2013 Midterm Solutions
P HIL 210: Brief Solutions to Mid-Term Exam, due 30 October, 2013. 1. (a) 1. f (h(g(e), g(b))) 6= a
3. Q(a) ∗ (R( f (c), g(b)) ∧ g(a) = f (b))
None of them: it’s true that g(e), g(b), h(g(e), g(b)), f (h(g(e), g(b))) and a are all terms in the language, but 6= is not allowed, since it is shorthand for ¬ = (x, y), and ¬ is not a connective in our language. Right number of brackets. 2. P(h(e, b) ∗ P(c)) None of them: P(x) is a unary predicate symbol which should take a term as an argument. However, what?s in the argument place in not a term. Right number of brackets. 1. (b)
None of them: ∗ is a binary connective which should take two sentences as arguments. However, the right-hand side of ∗ is not a sentence since ∧ is not a symbol in our language. Right number of brackets. 4. ((h(a, b) = f (c)) ∗ (b = c ∗ b = c)) ∗ (Q(e) ∗ P(g(b))) Non-atomic sentence, right number of brackets. 5. d = c ∗ (P(e) ∗ (Q(b))) Non-atomic sentence, right number of brackets.
Explain carefully why, if the inference from the sentences P1 , . . . , Pn to Q is valid, then the set of sentences {P1 , . . . , Pn , ¬Q} must be inconsistent. Suppose, for contradiction, that the inference from the sentences P1 , . . . , Pn to Q is valid, but the sentences {P1 , . . . , Pn , ¬Q} are not inconsistent, i.e., consistent. This means that there is an interpretation in which all of P1 , . . . , Pn , ¬Q are true. In this interpretation, all of P1 , . . . , Pn are true, but Q is false. But this contradicts our assumption that the inference from the sentences P1 , . . . , Pn to Q is valid, since what it means to say that this inference is valid is to say that there is no interpretation in which all of P1 , . . . , Pn are true and Q is false. So our initial supposition must be false. Hence, if the inference from the sentences P1 , . . . , Pn to Q is valid, the sentences {P1 , . . . , Pn , ¬Q} are inconsistent.
2.
Various questions based on the assumption: Suppose we have a truth-functional connective ‘∗’, which is such that P ∗ Q is true exactly when either P is false or Q is false.
1
The basic truth-table is: P T T F F
Q T F T F
P∗Q F T T T
Notice that this is the same connective as is discussed in exercise 7.29 (p. 197 of the textbook); there it is marked as ‘|’, the so-called Sheffer Stroke. The equivalences asked for in (b) can be summarised in the following table: P T T F F
Q T F T F
P∗Q F T T T
¬P F F T T
P∗P F F T T
P∨Q T T T F
(P ∗ P) F F T T
∗ T T T F
(Q ∗ Q) F T F T
2. (c)
A reasonable Fitch-style ‘elimination’ rule for P ∗ Q would be this: If P ∗ Q and P [resp. Q] appear in a proof, enter Q ∗ Q [resp. P ∗ P]. Note that this rule is based on Disjunctive Syllogism, on the grounds that P ∗ Q really ‘says’ ¬P ∨ ¬Q. (You could design a rule based on the ordinary rule of ∨-Elim if you wished.) A reasonable ‘introduction’ rule would be: If P ∗ P [resp. Q ∗ Q] appears in a proof under a given line, you can add P ∗ Q to that line.
2. (d)
This is 4.41 from the textbook: ⇔ ⇔ ⇔ ⇔ ⇔
3. (a)
C ∧ (A ∨ (B ∧C)) (C ∧ A) ∨ (C ∧ (B ∧C)) (C ∧ A) ∨ (C ∧ B ∧C) (C ∧ A) ∨ (C ∧C ∧ B) (C ∧ A) ∨ ((C ∧C) ∧ B) (C ∧ A) ∨ (C ∧ B)
Distribution of ‘∧’ over ‘∨’. Associativity of ‘∧’. Commutativity of ‘∧’. Associativity of ‘∧’. Idempotence of ‘∧’.
