# Solutions to Exercises from Session 2 and some new Puzzles

## Exercise from Session 2

Anderson proposed to utilize a violation constant $$\mathbf{v}$$ to express or define an obligation operator:1

• $$\mathsf{O} A$$ is defined by $$\neg A \Rightarrow \mathbf{v}$$: $$A$$ is obliged if $$\neg A$$ leads to violation.

We have not given a precise meaning to $$\Rightarrow$$ yet. Anderson had in mind an implication from the family of relevance logics. In the literature it has often been proposed to use strict implication instead (also since it has a comparably simple logical model):

• $$\mathsf{O} A$$ is defined by $$\Box(\neg A \rightarrow \mathbf{v})$$: $$A$$ is obliged if $$\neg A$$ necessarily leads to violation.

Permission can then be defined by:

• $$\mathsf{P}A$$ is defined by $$\Diamond(A \wedge \neg \mathbf{v})$$: $$A$$ is permissible if it is possible that $$A$$ and at the same time no violation takes place.

The necessity modality $$\Box$$ may be the modality from the modal logic K or T. In both logics, just like in SDL, a model is a quadruple $$M = \langle W, R, @, v \rangle$$ with a non-empty set of worlds $$W$$, an accessibility relation $$R$$ among worlds, an actual world $$@ \in W$$ and an interpretation of atoms at worlds $$v$$. For K there are no restrictions on $$R$$, while for T, $$R$$ is reflexive, i.e., for all worlds $$w \in W$$, $$wRw$$. We have one special requirement for $$\mathbf{v}$$ that every model and every world has to fulfill: there is always an accessible world without violation.

• For all $$w \in W$$ there is a $$w^{\prime} \in W$$ for which $$wRw^{\prime}$$ and $$M,w^{\prime} \models \neg \mathbf{v}$$.

Note that this requirement implies that the accessibility relation will be serial, just like in SDL.

Formulas at worlds are interpreted just as in SDL, where

• $$M,w \models \Box A$$ (“it is necessary that $$A$$“) iff for all $$w^{\prime}$$ for which $$wRw^{\prime}$$, $$M,w^{\prime} \models A$$.
• $$M,w \models \Diamond A$$ (“it is possibly that $$A$$“) iff there is an $$w^{\prime}$$ for which $$w R w^{\prime}$$ and $$M,w^{\prime} \models A$$.

Try to verify that the following principles of SDL are also valid in Anderson’s systems based on K and T (at every world in every model):

1. $$\mathsf{O}p \rightarrow \mathsf{P} p$$
2. $$(\mathsf{O}p \wedge \mathsf{O}q) \rightarrow \mathsf{O}(p \wedge q)$$

1. $$\Box p \rightarrow \mathsf{O}p$$
2. $$\Box(p \rightarrow q) \rightarrow (\mathsf{O}p \rightarrow \mathsf{O} q)$$ (Note the similarly to O-Inh.)
3. and where $$R$$ is reflexive: $$\mathsf{O}(\mathsf{O}p \rightarrow p)$$.

Extra question: in your opinion, in what sense (if any) does Anderson’s system give a “reduction” of deontic logic to alethic modal logic, and in what sense (if any) not?

## Solution to the Exercise from Session 2

### $$\mathsf{O} p \rightarrow \mathsf{P} p$$

The formula holds in every model of Anderson’s system (in every world). To see this suppose $$M = \langle W, R, @, v \rangle$$ is an abitrary model and $$w \in W$$. To show that $$\mathsf{O} p \rightarrow \mathsf{P}p$$ we consider two cases:

1. $$M,w \models \mathsf{O}p$$. Thus, $$M,w \models \Box(\neg p \rightarrow \mathbf{v})$$. Note that since $$\Diamond \neg \mathbf{v}$$ holds in every world, $$M,w \models \Diamond \neg \mathbf{v}$$. Hence, there is a world $$w^{\prime}$$ for which $$wRw^{\prime}$$ and $$M,w^{\prime} \models \neg \mathbf{v}$$. Since $$M,w \models \Box(\neg p \rightarrow \mathbf{v})$$ and $$wRw^{\prime}$$, also $$M,w^{\prime} \models \neg p \rightarrow \mathbf{v}$$. By Modus Tollens, $$M,w^{\prime} \models p$$. Hence, $$M,w \models \Diamond (p \wedge \neg \mathbf{v})$$ and hence $$M,w \models \mathsf{P}p$$. Altogether, $$M,w \models \mathsf{O}p \rightarrow \mathsf{P}p$$.
2. $$M,w \models \neg \mathsf{O}p$$. But then $$M,w \models \mathsf{O}p \rightarrow \mathsf{P}p$$.

