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Semantic Paradoxes Continued. RECAP The Barber Paradox Once upon a time there was a village, and in this village lived a barber named B. B shaved all.

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Presentation on theme: "Semantic Paradoxes Continued. RECAP The Barber Paradox Once upon a time there was a village, and in this village lived a barber named B. B shaved all."— Presentation transcript:

1 Semantic Paradoxes Continued

2 RECAP

3 The Barber Paradox Once upon a time there was a village, and in this village lived a barber named B. B shaved all the villagers who did not shave themselves, And B shaved none of the villagers who did shave themselves. Question, did B shave B, or not?

4 Suppose B Shaved B 1. B shaved BAssumption 2. B did not shave any villager X where X shaved X Assumption 3. B did not shave B1,2 Logic

5 Suppose B Did Not Shave B 1. B did not shave BAssumption 2. B shaved every villager X where X did not shave X Assumption 3. B shaved B1,2 Logic

6 The Law of Excluded Middle Everything is either true or not true. Either P or not-P, for any P. Either B shaved B or B did not shave B, there is not third option.

7 Disjunction Elimination A or B A implies C B implies C Therefore, C

8 Contradiction, No Assumptions B shaves B or B does not shave B [Law of Excluded Middle] If B shaves B, contradiction. If B does not shave B, contradiction. Therefore, contradiction

9 Contradictions Whenever we are confronted with a contradiction, we need to give up something that led us into the contradiction.

10 No Barber In this instance, however, it makes more sense to give up our initial acquiescence to the story: We assumed that there was a village with a barber who shaved all and only the villagers who did not shave themselves. The paradox shows us that there is no such barber, and that there cannot be.

11 Disquotation To say P is the same thing as saying P is true. This is the disquotation principle: P = P is true

12 Liar Sentence L = L is not true

13 L is true 1. L is trueAssumption 2. L 1, Disquotation 3. L is not true2, Def of L 1 & 3 form a contradiction

14 L is not true 1. L is not trueAssumption 2. L1, Def of L 3. L is true2, Disquotation 1 & 3 form a contradiction

15 Contradiction L is true or L is not true [Law of Excluded Middle] If L is true, then L is true and not true. If L is not true, then L is true and not true. Therefore, L is true and not true.

16 Solutions 1.Give up excluded middle 2.Give up disjunction elimination 3.Give up disquotation 4.Disallow self-reference 5.Accept that some contradictions are true

17 The Liars Lesson? There are lots of very complicated solutions to the liar, all of which do one of two things: abandon classical logic or abandon disquotation. Its clear we have to do one of these things, but neither is very satisfying, and there are no solutions to the liar that everyone likes.

18 Grellings Paradox Grellings Paradox or the paradox of heterological terms is very similar to the liar. To begin with, lets consider a principle like Disquotation, which Ill just call D2: F applies to x = x is F

19 Autological and Heterological The analogue of L in Grellings paradox is the new term heterological defined as follows: x is heterological = x does not apply to x We can also define autological, as follows: x is autological = x does apply to x

20 Question: Does heterological apply to heterological?

21 Yes? 1. H applies to HAssumption 2. H is H1 D2 3. H does not apply to H2 Def H

22 No? 1. H does not apply to HAssumption 2. H is H1 Def H 3. H applies to H2 D2

23 Contradiction Just like the liar, were led into a contradiction if we assume: D2: F applies to x = x is F Law of excluded middle: heterological either does or does not apply to itself. A or B, if A then C, if B then C; Therefore, C

24 RUSSELLS PARADOX

25 Sets There are dogs and cats and couches and mountains and countries and planets. According to Set Theory there are also sets. The set of dogs includes all the dogs as members, and all the members of the set of dogs are dogs. Likewise for the set of mountains, and the set of planets.

26 Notation We write out sets by putting names of their members between brackets. So the set of full professors in Lingnan philosophy is: {Darell, Neven, Paisley}

27 Notation We can also write the set using a condition: {x: x is a full professor in Lingnan philosophy} This is the same as the set {Darell, Neven, Paisley}. We might introduce a name for this set: F = {x: x is a full professor in Lingnan philosophy}

28 Membership The fundamental relation in set theory is membership, or being in. Members of a set are in the set, and non- members are not. Mt. Everest is in {x: x is a mountain}, Michael Jordan is not in {x: x is a mountain}.

