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1.5 (odds) - 30 Chapter 1 The Foundations Logic and Proofs...

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Unformatted text preview: 30 Chapter 1 The Foundations: Logic and Proofs SECTION 1.5 Rules of Inference 1. This section lays the groundwork for understanding proofs. You are asked to understand the logical rules of inference behind valid arguments, and you are asked to construct some highly stylized proofs using these rules. The proofs will become more informal in the next section and throughout the remainder of this book (and your mathematical studies). This is modus ponens. The first statement. is 'p —> q, where p is “Socrates is human” and q is “Socrates is mortal.” The second statement is p. The third is g. Modus poneus is valid. We can therefore conclude that the conclusion of the argument (third statement) is true, because the hypotheses (the first two statements) are true. . a) This is the addition rule. W's are concluding from p that p V q must be true, where ,u is “Alice is a mathematics major” and q is “Alice is a computer science major.” b) This is the simplification rule. We are concluding from p A q that 30 must be true, where p is “Jerry is a mathematics major” and q is “Jerry is a computer science major”. c) This is modus ponens. We are concluding from p —> q and ,0 that (1 must be true, where p is “it. is rainy” and q is “the pool will be closed.” (1) This is modus tollens. We are concluding from p A q and at] that up must be true, where p is “it will snow today” and q is “the university will close today.” e) This is hypothetical syllogism. We are concluding from p —i q and q —» r that p —> r must be true, where 5 p is “I will go swimming," (1 is “I will stay in the sun too long,” and r is “I will sunburn.” . Let in be the proposition “Randy works hard,” let (1 be the proposition “Randy is a dull boy,” and let j be the proposition “Randy will get the job.” We are given premises w, w —> d, and d -> fij. We want to conclude -u_;i . We set up the proof in two columns, with reasons, as in Example 6. Step Reason 1. w Hypothesis '2. w —) d Hypothesis 3. d Modus ponens using (2) and (3) 4. d —> 77' Hypothesis 5. -Ij Modus ponens using (3) and (4) First we use universal instantiation to conclude from “For all (i: if a: is a man then :1: is mortal” the Siecial 7 7 case of interest, “If Socrates is a man, then Socrates is mortal.” Then we use modus pouens to conclude that Socrates is mortal. a) Because it was sunny on Tuesday, we assume that it did not rain or snow on Tuesday (otherwise we cannot do anything with this problem). If we use modus tollens on the universal instantiation of the given conditional statement applied to Tuesday, then we conclude that I did not take Tuesday off. If we now apply disjunctive syllogism to the disjunction in light of this conclusion, we see that I took Thursday off. Now use modus ponens on the universal instantiation of the given conditional statement applied to Thursday; we conclude that it rained or snowed on Thursday. One more application of disjunctive syllogisin tells us that it rained on Thursday. b) Using modus tollens we conclude two things—that I did not eat spicy food and that it did not thunder. Therefore by the conjunction rule of inference (Table 1), we conclude “I did not eat spicy food and it did not thunder.” c) By disjunctive syllogism from the first two hypotheses we conclude that I am clever. The third hypothesis gives us no useful information. Section 1 5 Rules of Inference 31 11. 