Altogether, in any case, $$M,w \models \mathsf{O}p \rightarrow \mathsf{P} p$$. Since $$M$$ and $$w$$ were arbitrary, our proof is finished.

### $$(\mathsf{O} p \wedge \mathsf{O} q) \rightarrow \mathsf{O}(p \wedge q)$$

The formula holds in every model of Anderson’s system (in every world). To see this suppose $$M = \langle W, R, @, v \rangle$$ is an abitrary model and $$w \in W$$. We again consider two cases:

1. $$M,w \models \mathsf{O}p \wedge \mathsf{O}q$$. Thus, $$M,w \models \Box(\neg p \rightarrow \mathbf{v})$$ and $$M,w \models \Box(\neg q \rightarrow \mathbf{v})$$. Now let $$w^{\prime}$$ be any world for which $$wRw^{\prime}$$. Thus, $$M,w^{\prime} \models \neg p \rightarrow \mathbf{v}$$ and $$M, w^{\prime} \models \neg q \rightarrow \mathbf{v}$$. We distinguish two cases.

1. $$M,w ^{\prime} \models \mathbf{v}$$. Then $$M,w^{\prime} \models \neg (p \wedge q) \rightarrow \mathbf{v}$$.
2. $$M,w^{\prime} \models \neg \mathbf{v}$$. Then $$M,w^{\prime} \models p$$ and $$M,w^{\prime} \models q$$ (by Modus Tollens). Thus, $$M,w^{\prime} \models (p \wedge q)$$ and so $$M,w^{\prime} \models \neg(p \wedge q) \rightarrow \mathbf{v}$$.

Since $$w^{\prime}$$ was an arbitrary accessible world from $$w$$, $$M,w \models \Box(\neg(p \wedge q) \rightarrow \mathbf{v})$$ and so $$M,w \models \mathsf{O}(p \wedge q)$$. Thus, $$M,w \models (\mathsf{O}p \wedge \mathsf{O}q) \rightarrow \mathsf{O}(p \wedge q)$$.

2. $$M,w \models \neg(\mathsf{O}p \wedge \mathsf{O}q)$$. But then $$M,w \models (\mathsf{O}p \wedge \mathsf{O}q) \rightarrow \mathsf{O}(p \wedge q)$$.

Thus, in any case, $$M,w \models (\mathsf{O}p \wedge \mathsf{O}q) \rightarrow \mathsf{O}(p \wedge q)$$. Since $$M$$ and $$w$$ were arbitrary, our proof is finished.

### $$\Box p \rightarrow \mathsf{O} p$$

The formula holds in every model of Anderson’s system (in every world). To see this suppose $$M = \langle W, R, @, v \rangle$$ is an abitrary model and $$w \in W$$. We consider two cases:

1. $$M,w \models \Box p$$. Thus, for all $$w^{\prime}$$ for which $$wRw^{\prime}$$, $$M,w^{\prime} \models p$$. Thus, $$M,w^{\prime} \models \neg p \rightarrow \mathbf{v}$$. And so, $$M,w \models \Box(\neg p \rightarrow \mathbf{v})$$, which is to say, $$M,w \models \mathsf{O} p$$. Thus, $$M,w \models \Box p \rightarrow \mathsf{O} p$$.
2. $$M,w \models \neg \Box p$$. But then $$M,w \models \Box p \rightarrow \mathsf{O} p$$.

Thus, in any case, $$M,w \models \Box p \rightarrow \mathsf{O} p$$. Since $$M$$ and $$w$$ were arbitrary, our proof is finished.