29 Set Theoretic Rules Reduction: a is in {x: COND(x)} Therefore, COND(a) Abstraction: COND(a) Therefore, a is in {x: COND(x)}

30 Examples Reduction: Mt. Everest is in {x: x is a mountain} Therefore, Mt. Everest is a mountain. Abstraction: Mt. Everest is a mountain. Therefore, Mt. Everest is in {x: x is a mountain}

31 Self-Membered Sets Its possible that some sets are members of themselves. Let S = {x: x is a set}. Since S is a set, S is in {x: x is a set} (by abstraction), and thus S is in S (by Def of S). Or consider H = {x: Michael hates x}. Maybe I even hate the set of things I hate. So H is in H.

32 Russells Paradox Set Most sets are non-self-membered. The set of mountains is not a mountain; the set of planets is not a planet; and so on. Define: R = {x: x is not in x}

33 Is R in R? 1. R is in RYes? 2. R is in {x: x is not in x}1, Def of R 3. R is not in R2, Reduction 4. R is not in RNo? 5. R is in {x: x is not in x}4, Abstraction 6. R is in R5, Def of R

34 Historical Importance Russells paradox was what caused Frege to stop doing mathematics and do philosophy of language instead.

35 Comparison with the Liar Russell thought that his paradox was of a kind with the liar, and that any solution to one should be a solution to the other. Basically, he saw both as arising from a sort of vicious circularity.

36 Comparison with the Liar If this is right the semantic paradoxes may not be properly semantic at all, but arise from a structural feature that many non- semantic things (like sets) also have.

37 The von Neumann Heirarchy

38 CURRYS PARADOX

39 Haskell Brooks Curry Mathematician who worked on combinatory logic. Has three computer languages named after him: Haskell, Brooks, and Curry. Devised a semantic paradox.

40 Conditional Proof Suppose you want to prove a conditional (if- then) statement. For example, suppose you want to show that if the accuser is telling the truth, then the accused should go to jail.

41 The Accusation Michael kicked me.

42 Assuming for the Sake of Argument First, you would assume for the sake of argument that the accuser is telling the truth. Assume that Michael did in fact kick the puppy. (Even though of course hes innocent.)

43 Conditional Proof Then you would use that assumption to show that Michael belonged in jail. You would argue that since kicking puppies violates article 2, section 6, paragraph 3 of the criminal code, Michael belongs in jail.

44 Conditional Proof Finally, you would stop assuming that Michael did actually kick the puppy and conclude: If the accuser is telling the truth, then Michael belongs in jail.

45 Modus Ponens Theres one other rule of logic that involves conditionals. This rule is used when we already know a conditional is true: Premise: if A, then B Premise: A Conclusion: B

46 Currys Paradox Define the Curry sentence C as follows: C = If C is true, then Michael is God.

47 Proving C To prove C we do a conditional proof: C = If C is true, then Michael is God.

48 Proving C To prove C we do a conditional proof: C = If C is true, then Michael is God. Assume this.

49 Proving C To prove C we do a conditional proof: C = If C is true, then Michael is God. Prove this.

50 Proving C To prove C we do a conditional proof: C = If C is true, then Michael is God. Throw out assumption and conclude this.

51 Proof of C 1.C is trueAssumption 2.C1, Disquotation 3.C is true, then Michael is God 2, Definition of C 4. Michael is God1, 3, Modus Ponens

52 Proof of C 1.C is trueAssumption 2.C1, Disquotation 3.C is true, then Michael is God 2, Definition of C 4. Michael is God1, 3, Modus Ponens 5. If C is true, then Michael is God 1, 4 Cond. Proof

53 No Paradox Yet Now we have proven C. This in itself should not bother us. We are not committed to saying Michael is God, only that if C is true, then Michael is God.

54 Proof that Michael is God 1. If C is true, then Michael is God. Previous Proof 2. C1, Definition of C 3. C is true2, Disquotation 4. Michael is God1, 3, Modus Ponens

55 Set Theory Version Currys Paradox also comes in a set theoretic flavor. The paradoxical set is this one: M = {x: if x is in x, then Michael is God}

56 Requirements Currys Paradox leads to contradiction (or absurdity, or anything you like) and all it requires is: Self-reference Disquotation Conditional Proof Modus Ponens

57 Important Feature Currys Paradox uses different logical rules (modus ponens, conditional proof) than the liar paradox (excluded middle, disjunction elimination). This suggests that its probably not the logic, but instead self-reference and/ or disquotation that is the problem.

58 Important Feature Recall that paraconsistent logics can avoid the Liar Paradox by assuming that some sentences can be both true and false while avoiding explosion. Even with such an assumption, however, Currys Paradox still works, and leads to explosion, since you can prove anything with it.

