## Exercises for Session 6 (Supererogation)

In the last session Laura and Thomas presented McNamara’s account of supererogation, as presented in his 1996 article in Mind. In order to get a better grip on his semantic framework, we investigate some properties that may hold or may not hold in his system. Consider always first what your pre-analytic intuitions tell you about the property and decide for yourself whether the formal framework is useful to get a more precise understanding of the underlying concepts. [Read More]

## Session 5: More on SDDL and Solution to Exercises

In the last session we had the following exercises.

## Exercise 1

How does an SDDL-model of the Chisholm scenario look like? Recall, the Chisholm scenario consists of the following norms:

1. $$\mathsf{O}(g \mid \top)$$
2. $$\mathsf{O}(t \mid g)$$
3. $$\mathsf{O}(\neg t \mid \neg g)$$

Using the semantics clauses for the $$\mathsf{O}$$-operator, we know that in any model of our given norms, there have to be three worlds that serve as cutoff points:

• $$w_1$$ for which $$M,w_1 \models g \wedge \top$$ (which means: $$M,w_1 \models g$$) and for all $$w \ge w_1$$, $$M,w \models \top \rightarrow g$$ (which means: $$M,w \models g$$);
• $$w_2$$ for which $$M,w_2 \models g \wedge t$$ and for all $$w \ge w_2$$, $$M,w \models g \rightarrow t$$;
• $$w_3$$ for which $$M,w_3 \models \neg g \wedge \neg t$$ and for all $$w \ge w_3$$, $$M,w_3 \models \neg g \rightarrow \neg t$$.

A model of the Chisholm scenario may look, for instance, as follows:

## Exercise 2

Do the following principles hold in SDDL (that is: are they validated in each SDDL model at each world)?

1. $$(\mathsf{O}(p \mid q) \wedge \mathsf{O}(s \mid q)) \rightarrow \mathsf{O}(p \wedge s \mid q)$$
2. $$\mathsf{O}(p \mid q) \rightarrow \mathsf{O}(p \mid q \wedge s)$$
3. $$\mathsf{O}(p \mid q) \rightarrow \mathsf{O}(p \wedge s \mid q)$$
4. $$\mathsf{O}(p \mid q) \rightarrow \mathsf{O}(p \wedge q \mid q)$$
5. $$\mathsf{O}(p \mid q) \wedge \mathsf{O}(p \mid s) \rightarrow \mathsf{O}(p \mid q \wedge s)$$
6. $$\mathsf{O}(p \mid q) \rightarrow \mathsf{P}(p \mid q)$$
7. $$\mathsf{O}(p \mid q) \rightarrow \mathsf{O}(q \rightarrow p \mid \top)$$
8. $$\mathsf{O}(p \rightarrow q \mid s) \rightarrow (\mathsf{O}(p \mid s) \rightarrow \mathsf{O}(q \mid s))$$
9. $$(\mathsf{O}(p \mid q) \wedge \mathsf{O}(s \mid q)) \rightarrow \mathsf{O}(p \mid q \wedge s)$$

I have demonstrated a proof of property 1 in details in the last blog entry.

I have given a counter-example in the last blog entry.

This property doesn’t hold. Here is a counter-example.

Our model $$M = \langle \{w\}, \{(w,w)\}, v \rangle$$ has exactly one world for which

world $$p$$ $$q$$ $$s$$
$$w$$ 1 1 0

Note that $$M \models \mathsf{O}(p \mid q)$$ while it is not the case that $$M \models \mathsf{O}(p \wedge s \mid q)$$ since there is no potential cutoff point, i.e., there is no world where $$p \wedge s \wedge q$$ is true.

This holds. Suppose we have a model $$M = \langle W, \ge, v \rangle$$ where $$M \models \mathsf{O}(p \mid q)$$. Thus, there is a $$w_1 \in W$$ for which

1. $$M,w_1 \models p \wedge q$$ and
2. for all $$w \ge w_1$$, $$M,w \models q \rightarrow p$$.

Since for all worlds $$w$$, $$M,w \models q \rightarrow q$$ we get with item 2,

$$3$$. $$M,w \models q \rightarrow (p \wedge q)$$ for all worlds $$w \ge w_1$$.

So, with items 1 and 3, $$M \models \mathsf{O}(p \wedge q \mid q)$$.

This does not hold in general. We give a counter-example, i.e., a model $$M = \langle W, \ge, v \rangle$$ for which (i) $$M \models \mathsf{O}(p \mid q) \wedge \mathsf{O}(p \mid s)$$ while (ii) it is not the case that $$M \models \mathsf{O}(p \mid q \wedge s)$$.

Let $$W = \{w_1, w_2\}$$ where the assignment is as follows:

world $$p$$ $$q$$ $$s$$
$$w_1$$ 1 1 0
$$w_2$$ 1 0 1

Let $${\ge} = \{(w_1,w_1), (w_2,w_2), (w_2,w_1)\}$$. It is easy to see that $$M$$ fulfills our desiderata.

This holds in SDDL. Recall that $$\mathsf{P}(p \mid q)$$ is defined by $$\neg \mathsf{O}(\neg p \mid q)$$. Suppose we have an arbitrary model $$M = \langle W, \ge, v \rangle$$ in which $$\mathsf{O}(p \mid q)$$ holds. (If it doesn’t then obviously $$M \models \mathsf{O}(p \mid q) \rightarrow \mathsf{P}(p \mid q)$$.) Thus, there is a $$w_1 \in W$$ for which $$M \models p \wedge q$$ and for all $$w \ge w_1$$, $$M,w \models q \rightarrow p$$.

We have to show that $$M \models \neg \mathsf{O}(\neg p \mid q)$$. That means, for every potential cutoff world $$w_2$$, i.e., every world $$w_2$$ for which $$M,w_2 \models q \wedge \neg p$$, there must be a world $$w_2^{\prime} \ge w_2$$ for which $$M,w_2^{\prime} \models \neg (q \rightarrow \neg p)$$ (and hence, $$M,w_2^{\prime} \models q \wedge p$$).

• If there is no potential cutoff point $$w_2$$, then clearly $$M \models \neg \mathsf{O}(\neg p \mid q)$$.
• Else, let $$w_2$$ be such a world for which $$M,w_2 \models q \wedge \neg p$$. Note that $$w_1 \ge w_2$$, since for all $$w \ge w_1$$, $$M,w \models \neg ( q \wedge \neg p)$$ (since $$M,w \models q \rightarrow p$$). Also recall that $$M,w_1 \models \neg(q \rightarrow \neg p)$$.

This concludes our proof.

This holds. To show this we again assume an arbitrary model $$M = \langle W, \ge, v \rangle$$ for which $$M \models \mathsf{O}(p \mid q)$$.

Thus, there is a $$w_1 \in W$$ for which $$M,w_1 \models p \wedge q$$ and for all $$w \ge w_1$$, $$M,w \models q \rightarrow p$$. Clearly, $$M,w_1 \models \top \wedge (q \rightarrow p)$$ and for all $$w \ge w_1$$, $$M,w \models \top \rightarrow (q \rightarrow p)$$. Thus, $$M \models \mathsf{O}(q \rightarrow p \mid \top)$$.

This holds. Let $$M = \langle W, \ge, v \rangle$$ be a model for which $$M \models \mathsf{O}(p \rightarrow q \mid s)$$. Thus, there is a $$w_1 \in W$$ for which $$M,w_1 \models s \wedge (p \rightarrow q)$$ and for all $$w \ge w_1$$, $$M,w \models s \rightarrow (p \rightarrow q)$$.

Now if it isn’t the case that $$M \models \mathsf{O}(p \mid s)$$ then $$M \models \neg \mathsf{O}(p \mid s)$$ and therefore $$M \models \mathsf{O}(p \mid s) \rightarrow \mathsf{O}(q\mid s)$$. Thus, $$M \models \mathsf{O}(p \rightarrow q \mid s) \rightarrow (\mathsf{O}(p \mid s) \rightarrow \mathsf{O}(q \mid s))$$.

