11Answers to the Exercises
Chapter 2
Chapter 3
Chapter 4
Chapter 5
Chapter 6
Chapter 7
Chapter 8
Chapter 9
Chapter 10
Chapter 1
(c) and (g) are truthfunctional; (a), (b), (d), and (e) are not truthfunctional.
(f) is truthfunctional if God is omniscient (and infallible); it is also truthfunctional if God doesn’t exist, or if God believes all and only false things; otherwise (f) is not truthfunctional.
 (a)
 \(\Diamond p\) \(p\): I offended the principal.
 (b)
 \(\neg \Diamond p\) \(p\): It is raining.
 (c)
 \(\Diamond p\) \(p\): There is life on Mars.
 (d)
 \(\Box (p \to q)\) \(p\): The murderer escaped through the window; \(q\): There are traces on the ground.
 (e)
 \(\Diamond (p \land q)\) \(p\): The murderer escaped through the window; \(q\): There are traces on the ground.
 (a)
 \(\Diamond p\) \(p\): I study architecture.
 (b)
 \(\Diamond p\) \(p\): The bridge collapses.
 (c)
 \(\neg \Diamond (p \land q)\) \(p\): You are talking to me from the kitchen; \(q\): I hear you.
 (d)
 \(p \to \Diamond q\) \(p\): You have a smartphone; \(q\): You use an electronic ticket.
 (a)
 \begin {alignat*} {2} 1.\quad & \Box p \to p &\quad & \text {(T)}\\ 2.\quad & \Box (\Box p \to p) &\quad & \text {(1, Nec)} \end {alignat*}
 (b)
 \begin {alignat*} {2} 1.\quad & p \to (q \to (p \land q)) &\quad & \text {(CPL)}\\ 2.\quad & \Box (p \to (q \to (p \land q))) &\quad & \text {(1, Nec)}\\ 3.\quad & \Box (p \to (q \to (p \land q))) \to (\Box p \to \Box (q \to (p \land q))) &\quad & \text {(K)}\\ 4.\quad & \Box p \to \Box (q \to (p \land q))) &\quad & \text {(2, 3, CPL)}\\ 5.\quad & \Box (q \to (p \land q))) \to (\Box q \to \Box (p\land q)) &\quad & \text {(K)}\\ 6.\quad & \Box p \to (\Box q \to \Box (p\land q)) &\quad & \text {(4, 5, CPL)}\\ 7.\quad & (\Box q \land \Box q) \to \Box (p\land q) &\quad & \text {(6, CPL)} \end {alignat*}
 (c)
 \begin {alignat*} {2} 1.\quad & \neg \Diamond \neg p \leftrightarrow \Box \neg \neg p &\quad & \text {(Dual)}\\ 2.\quad & \neg \neg \Diamond \neg p \leftrightarrow \neg \Box \neg \neg p &\quad & \text {(1, CPL)}\\ 3.\quad & \Diamond \neg p \leftrightarrow \neg \Box \neg \neg p &\quad & \text {(2, CPL)}\\ 4.\quad & \neg \neg p \to p &\quad & \text {(CPL)}\\ 5.\quad & \Box (\neg \neg p \to p) &\quad & \text {(4, Nec)}\\ 6.\quad & \Box (\neg \neg p \to p) \to (\Box \neg \neg p \to \Box p)&\quad & \text {(K)}\\ 7.\quad & \Box \neg \neg p \to \Box p&\quad & \text {(5, 6, CPL)}\\ 8.\quad & p \to \neg \neg p &\quad & \text {(CPL)}\\ 9.\quad & \Box (p \to \neg \neg p) &\quad & \text {(8, Nec)}\\ 10.\quad & \Box (p \to \neg \neg p) \to (\Box p \to \Box \neg \neg p)&\quad & \text {(K)}\\ 11.\quad & \Box p \to \Box \neg \neg p &\quad & \text {(9, 10, CPL)}\\ 12.\quad & \Box \neg \neg p \leftrightarrow \Box p &\quad & \text {(7, 11, CPL)}\\ 13.\quad & \neg \Box \neg \neg p \leftrightarrow \neg \Box p &\quad & \text {(12, CPL)}\\ 14.\quad & \Diamond \neg p \leftrightarrow \neg \Box p &\quad & \text {(3, 13, CPL)} \end {alignat*}
The rules of the axiomatic calculus cannot be used to directly derive assumptions from arbitrary premises. To show that \(A\) entails \(B\), you have to prove \(A \to B\).