13. d) We can apply universal instantiation to the conditional statement and conclude that if Ralph (respectively, Ann) is a CS major, then he (she) has a PC. Now modus tollens tells us that Ralph is not a CS major. There are no conclusions to be drawn about Ann. 3) The first two conditional statements can be phrased as “If .19 is good for corporations, then a: is good for the US.” and “If at is good for the U.S., then .1“ is good for you.” If we now apply universal instantiation with .13 being “for you to buy lots of stuff,” then we can conclude using modus ponens twice that for you to buy lots of stuff is good for the US. and is good for you. t) The given conditional statement is “For all .1", if .7: is a rodent, then :1 gnaws its food.” We can form the universal instantiation of this with a being a mouse, a rabbit, and a bat. Then modus ponens allows us to conclude that mice gnaw their food; and modns tollens allows us to conclude that rabbits are not rodents. We can conclude nothing about bats. we are asked to Show that whenever p1, p2 , , 1),, are true, then (1 a 1’ must be true, given that we know that whenever p1, pg, 1),, and q are true, then 7° must be true. So suppose that m, p2, . . ., 3),, are true. we want to establish that q —> r is true. If q is false, then we are clone, vacuously. Otherwise, q is true, so by the validity of the given argument. form, we know that r is true. In each case we set rip the proof in two columns with reasons, as in Example 6. a) Let c(:.: 7;) be “:c is in this class, ’ let j(.17) be‘ :1: knows how to write programs in .,"’]AVA and let 11(1) be “.1" can get a high paying job.’ We are given p1emises C(Doug), j(Doug), and V$(j(rc) *1 11(1 ',)) and we want to conclude Elm(c(3:) A h(:t)). Step Reason 1. ‘1'!ch (:13) h(:r)) Hypothesis 2. j(Doug) —-r h.(Doug) Universal instantiation using (1) 3. j(Do11g) Hypothesis 4. [1(Doug) Modus ponens using (2) and (3) 5. C(Doug) Hypothesis 6. c(Doug) /\ h(Doug ) Conjunction using (4) and (5) 7. 31(c(a) A 11(3)) Existential generalization using (6) b) Let C(33) be “:1: is in this class,” let 111(31) be “a: enjoys whale watching,” and let 13(3)) be “:1; cares about ocean pollution.” We are given premises 3:5(c(:r ) A 111(.r)) and V3;(111(;r) —+ p(.1: )), and we want to conclude 39;(c(:r) A p(.r)). In our proof, :1) represents an unspecified particular person. Step Reason 1. Elrc(c(rc) /\ w(;t)) Hypothesis 2. 13(3)) A 111(y) Existential instantiation using (1) 3. 111(3)) Simplification using ('2) 4. C(31) Simplification using (2) Va:(w(:1:) —+ p(;L')) Hypothesis 111(3)) a p(y) Universal instantiation using (5) Modus ponens using (3) and (6) C(31) /\ 11(1)) Conjunction using (4) and (7 ) 59907453"? T: x—-. C“. v .:1:El:c(c() )/\ p (1')) Existential generalization using (8) c) Let c(3:) be‘ r is in this class," let [1(a) be“ .1: owns a PC,’ and let 111(3" ) be “a can use a word processing program.” We a1e given p1emises C(Zeke), ‘v’rr(c (.1) #1 11(1)), and \zt’rt(p(r1:) —1 111(1)), and we want to conclnde ’w(Zeke). 32 l5. 17. 19. 21. 23. 25. Chapter 1 The Foundations: Logic and Proofs Step Reason 1. ‘t/:r(c (m) —> p(1:)) Hypothesis 2.c (Zeke) —> p(Zeke) Universal instantiation using (1) 3. C(Zeke) Hypothesis 4. p(Zeke) Modus ponens using (2) and (3) 5. V:,r(p(:c) ~> 111(1)) Hypothesis 6. p(Zeke) —> 111(Zeke) Universal instantiation using (5) T. w(Zeke) Modus poncns using (4) and (6) H d) Let j(:i;) be “:r is in New Jersey,” let f(:r) be “:r lives within fifty miles of the ocean, and let s(:1;) be “at has seen the ocean.” We are given premises V:1:(j (I) —> f(3:)) and 3m(j(:r.) /\ fis(_;r)), and we want to conclude 3:1;(f(:r) A —us(;i:)). In our proof, 3,: represents an unspecified particular person. Step Reason 1. 3:};(j(.'ti) As(.r)) Hypothesis 2. fly) /\ sy( ) Existential instantiation using (1) 3. j(y) Simplification. using (2) 4. Va;(j(:r) —1 f(:t)) Hypothesis 5. j(y) ~> fly) Universal instantiation using (4) 6. fly) Modus ponens using ('3) and (5 ) 7. ns(y) Simplification using (2) 8. [(31) /\ n.3(y) Conjunction using (6) and (7) 9. 