Suppose now that $$M \models \mathsf{O}(p \mid s)$$. Thus, there is a $$w_2 \in W$$, for which $$M,w_2 \models s \wedge p$$ and for all $$w \ge w_2$$, $$M,w \models s \rightarrow p$$. We need to show that $$M \models \mathsf{O}(q\mid s)$$.

We have two possible scenarios. Either $$w_1 \ge w_2$$

or $$w_2 \ge w_1$$:

We discuss the former case. (The latter case is similar.) In order to show that $$M \models \mathsf{O}(q \mid s)$$ we need to locate a world $$w_3$$ for which $$M,w_3 \models q \wedge s$$ and for all $$w \ge w_3$$ ,$$M,w \models s \rightarrow q$$. Let for this $$w_3 = w_1$$. Note that $$M,w_1 \models s \wedge (p \rightarrow q)$$ and since $$w_1 \ge w_2$$, $$M,w_1 \models s \rightarrow p$$. Thus, $$M,w_1 \models s \rightarrow q$$ (by classical propositional logic) and since $$M,w_1 \models s$$ also $$M,w_1 \models q$$ (by Modus Ponens). Altogether, $$M,w_1 \models s \wedge q$$. Now, since for all $$w \ge w_1$$ we have both $$M,w \models s \rightarrow p$$ and $$M,w \models s \rightarrow (p \rightarrow q)$$ we also have $$M,w \models s \rightarrow q$$.

This principle holds. To prove it we consider a model $$M = \langle W, \ge, v \rangle$$ for which $$M \models \mathsf{O}(p \mid q) \wedge \mathsf{O}(s \mid q)$$ and show that $$M \models \mathsf{O}(p \mid q \wedge s)$$.

• Since $$M \models \mathsf{O}(p \mid q)$$ there is a $$w_1 \in W$$ for which $$M,w_1 \models p \wedge q$$ and for all $$w \ge w_1$$, $$M,w \models q \rightarrow p$$.
• Since $$M \models \mathsf{O}(s \mid q)$$ there is a $$w_2 \in W$$ for which $$M, w_2 \models q \wedge s$$ and for all $$w \ge w_2$$, $$M,w \models q \rightarrow s$$.

We have two possible scenarios: $$w_1 \ge w_2$$ and $$w_2 \ge w_1$$. We discuss the former (the latter is analogous).

In order to verify that $$M \models \mathsf{O}(p \mid q \wedge s)$$ we need to find a world $$w_3$$ for which $$M,w_3 \models p \wedge q \wedge s$$ and for all $$w \ge w_3$$, $$M,w \models (q \wedge s) \rightarrow p$$. Let $$w_3 = w_1$$. Then $$M,w_3 \models p \wedge q \wedge s$$ since $$M,w_1 \models p \wedge q$$ and since $$w_1 \ge w_2$$, $$M,w_1 \models q \rightarrow s$$ and so by Modus Ponens also $$M,w_1 \models s$$. Furthermore, for all $$w \ge w_3$$, $$M,w \models (q \wedge s) \rightarrow p$$ since $$M,w \models q \rightarrow p$$ and $$M,w \models q \rightarrow s$$.

## Some Potential Problems with SDDL

We spent the last minutes of our session discussion two potential problems with SDDL.

### The problem with Rational Monotony

One is that from $$\mathsf{O}(\neg f \mid \top)$$ and $$\mathsf{P}(a \mid \top)$$ follows in SDDL $$\mathsf{O}(\neg f \mid a)$$. In other words, the following is validated in SDDL: $$(\mathsf{O}(\neg f \mid \top) \wedge \mathsf{P}(a \mid \top) ) \rightarrow \mathsf{O}(\neg f \mid a)$$.

This is problematic for specificity-like scenarios such as Horty’s asparagus. After all, the following setting seems perfectly consistent:

1. In general, you ought not to eat with your fingers. $$\mathsf{O}(\neg f \mid \top)$$
2. However, if you’re being served asparagus, you are allowed to eat with your fingers. $$\mathsf{P}(f \mid a)$$.

Note that item 2 is equivalent to $$\neg \mathsf{O}(\neg f \mid a)$$. So, if we, additionally to our norms 1 and 2, have also $$\mathsf{P}(a \mid \top)$$ (“In general you’re allowed to eat asparagus.”), we get an inconsistent set of norms from which anything follows. This seems not satisfactory.

Since in classical propositional logic we have $$(A \wedge B) \rightarrow C$$ is equivalent to $$(A \wedge \neg C) \rightarrow \neg B$$ we can transform the principle above to: $$(\mathsf{O}(\neg f \mid \top) \wedge P(f \mid a)) \rightarrow \mathsf{O}(\neg a \mid \top)$$. This seems similarly counter-intuitive.

The deeper reason behind these kind of problems is that Rational Monotony is a principle underlying SDDL:

• Rational Monotony: From $$\mathsf{O}(A \mid B)$$ and $$\mathsf{P}(C \mid B)$$ follows $$\mathsf{O}(A \mid B \wedge C)$$.

Exercise: Try to show that Rational Monotony holds in SDDL.

### The problem with factual detachment

In SDDL, as we presented it, we have no way of validating Factual Detachment, the principle which says that if $$A$$ and $$\mathsf{O}(B \mid A)$$ then $$\mathsf{O}B$$.

Clearly, we would like to derive from conditional obligations and the given fact, the actual unconditional obligations that guide the actions of an agent. This is in general tricky, since we need to take into account exceptions and contrary-to-duty obligations. E.g., in the asparagus example, if we’re served asparagus the more general obligation to not eat with your fingers should be blocked.

One possible way to equip SDDL with factual detachment is by making use of an actual world. Then models are quadruples $$M = \langle W, \ge, v, @ \rangle$$ where $$@ \in W$$ is the “actual world”. We define $$M \models A$$ iff $$M,@ \models A$$. Now, suppose we have the premise set $$\{p, \mathsf{O}(q \mid p)\}$$. A model may look as follows:

Now, if our actual world $$@$$ is at least as ideal as $$w_1$$, we have $$M,@ \models q$$. However, if we are in a less ideal world than $$w_1$$ we might have to deal with a world for which $$M,@ \models \neg q$$. So,

• Problem 1: How to situate the actual world in the “right spot”? The right spot seems to be as ideal as the given premises allow for.

Another problem is that even if $$@ = w_3$$, we get $$M,@ \models q$$ while we expect $$M,@ \models \mathsf{O}q$$. Indeed we don’t want to derive new facts but rather unconditional obligations.

• Problem 2: How to model unconditional obligations in SDDL?

The problem of factual detachment will still concern us some more in future sessions.

## Session 5: More on SDDL and Solution to Exercises

In the last session we had the following exercises.

## Exercise 1

How does an SDDL-model of the Chisholm scenario look like? Recall, the Chisholm scenario consists of the following norms:

1. $$\mathsf{O}(g \mid \top)$$
2. $$\mathsf{O}(t \mid g)$$
3. $$\mathsf{O}(\neg t \mid \neg g)$$

Using the semantics clauses for the $$\mathsf{O}$$-operator, we know that in any model of our given norms, there have to be three worlds that serve as cutoff points:

• $$w_1$$ for which $$M,w_1 \models g \wedge \top$$ (which means: $$M,w_1 \models g$$) and for all $$w \ge w_1$$, $$M,w \models \top \rightarrow g$$ (which means: $$M,w \models g$$);
• $$w_2$$ for which $$M,w_2 \models g \wedge t$$ and for all $$w \ge w_2$$, $$M,w \models g \rightarrow t$$;
• $$w_3$$ for which $$M,w_3 \models \neg g \wedge \neg t$$ and for all $$w \ge w_3$$, $$M,w_3 \models \neg g \rightarrow \neg t$$.