Chapter 2
This shows that the truth of \(p\) (at a world in a model) does not entail the truth of \(\Box p\) (at the world in the model), even though the validity of \(p\) entails the validity of \(\Box p\), as per the previous exercise.
 (a)
 Target: \(p\to q\)
Countermodel: \(W = \{ w \}, V(p) = \{ w \}, V(q) = \emptyset \).
 (b)
 Target: \(p \to \Box (p \lor q)\)
Countermodel: \(W = \{ w, v \}, V(p) = \{ w \}, V(q) = \emptyset \).
 (c)
 Target: \(\Box p \lor \Box \neg p\)
Countermodel: \(W = \{ w,v,u \}, V(p) = \{ u \}\).
 (d)
 Target: \(\Diamond (p \to q) \to (\Diamond p \to \Diamond q)\)
Countermodel: \(W = \{ w,v,u \}, V(p) = \{ u \}, V(q) = \emptyset \).
 (e)
 \(\Box \Diamond p \to p\)
tree grows forever. The target sentence isn’t valid, but the tree method only gives us an infinite countermodel. In such a case, it may be useful to read off a model from an incomplete version of the tree and manually check whether it is a genuine countermodel. The model determined by the first five nodes of the present tree is \(W = \{ w,v \}, V(p)=\{ v \}\), and you can confirm that it is a countermodel to the target sentence.
If you read off a model from an incomplete tree, you can’t be sure that it is a countermodel for the target sentence. You must always doublecheck!
(d) and (f) are invalid. Here is a tree for (d):
can choose either of the open branches to read off a countermodel. In fact, here we get the same countermodel no matter which open branch we choose: \(W = \{ w,v,u \}, V(p)=\{ v\}, V(q)=\{u\}\).
A tree for (e) might begin like this:
tree grows forever. The model determined by the first seven nodes of the present tree is \(W = \{ w,v,u \}, V(p)=\{ v \}\). It is a countermodel to the target sentence.
To show that \(A\) and \(B\) are equivalent, we can draw a tree for \(A \leftrightarrow B\).
Chapter 3
 (a)
 One answer: \(A\) is physically necessary iff \(A\) is true at all worlds that are compatible with the laws of physics. A possibly better answer: \(A\) is physically necessary iff \(A\) is true at all worlds that are compatible with the laws of physics and the current state of the universe.
 (b)
 A simple, if somewhat uninformative answer: We know that \(A\) iff \(A\) is true at all worlds that are compatible which our knowledge.
 (c)
 It is true that \(A\) iff \(A\) is true at all worlds that are identical to the actual world.
If this seems strange, remember that \(\Box A\) is equivalent to \(\neg \Diamond \neg A\). And \(\Diamond \neg A\) means that there’s an accessible world where \(\neg A\) is true. If there are no accessible worlds, then this is false. So \(\neg \Diamond \neg A\) is true.
 (a)
 For example: \(W = \{ w,v \}\), \(R = \{ (w,v), (v,w) \}\), \(V(p) = \{ v \}\). \(\Box p \to \Box \Box p\) is false at \(w\). (‘\(R = \{ (w,v), (v,w) \}\)’ means that \(R\) relates \(w\) to \(v\) and \(v\) to \(w\) and nothing else to anything else.)
 (b)
 For example: \(W = \{ w,v \}\), \(R = \{ (w,w), (w,v) \}\), \(V(p) = \{ w \}\). \(\Diamond p \to \Box \Diamond p\) is false at \(w\).
 (a)
 Suppose \(R\) is symmetric and transitive, and that \(xRy\) and \(xRz\). By symmetry, \(yRx\). By transitivity, \(yRz\).
 (b)
 Suppose \(R\) is symmetric and euclidean, and that \(xRy\) and \(yRz\). By symmetry, \(yRx\). By euclidity, \(xRz\).
 (c)
 Suppose \(R\) is reflexive and euclidean, and that \(xRy\). By reflexivity, \(xRx\). By euclidity, \(yRx\).
Chapter 4
Method B is sound, but not complete. Since every instance of \(\Box (A \lor \neg A)\) is Kvalid, everything that is provable with method B is Kvalid. But many Kvalid sentences (e.g., \(p \to p\)) aren’t provable with method B.
