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Interpretations and structures

  1. Mar 15, 2005 #1

    honestrosewater

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    A few lingering questions...
    Suppose an interpretation is an assignment of capital letters to declarative sentences. I see that assigning the same letter to more than one sentence limits the usefulness of your interpretation, but is it ever useful or necessary to assign more than one letter to the same sentence?

    Suppose a set E of English sentences is translated, by a suitable interpretation, into a set X of formulas of formal language L. The information contained within each sentence is lost in the translation. Because of this, X may be consistent while E is inconsistent (the inconsistencies could lie in the lost information). However, if X is inconsistent, E is inconsistent because [?blank?]. I can't say because inconsistency in L is only a relationship between formula and if a relationship holds between formula, it holds between sentences- since that would also be true of consistency. There's some difference between consistency and inconsistency that I'm missing.

    Similarly, suppose an argument (E/c), where E is as before, and c is an English sentence and the conclusion, is translated into a sequent (X l= p), where X is as before and p is formula of L. If (X l= p) is correct, (E/c) is a valid argument because 1) if E inconsistent, the argument is still valid, 2) if c is self-contradictory, E must be inconsistent- right?, 3) if c is tautologous, (E/c) is valid, and 4) are there other cases to consider?
    If (X l= p) is incorrect, (E/c) is [?blank?]. I don't think this tells us anything about (E/c). (E/c) could easily be invalid, and, if, for instance, M: "John died on Monday" and T: "John died on the day before Tuesday" are in EMT, this lost information could make several (EMT/c)s valid even if (XMT l= p) is incorrect.
    BTW, I see that (E/c) could be circular even if (X l= p) is not, since 1) we could have used more than one letter for at least one sentence in E, or 2) two or more sentences in E could be equivalent, as in EMT. I would love any tidbits such as that.
    I don't need everything spelled out for me; Hints would be enough and appreciated.
     
  2. jcsd
  3. Mar 15, 2005 #2

    AKG

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    When formalizing a sentence, some information is taken away. Now, if the remaining information is inconsistent, then adding the lost information won't remove the inconsistency. It could possibly add more inconsistency (which is why a consistent X doesn't ensure a consistent E), but it can't possibly remove the original inconsistency.

    Now, I assume you mean to say that p is the formalization of c? I don't know whether you've covered the case where E is consistent, or where c is neither tautologous nor self-contradictory. At any rate, depending on what information is lost, I think it is possible that (X |= p) is valid while all sentences of E are true and c is false. For the other way around (when X |= p is incorrect) I think it still depends on the lost information.

    That's is my take on it...
     
  4. Mar 15, 2005 #3

    honestrosewater

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    Oh, thanks, it's so obvious now.
    Yes, I meant to add that after I realized p could be compound.
    Okay, if (X |= p) is valid and X is inconsistent, then E is inconsistent and (E/c) is valid. If (X |= p) is valid and X is consistent and
    1) E is inconsistent, then (E/c) is valid.
    2) E is consistent, then all sentences in E can be true in some structure, and all the relationships among (E/c) still hold as they do among (X l= p), right? That is, E still entails c. [Edit: Actually if E entails c, (E/c) can't be invalid, but I don't know which thoughts are correct] So the only way for (E/c) to be invalid is if c is self-contradictory, right? But there is no set of consistent sentences that can entail a self-contradiction, is there?
    My book says if (X l= p) is valid, (C/n) is valid, so I'm quite sure it is- I just want to make sure I know why.
     
    Last edited: Mar 15, 2005
  5. Mar 16, 2005 #4

    AKG

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    E is consistent, then all sentences in E can be true in some structure, and all the relationships among (E/c) still hold as they do among (X l= p), right?

    I think so. In order to have E/c, we simply need that whenever all of E's elements are true, c is true. If all of E are true, then I believe it is safe to say that all of X are true, since it seems to me E is just X with some additional information. If X are all true, then by (X |= p), p is true. But from this, can we conclude c is true? Note that we're going backwards this time, so p is like c, but with less information. p may be like a "part" of c, and this part of c is true, but the whole of c may not be.

    But there is no set of consistent sentences that can entail a self-contradiction, is there?

    If a set of sentences is consistent, then all the sentences are true for some (not necessarily all) truth value assignment of its atomic parts. This truth value assignment is, from what I can tell, what you're calling a 'structure.' So, for some truth value assignment E are all true, but for all truth value assignments, c is false, so there is a truth value assignment where all of E are true but c is false, therefore E/c is not possible.
     