A model of the Chisholm scenario may look, for instance, as follows:

## Exercise 2

Do the following principles hold in SDDL (that is: are they validated in each SDDL model at each world)?

1. $$(\mathsf{O}(p \mid q) \wedge \mathsf{O}(s \mid q)) \rightarrow \mathsf{O}(p \wedge s \mid q)$$
2. $$\mathsf{O}(p \mid q) \rightarrow \mathsf{O}(p \mid q \wedge s)$$
3. $$\mathsf{O}(p \mid q) \rightarrow \mathsf{O}(p \wedge s \mid q)$$
4. $$\mathsf{O}(p \mid q) \rightarrow \mathsf{O}(p \wedge q \mid q)$$
5. $$\mathsf{O}(p \mid q) \wedge \mathsf{O}(p \mid s) \rightarrow \mathsf{O}(p \mid q \wedge s)$$
6. $$\mathsf{O}(p \mid q) \rightarrow \mathsf{P}(p \mid q)$$
7. $$\mathsf{O}(p \mid q) \rightarrow \mathsf{O}(q \rightarrow p \mid \top)$$
8. $$\mathsf{O}(p \rightarrow q \mid s) \rightarrow (\mathsf{O}(p \mid s) \rightarrow \mathsf{O}(q \mid s))$$
9. $$(\mathsf{O}(p \mid q) \wedge \mathsf{O}(s \mid q)) \rightarrow \mathsf{O}(p \mid q \wedge s)$$

I have demonstrated a proof of property 1 in details in the last blog entry.

I have given a counter-example in the last blog entry.

This property doesn’t hold. Here is a counter-example.

Our model $$M = \langle \{w\}, \{(w,w)\}, v \rangle$$ has exactly one world for which

world $$p$$ $$q$$ $$s$$
$$w$$ 1 1 0

Note that $$M \models \mathsf{O}(p \mid q)$$ while it is not the case that $$M \models \mathsf{O}(p \wedge s \mid q)$$ since there is no potential cutoff point, i.e., there is no world where $$p \wedge s \wedge q$$ is true.

This holds. Suppose we have a model $$M = \langle W, \ge, v \rangle$$ where $$M \models \mathsf{O}(p \mid q)$$. Thus, there is a $$w_1 \in W$$ for which

1. $$M,w_1 \models p \wedge q$$ and
2. for all $$w \ge w_1$$, $$M,w \models q \rightarrow p$$.

Since for all worlds $$w$$, $$M,w \models q \rightarrow q$$ we get with item 2,

$$3$$. $$M,w \models q \rightarrow (p \wedge q)$$ for all worlds $$w \ge w_1$$.

So, with items 1 and 3, $$M \models \mathsf{O}(p \wedge q \mid q)$$.

This does not hold in general. We give a counter-example, i.e., a model $$M = \langle W, \ge, v \rangle$$ for which (i) $$M \models \mathsf{O}(p \mid q) \wedge \mathsf{O}(p \mid s)$$ while (ii) it is not the case that $$M \models \mathsf{O}(p \mid q \wedge s)$$.

Let $$W = \{w_1, w_2\}$$ where the assignment is as follows:

world $$p$$ $$q$$ $$s$$
$$w_1$$ 1 1 0
$$w_2$$ 1 0 1

Let $${\ge} = \{(w_1,w_1), (w_2,w_2), (w_2,w_1)\}$$. It is easy to see that $$M$$ fulfills our desiderata.

This holds in SDDL. Recall that $$\mathsf{P}(p \mid q)$$ is defined by $$\neg \mathsf{O}(\neg p \mid q)$$. Suppose we have an arbitrary model $$M = \langle W, \ge, v \rangle$$ in which $$\mathsf{O}(p \mid q)$$ holds. (If it doesn’t then obviously $$M \models \mathsf{O}(p \mid q) \rightarrow \mathsf{P}(p \mid q)$$.) Thus, there is a $$w_1 \in W$$ for which $$M \models p \wedge q$$ and for all $$w \ge w_1$$, $$M,w \models q \rightarrow p$$.

We have to show that $$M \models \neg \mathsf{O}(\neg p \mid q)$$. That means, for every potential cutoff world $$w_2$$, i.e., every world $$w_2$$ for which $$M,w_2 \models q \wedge \neg p$$, there must be a world $$w_2^{\prime} \ge w_2$$ for which $$M,w_2^{\prime} \models \neg (q \rightarrow \neg p)$$ (and hence, $$M,w_2^{\prime} \models q \wedge p$$).

• If there is no potential cutoff point $$w_2$$, then clearly $$M \models \neg \mathsf{O}(\neg p \mid q)$$.
• Else, let $$w_2$$ be such a world for which $$M,w_2 \models q \wedge \neg p$$. Note that $$w_1 \ge w_2$$, since for all $$w \ge w_1$$, $$M,w \models \neg ( q \wedge \neg p)$$ (since $$M,w \models q \rightarrow p$$). Also recall that $$M,w_1 \models \neg(q \rightarrow \neg p)$$.

This concludes our proof.

This holds. To show this we again assume an arbitrary model $$M = \langle W, \ge, v \rangle$$ for which $$M \models \mathsf{O}(p \mid q)$$.

Thus, there is a $$w_1 \in W$$ for which $$M,w_1 \models p \wedge q$$ and for all $$w \ge w_1$$, $$M,w \models q \rightarrow p$$. Clearly, $$M,w_1 \models \top \wedge (q \rightarrow p)$$ and for all $$w \ge w_1$$, $$M,w \models \top \rightarrow (q \rightarrow p)$$. Thus, $$M \models \mathsf{O}(q \rightarrow p \mid \top)$$.

This holds. Let $$M = \langle W, \ge, v \rangle$$ be a model for which $$M \models \mathsf{O}(p \rightarrow q \mid s)$$. Thus, there is a $$w_1 \in W$$ for which $$M,w_1 \models s \wedge (p \rightarrow q)$$ and for all $$w \ge w_1$$, $$M,w \models s \rightarrow (p \rightarrow q)$$.

Now if it isn’t the case that $$M \models \mathsf{O}(p \mid s)$$ then $$M \models \neg \mathsf{O}(p \mid s)$$ and therefore $$M \models \mathsf{O}(p \mid s) \rightarrow \mathsf{O}(q\mid s)$$. Thus, $$M \models \mathsf{O}(p \rightarrow q \mid s) \rightarrow (\mathsf{O}(p \mid s) \rightarrow \mathsf{O}(q \mid s))$$.

Suppose now that $$M \models \mathsf{O}(p \mid s)$$. Thus, there is a $$w_2 \in W$$, for which $$M,w_2 \models s \wedge p$$ and for all $$w \ge w_2$$, $$M,w \models s \rightarrow p$$. We need to show that $$M \models \mathsf{O}(q\mid s)$$.

We have two possible scenarios. Either $$w_1 \ge w_2$$

or $$w_2 \ge w_1$$:

We discuss the former case. (The latter case is similar.) In order to show that $$M \models \mathsf{O}(q \mid s)$$ we need to locate a world $$w_3$$ for which $$M,w_3 \models q \wedge s$$ and for all $$w \ge w_3$$ ,$$M,w \models s \rightarrow q$$. Let for this $$w_3 = w_1$$. Note that $$M,w_1 \models s \wedge (p \rightarrow q)$$ and since $$w_1 \ge w_2$$, $$M,w_1 \models s \rightarrow p$$. Thus, $$M,w_1 \models s \rightarrow q$$ (by classical propositional logic) and since $$M,w_1 \models s$$ also $$M,w_1 \models q$$ (by Modus Ponens). Altogether, $$M,w_1 \models s \wedge q$$. Now, since for all $$w \ge w_1$$ we have both $$M,w \models s \rightarrow p$$ and $$M,w \models s \rightarrow (p \rightarrow q)$$ we also have $$M,w \models s \rightarrow q$$.

