Solutions Q1, Q3, Q4.(a), Q5, Q6 to INTLOGS16 Test 1

Transcription

1 Solutions Q1, Q3, Q4.(a), Q5, Q6 to INTLOGS16 Test 1 Prof S Bringsjord NY Contents I Problems 1 II Solutions 3 Solution to Q1 3 Solutions to Q3 4 Solutions to Q4.(a) (i) 4 Solution to Q4.(a) Solution to Q5 5 Solution to Q6 5

2 Part I Problems Q1 Prove that E N. (This is of course to be an informal proof, not a formal proof in Slate. By informal proof we mean the style of proof that is expressed in a mixture of English and formal symbols.) In addition, we stipulate that E here denotes the positive even numbers; i.e., the set {2, 4, 6,...}, a definition of E that differs slightly from what is in the LAMA-BDL book.) Q2 Suppose that you are presented with a version of the Wason Selection Task exactly the same as the one we considered in class, except the four cards in front of you have the following appearance. Which card or cards should you turn over? Prove that your answer is correct. (While you can use Slate to help you, your proof should be an informal one.) G E 7 4 Q3 The following four statements are either all true, or all false. Given this, can we deduce that Lola lies? Prove that your answer is correct. (While you can use Slate to help you, your proof should be an informal one.) 1. If Lucy lies, then so does Larry. 2. If Larry lies, then so does Linda. 3. If Linda lies, then Lola does as well. 4. Lucy lies. Q4 For each of the following formulae, first (i) write down the representation of the formula as an S-expression, then (ii) give a yes or no answer as to whether it s a theorem in the propositional calculus. In addition, (iii) for each of your affirmative verdicts, provide a clear, informal proof that confirms your verdict; and for each of your negative verdicts, provide a confirming counter-example (in the form of a set of assignments of true or false to the relevant atomic formulae (or to in Slate s on-screen terminology the relevant literals). (a) P (P Q) (b) (P Q) ( Q P) (c) ((P Q) Q) P (d) P (Q P) (e) (Z Z) P (f) (P (Q R)) ((P Q) (P R)) (g) ((P Q) P) Q (h) P P (i) (D E) ( D E) Q5 In a paragraph or two, explain cogently why the Entscheidungsproblem is relevant to claims that The Singularity will soon occur. 1

3 Q6 Suppose that E is a string built by concatenating the symbols that allow well-formed formulas of the propositional calculus to be built. Such symbols include, of course, the five truthfunctional connectives (,,,, ), and atomic formulas like P, Q, etc. Can you program a Raven-machine to decide whether or not E is in fact a well-formed formula of the propositional calculus? Rigorously justify your answer. 2

4 Part II Solutions Solution to Q1 Proof: By the definition of equinumerosity, we are done if we can establish that there is a bijection b from E to N. We define: b(n) = (n 2) 1, where n E (1) Applying the function b to the smallest element of E, and then size-wise the next two elements, gives: b(2) = 0, b(4) = 1, b(6) = 2 To show that b is bijective, it suffices to show that it s both injective and surjective. Consider injectivity first, and suppose that b is not injective. Then two distinct elements of the domain map to the same natural number k. That is, we have this situaion: where n m. This yields b(n) = k and b(m) = k, n 2 1 = m 2 1, which in turn by adding 1 to both sides gives n 2 = m 2, which in turn by multiplying both sides by 2 gives n = m, and we have reached a contradiction. Our supposition is hence false: b is injective. As to surjectivity: (We will grant full credit to any cogent rationale here (as we have not covered yet the informal proof technique mathematical induction, which is the natural tool to use in order to demonstrate surjectivity in the present case). Clearly, there can t possibly be a natural number that isn t mapped-to from some even numbers, because the pattern continues forever, with one natural number at a time hit. Hence, for any natural number j N, j will eventually be reached in this progression. This means that the function b is surjective. QED 3

5 Solution to Q3 Theorem: Yes. Proof: There are two exhaustive cases to consider: one in which all four statements hold, and one in which they are all false. The first case is transparent: here, we can deduce that Lola lies by repeated application of modus ponens. In the second case, the first two statements are false; i.e., 1n. (LucyLies LarryLies) 2n. (LarryLies LindaLies) But as we have noted repeatedly from the very outset of the course, there is one and only one way a material conditional can be false: viz., when the antecedent holds, but the consequent doesn t. Therefore we can deduce from 1n. and 2n., respectively, this: (i) LucyLies LarryLies (ii) LarryLies LindaLies) But (i) and (ii) together immediately yield a contradiction that now stares us in the face: We have on the one hand LarryLies, and on the other LarryLies. Since eveything follows from a contradiction, we can immediately deduce LolaLies. QED Solutions to Q4.(a) (i) Solution to Q4.(a) (i) (if P (or P Q)) Note: I verified this in Slate to be sure. (ii) yes Note: I verified this in Slate to be sure. (iii) Proof: Suppose that P is true. A disjunction φ ψ in the propositional calculus holds provided that at least one of the disjuncts is true. But since by supposition P is true, it the follows immediately that the specific disjunction P Q is true. We have thus shown that if P, then P Q; that is, that P (P Q). QED Note: I built an oracular proof in Slate to be sure of my reasoning. 4

6 Solution to Q5 The easy version of the Entscheidungsproblem, i.e. the problem of deciding whether a given formula in the propositional caclulus is a theorem of that calculus, can be be solved by a standard algorithm (= by a Turing machine, by a program written in a standard programming language like Python, etc.). In terms of the slides/lectures, an algorithm to compute THEOREM P C exists, and can be engineered by us. (Indeed, Q4 repeatedly asks whether the test-taker can match wits with a machine able to compute THEOREM P C.) However, the Entscheidungsproblem is a harder problem, and it cannot be solved by a standard algorithm. In light of this, when people like Kurzweil glibly assert that computing machines of the future will be phenomenally smart, a key question to ask is whether these machines will be so smart that they can solve the Entscheidungsproblem. If the response is that they can solve the problem, then there is a major problem, because a computing machine unable to solve the Entscheidungsproblem cannot build one that does, so how would these Entscheidungsproblem-solving machines arrive in the first place? Suppose on the other hand that the response is a negative one. In this case, the only fundamental source of the intelligence of these super-smart future machines is that they process information faster than the machines (and humans) that come before them. But what s so special about mere speed of processing? After all, our calculators have long been much, much faster than us but does that make them more intelligent than us? Solution to Q6 Yes. 1 A proof isn t required here, only a rigrous justification for the affirmative. Here s an argument that fits the bill: First, Roger needs to be able to answer with a yes or no upon receiving E, and he needs to be invariably correct. In order to do this, he can just run an algorithm based on the Sept. 11 slide deck, which emphasized the grammar of the proposition calculus from the book, by centering around a table with six rows, where each row is a kind of wff for the propositional calculus. So here s an algorithm for Roger: 1. Check if E matches one of the forms in the six rows. (a) If not, set answer to no and halt. (b) If E is an atomic formula, set answer to yes. (c) If E is a negation φ, set E to φ and go to 1. (d) If E is a conjunction (disjunction), for as long as you don t get a command to halt, try each conjunct (disjunct) in φ i in turn as E and then go to 1. (e) If E is a conditional or biconditional φ ψ (where is either or ), first for φ set to E and then if you don t get a halt command for ψ set to E, go to Print answer and halt. 1 This question isn t asking whether a particular person can program Roger to be a good decider; it s asking whether, in general, one can program Roger to be a good decider. 5