## Session 9: Seeing to it that: deliberative stit, refraining, and group actions

## Session 8: Seeing to it that …

## Session 7: Free Choice Permission and Truthmaker Semantics

In this session, Tim Forche gave an excellent presentation on the very interesting solution to the problem of Free Choice Permission based on truth-maker semantics proposed by Anglberger, Korbmacher and Faroldi in their DEON 2016 paper.

[Under construction.]

## Session 6: McNamara on Supererogation (will be contributed by Laura and Thomas)

This will be an entry by Laura and Thomas. It is not finished yet.

## Exercises for Session 6 (Supererogation)

## 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:

- \(\mathsf{O}(g \mid \top)\)
- \(\mathsf{O}(t \mid g)\)
- \(\mathsf{O}(\neg t \mid \neg g)\)

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)?

- \((\mathsf{O}(p \mid q) \wedge \mathsf{O}(s \mid q)) \rightarrow \mathsf{O}(p \wedge s \mid q)\)
- \(\mathsf{O}(p \mid q) \rightarrow \mathsf{O}(p \mid q \wedge s)\)
- \(\mathsf{O}(p \mid q) \rightarrow \mathsf{O}(p \wedge s \mid q)\)
- \(\mathsf{O}(p \mid q) \rightarrow \mathsf{O}(p \wedge q \mid q)\)
- \(\mathsf{O}(p \mid q) \wedge \mathsf{O}(p \mid s) \rightarrow \mathsf{O}(p \mid q \wedge s)\)
- \(\mathsf{O}(p \mid q) \rightarrow \mathsf{P}(p \mid q)\)
- \(\mathsf{O}(p \mid q) \rightarrow \mathsf{O}(q \rightarrow p \mid \top)\)
- \(\mathsf{O}(p \rightarrow q \mid s) \rightarrow (\mathsf{O}(p \mid s) \rightarrow \mathsf{O}(q \mid s))\)
- \((\mathsf{O}(p \mid q) \wedge \mathsf{O}(s \mid q)) \rightarrow \mathsf{O}(p \mid q \wedge s)\)

### Ad 1

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

### Ad 2

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

### Ad 3

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.

### Ad 4

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

- \(M,w_1 \models p \wedge q\) and
- 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, \(M \models \mathsf{O}(p \wedge q \mid q)\).

### Ad 5

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.

### Ad 6

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.

### Ad 7

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, \(M \models \mathsf{O}(q \rightarrow p \mid \top)\).

### Ad 8

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 \(M,w \models s \rightarrow q\).

### Ad 9

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:

- In general, you ought not to eat with your fingers. \(\mathsf{O}(\neg f \mid \top)\)
- 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:

- \(\mathsf{O}(g \mid \top)\)
- \(\mathsf{O}(t \mid g)\)
- \(\mathsf{O}(\neg t \mid \neg g)\)

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)?

- \((\mathsf{O}(p \mid q) \wedge \mathsf{O}(s \mid q)) \rightarrow \mathsf{O}(p \wedge s \mid q)\)
- \(\mathsf{O}(p \mid q) \rightarrow \mathsf{O}(p \mid q \wedge s)\)
- \(\mathsf{O}(p \mid q) \rightarrow \mathsf{O}(p \wedge s \mid q)\)
- \(\mathsf{O}(p \mid q) \rightarrow \mathsf{O}(p \wedge q \mid q)\)
- \(\mathsf{O}(p \mid q) \wedge \mathsf{O}(p \mid s) \rightarrow \mathsf{O}(p \mid q \wedge s)\)
- \(\mathsf{O}(p \mid q) \rightarrow \mathsf{P}(p \mid q)\)
- \(\mathsf{O}(p \mid q) \rightarrow \mathsf{O}(q \rightarrow p \mid \top)\)
- \(\mathsf{O}(p \rightarrow q \mid s) \rightarrow (\mathsf{O}(p \mid s) \rightarrow \mathsf{O}(q \mid s))\)
- \((\mathsf{O}(p \mid q) \wedge \mathsf{O}(s \mid q)) \rightarrow \mathsf{O}(p \mid q \wedge s)\)

### Ad 1

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

### Ad 2

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

### Ad 3

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.

### Ad 4

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

- \(M,w_1 \models p \wedge q\) and
- 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, \(M \models \mathsf{O}(p \wedge q \mid q)\).

### Ad 5

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.

### Ad 6

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.

### Ad 7

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, \(M \models \mathsf{O}(q \rightarrow p \mid \top)\).

### Ad 8

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 \(M,w \models s \rightarrow q\).

### Ad 9

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:

- In general, you ought not to eat with your fingers. \(\mathsf{O}(\neg f \mid \top)\)
- 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.

## Solutions to the Exercises from Session 3

## Exercises

The following are a bit harder puzzles for the courageous.

- 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.
- What about \(\mathsf{O}\mathsf{O}p \rightarrow \mathsf{O}p\)?
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?

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

### Ad 1

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 \(w^{\prime} R w^{\prime}\).

### Ad 2

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 \(w^{\prime\prime} R w^{\prime}\).

### Ad 3

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:

- \(\mathsf{O}(\mathsf{O}p \rightarrow p) \rightarrow ( \mathsf{O} \mathsf{O} p \rightarrow \mathsf{O}p)\).

### Ad 4

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:

- You ought to go to help your friend.
- If you go, call.
- If you don’t go, don’t call.
- 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:

- 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 \(A \rightarrow B\).)
- 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.
- 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.
- 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):

- It should handle contrary-to-duty obligations without generating explosion or other oddities.
- It should be able to distinguish specificity cases from contrary-to-duty cases.
- 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).

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