This principle holds. To prove it we consider a model $$M = \langle W, \ge, v \rangle$$ for which $$M \models \mathsf{O}(p \mid q) \wedge \mathsf{O}(s \mid q)$$ and show that $$M \models \mathsf{O}(p \mid q \wedge s)$$.

• Since $$M \models \mathsf{O}(p \mid q)$$ there is a $$w_1 \in W$$ for which $$M,w_1 \models p \wedge q$$ and for all $$w \ge w_1$$, $$M,w \models q \rightarrow p$$.
• Since $$M \models \mathsf{O}(s \mid q)$$ there is a $$w_2 \in W$$ for which $$M, w_2 \models q \wedge s$$ and for all $$w \ge w_2$$, $$M,w \models q \rightarrow s$$.

We have two possible scenarios: $$w_1 \ge w_2$$ and $$w_2 \ge w_1$$. We discuss the former (the latter is analogous).

In order to verify that $$M \models \mathsf{O}(p \mid q \wedge s)$$ we need to find a world $$w_3$$ for which $$M,w_3 \models p \wedge q \wedge s$$ and for all $$w \ge w_3$$, $$M,w \models (q \wedge s) \rightarrow p$$. Let $$w_3 = w_1$$. Then $$M,w_3 \models p \wedge q \wedge s$$ since $$M,w_1 \models p \wedge q$$ and since $$w_1 \ge w_2$$, $$M,w_1 \models q \rightarrow s$$ and so by Modus Ponens also $$M,w_1 \models s$$. Furthermore, for all $$w \ge w_3$$, $$M,w \models (q \wedge s) \rightarrow p$$ since $$M,w \models q \rightarrow p$$ and $$M,w \models q \rightarrow s$$.

## Some Potential Problems with SDDL

We spent the last minutes of our session discussion two potential problems with SDDL.

### The problem with Rational Monotony

One is that from $$\mathsf{O}(\neg f \mid \top)$$ and $$\mathsf{P}(a \mid \top)$$ follows in SDDL $$\mathsf{O}(\neg f \mid a)$$. In other words, the following is validated in SDDL: $$(\mathsf{O}(\neg f \mid \top) \wedge \mathsf{P}(a \mid \top) ) \rightarrow \mathsf{O}(\neg f \mid a)$$.

This is problematic for specificity-like scenarios such as Horty’s asparagus. After all, the following setting seems perfectly consistent:

1. In general, you ought not to eat with your fingers. $$\mathsf{O}(\neg f \mid \top)$$
2. However, if you’re being served asparagus, you are allowed to eat with your fingers. $$\mathsf{P}(f \mid a)$$.

Note that item 2 is equivalent to $$\neg \mathsf{O}(\neg f \mid a)$$. So, if we, additionally to our norms 1 and 2, have also $$\mathsf{P}(a \mid \top)$$ (“In general you’re allowed to eat asparagus.”), we get an inconsistent set of norms from which anything follows. This seems not satisfactory.

Since in classical propositional logic we have $$(A \wedge B) \rightarrow C$$ is equivalent to $$(A \wedge \neg C) \rightarrow \neg B$$ we can transform the principle above to: $$(\mathsf{O}(\neg f \mid \top) \wedge P(f \mid a)) \rightarrow \mathsf{O}(\neg a \mid \top)$$. This seems similarly counter-intuitive.

The deeper reason behind these kind of problems is that Rational Monotony is a principle underlying SDDL:

• Rational Monotony: From $$\mathsf{O}(A \mid B)$$ and $$\mathsf{P}(C \mid B)$$ follows $$\mathsf{O}(A \mid B \wedge C)$$.

Exercise: Try to show that Rational Monotony holds in SDDL.

### The problem with factual detachment

In SDDL, as we presented it, we have no way of validating Factual Detachment, the principle which says that if $$A$$ and $$\mathsf{O}(B \mid A)$$ then $$\mathsf{O}B$$.

Clearly, we would like to derive from conditional obligations and the given fact, the actual unconditional obligations that guide the actions of an agent. This is in general tricky, since we need to take into account exceptions and contrary-to-duty obligations. E.g., in the asparagus example, if we’re served asparagus the more general obligation to not eat with your fingers should be blocked.

One possible way to equip SDDL with factual detachment is by making use of an actual world. Then models are quadruples $$M = \langle W, \ge, v, @ \rangle$$ where $$@ \in W$$ is the “actual world”. We define $$M \models A$$ iff $$M,@ \models A$$. Now, suppose we have the premise set $$\{p, \mathsf{O}(q \mid p)\}$$. A model may look as follows:

Now, if our actual world $$@$$ is at least as ideal as $$w_1$$, we have $$M,@ \models q$$. However, if we are in a less ideal world than $$w_1$$ we might have to deal with a world for which $$M,@ \models \neg q$$. So,

• Problem 1: How to situate the actual world in the “right spot”? The right spot seems to be as ideal as the given premises allow for.

Another problem is that even if $$@ = w_3$$, we get $$M,@ \models q$$ while we expect $$M,@ \models \mathsf{O}q$$. Indeed we don’t want to derive new facts but rather unconditional obligations.

• Problem 2: How to model unconditional obligations in SDDL?

The problem of factual detachment will still concern us some more in future sessions.

## 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 $$M, w_2 \models \mathsf{O}p \wedge \neg p$$.

## Session 3: More Problems with Normative Conditionals

In Chisholm (1963) we find the following example:

2. If you go, call.
3. If you don’t go, don’t call.
4. You don’t go.

We now present one possible analysis of this example in SDL:

We have translated the first premise by $$\mathsf{O} g$$, the second by $$\mathsf{O}(g \rightarrow t)$$ via wide-scoping, the third by $$\neg g \rightarrow \mathsf{O}\neg t$$ via narrow-scoping and the fourth by $$\neg g$$. As we can see, this leads to catastrophe in SDL: anything follows. Here,

• DDet (deontic detachment) is the principle according to which from $$\mathsf{O} A$$ and $$\mathsf{O}(A \rightarrow B)$$ follows $$\mathsf{O}B$$. (Easy exercise: show that it holds in SDL.)
• FDet (factual detachment) is, in our context (we’ll come back to this), the principle according to which from $$A$$ and $$A \rightarrow \mathsf{O} B$$ follows $$\mathsf{O} B$$. This clearly holds for SDL since it is just an instance of Modus Ponens.
• LogExp is the principle of Deontic Explosion: in SDL if we have $$\mathsf{O} A$$ and $$\mathsf{O} \neg A$$, anything follows. The reason is that there are no SDL-models where $$\mathsf{O} A$$ and $$\mathsf{O} \neg A$$ hold at some world. (Easy exercise: show this! Note that you need to use the fact that the accessibility relation is serial!)

What do we learn from that? Clearly, the formalization of our natural language example leads to trouble in SDL. There is still hope, though. Maybe we can just change our formalization: we can for instance narrow scope premise 2, or/and wide-scope premise 3.

The following table provides an overview and indicates problems we get with the other formalizations.

