Appendix A Answers and Hints to Selected Exercises Exercise 1.1a “‘P P’ is a logical truth” is a sentence of the metalanguage, and (I would say) is false.∨∼ ‘P P’ contains the meaningless letter ‘P’, so it isn’t a logical truth. Rather, it represents∨∼ logical truths (assuming the law of the excluded middle is correct! See chapter 3.) Exercise 1.1b ‘(P Q) (Q P)’ is a sentence of the object language. Since it contains meaningless_ expressions! _ (‘P’, ‘Q’), it isn’t true. (Not that it’s false!) Exercise 1.1c This is a bit of a trick question. “‘Frank and Joe are brothers’ logically implies ‘Frank and Joe are siblings’” is a sentence of English, which is talking about further sentences of English. So English is functioning here both as the object language and as the metalanguage. As for whether the sentence is true, I would say no, since the implication is not “formal”. Exercise 1.2a ‘Attorney and lawyer are synonyms’ confuses use and mention; inserting quotation marks thus xes the problem: ‘Attorney’ and ‘lawyer’ are synonyms. Exercise 1.2b How can we insert quotation marks to remove the use-mention confusion in ‘If S1 is an English sentence and S2 is another English sentence, then the string S1 and S2 is also an English sentence’? This is again a bit of a trick question. You might think to do it this way: 335 APPENDIX A. ANSWERS AND HINTS 336 If S1 is an English sentence and S2 is another English sentence, then the string ‘S1 and S2’ is also an English sentence. But this isn’t right. It makes the (false) claim that the string of letters ‘S1 and S2’ (a string that contains the variables ‘S1’ and ‘S2’) is an English sentence, whereas the intention of the original sentence was to say that strings like ‘Snow is white and grass is green’ and ‘Roses are red and violets are blue’ are English sentences. Really, what we want is something like this: If S1 is an English sentence and S2 is another English sentence, then the string consisting of S1, followed by ‘and’, followed by S2, is also an English sentence. Quine (1940, 36) invented a device for saying such things more concisely. In his notation, we could write instead: If S1 is an English sentence and S2 is another English sentence, then pS1 and S2q is also an English sentence. His “corner quotes”, ‘p’ and ‘q’, work like regular quotation marks, except when it comes to variables of the metalanguage such as ‘S1’ and ‘S2’. Expressions other than such variables simply refer to themselves within corner quotes, just as all expressions do within regular quotation marks. But metalanguage variables refer to their values—i.e., the linguistic expressions they stand for—rather than themselves, within Quine’s corner quotes. Thus, pS1 and S2q means the same as: the string consisting of S1, followed by ‘and’, followed by S2 Exercise 1.3 Let sentence S1 be ‘There exists an x such that x and x are identical’, and let S2 be ‘There exists an x such that there exists a y such that x and y are not identical’. Does S1 logically imply S2 according to the modal criterion? Well, that depends. It depends on what is possible. You might think that there could have existed only a single thing, in which case S1 would be true and S2 would be APPENDIX A. ANSWERS AND HINTS 337 false. If this is indeed possible, then S1 doesn’t logically imply S2 (given the modal criterion). But some people think that numbers exist necessarily, and in particular that it’s necessarily true that the numbers 0 and 1 exist and are not identical. If this is correct, then it wouldn’t be possible for S1 to be true while S2 is false (since it wouldn’t be possible for S2 to be false.) And so, S1 would logically imply S2, given the modal criterion. How about according to Quine’s criterion? Again, it depends—in this case on which expressions are logical expressions. If (as is commonly supposed) ‘there exists an x such that’, ‘there exists a y such that’, ‘not’, and ‘are identical’ are all logical expressions, then all expressions in S1 and S2 are logical expressions. So, since each sentence is in fact true, there’s no way to substitute nonlogical expressions to make S1 true and S2 false. So S1 logically implies S2 (according to Quine’s criterion). But suppose ‘are identical’ is not a logical expression. Then S1 would not logically imply S2, according to Quine’s criterion. For consider the result of substituting the