### $$\Box(p \rightarrow q) \rightarrow (\mathsf{O}p \rightarrow \mathsf{O}q)$$

The formula holds in every model of Anderson’s system (in every world). To see this suppose $$M = \langle W, R, @, v \rangle$$ is an abitrary model and $$w \in W$$. We consider two cases:

1. $$M,w \models \Box(p \rightarrow q)$$. Let $$w^{\prime}$$ be an arbitrary world for which $$w R w^{\prime}$$. Thus, $$M,w ^{\prime} \models p \rightarrow q$$. We have to show that $$M,w \models \mathsf{O}p \rightarrow \mathsf{O}q$$. If $$M,w \models \neg \mathsf{O}p$$ we immediately get $$M,w \models \mathsf{O}p \rightarrow \mathsf{O}q$$. Suppose $$M,w \models \mathsf{O}p$$. Thus, $$M,w \models \Box(\neg p \rightarrow \mathbf{v})$$. Thus, $$M, w^{\prime} \models \neg p \rightarrow \mathbf{v}$$. We consider two cases:
1. $$M,w^{\prime} \models p$$. Thus, by Modus Ponens, $$M,w^{\prime} \models q$$ and thus $$M,w^{\prime} \models \neg q \rightarrow \mathbf{v}$$.
2. $$M,w^{\prime} \models \neg p$$. Thus, by Modus Ponens, $$M,w^{\prime} \models \mathbf{v}$$ and thus $$M,w^{\prime} \models \neg q \rightarrow \mathbf{v}$$.
2. $$M,w \models \neg \Box(p \rightarrow q)$$. But then $$M,w \models \Box(p \rightarrow q) \rightarrow (\mathsf{O}p \rightarrow \mathsf{O}q)$$.

Thus, in any case, $$M,w \models \Box (p \rightarrow q) \rightarrow (\mathsf{O}p \rightarrow \mathsf{O}q)$$. Since $$M$$ and $$w$$ were arbitrary, our proof is finished.

### $$\mathsf{O}(\mathsf{O}p \rightarrow p)$$ (where $$R$$ is reflexive)

The formula holds in every model of Anderson’s system (in every world) with reflexive accessibility relation. To see this suppose $$M = \langle W, R, @, v \rangle$$ is an abitrary model and $$w \in W$$.

Assume for a contradiction that $$M,w \models \neg \mathsf{O}(\mathsf{O}p \rightarrow p)$$.

1. Thus, $$M,w \models \neg \Box(\neg(\mathsf{O}p \rightarrow p) \rightarrow \mathbf{v})$$.
2. Thus, there is a world $$w^{\prime}$$ for which $$w R w^{\prime}$$ and $$M,w^{\prime} \models \neg(\neg(\mathsf{O}p \rightarrow p) \rightarrow \mathbf{v})$$.
3. Hence, $$M,w^{\prime} \models \neg(\mathsf{O}p \rightarrow p)$$ and $$M,w^{\prime} \models \neg \mathbf{v}$$.
4. So, $$M,w^{\prime} \models \mathsf{O}p$$ and $$M, w^{\prime} \models \neg p$$.
5. So, $$M,w^{\prime} \models \Box(\neg p \rightarrow \mathbf{v})$$.
6. Since $$R$$ is reflexive, $$w^{\prime} R w^{\prime}$$ and so $$M,w^{\prime} \models \neg p \rightarrow \mathbf{v}$$.
7. By Modus Ponens, $$M,w^{\prime} \models \mathbf{v}$$, but this is a contradiction (with line 3).

So for all worlds $$w \in W$$, $$M,w \models \mathsf{O}(\mathsf{O}p \rightarrow p)$$.

## New Puzzles

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?

1. A dual approach, not with a violation constant, but with an ideality constant, has been proposed by Stig Kanger in New foundations for ethical theory. In R. Hilpinen (Eds.), Deontic Logic: Introductory and Systematic Readings (pp. 36–58). D. Reidel Company, Dordrecht. Anderson’s approach can be found in A reduction of deontic logic to alethic modal logic. Mind, 67(265), 100–103, 1958. [return]