Norm F1 F2 F3 F4
1 Go! $$\mathsf{O}g$$ $$\mathsf{O}g$$ $$\mathsf{O}g$$ $$\mathsf{O}g$$
2 If you go, tell! $$g \rightarrow \mathsf{O}t$$ $$\mathsf{O}(g \rightarrow t)$$ $$g \rightarrow \mathsf{O}t$$ $$\mathsf{O}(g \rightarrow t)$$
3 If you don’t go, don’t tell! $$\neg g \rightarrow \mathsf{O} \neg t$$ $$\neg g \rightarrow \mathsf{O} \neg t$$ $$\mathsf{O}(\neg g \rightarrow \neg t)$$ $$\mathsf{O}(\neg g \rightarrow \neg t)$$
4 You don’t go. $$\neg g$$ $$\neg g$$ $$\neg g$$ $$\neg g$$
Dependence $$4 \rightarrow 2$$ $$4 \rightarrow 2$$ $$1 \rightarrow 3$$
Asymmetry 2 + 3 2 + 3
Explosion Yes
Oddities $$g \rightarrow \mathsf{O} \neg t$$ $$g \rightarrow \mathsf{O}\neg t$$ $$\mathsf{O}(\neg g \rightarrow t)$$

We indicate 4 types of problems:

1. Dependence: intuitively speaking the different premises are logically independent which means that neither should be derivable from another. This, however, is violated in F1, F3 and F4. For instance, in F1, the fourth premise implies the second one. (This is an instance of the classical inference: $$\neg A$$ implies $$A \rightarrow B$$.)
2. We have already talked about the problem of explosion: clearly, intuitively speaking our scenario describes a consistent setting which should not be rendered inconsistent by our logic.
3. Asymmetry: one may argue that conditional obligations should be modelled in analogous ways in the examples to avoid ad hoc adjustments to obtain the wanted outcome. Note, however, that there are subtle differences between wide-scoping and narrow-scoping. It is one thing to say that $$A \rightarrow B$$ should ideally be the case and another one to say that if $$A$$, then $$B$$ should ideally be the case.
4. Oddities: A deontic logic should not allow to derive formulas which are arguably non-sequiturs for the given scenario. For instance, in F1, we get $$g \rightarrow \mathsf{O} \neg t$$. This seems to be dubious if we intend to express conditional obligations via $$A \rightarrow \mathsf{O} B$$ as is done for premise 3.

Altogether, neither formalization is close to being satisfactory. Chisholm’s and Forrester’s examples are usually considered as the deathblow to SDL. On the positive side, these problematic examples helped establishing Deontic Logic as a research program with its own specific (hard) problems and techniques, far from being a simple instantiation of orthodox systems of modal logic. This lead to lots of innovation and also cross-fertilization (for instance with the field of non-monotonic logic). We will see more of this in future sessions of this seminar … in a sense we only get really started now.

We used the rest of the time in the seminar to talk a bit more about factual detachment. As we have seen last time, Forrester claimed that it doesn’t always make sense, but often it does. Here is an illustrative example (which goes back to John Horty). Since SDL is dead, we will use a new notation for conditional obligations from now on: $$A \Rightarrow_{\mathsf{O}} B$$ expressing that $$A$$ commits us/a given agent/etc. to bring about $$B$$. Often deontic logicians also write $$\mathsf{O}(B \mid A)$$ for this. For novices $$\Rightarrow_{\mathsf{O}}$$ is easier to read, so we will stick with it for now.

• Being served a meal, you ought to not eat with your fingers. $$m \Rightarrow_{\mathsf{O}} \neg f$$
• However, if you’re being served asparagus, you ought to eat with your fingers. $$(m \wedge a) \Rightarrow_{\mathsf{O}} f$$.
• Your being served asparagus. $$m \wedge a$$

Now if we were to rigorously apply factual detachment,

• Factual Detachment: From $$A$$ and $$A \Rightarrow_{\mathsf{O}}B$$ follows $$\mathsf{O} B$$.

we would be able to infer both $$\mathsf{O} \neg f$$ and $$\mathsf{O} f$$. This is not intuitive, to say the least.

In cases where a more specific obligation contradicts a more general one, it often makes sense to prioritize the more specific obligation. Logicians call such scenarios cases of Specificity. In these cases, the general obligation is excepted or cancelled.

One has to be careful, though. As we have seen in the Forrester case, contrary-to-duty obligations are tricky.

• In general you ought not to kill. $$\top \Rightarrow_{\mathsf{O}} \neg k$$
• However, if you kill, you ought to kill gently. $$k \Rightarrow_{\mathsf{O}} g$$.
• You kill.

In this case it seems that the general obligation, not to kill, is not really cancelled: it is still in place and violated.1 Our agent should be held accountable for it!

The above discussion gives us several desiderata for a deontic logic (there are many more, of course):

1. It should handle contrary-to-duty obligations without generating explosion or other oddities.
2. It should be able to distinguish specificity cases from contrary-to-duty cases.
3. It should handle factual detachment in such a way that it is sensitive to exceptional situations.

Let’s keep these in mind when investigating systems from the literature in future sessions of this seminar.

## References

• Chisholm, R. M. (1963). Contrary-to-duty imperatives and deontic logic. Analysis, 24, 33–36.
• Forrester, J. . (1984). Gentle murder, or the adverbial samaritan. Journal of Philosophy, 81, 193–197.
• Straßer, C. (2011). A deontic logic framework allowing for factual detachment. Journal of Applied Logic, 9(1), 61–80.
• Torre, L. V. D., & Tan, Y. (1995). Cancelling and overshadowing: two types of defeasibility in defeasible deontic logic. In , In Proceedings of the Fourteenth International Joint Conference on Artificial Intelligence (IJCAI95) (pp. 1525–1532).

1. More on this distinctions can be found in, for instance, Torre&Tan (1995) and Straßer (2011). [return]

## 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]

## Session 2: The limitations of Standard Deontic Logic

In Standard Deontic Logic (henceforth, SDL), the following is a valid principle:

• If $$\vdash A \rightarrow B$$ then also $$\vdash \mathsf{O} A \rightarrow \mathsf{O} B$$.

This principle is sometimes called modal inheritance (for the $$\mathsf{O}$$ operator). We will abbreviate it with O-Inh. It means, if $$A$$ logically implies $$B$$ then also $$\mathsf{O}A$$ implies $$\mathsf{O} B$$.

Several problems with this principle have been pointed out in the literature, some of which are considered as core problems of SDL. Consider the following obligation:

• It ought to be that Peter, who has been robbed, is helped. (S1)

Now, if we interpret this in the following way

• It ought to be that (Peter who has been robbed is helped). (SWC1)

we face the problem that from

• “Peter who has been robbed is helped”

follows

• “Peter has been robbed.”

So with O-Inh we get:

• It ought to be that (Peter has been robbed).

This is clearly not intuitive.

Forrester (1984) points out that in such cases we also have also the option to interpret S1 differently:

• Peter has been robbed. It ought to be that Peter is helped. (SNS1)

The difference between SWC1 and SNS1 is a matter of scope: in SWC1 the scope of the $$\mathsf{O}$$-operator (i.e., the part of the sentence to which it applies) is the full statement “Peter who has been robbed is helped” while in SNS1 it is restricted to “Peter is helped”. In the first case the scope is wide, in the second narrow.

Cases like our example have been dubbed Good Samaritan in the literature. 1

Aqvist (1967) proposed an epistemic version of it:

1. It ought to be that Smith refrains from robbing Jones.
2. I ought to know that Smith robs Jones. 2

Following what most epistemologists tell us about the nature of knowledge, from knowing that $$A$$ follows that $$A$$. Formally,

• $$\vdash \mathsf{K}A \rightarrow A$$. Here $$\mathsf{K}$$ is a modal operator that expresses knowledge (of a given agent) in its argument.

Now, if we plug in O-Inh we get from 2 (formalized as $$\mathsf{O} \mathsf{K} p$$):

• It ought to be that Smith robs Jones. (Formally, $$\mathsf{O} p$$.) 3

In this example a solution such as above with simply changes the scope of $$\mathsf{O}$$, seems to not be available since it essentially concerns the interaction of the deontic modality $$\mathsf{O}$$ with the epistemic modality $$\mathsf{K}$$.

Forrester in his article investigates whether there are cases where (a) the scoping solution does not work and (b) that are not related to the nesting of (different types of) modalities (like Aqvist’s case). For this, he proposes the following example:

1. In general you ought not to kill.
2. However, if you kill, you better do so in a gentle way.

The second obligation is a so-called contrary-to-duty obligation. Such obligations help us to express obligations that are triggered in case other, more general, obligations are violated. In many scenarios it seems reasonable to differentiate between ideal, sub-ideal and bad situations. E.g., best if no killing happens, worse is if a killing happens, but if so it is better if the killing is at least gently. We differentiate often between better and worse ways of violating obligations when we assign blame, responsibility, or punishment to agents. The punishment and blame for a violent killing may be in many situations less severe than for a gentle killing, to utilize Forrester’s morbid example a bit more.