predicate ‘are both existent’ for ‘are identical’. S1 then becomes true: ‘There exists an x such that x and x are both existent’, whereas S2 becomes false: ‘There exists an x such that there exists a y such that x and y are not both existent’. Exercise 1.4 Here is the de nition of the powerset of A: u : u A . The f ⊆ g powerset of 2,4,6 is ?, 2 , 4 , 6 , 2,4 , 2,6 , 4,6 , 2,4,6 . Notice that the powersetf of a setg alwaysf f containsg f g f g bothf theg f nullg setf andg f the setgg itself (look at the de nition of ‘subset’ to see why this is so.) Exercise 1.5 N and Z are equinumerous, because of the following function f : f (0) = 0, f (1) = 1, f (2) = 1, f (3) = 2, f (4) = 2, f (5) = 3, f (6) = 3,.... This function can be de ned− more rigorously as follows:− − ( n if n is even f n − 2 (for any n N) ( ) = n+1 2 if n is odd 2 Exercise 2.8 Hint: instead of trying to show directly that every wff without repetition of sentence letters has the feature of PL-invalidity, nd some feature F that is stronger than PL-invalidity (i.e., some feature F from which PL- invalidity follows), and show by induction that every wff without repeated sentence letters has this feature F ; and then, nally, conclude that every wff without repeated sentence letters is PL-invalid. APPENDIX A. ANSWERS AND HINTS 338 Exercise 2.10 Hint: call a sequent Γ φ valid iff Γ φ; prove by induction that every provable sequent is a valid sequent.) Exercise 3.7 We’re to show that there are no valid formulas in Kleene’s system. Consider the trivalent interpretation that assigns # to every sentence I letter. If there existed any Kleene-valid formula φ then KV (φ) would need to be 1, whereas we can show by induction that KV ( ) = # forI every wff . Base case: all the sentence letters are obviously # in I. Inductive step: assume that φ and are both # in . We need now to showI that φ , φ , and φ are all # in . But that’sI easy—just look at the truth tables^ for _, and . !# # is #, # # is I#, and # # is #. ^ _ ! ^ _ ! Exercise 3.8 Hint: use induction. Exercise 3.11 Hint: in each direction, prove the contrapositive. Exercise 3.8 might come in handy. Exercise 3.15 Hint: this isn’t hard, but it’s a bit tricky. It might help to note that every classical (bivalent) interpretation also counts as a trivalent interpretation, with itself as its only precisi cation. Exercise 3.16 We’re to argue that contraposition and reductio should fail, given a supervaluational semantics for (assuming the identi cation of truth with truth-in-all-sharpenings). Contraposition:4 as argued in the text, for all φ, φ logically implies “de nitely, φ”. So ‘Middling Mary is rich’ logically implies ‘Middling Mary is de nitely rich’. But ‘not: de nitely, Middling Mary is rich’ doesn’t logically imply ‘not: Middling Mary is rich’, since if Mary is a (de nite) borderline case of being rich, the rst is true on all sharpenings and hence is true, while the second is false under some sharpenings and so is not true. So to model these results, it should turn out under the supervaluationist semantics that P P but P 2 P. As for4 reductio,∼4 “Mary∼ is rich and Mary is not de nitely rich” cannot be true (on logical grounds), and so vacuously implies anything at all. (If it were true, then it would be true on all sharpenings; but then ‘Mary is rich’ would be true on all sharpenings; but then ‘Mary is not de nitely rich’ would be false.) So in particular, it logically implies both ‘Snow is white’ and ‘Snow is not white’ (say). But, contrary to reductio, ‘not: Mary is rich and Mary is not de nitely rich’ is not a logical truth, since it isn’t true. For there are sharpenings in which APPENDIX A. ANSWERS AND HINTS 339 both ‘Mary is rich’ and ‘Mary is not de nitely rich’ are true. Exercise 3.17 For the systems of Łukasiewicz, Kleene, and Priest, we are to nd intuitionistically provable sequents whose premises do not semantically imply their conclusions. Let’s begin with Kleene’s system. We showed in exercise 3.7 that there are no Kleene-valid wffs. Thus, ? 2K P P. But the following is an intuitionistically acceptable proof of the sequent ! P P: ? ) ! 1. P P RA (for conditional proof) ) 2. P P 1, I ? ) ! ! Next, ŁV ( (P P)) = # for any trivalent assignment in which P is #, I ∼ ∧∼ I so (P P) is Łukasiewicz-invalid. But ? (P P) is intuitionistically provable:∼ ∧∼ ) ∼ ∧∼ 1. P P P P RA (for reductio) ∧∼ ) ∧∼ 2.