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?