Given this motivation or justification of contrary-to-duty norms, we should consider the example above consistent. Especially, a deontic logic when applied to it should not render it inconsistent. Let us therefore check what happens when we formalize the example in SDL. I first show a proof tree and then comment on the various steps:

In blue we introduce different types of fact which characterize this example.

• Two facts are independent from deontic information: first, the killing takes place ($$k$$), second, killing gently implies killing ($$\vdash g \rightarrow k$$). We write this as a logical axiom since we consider this a conceptual truth which should hold in every possible world.
• We have two norms:
1. The categorical obligation not to kill ($$\mathsf{O} \neg k$$).
2. The conditional obligation that if you kill you ought to kill gently ($$\mathsf{O}(k \rightarrow g)$$).

FDet (in red) is a principle that expresses that conditional obligations should be “detachable”: if the condition of the obligation is true, the obligation applies. In our example, if you really do kill, you have to do it gently ($$k \rightarrow \mathsf{O}g$$). This is not a general principle of SDL, but rather stipulated by Forrester.

So, applying Modus Ponens (MP) to the fact $$k$$ and $$k \rightarrow \mathsf{O}g$$ we obtain $$\mathsf{O}g$$ (in orange).

O-Inh was discussed above and is used in our example to derive $$\mathsf{O}g \rightarrow \mathsf{O} k$$ (in purple).

Now, in SDL we have the principle O-Cons expressing that there are no deontic conflicts, that is, situation where $$\mathsf{O}A$$ and $$\mathsf{O}\neg A$$ holds for some formula $$A$$. So, if $$\mathsf{O}\neg k$$ holds, $$\mathsf{O}k$$ cannot also hold and therefore $$\neg \mathsf{O}k$$ holds (in light blue). If you want to see how this plays out in the semantics of SDL, we’ll address this in a second.

From $$\mathsf{O}g$$ and $$\mathsf{O}g \rightarrow \mathsf{O}k$$ we can, via Modus Ponens, derive $$\mathsf{O}k$$ (in a box, because this is really bad! – if you don’t see why immediately, shortly think about it …).

Now, classical logic gives us an additional problem, namely that from a contradiction, like having $$\mathsf{O}k$$ and $$\neg \mathsf{O}k$$, anything follows. We write $$\bot$$ for the falsity symbol, which is always interpreted as false, as the name suggests.

So why is this really bad? It means SDL renders a scenario as inconsistent that is intuitively speaking perfectly reasonable. And even if you would block the last inference (logical explosion) by choosing, for instance, a paraconsistent logic, we’d still be left with $$\mathsf{O}k$$ which seems not a reasonable inference to make in the given context (or ever?).

Again, the culprit seems to be O-Inh: it seems not reasonable to infer from $$\mathsf{O}g$$ the obligation to kill. As you can see, now we have no way out by simply scoping our $$\mathsf{O}$$ in a different way, as in the example in the beginning of this entry. Similarly, the solution cannot lie with changing the way $$\mathsf{O}$$ interacts with the epistemic modality $$\mathsf{K}$$ as in Aqvist’s example. How will we get out of this misery! Stay tuned … for future sessions.

Let us in the end shortly see why the principles used in Forrester’s proof are indeed valid in SDL.

We first consider O-Inh. Suppose $$\vdash g \rightarrow k$$ and so $$g \rightarrow k$$ holds in every model at every world. We want to show that then also $$\mathsf{O}g \rightarrow \mathsf{O}k$$ holds in every model at every world. So let $$M = \langle W, R, @, v \rangle$$ be an arbitrary SDL-model and $$w \in W$$ be an arbitrary world. We want to show that $$M,w \models \mathsf{O}g \rightarrow \mathsf{O}k$$. We show that if $$M,w \models \mathsf{O}g$$ then $$M,w \models \mathsf{O}k$$. (Check the truth-table for $$\rightarrow$$ to see why this is sufficient.)

• Suppose $$M,w \models \mathsf{O}g$$.
• Thus, for all $$w^{\prime} \in W$$ for which $$w R w^{\prime}$$, $$M,w^{\prime} \models g$$.
• Since $$g \rightarrow k$$ is a logical axiom, it holds in every world.
• In particular, for all $$w^{\prime} \in W$$ for which $$w R w^{\prime}$$, $$M, w^{\prime} \models g \rightarrow k$$.
• Since our worlds obey the rules of classical logic, in particular Modus Ponens is valid in them. So, for all $$w^{\prime} \in W$$ for which $$w R w^{\prime}$$, $$M, w^{\prime} \models k$$.
• Thus, $$M,w \models \mathsf{O}k$$.

It is an exercise to show that the principle O-Cons holds in all SDL-models.

## Exercise: The Andersonian Reduction

Anderson proposed to utilize a violation constant $$\mathbf{v}$$ to express or define an obligation operator:4

• $$\mathsf{O} A$$ 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?

## Solutions to the exercises from last time

### Exercise 1

We have the following model $$M = (W, R, @, v)$$ where

• $$W = \{@, w\}$$
• $$R = \{(@,w), (w,w)\}$$

and $$v$$ is given by (to keep things simple we work with a language with only two atoms $$p,q$$):

$$@$$ $$w$$
$$p$$ 1 1
$$q$$ 0 0

Which of the following statements is true:

1. $$M,@ \models \mathsf{O}p$$
2. $$M,@ \models \mathsf{O} \mathsf{O}p$$
3. $$M,@ \models \mathsf{P} q$$
4. $$M,@ \models \mathsf{O}p \rightarrow q$$.
5. $$M,@ \models \mathsf{O}p \rightarrow p$$.

Item Yes/No
1. Yes
2. Yes
3. No
4. Yes
5. Yes

### Exercise 2

Can you find a model with only two worlds and a serial accessibility relation for which it doesn’t hold that $$M,@ \models \mathsf{O} \mathsf{O} p \rightarrow \mathsf{O}p$$?

#### Solution

Take

where

world p
@ 1
w 0

In this case we have:

• $$M,@ \models \mathsf{O} \mathsf{O} p$$ but
• $$M,@ \models \neg \mathsf{O}p$$.

### Exercise 3

How many worlds does a model with serial accessibility relation have, whose accessibility relation is transitive and irreflexive. Recall:

• transitive means that for all $$w_1, w_2, w_3$$ we have that if $$w_1 R w_2$$ and $$w_2 R w_3$$ then $$w_1 R w_3$$.
• irreflexive means that for all $$w$$ we don’t have $$wRw$$.

#### Solution

There is no model with a finite set of worlds that fulfills our requirements. Assume for a contradiction there is such a model $$M= \langle W, R, @, v \rangle$$ where $$W = \{ @, w_1, \dotsc, w_n\}$$ for some natural number $$n$$.

• Since $$R$$ is serial and irreflexive, there must be a $$w \neq @$$ in $$W$$ such that $$@Rw$$. Without loss of generality assume that $$w = w_1$$.
• For the same reason there must be a $$w \neq w_1$$ such that $$w_1Rw$$. It cannot be that $$w = @$$ since otherwise by transitivity $$w_1Rw_1$$. So, $$w \in W \setminus \{@,w_1\}$$. Without loss of generality assume that $$w = w_2$$.
• Again, following the same line of reasoning there must be a $$w \neq w_2$$ such that $$w_2Rw$$. It cannot be that $$w \in \{@,w_1\}$$ since otherwise by transitivity $$w_2Rw_2$$. So $$w \in W \setminus \{@,w_1,w_2\}$$. Without loss of generality assume that $$w =w_3$$.
• etc.
• Following the same line of reasoning there must be a $$w \neq w_n$$ such that $$w_nRw$$. It cannot be that $$w \in \{@,w_1, \ldots, w_{n-1}\}$$ since otherwise by transitivity $$w_n R w_n$$. So $$w \in W$$ which is a contradiction.