Method C is neither sound nor complete. It is not sound because many Kinvalid sentences figure in logic textbooks. It is not complete because there are infinitely many Kvalid sentences almost all of which don’t occur in any textbooks.
For \(\neg \Diamond A\): Suppose \(\beta \) contains nodes of the form \(\neg \Diamond A \;(\omega )\) and \(\omega R \upsilon \), and the branch is extended by adding \(\neg A \;(\upsilon )\). Since \(\neg \Diamond A \;(\omega )\) and \(\omega R \upsilon \) are correct statement about \(M\) under \(f\), we have \(M,f(\omega ) \models \neg \Diamond A\) and \(f(\omega )Rf(\upsilon )\). By clause (b) of definition 3.2, \(M,f(\omega ) \models \neg \Diamond A\) implies \(M,f(\omega ) \not \models \Diamond A\). By clause (h), it follows that \(M,f(\upsilon ) \models \neg A\). So the extended branch correctly describes \(M\).
For \(\neg \Diamond B\): If \(A\) is a negated diamond sentence \(\neg \Diamond B\), then \(\beta \) contains a node \(\neg B \;(\upsilon )\) for each world variable \(\upsilon \) for which \(\omega R \upsilon \) is on \(\beta \) (because the tree is fully developed). By induction hypothesis, \(M, \upsilon \models \neg B\), for each such \(\upsilon \). By definition 4.2, it follows that \(M,\upsilon \models \neg B\) for all worlds \(\upsilon \) such that \(\omega R \upsilon \). By clauses (b) and (g) of definition 3.2, it follows that \(M, \omega \models A\).
By the Completeness Lemma, the target sentence \(\neg \neg A\) is true at world \(w\) in \(M\). So \(A\) is true at \(w\) in \(M\). So \(A\) is true at some world in some acyclical model.
Suppose \(A\) is not S5provable. Then \(\{ \neg A \}\) is S5consistent. It follows by Lindenbaum’s Lemma that \(\{ \neg A \}\) is included in some maximal S5consistent set \(\Gamma \). By definition of canonical models, this set is a world in the canonical model \(M_{S5}\) for S5. Since \(\neg A\) is in \(\Gamma \), it follows from the Canonical Model Lemma that \(M_{S5},\Gamma \models \neg A\). So \(M_{S5},S \not \models A\).
It remains to show that the accessibility relation in \(M_{S5}\) has the right formal properties. We know that a sentence is S5validity iff it is valid in the class of Kripke models whose accessibility relation is an equivalence relation. So we will show that the accessibility relation in \(M_{S5}\) is reflexive, transitive, and symmetric.
By definition, a world \(v\) in a canonical model is accessible from \(w\) iff whenever \(\Box A \in w\) then \(A \in v\). Since the worlds in \(M_{S5}\) are maximal S5consistent sets of sentences, and every such set contains every instance of the (T)schema \(\Box A \to A\), there is no world in \(M_{S5}\) that contains \(\Box A\) but not \(A\). So every world in \(M_{S5}\) has access to itself.
For transitivity, suppose for some worlds \(w,v,u\) in \(M_{S5}\) we have \(wRv\) and \(vRu\). We need to show that \(wRu\). Given how \(R\) is defined in \(M_{S5}\), we have to show that \(u\) contains all sentences \(A\) for which \(w\) contains \(\Box A\). So let \(A\) be an arbitrary sentence for which \(w\) contains \(\Box A\). Since every world in \(M_{S5}\) contains every instance of \(\Box A \to \Box \Box A\), we know that \(w\) also contains \(\Box \Box A\). From \(wRv\), we can infer that \(v\) contains \(\Box A\). And from \(vRu\), we can infer that \(u\) contains \(A\).
For symmetry, suppose for some worlds \(w,v\) in \(M_{S5}\) we have \(wRv\) and not \(vRw\). Given how \(R\) is defined, this means that there is some sentence \(A\) for which \(\Box A\) is in \(v\) but \(\neg A\) is in \(w\). Since \(w\) contains the Tprovable sentence \(\neg A \to \Diamond \neg A\) and the (5)instance \(\Diamond \neg A \to \Box \Diamond \neg A\), it also contains \(\Box \Diamond \neg A\). So \(v\) contains \(\Diamond \neg A\). This contradicts the assumption that \(v\) is S5consistent, given that \(v\) contains \(\Box A\).