The problem with the purported proof on p. 168 is line 8: it derives Tet(a) ∧ Dodec(b) from Tet(a) (line 7) and Dodec(b) (line 5) by conjunction introduction. However, the claim that Dodec(b) (line 5) only follows given the assumption that Tet(a) ∧ Dodec(b) at line 4, which has been discharged at line 7, i.e., Dodec(b) doesn’t follow from the premise alone, and so it cannot be used to establish the desired conclusion stated at line 8. Consider the following counterexample, depicted below. Here, we see that a is a tetrahedron and c is large, which makes the premise true. But b is a cube, which makes the conclusion false. false.
!
2
3. (b)
[This is 6.40 in the textbook.] This is a valid argument; here’s a proof, generated in the Fitch program:
3. (c)
[This is 8.47 in the textbook.] This is an invalid argument. Consider the counterexample depicted below. Here, we see that a is small and b is medium. So the first premise is true. We also see that a is in front of b. So the second premise is true, too. Since c is a small cube, the third premise is also true. But the conclusion is false: c is not a tetrahedron, although it is not in front of b:
3. (d)
[This is 8.48 in the textbook.] The argument is valid; the Fitch program generates the following proof: This is a valid argument. Here’s a proof:
4.
Assume Harry is a Knave. So what he says is false, which means (since it’s a conditional) that Hermione is a Knight and it is not the case that Snape is not on the island, i.e., Snape is on the island. Since it follows that Hermione is a Knight, then what she says is true, viz.: If Harry is a Knave, then Snape is not on the island. Since we?re assuming that Harry is a Knave, it follows that Snape is not on the island. Now we’ve reached a contradiction: Snape both is and is not on the island. This means that our initial assumption must be false, and Harry must be a Knight. Now suppose Hermione is a Knave, and so what she says is false. So, since this is a conditional, it follows that Harry is a Knave and it is not the case that Snape is not on the island. But this contradicts the fact that Harry’s a Knight. So, our assumption must be wrong, and Hermione must be a Knight, too.
3
But, given that Harry (a Knight) utters a true conditional, and given that Hermione is a Knight (true antecedent), it must follow (by modus ponens) that Snape is not on the island. So Harry and Hermione are both Knights, and Snape is not on the island.
4
3. Q(a) ∗ (R( f (c), g(b)) ∧ g(a) = f (b))
None of them: it’s true that g(e), g(b), h(g(e), g(b)), f (h(g(e), g(b))) and a are all terms in the language, but 6= is not allowed, since it is shorthand for ¬ = (x, y), and ¬ is not a connective in our language. Right number of brackets. 2. P(h(e, b) ∗ P(c)) None of them: P(x) is a unary predicate symbol which should take a term as an argument. However, what?s in the argument place in not a term. Right number of brackets. 1. (b)
None of them: ∗ is a binary connective which should take two sentences as arguments. However, the right-hand side of ∗ is not a sentence since ∧ is not a symbol in our language. Right number of brackets. 4. ((h(a, b) = f (c)) ∗ (b = c ∗ b = c)) ∗ (Q(e) ∗ P(g(b))) Non-atomic sentence, right number of brackets. 5. d = c ∗ (P(e) ∗ (Q(b))) Non-atomic sentence, right number of brackets.
Explain carefully why, if the inference from the sentences P1 , . . . , Pn to Q is valid, then the set of sentences {P1 , . . . , Pn , ¬Q} must be inconsistent. Suppose, for contradiction, that the inference from the sentences P1 , . . . , Pn to Q is valid, but the sentences {P1 , . . . , Pn , ¬Q} are not inconsistent, i.e., consistent. This means that there is an interpretation in which all of P1 , . . . , Pn , ¬Q are true. In this interpretation, all of P1 , . . . , Pn are true, but Q is false. But this contradicts our assumption that the inference from the sentences P1 , . . . , Pn to Q is valid, since what it means to say that this inference is valid is to say that there is no interpretation in which all of P1 , . . . , Pn are true and Q is false. So our initial supposition must be false. Hence, if the inference from the sentences P1 , . . . , Pn to Q is valid, the sentences {P1 , . . . , Pn , ¬Q} are inconsistent.
2.