### Exercise 4

Construct a model $$M = (W, R, @, v)$$ in which the following are true:

1. $$M,@ \models \mathsf{O}(p \vee q)$$
2. $$M,@ \models \neg\mathsf{O}p \wedge \neg \mathsf{O}q$$
3. $$M,@ \models \mathsf{P}p$$
4. $$M,@ \models \mathsf{P}q$$

#### Solution

Let the model be given by

where

world p q
@ 1 0
w 0 1

## References

• Anderson, A. R. (1958). A reduction of deontic logic to alethic modal logic. Mind, 67(265), 100–103.
• Aqvist, L. (1967). Good samaritans, contrary-to-duty imperatives, and epistemic obligations. Nous, 1(4), 361–379.
• Chisholm, R. M. (1963). Contrary-to-duty imperatives and deontic logic. Analysis, 24(), 33–36.
• Forrester, J. . (1984). Gentle murder, or the adverbial samaritan. Journal of Philosophy, 81, 193–197.
• Kanger, S. (1971). New foundations for ethical theory. In R. Hilpinen (Eds.), Deontic Logic: Introductory and Systematic Readings (pp. 36–58). D. Reidel Company, Dordrecht.

1. Parable of the Good Samaritan - Wikipedia. [return]
2. Aqvist doesn’t give us much of context as to why I ought to know that. All that matters to him is that the two obligations are intuitively speaking consistent. [return]
3. If you are complaining that we happily jump from ought-to-be to ought-to-do and back, you are right. I am following Aqvist here: one could be more hygienic and transform the second obligation into an ought-to-be: It ought to be that I know that … [return]
4. 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]

## Session 1: Introduction

Today’s session served as a general introduction to basic ideas and concepts underlying deontic logic. What is deontic logic? To answer this question it was useful to remind ourselves of the general aim of formal logic. It is to provide theories (or the (?) theory) of valid inference. It helps us to understand rules underlying correct inferences and to identify faulty reasoning. Take the following example:

• Either he is at the Mensa or he is at Q-West.
• He’s not at the Mensa.
• Thus, he is at Q-West.

This inference follows a general rule, disjunctive syllogism. In order to phrase general laws we have to abstract from the concrete content of inferences and reduce them to their logical form. For this, in turn, we need an abstract, formal language. The language of propositional logic, which will accompany us in this course, has the following ingredients:

• logical constants: $$\wedge$$ (“and”: conjunction), $$\vee$$ (“or”: disjunction), $$\neg$$ (“not”: negation), $$\rightarrow$$ (“implies”: implication), and $$\equiv$$ (“equivalent/co-implies”: equivalence/co-implication).
• atoms (or “sentential letters”): $$p, q, s, r$$ …

Having this linguistic tools in our repertoire, we can phrase the rule of disjunctive syllogism in its general form:

• $$p \vee q$$
• $$\neg p$$
• Thus, $$q$$.

Our concrete inference above is an instance of this if we interpret $$p$$ with “He is at the Mensa.” and $$q$$ with “He is at Q-West.”.

Now, deontic logic has it’s focus on a specific kind of inference for which the vocabulary of propositional logic is too poor: reasoning with norms. In particular this concerns inferences featuring the following norm types (which come with typical verbs and which will be formally expressed with logical constants on the right):

norm type typical verbs logical constant
obligations ought, must $$\mathsf{O}$$
permission may, permitted to, can $$\mathsf{P}$$
prohibition forbidden, prohibited $$\mathsf{F}$$

(We will at some point discuss others, such as supererogation, etc.)

Here are some examples of statements featuring obligations, that will help us to make some more fine-grained distinctions:

1. There ought to be an exam at the end of the term.
2. You ought to bring about that the dishes are washed.
3. Peter ought to wash the dishes.
4. Wash the dishes!
5. There is an obligation to write an exam at the end of the term.

When phrasing normative statements with ought to be (statement 1), the ought applies to a proposition: it ought to be that $$A$$ is true (like in 1 that “there is an exam at the end of the term”). In contrast, when phrasing a normative statement with ought to do (statement 3), our obligation applies to an action and it concerns an agent. Statement 2 is interesting in that it seems to offer a compromise between the two: while the ought still applies to a —rather generic— action, namely to bring something about, what is to be brought about is a state of affairs which is described by a proposition (“The dishes are washed”). So, if we let $$\mathsf{O}$$ be interpreted as ought-to-bring-about or as ought-to-be, it takes as arguments propositions, which we are able to express in propositional logic. However, if we let it express ought-to-do, we need to enrich our language with ways to express atomic and complex actions, such as washing the dishes. There are ways to do so, but in the first part of the course we will stick to interpretations of $$\mathsf{O}$$ that apply to propositions.

Statement 4 is an imperative: by uttering it, a norm is “brought to life” (i.e., if certain contextual factors are fulfilled such as: the one uttering the imperative has an authority, is not acting or joking, etc.). Imperatives are prescriptive statements: they express norms and do not just report on their existence. Prescriptive statements need not take the form of an imperative, though. E.g., statement 2 may have a prescriptive reading under which you are given the obligation to wash the dishes. In contrast, there are also descriptive statements reporting on the existence of norms. Clearly, statement 5 falls into this category: it does not itself express the norm or bring it into existence, but rather reports on it. Often, without contextual information, statements are ambiguous: they can be read prescriptively or descriptively. (E.g., take statement 1.) It is not clear, and even quite dubious, whether prescriptive statements have truth values. Given the centrality of truth-values in logical theories of valid inference, this has been considered a problem. Can we not have a logical theory about reasoning with “norms” (so, prescriptive statements), but only about “normative propositions” (so, descriptive statements about norms)?

Now, is there at all some logical regularity governing normative reasoning? If not, deontic logic may be an idle business. Let us look at some examples of intuitively sound normative inferences:

Anne may write an exam or give a talk. $$\mathsf{P}(p \vee q)$$
Thus, Anne may write an exam or she may give a talk. $$\mathsf{P} p \vee \mathsf{P} q$$

One may even argue we can conclude $$\mathsf{P}p \wedge \mathsf{P}q$$ from $$\mathsf{P}(p \vee q)$$. This is called “free choice permission”: if you are permitted to $$p$$ or $$q$$, whatever your choice, $$p$$ or $$q$$, you have the permission to do so. This principle is surprisingly hard to model in deontic logic and will be discussed thoroughly in another session.

Anne is obliged to write an exam $$\mathsf{O} p$$
Thus, Anne is permitted to write an exam. $$\mathsf{P}p$$

It seems reasonable that if you’re obliged to bring about $$p$$, you’re also permitted to do so.

Now here are two faulty inferences:

Anne is obliged to write an exam or give a talk. $$\mathsf{O}(p \vee q)$$
* Thus, Anne is obliged to write an exam or she is obliged to give a talk. $$\mathsf{O} p \vee \mathsf{O} q$$

and

Anne is obliged to write an exam or give a talk. $$\mathsf{O}(p \vee q)$$
She’s allowed to not write an exam. $$\mathsf{P}\neg p$$
* Thus, she’s obliged to give a talk. $$\mathsf{O} q$$

Clearly, we expect deontic logic to give us an understanding of what goes wrong with these inferences and to help us identify other faulty inferences.

So, what if we replace $$\mathsf{P}\neg p$$ in the last inference with $$\mathsf{O}\neg p$$? If this doesn’t make you scratch your head yet, things can get arbitrarily complicated. A deontic logic should give us a precise framework and methods to solve such puzzles.