Chapter 5
 (a)
 \(\mathsf {K}(r \lor s)\)
\(r\): It is raining; \(s\): It is snowing  (b)
 \(\mathsf {K} r \lor \mathsf {K} s\)
\(r\): It is raining; \(s\): It is snowing  (c)
 \(\mathsf {K} r \lor \mathsf {K} \neg r\)
\(r\): It is raining  (d)
 This sentence is ambiguous. On one reading, it could be translated as \(\mathsf {M} g\to \mathsf {K} g\), on
the other as \(\mathsf {K}(\mathsf {M} g \to g)\)
\(g\): You are guilty
The tree for (c) doesn’t close:
We could add a few more applications of Reflexivity, but the tree would remain open. It also gives us a countermodel: let \(W\) = \(\{ w,v,u \}\); \(w\) has 1access to \(v\) and \(u\); each world has 1 and 2access to itself; \(V(p) = \{ v \}\). In this model, at world \(w\), \(\mathsf {M}_1\mathsf {K}_2 p\) is true while \(\mathsf {M}_2\mathsf {K}_1 p\) is false.
By (PI), \(\mathsf {B} A \to \mathsf {K}\mathsf {B} A\) is valid. By (KB), so is \(\mathsf {K}\mathsf {B} A \to \mathsf {B}\mathsf {B} A\). By propositional logic, these entail \(\mathsf {B} A \to \mathsf {B}\mathsf {B} A\).
By (NI), \(\neg \mathsf {B} \neg A \to \mathsf {K}\neg \mathsf {B} \neg A\) is valid. By (KB), so is \(\mathsf {K}\neg \mathsf {B} \neg A \to \mathsf {B}\neg \mathsf {B} \neg A\). By propositional logic, these entail \(\neg \mathsf {B} \neg A \to \mathsf {B}\neg \mathsf {B} \neg A\).
Chapter 6
 (a)
 \(\mathsf {O} \neg p\); \(p\): You go into the garden.
 (b)
 \(\mathsf {O} \neg p\); \(p\): You go into the garden.
 (c)
 \(\mathsf {O} p\); \(p\): Jones helps his neighbours.
 (d)
 \(\mathsf {O} (p \to q)\); \(p\): Jones helps his neighbours, \(q\): Jones tells his neighbours that he’s coming.
 (e)
 You might try \(\mathsf {O} (\neg p \to \neg q)\) or \(\neg p \to \mathsf {O} \neg q\) \(p\): Jones helps his neighbours, \(q\): Jones tells his neighbours that he’s coming.
See section 6.3, especially exercise 6.14, for why neither translation of (e) is fully satisfactory.
(Dual2) says that \(\neg \Box A\) is equivalent to \(\Diamond \neg A\). If nothing is forbidden then \(\neg \Box A\) is true for all \(A\). But if nothing is permitted then \(\Diamond \neg A\) is false for all \(A\).
Note that the absolutist approach validates schemas like \(\mathsf {O} A \to \mathsf {O}\mathsf {O} A\) without assuming that the norms say anything about whether they should be in force.
Factual detachment is invalid. A counterexample is the “gentle murder puzzle”. Suppose John is determined to kill his grandmother. If he will go ahead and kill her, he ought to do so gently. Can we conclude that John ought to gently kill his grandmother? Arguably not. He shouldn’t kill her at all! We have \(k\) and \(\mathsf {O}(g/k)\), but not \(\mathsf {O}(g)\). Formally, \(g\) is true at the best of the accessible \(k\)worlds, but since all the \(k\)worlds are quite bad, \(g\) is not true at the best of the accessible worlds.
You probably translated (d) as either \(p \to \mathsf {O} q\) or as \(\mathsf {O}(p \to q)\). \(p \to \mathsf {O} q\) is entailed by (f), so it violates the nonentailment condition. Assume then that (d) is translated as \(\mathsf {O}(p \to q)\).
The most obvious translations for (e) are \(\neg p \to \mathsf {O}\neg q\) and \(\mathsf {O}(\neg p \to \neg q)\). The latter is entailed by (c). If we choose the former, then (c)–(f) constitute a deontic dilemma: (e) and (f) would entail \(\mathsf {O} \neg t\), but (c) and (d) would entail \(\mathsf {O} t\).
The Paradox of Free Choice: ‘Alice might be in the office or in the library’ seems to imply that Alice might be in the office and that she might be in the library.
