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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 the villagers who did not shave themselves, And B shaved none of the villagers who did shave themselves.

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Semantic Paradoxes Continued


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    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 B Assumption 2. B did not shave any villager X where X shaved X Assumption 3. B did not shave B 1,2 Logic

    5. Suppose B Did Not Shave B 1.B did not shave B Assumption 2. B shaved every villager X where X did not shave X Assumption 3. B shaved B 1,2 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 true Assumption 2. L 1, Disquotation 3. ‘L’ is not true 2, Def of L 1 & 3 form a contradiction

    14. “‘L’ is not true” 1. ‘L’ is not true Assumption 2. L 1, Def of L 3. ‘L’ is true 2, 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 • Give up excluded middle • Give up disjunction elimination • Give up disquotation • Disallow self-reference • Accept that some contradictions are true

    17. The Liar’s 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. It’s 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. Grelling’s Paradox Grelling’s Paradox or the paradox of heterological terms is very similar to the liar. To begin with, let’s consider a principle like Disquotation, which I’ll just call D2: ‘F’ applies to x = x is F

    19. Autological and Heterological The analogue of ‘L’ in Grelling’s 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 ‘H’ Assumption 2. ‘H’ is H 1 D2 3. ‘H’ does not apply to ‘H’ 2 Def H

    22. No? 1. ‘H’ does not apply to ‘H’ Assumption 2. ‘H’ is H 1 Def H 3. ‘H’ applies to ‘H’ 2 D2

    23. Contradiction Just like the liar, we’re 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. Russell’s 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 It’s 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. Russell’s 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 R Yes? 2. R is in {x: x is not in x} 1, Def of R 3. R is not in R 2, Reduction 4. R is not in R No? 5. R is in {x: x is not in x} 4, Abstraction 6. R is in R 5, Def of R

    34. Historical Importance Russell’s 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. Curry’s 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 he’s 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 There’s 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. Curry’s 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.