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

Show that additionally we get:

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?

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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. ↩︎