Chapter 7
 (a)
 \(\mathsf {H} \neg p\)
\(p\): It is warm  (b)
 \(\mathsf {F} p\)
\(p\): There is a sea battle  (c)
 \(\neg \mathsf {F} \mathsf {P} p\) or, perhaps, \(\mathsf {F}\neg \mathsf {P} p\)
\(p\): There is a sea battle  (d)
 \(\mathsf {F}(p \lor \mathsf {P} q)\) or \(\mathsf {F}(\mathsf {F} p \lor \mathsf {F}\mathsf {P} q)\)
\(p\): It is warm  (e)
 \(\neg \mathsf {P} p \to \neg \mathsf {F} q\) or \(\mathsf {G}(\neg \mathsf {P} p \to \neg q)\)
\(p\): You study, \(q\): you pass the exam  (f)
 \(\mathsf {P}(p \land q)\)
\(p\): I am having tea, \(q\): the door bell rings
 (a)
 For example, \(\mathsf {G} A \to \mathsf {F} A\).
 (b)
 For example, \(\mathsf {H} A \to \mathsf {P} A\).
 (c)
 No schema corresponds to the class of frames with a last time. If we also assume linearity, \(\mathsf {G}(A \land \neg A) \lor \mathsf {F}\mathsf {G}(A \land \neg A)\) works.
 (d)
 No schema corresponds to the class of frames with a first time. If we assume linearity, then \(\mathsf {H}(A \land \neg A) \lor \mathsf {P}\mathsf {H}(A \land \neg A)\) works.
In the other direction, we have to show that if a frame isn’t dense then some instance of \(\mathsf {F} A \to \mathsf {F}\mathsf {F} A\) is false at some point \(t\) in some model \(M\) based on that frame. We take the simplest instance \(\mathsf {F} p \to \mathsf {F}\mathsf {F} p\). If a frame isn’t dense then there are points \(t,s\) such that \(t<s\) and no point lies in between \(t\) and \(s\). Let \(V\) be an interpretation function that makes \(p\) true at \(s\) and false everywhere else. Then \(\mathsf {F} p\) is true at \(t\) but \(\mathsf {F}\mathsf {F} p\) is false. So \(\mathsf {F} p \to \mathsf {F}\mathsf {F} p\) is false at \(t\).
To show that a schema is valid, assume for reductio that there is some time \(t\) on some history \(H\) in some model \(M\) at which the schema is false. Then (repeatedly) use definition 7.3 to derive a contradiction.
For (e), consider a model with three times \(t,s,r\) such that \(s < t\), \(r< t\), and neither \(s < r\) nor \(r< s\). Let \(q\) be true at \(s\) and false at the other two times. \(\mathsf {P} q \to \Box \mathsf {P} \Diamond q\) is false at \(t\) on the history \(\langle s,t \rangle \).
Suppose \(A\) Ockhamentails \(B\). Let \(t\) be any time in any temporal model at which \(A\) is true, i.e.: true relative to all histories through \(t\). Since \(A\) Ockhamentails \(B\), \(B\) is true at \(t\) relative to all histories through \(t\). So \(A\) superentails \(B\).
But suppose \(A\) superentails \(B\). Let \(t\) be any time on any history \(h\) in any temporal model at which \(A\) is true. We can’t infer that \(B\) is true at \(t\) on \(h\), for \(A\) may be false at \(t\) relative to other histories \(h'\). So we can’t infer that \(A\) Ockhamentails \(B\). Indeed, \(\mathsf {F} p\) superentails \(\Box \mathsf {F} p\), but \(\mathsf {F} p\) does not Ockhamentail \(\Box \mathsf {F} p\).
Chapter 8

If our opponents had been cheating, we would never have found out. Therefore: If we had found out that our opponents are cheating, then they wouldn’t have been cheating.

If you had added sugar to your coffee, it would have tasted good. Therefore: If you had added sugar and vinegar to your coffee, it would have tasted good.
Both of these inferences are valid if subjunctive conditionals are strict conditionals. But they don’t sound good.
The inference might nonetheless look reasonable because it would normally be inappropriate to assert a disjunction ‘\(p\) or \(q\)’ unless the disjunction is known – unless it is true at all epistemically accessible worlds. And if \(p \lor q\) is true at all epistemically accessible worlds then \(\neg p \to q\) is also true at all epistemically accessible worlds, and so \(\Box (\neg p \to q)\) is true. Thus the conclusion of ortoif is true in any situation in which the premise is assertable. If the logic of knowledge validates the (4)schema, we can go further and say that the conclusion is assertable in any situation in which the premise is assertable.