Various questions based on the assumption: Suppose we have a truth-functional connective ‘∗’, which is such that P ∗ Q is true exactly when either P is false or Q is false.
1
The basic truth-table is: P T T F F
Q T F T F
P∗Q F T T T
Notice that this is the same connective as is discussed in exercise 7.29 (p. 197 of the textbook); there it is marked as ‘|’, the so-called Sheffer Stroke. The equivalences asked for in (b) can be summarised in the following table: P T T F F
Q T F T F
P∗Q F T T T
¬P F F T T
P∗P F F T T
P∨Q T T T F
(P ∗ P) F F T T
∗ T T T F
(Q ∗ Q) F T F T
2. (c)
A reasonable Fitch-style ‘elimination’ rule for P ∗ Q would be this: If P ∗ Q and P [resp. Q] appear in a proof, enter Q ∗ Q [resp. P ∗ P]. Note that this rule is based on Disjunctive Syllogism, on the grounds that P ∗ Q really ‘says’ ¬P ∨ ¬Q. (You could design a rule based on the ordinary rule of ∨-Elim if you wished.) A reasonable ‘introduction’ rule would be: If P ∗ P [resp. Q ∗ Q] appears in a proof under a given line, you can add P ∗ Q to that line.
2. (d)
This is 4.41 from the textbook: ⇔ ⇔ ⇔ ⇔ ⇔
3. (a)
C ∧ (A ∨ (B ∧C)) (C ∧ A) ∨ (C ∧ (B ∧C)) (C ∧ A) ∨ (C ∧ B ∧C) (C ∧ A) ∨ (C ∧C ∧ B) (C ∧ A) ∨ ((C ∧C) ∧ B) (C ∧ A) ∨ (C ∧ B)
Distribution of ‘∧’ over ‘∨’. Associativity of ‘∧’. Commutativity of ‘∧’. Associativity of ‘∧’. Idempotence of ‘∧’.
The problem with the purported proof on p. 168 is line 8: it derives Tet(a) ∧ Dodec(b) from Tet(a) (line 7) and Dodec(b) (line 5) by conjunction introduction. However, the claim that Dodec(b) (line 5) only follows given the assumption that Tet(a) ∧ Dodec(b) at line 4, which has been discharged at line 7, i.e., Dodec(b) doesn’t follow from the premise alone, and so it cannot be used to establish the desired conclusion stated at line 8. Consider the following counterexample, depicted below. Here, we see that a is a tetrahedron and c is large, which makes the premise true. But b is a cube, which makes the conclusion false. false.
!
2
3. (b)
[This is 6.40 in the textbook.] This is a valid argument; here’s a proof, generated in the Fitch program:
3. (c)
[This is 8.47 in the textbook.] This is an invalid argument. Consider the counterexample depicted below. Here, we see that a is small and b is medium. So the first premise is true. We also see that a is in front of b. So the second premise is true, too. Since c is a small cube, the third premise is also true. But the conclusion is false: c is not a tetrahedron, although it is not in front of b:
3. (d)
[This is 8.48 in the textbook.] The argument is valid; the Fitch program generates the following proof: This is a valid argument. Here’s a proof:
4.
Assume Harry is a Knave. So what he says is false, which means (since it’s a conditional) that Hermione is a Knight and it is not the case that Snape is not on the island, i.e., Snape is on the island. Since it follows that Hermione is a Knight, then what she says is true, viz.: If Harry is a Knave, then Snape is not on the island. Since we?re assuming that Harry is a Knave, it follows that Snape is not on the island. Now we’ve reached a contradiction: Snape both is and is not on the island. This means that our initial assumption must be false, and Harry must be a Knight. Now suppose Hermione is a Knave, and so what she says is false. So, since this is a conditional, it follows that Harry is a Knave and it is not the case that Snape is not on the island. But this contradicts the fact that Harry’s a Knight. So, our assumption must be wrong, and Hermione must be a Knight, too.
3
But, given that Harry (a Knight) utters a true conditional, and given that Hermione is a Knight (true antecedent), it must follow (by modus ponens) that Snape is not on the island. So Harry and Hermione are both Knights, and Snape is not on the island.
4