But how to do so? So far we only learned how to formalize our language, but how to identify general laws of reasoning? Let us recall one way to do so in propositional logic. There, it was useful to formulate a theory of the meaning of logical constants. Such a theory was given in terms of truth tables, such as

$$\wedge$$ 0 1 $$\vee$$ 0 1 $$\neg$$
0 0 0 0 1 1
1 0 1 1 1 0

A formula was tautological, if it was true in all possible interpretations of the atoms occurring in it. E.g., in the example at the beginning of this entry we have:

$$p$$ $$q$$ $$((p \vee q))$$ $$\wedge$$ $$\neg p)$$ $$\rightarrow q$$
0 0 0 0 1 1
0 1 1 1 1 1
1 0 1 0 0 1
1 1 1 0 0 1

So, a natural question is, whether truth tables help us also to express the meaning of $$\mathsf{O}$$ and $$\mathsf{P}$$. Unfortunately, this approach doesn’t work. The reason is that the truth-value of $$\mathsf{O}p$$ is independent of the truth-value of $$p$$. ($$\mathsf{O}$$ and $$\mathsf{P}$$ are not truth-functional, unlike the propositional connectives $$\wedge, \neg, \vee, \rightarrow$$ and $$\equiv$$.)

One way to think about the meaning of obligations is that their arguments should be true in ideal worlds: after all, ideally we fulfill our obligations. So, it is not sufficient to only observe the current factual situation and what is true in it (given by the truth-values of atoms), but rather we need to consider ideal situations.

This idea is made formally precise in so-called possible worlds semantics. A logical model $$M$$ is given by a quadruple $$( W, R, @, v)$$ where

• $$W$$ is a set of worlds
• $$R$$ is an accessibility relation between worlds
• $$@$$ is the actual world, our reference point
• $$v$$ is a function that assigns truth values 0 and 1 to pairs $$(w, p)$$ of worlds in $$W$$ and atoms. It expresses whether $$p$$ holds at world $$w$$ or not.

The basic understanding is that worlds reachable from the actual world $$@$$ via the accessibility relation $$R$$ are ideal counterparts of $$@$$. We expect all of the given obligations to be fulfilled in these ideal worlds. Also other worlds than $$@$$ may have ideal counterparts. (If these are really “ideal” worlds, you may expect them to be ideal counterparts of themselves, but we don’t demand this at this point.)

The model is not informative yet, since we still need to determine when a formula is valid at a given world in the model. For this we use our assignment function $$v$$ in a bottom-up way. Where $$w \in W$$, we have:

• $$M,w \models p$$ iff $$v(w,p) = 1$$.
• $$M,w \models \neg A$$ iff $$M, w \not\models A$$.
• $$M,w \models A \wedge B$$ iff $$M,w \models A$$ and $$M,w \models B$$.
• $$M,w \models A \vee B$$ iff $$M,w \models A$$ or $$M,w \models B$$.
• $$M,w \models A \rightarrow B$$ iff $$M,w \not\models A$$ or $$M,w\models B$$.
• $$M,w \models A \equiv B$$ iff ($$M,w \models A$$ iff $$M,w \models B$$).

Finally, and most interestingly, when is a formula $$\mathsf{O}A$$ true at some world $$w$$. Well, we follow our initial motivation: exactly when $$A$$ holds in all ideal counterparts of $$w’$$!

• $$M,w \models \mathsf{O}A$$ iff for all $$w’$$ which are $$R$$-reachable from $$w$$, $$M,w’ \models A$$.
• $$M,w \models \mathsf{P}A$$ iff there is a $$w’$$ which is $$R$$-reachable from $$w$$ and $$M,w’ \models A$$.

For permission, it is sufficient to find one ideal counterpart of the given world where the argument of $$\mathsf{P}$$ holds.

Let us test this approach with our previous example:

Anne may write an exam or give a talk. $$\mathsf{P}(p \vee q)$$
Thus, Anne may write an exam or the may give a talk. $$\mathsf{P} p \vee \mathsf{P} q$$

If this is a valid inference, it should be true in any model in any world. Is this so. Let’s see:

• Suppose $$M,w \models \mathsf{P}(p \vee q)$$
• Thus, there must be an ideal counterpart $$w’$$ of $$w$$ for which $$M,w’ \models p \vee q$$.
• Thus, $$M,w’ \models p$$ or $$M,w’ \models q$$.
• Hence, $$M,w \models \mathsf{P}p$$ or $$M,w \models \mathsf{P}q$$.
• Thus, $$M,w \models \mathsf{P}p \vee \mathsf{P}q$$.

We have to be careful however with the inference from $$\mathsf{O}p$$ to $$\mathsf{P}p$$. According to our definition of a model above, it could happen that a world has no ideal counterpart! In such a world for any formula $$A$$, $$\mathsf{O} A$$ and $$\neg \mathsf{P} A$$ will hold. (See for yourself why this is so!)

In order to disallow such situations, we require that every world has at least one ideal counterpart. In order words, the accessibility relation should be serial:

• for all $$w \in W$$ it holds that $$wRw’$$ for some $$w’ \in W$$. (This $$w’$$ could be but need not be identical to $$w$$.)

The logic which uses possible world models with serial accessibility relations is called KD or SDL (for Standard Deontic Logic). We will see that it has serious limitations, but it is a good starting point for gaining a precise formal understanding of normative reasoning.

In SDL also our other example

Anne is obliged to write an exam $$\mathsf{O} p$$
Anne is permitted to write an exam. $$\mathsf{P}p$$

becomes a valid inference. Let us show this:

• Suppose for some world $$w$$ in a model $$M$$ (with serial accessibility relation $$R$$) we have $$M,w \models \mathsf{O}p$$.
• Hence, in all ideal counterparts $$w’$$ of $$w$$ we have $$M,w’ \models p$$.
• Since in view of the seriality of $R$e there is at least one such $$w’$$, we also have $$M,w\models \mathsf{P} p$$.

## Exercises

### Exercise 1

We have the following model $$M = (W, R, @, v)$$ where

• $$W = \{@, w\}$$
• $$R = \{(@,w), (w,w)\}$$

and $$v$$ is given by (to keep things simple we work with a language with only two atoms $$p,q$$):

$$@$$ $$w$$
$$p$$ 1 1
$$q$$ 0 0

Which of the following statements is true:

1. $$M,@ \models \mathsf{O}p$$
2. $$M,@ \models \mathsf{O} \mathsf{O}p$$
3. $$M,@ \models \mathsf{P} q$$
4. $$M,@ \models \mathsf{O}p \rightarrow q$$.
5. $$M,@ \models \mathsf{O}p \rightarrow p$$.

### Exercise 2

Can you find a model with only two worlds and a serial accessibility relation for which it doesn’t hold that $$M,@ \models \mathsf{O} \mathsf{O} p \rightarrow \mathsf{O}p$$?

### Exercise 3

How many worlds does a model with serial accessibility relation have, whose accessibility relation is transitive and irreflexive. Recall:

• transitive means that for all $$w_1, w_2, w_3$$ we have that if $$w_1 R w_2$$ and $$w_2 R w_3$$ then $$w_1 R w_3$$.
• irreflexive means that for all $$w$$ we don’t have $$wRw$$.

### Exercise 4

Construct a model $$M = (W, R, @, v)$$ in which the following are true:

1. $$M,@ \models \mathsf{O}(p \vee q)$$
2. $$M,@ \models \neg\mathsf{O}p \wedge \neg \mathsf{O}q$$
3. $$M,@ \models \mathsf{P}p$$
4. $$M,@ \models \mathsf{P}q$$

## Ongoing List of Topics and Literature

This entry contains a list of topics and literature. I will from time to time add stuff here. Feel free to suggest topics and papers you are interested in! Possible worlds and SDL (session #1) How do possible world semantics in modal logic work? Standard Deontic Logic: what are the main ideas. Seminal work in deontic logic: the limits of SDL Von Wright’s seminal paper on deontic logic Von Wright, G. [Read More]