Asymmetry thereby turns into: if \(u \npreceq _{w} v\) then \(v \preceq _{w} u\). Equivalently: either \(u \preceq _{w} v\) or \(v \preceq _{w} u\). This property of relations is called completeness. Notice that it entails reflexivity.
Quasiconnectedness turns into: if \(u \npreceq _{w} v\) then for all \(t\), either \(t \npreceq _{w} v\) or \(u \npreceq _{w} t\). This is equivalent to transitivity for \(\preceq \).
The Limit Assumption turns into: for any nonempty set of worlds \(X\) and world \(w\) there is a \(v\in X\) such that there is no \(u \in X\) with \(v \npreceq _{w} u\). Equivalently, for any nonempty set of worlds \(X\) and world \(w\) there is a \(v\in X\) such that \(v \preceq _{w} u\) for all \(u\in X\).
This model also shows that (E2) and (E3) are invalid. (E2) is an inference from \(\neg p\) to \(p \mathrel {\mathop \Box }\mathrel {\mkern 2.5mu}\rightarrow q\). In the model, \(\neg p\) is true at \(w\) but \(p \mathrel {\mathop \Box }\mathrel {\mkern 2.5mu}\rightarrow q\) is false. (E3) is an inference from \(\neg (p \mathrel {\mathop \Box }\mathrel {\mkern 2.5mu}\rightarrow q)\) to \(p\). In the model, \(\neg (p \mathrel {\mathop \Box }\mathrel {\mkern 2.5mu}\rightarrow q)\) is true at \(w\) but \(p\) is false.
(E4) is an inference from \(p \mathrel {\mathop \Box }\mathrel {\mkern 2.5mu}\rightarrow q\) to \(\neg q \mathrel {\mathop \Box }\mathrel {\mkern 2.5mu}\rightarrow \neg p\). In the following model, the premise is true at \(w\) and the conclusion false.
The same style of example works for indicative conditionals.
 (a)
 Assume \(A\land B\) is true at some world \(w\) in some model \(M\). By Centring, \(w\) is among the closest \(A\)worlds to \(w\). By connectedness, \(w\) is the unique closest \(A\)world to \(w\). So \(B\) is true at all closest \(A\)worlds to \(w\).
 (b)
 Assume \(A \mathrel {\mathop \Box }\mathrel {\mkern 2.5mu}\rightarrow (B\lor C)\) is true at some world \(w\) in some model \(M\). So all the closest \(A\)worlds to \(w\) are \((B\lor C)\)worlds. If there are no \(A\)worlds then \(A \mathrel {\mathop \Box }\mathrel {\mkern 2.5mu}\rightarrow B\) and \(A \mathrel {\mathop \Box }\mathrel {\mkern 2.5mu}\rightarrow C\) are both true. If there are \(A\)worlds then Stalnaker’s semantics implies that there is a unique closest \(A\)world \(v\) to \(w\). Since \(B\lor C\) is true at \(v\), either \(B\) or \(C\) must be true at \(v\). So either \(B\) is true at all closest \(A\)worlds to \(w\) or \(C\) is true at all closest \(A\)worlds to \(v\).
Chapter 9
 (a)
 \(Srj \land Skj\); \(r\): Keren, \(k\): Keziah, \(j\): Jemima, \(S\): – is a sister of –
 (b)
 \(\forall x (Mx \to Ox)\); \(M\): – is a myriapod, \(O\): – is oviparous
 (c)
 \(\exists x (Cx \land Nx \land Hfx)\); \(f\): Fred, \(C\): – is a car, \(N\): – is new, \(H\): – has –
 (d)
 \(\neg \forall x (Sx \to Lxl)\); \(l\): logic; \(S\): – is a student, \(L\): – loves –
 (e)
 \(\forall x ((Sx \land Lxl) \to \exists y Lxy)\); \(l\): logic; \(S\): – is a student, \(L\): – loves –
(b) Let \(M_1\) be a model with a single world that can see itself. Let \(M_{2}\) be a model with two worlds, each of which can see the other but not itself. In both models, all sentence letters are false at all worlds. The very same \(\mathfrak {L}_M\)sentences are true at all worlds in these models (as a simple proof by induction shows). But the first model is reflexive and the second isn’t. So there is no \(\mathfrak {L}_{M}\)question that is true at a world in a model iff the model’s accessibility relation is reflexive.
