}$$ then $$w^{\prime} R w^{\prime}$$.

In the previous model we have:

• $$M,w_1 \models \mathsf{O} \mathsf{O} p$$ but
• $$M,w_1 \models \neg \mathsf{O} p$$ since $$M,w_2 \models \neg p$$.

Additional information: the principle $$\mathsf{O}\mathsf{O}p \rightarrow \mathsf{O}p$$ can be warrented if we impose density on the accessibility relation. It says: for all $$w$$ and $$w^{\prime}$$, if $$w R w^{\prime}$$ then there is a $$w^{\prime\prime}$$ for which $$wRw^{\prime\prime}$$ and $$w^{\prime\prime} R w^{\prime}$$.

In SDL the distribution principle holds for $$\mathsf{O}$$. So, for any formulas $$A$$ and $$B$$,

• $$\mathsf{O}(A \rightarrow B) \rightarrow (\mathsf{O} A \rightarrow \mathsf{O} B)$$.

If we now set $$A = \mathsf{O}p$$ and $$B= p$$ then:

• $$\mathsf{O}(\mathsf{O}p \rightarrow p) \rightarrow ( \mathsf{O} \mathsf{O} p \rightarrow \mathsf{O}p)$$.

No it doesn’t. We give a counter-model $$M = \langle \{w_{1}, w_{2}, w_3\}, R, v, w_{1} \rangle$$. The accessibility relation is given in the following graph:

Our assignment is as follows:

world $$p$$
$$w_1$$ 1
$$w_2$$ 0
$$w_3$$ 1

We have:

• $$M, w_1 \models \mathsf{O}\mathsf{O}p \rightarrow \mathsf{O}p$$ since $$M,w_1 \models \neg \mathsf{O} \mathsf{O} p$$. However,
• $$M, w_1 \models \neg \mathsf{O}(\mathsf{O}p \rightarrow p)$$ since $$M, w_2 \models \mathsf{O}p \wedge \neg p$$.