3:1;(f(:1:) /\ —us(x)) Existential generalization using (8) a) This is correct, using universal instantiation and modus ponens. b) This is invalid. After applying universal instantiation, it contains the fallacy of affirming the conclusion. c) This is invalid. After applying universal instantiation, it contains the fallacy of denying the hypothesis. (1) This is valid by universal instantiation and modus tollens. We know that some :1: exists that. makes H(:1:) tr.ue but we cannot conclude that Lola is one such 3;. Alias-be only Suzanne is happy and everyone. else is not happy. Then 3:1; H(. ) is t1 no, but H(Lola) is false. “ a) This is the fallacy of affirming the conclusion, since it has the form p —> (1 and (1 implies p.” b) This reasoning is valid; it is modus tollens. c) This is the fallacy of denying the hypothesis, since it has the form “31 ~> q and 10 implies fig.” Let us give an argument justifying the conclusion. By the second premise, there is some lion that does not drink coffee. Let us call him Leo. Thus we know that Lee is a lion and that Lee does not drink coffee. By simplification this allows us to assert each of these. statements separately. The first premise says that all lions are fierce; in particular, if Leo is a lion, then. Leo is fierce. By modus ponens, we can conclude that Lee is fierce. Thus we conclude that Lee is fierce and Leo does not drink coffee. By the definition of the existential quantifier, this tells us that there exist fierce creatures that do not drink coffee; in other words, that some fierce creatures do not drink coffee. The error occurs in step (5), because we cannot assume, as is being done here, that the c that makes P true. is the same as the c that makes Q true. ') ~> Q(:L’)) and “(2(0) We want to Show nP((1). Suppose, to the contrary, \Ve are given the premises Vrr(P(:L 13(0) is true. Therefore by universal modus ponens, we have Q(a) But this that -1P(a) is not true. Then contradicts the given premise HQ (1). Therefore our supposition must have been wrong, and so -uP(a.) is true, as desired. Section 1.5 Rules of Inference 33 27. we can set this up in twocolumn format. Step Reason 1. V.77(P($) A H($)) Premise 2. PM) A R(a) Universal instantiation using (1) 3. P(a.) Simplification using (2) 4 V.’r:(P(1-) —> (Q(:c) A 5(1)» Premise QM) /\ 3(a) Universal modus ponens using (3) and (4) Simplification using (5) Simplification using (2) Rte.) /\ 8(a) Conjunction using (7) and (6) Vac-(Rte) /\ 8(3)) Universal generalization using (-5) $090749>m. to H D 29. We can set this up in twoeolumn format. The proof is rather long but, straightforward if we go one step at a 31. 33. 35. time. Step Reason 1. EmnPt’t) Premise ‘2. -uP(c) Existential instantiation using (1) 3. VII:(P(:c) V Q(:r)) Premise :1. P(c) V Q(c) Universal instantiation using (3) 5. Q(c) Disjunctive syllogism using (4) and (2) 6. VstW) V 5(1)) Premise 7. nQ(c) V S((:) Universal instantiation using (6) 8. 8(a) Disjunctive syllog'ism using (-5) and (7), since fifiQ(c) E Q(c) 9. VLE(R(1:) —> nS(:L')) Premise lO. R(c) a fiS(c) Universal instantiation using (9) ll. oli’.(c) Modus tollens using (8) and (10), since noble) E S(c) 12. flxnfific) Existential generalization using (11) Let p be “It is raining"; let (1 be “Yvette has her umbrella”; let 7" be “Yvette gets wet." Then our assumptions are op V q, sq V or, and 1) V or. Using resolution on the first. two assumptions gives us up V or. Using resolution on this and the third assumption gives us or, so Yvette does not get wet. Assume that this proposition is satisfiable. Using resolution on the first two clauses allows us to conclude q V q; in other words, we know that q has to be true. Using resolution on the last two clauses allows us to conclude fiqV fiq; in other words, we know that -‘q has to be true. This is a contradiction. So this proposition is not satisfiable. This argument is valid. We argue by contradiction. Assume that Superman does exist. Then he is not impotent, and he is not malevolent (this follows from the fourth sentence). Therefore by (the contrapositives of) the two parts of the second sentence, we conclude that he is able. to prevent evil, and he is willing to prevent. evil. By the first sentence, we therefore know that Superman does prevent evil. This contradicts the third sentence. Since we have arrived at a contradiction, our original assumption must. have been false= so we conclude finally that Superman does not exist. ...
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