 (a)
 \(\Box Fa\)
\(a\): John, \(F\): – is hungry.
(Might be classified as either de re or de dicto.)  (b)
 \(\Box \forall x(Fx \to Gx)\)
\(F\): – is a cyclist, \(G\): – has legs.
This is de dicto. Also correct (but different in meaning) is the de re translation \(\forall x (Fx \to \Box Gx)\). Close but incorrect (and de re): \(\forall x \Box (Fx \to Gx)\).  (c)
 \(\forall x (Fx \to \Diamond Gx)\)
\(F\): – is a day, \(G\): – is our last day.
Better: \(\forall x (Fx \to \Diamond (Hx \land \neg \exists y(Fy \land Lyx \land Hy)))\)
\(F\): – is a day, \(L\): – is later than –, \(H\): We are alive on –.Both de re. The English sentence could also be understood de dicto, as \(\Diamond \forall x (Fx \to Gx)\), but that would be a very strange thing to say.
 (d)
 I would translate this as \(\forall x \mathsf {O}(Fx \to Gx)\)
\(F\): – wants to leave early, \(G\): – leaves quietly.
Even better, if we can use the conditional obligation operator: \(\forall x \mathsf {O}(Gx / Fx)\). Also defensible are \(\forall x (Fx \to \mathsf {O} Gx)\) and \(\mathsf {O}\forall x(Fx \to Gx)\).All of these are de re.
 (e)

\(\forall x (\exists y (Fy \land Hxy) \to \mathsf {P} Gx)\)
\(F\): – is a ticket, \(G\): – enters, \(H\): – bought –.
Perhaps even better: \(\forall x \mathsf {P}(Gx/ \exists y (Fy \land Hxy))\). Both of these are de re.
You could translate ‘bought a ticket’ as a simple predicate here; you could also use a temporal operator to account for the past tense of ‘bought’ (but it’s confusing to use two different kinds of ‘\(\mathsf {P}\)’ in one sentence).
 (a)
 \begin {align*} 1. \quad & a=a &&\text { (SI)}\\ 2. \quad & \forall x\, x\not = a \to a\not = a &&\text { (UI)}\\ 3. \quad & \neg \forall x\, x\not =a &&\text { (1, 2, CPL)}\\ 4. \quad & \neg \exists x\, x=a \leftrightarrow \forall x \, x\not =a &&\text { ($\forall \exists $)}\\ 5. \quad & \exists x\, x=a &&\text { (3, 4, CPL)}\\ 6. \quad & \Box \exists x\, x=a &&\text { (5, Nec)} \end {align*}
 (b)
 There are many correct answers. For example: historians debate whether Homer
ever existed. If \(a\) translates ‘Homer’ then \(\exists x\, x=a\) is arguably false if Homer isn’t a real
person. Since the available evidence is compatible with \(\neg \exists x\, x=a\), the sentence \(\Box \exists x\, x=a\) is false on an
epistemic interpretation of the box.
Where does the proof go wrong? Each of steps 1, 2, and 6 might be blamed.
Chapter 10
 (a)
 \(W=\{ w \}\), \(wRw\), \(D = \{ \text {Alice} \}\), \(V(F,w) = \{ \text {Alice} \}\), \(V(G,w) = \emptyset \)
 (b)
 \(W=\{ w,v \}\), \(wRw\) and \(wRv\), \(D = \{ \text {Alice}, \text {Bob} \}\), \(V(F,w) = \{ \text {Alice} \}\), \(V(F,v) = \{ \text {Bob} \}\)
 (c)
 \(W=\{ w,v \}\), \(wRw\) and \(wRv\), \(D = \{ \text {Alice}, \text {Bob} \}\), \(V(F,w) = \{ \text {Alice} \}\), \(V(F,v) = \emptyset \)
 (d)
 \(W=\{ w,v \}\), \(wRw\) and \(wRv\), \(D = \{ \text {Alice}, \text {Bob} \}\), \(V(P,w) = \{ \text {Alice} \}\), \(V(P,v) = \emptyset \), \(V(Q,w) = \{ \text {Alice} \}\), \(V(Q,v) = \emptyset \)
Curiously, (4) seems to be equivalent to the Converse Barcan Formula: it, too, is valid on a frame iff the frame has increasing domains. It also rules out scenarios in which individuals at one world may fail to exist at an accessible world.