59 THE PARADOX OF THE HEAP

60 Sorites 1.1 grain of sand is not a heap. 2.For all numbers n: if n grains of sand are not a heap, then n + 1 grains of sand are not a heap. 3.Therefore, 200 trillion grains of sand are not a heap.

61 The Other Way trillion grains of sand makes a heap. 2.For all numbers n: if n + 1 grains of sand make a heap, then n grains of sand make a heap. 3.Therefore 1 grain of sand makes a heap.

62 Paradox Neither of these sorites arguments results in a contradiction… until you add in the obvious fact that the conclusion of each is false.

63 Borderline Cases The paradox seems to arise whenever we have a term that admits of borderline cases. There are some people that I dont know whether theyre rich out of uncertainty: because I dont know how much money they have. These are not borderline cases.

64 Borderline Cases The paradox seems to arise whenever we have a term that admits of borderline cases. But there are other people that I dont know whether they are rich even though I know exactly how much money they have. These are borderline cases.

65 Borderline Cases Most of our ordinary language admits of borderline cases: Big, tall, short, rich, fast, slow, smart, dumb, funny, long, flat, narrow… Also: mountain, car, tree, horse…

66 What To Do? Neither of these sorites arguments results in a contradiction… until you add in the obvious fact that the conclusion of each is false. To deny the conclusion, we need to deny either premise 1 or premise 2 or logic.

67 Denying Premise 1 In the first argument, premise 1 is: 1 grain of sand is not a heap. In the second its: 200 trillion grains of sand is a heap.

68 Denying Premise 2 Premise 2 (Argument 1) says: For all numbers n: if n grains of sand are not a heap, then n + 1 grains of sand are not a heap. The negation of this is: There exists a number n such that: n grains of sand are not a heap, but n + 1 grains of sand are a heap.

69 Denying Premise 2 Premise 2 (Argument 2) says: For all numbers n: if n + 1 grains of sand make a heap, then n grains of sand make a heap. The negation of this is: There is a number n such that: n + 1 grains of sand make a heap, but n grains of sand do not make a heap.

70 No Sharp Boundaries Premise 2 in both cases asserts No Sharp Boundaries. Its never true that one grain of sand makes the difference between a heap and not a heap.

71 No Sharp Boundaries One hair doesnt make the difference between being bald and not bald. One micrometer doesnt make the difference between being tall and not tall. $0.10HKD does not make the difference between being rich and not rich. One nanosecond does not make the difference between being old and not old.

72 Solutions 1.Accept Sharp Boundaries. 2.Introduce more truth-values.

73 Epistemicism One solution is to claim that there ARE sharp boundaries, but we can never know where they are. Acquiring $0.10 can make someone go from not rich to rich, but we cant ever know when this happens.

74 Epistemicism Basic problem: What determines the boundary if not how we use the words? What determines how we use the words if not what we (can) know?

75 Epistemicism Further problem: the epistemicist says we cant know where the Sharp Boundary is, but that it exists. However, he has to admit that we can: Guess where the Sharp Boundary is. Wonder where the Sharp Boundary is. Fear that we are crossing the Sharp Boundary (e.g. for getting old). But all these seem silly!

76 Many-Valued Logics Another solution is to introduce a new truth- value: True, False, and Undefined. Theres No Sharp Boundaries, because theres no point at which adding one hair moves someone from truly bald to falsely bald.

77 Many-Valued Logics More hairs tttttttttttttttttttttttuuuuuuuuuuufffffffffffffffffff

78 Higher-Order Vagueness The problem is that now there are sharp boundaries between being truly bald and undefinedly bald, and between being undefinedly bald, and falsely bald. Intuitively, adding one hair to a truly bald person cant make them undefinedly bald.

79 Many-Valued Logics More hairs tttttttttttttttttttttttuuuuuuuuuuufffffffffffffffffff Two sharp boundaries!

80 Fuzzy Logic Instead, we might try having infinitely many truth-values: 1 is fully true, 0 is fully false, and any number in between is less than fully true. More hairs …

81 Fuzzy Logic A fuzzy logician has to explain how to calculate the truth-values of complex expressions from the truth values of their parts. Common rules: The truth-value of ~P is 1 minus the truth- value of P The truth-value of P & Q is the lowest of the truth-values of P and Q. The truth-value of P or Q is the highest of the truth values of P and Q.

82 Problems P & ~P should always be fully false: 0. But if P = 0.5, then P & ~P = 0.5


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