  6. Mar 17, 2005 #5

    honestrosewater

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    Okay, I understand everything except when (X l= p) is valid and X & E are both consistent.
    1) E is consistent & X is consistent
    2) E is consistent & X is inconsistent [impossible]
    3) E is inconsistent & X is consistent
    4) E is inconsistent & X is inconsistent
    (1-4) exhaust the cases. We already know if X is inconsistent, then E is incosistent, so (2) is impossible. Thus if E is consistent, then X is consistent (but the converse fails).

    Anyway, there's a shorter way to do this, but I want to consider each case.
    I) (X l= p) is valid and E is consistent and
    -1) c is tautologous. (E/c) is valid; A tautology is entailed by every E.
    -2) c is self-contradictory. Impossible; E cannot be consistent and entail a contradiction.
    -3) c is contingent and
    --a) c is a sentence of E. Since c is a sentence of E and E is consistent, when all sentences of E are true, it is impossible for c to be false, so (E/c) is valid.
    --b) c is a proper subsentence of E. (Meaning c is a subsentence but not a sentence, ex., in (Pigs can fly v I'm a monkey), "(Pigs can fly v I'm a monkey)" is a sentence and "(Pigs can fly v I'm a monkey)", "Pigs can fly", and "I'm a monkey" are subsentences.) I don't know where to go with this. Translation preserves an argument's form. Since c is contingent and no inconsistencies were introduced in going from (X l= p) to (E/c), all the possible truth-values remain the same. I don't know what else to say. There doesn't seem to be any relevant difference between (X l= p) and (E/c).
    --c) c is not a subsentence of E. Impossible. Since c is contingent, p is contingent. If c is not a subsentence of E, then p is not a subformula of X. But since (X l= p) is valid and E is consistent and p is contingent, p must be a subformula of X (otherwise, (X l= p) would be invalid).
     
  7. Mar 17, 2005 #6

    AKG

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    Why must p be a subformula of X in the last case? Suppose q is a subformula of X and r is not, that both q and r are contingent, and that (X |= q) is valid. Then we can always use disjunction introduction and get that (X |= (q v r)) is valid, but the conclusion is not a subformula of X. Is this right?

    Also, in case 2), do you mean that if c is self-contradictory, then so is p, and because of that "c is self-contraditory" is impossible? Otherwise, all you have is that (E/c) would be invalid, and you're trying to show that it is valid, but you use the "fact" that it is valid to show this case impossible, i.e. your reasoning is circular.
     
  8. Mar 17, 2005 #7

    honestrosewater

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    Dab nabbit. Yes, I didn't consider compound p. (If p is a tautology, neither p nor any of p's subformulas need to be in X either.) It's basically the same idea.
    --c) c and E have no subsentences in common. Impossible. Since c is contingent, p is contingent. If c and E have no subsentences in common, then p and X have no subformula in common. But since (X l= p) is valid and E is consistent and p is contingent, p and X must have at least one subformula in common. Sadly, I haven't a clue how to prove this. If p is contingent and p and X are completely independent... duh? Will duh suffice? :wink:
    Okay, by definition, (X l= p) iff (X, ~p l= ). (X l= ) means there is no structure in which all formulas of X are defined and true, i.e., X is inconsistent. Let p be contingent, X be consistent, and (X l= p). Assume X and p have no subformulas in common. Since (X l= p), (X, ~p l= ). Thus p is a subformula of X, a contradiction. It's more obvious if you say (X, ~p) is inconsistent, but whatever. How's that?
    c being self-contradictory doesn't imply that p is contradictory. c could be "This sentence is twenty words long." Since c is atomic, p is atomic so cannot be contradictory. If c is self-contradictory, cannot (E/c) still be valid if E is inconsistent? It certainly works for formal logic; [P, ~P l= (P & ~P)]. I was using the fact that I had already stated the condition that E is consistent (in I). That's okay, isn't it?
    Edit: Oh, if E is consistent AND (E/c) is valid, c cannot be self-contradictory. Phewy. I'll look it over.
     
    Last edited: Mar 17, 2005
  9. Mar 17, 2005 #8

    AKG

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    I think so ;). If p and X are independent, that is p doesn't depend on any atomic component of X, then p is not contingent on any atomic component of X, and so p is not contingent, contradiction. I don't know if that's enough to say p is contingent, but it is contingent relative to the context. I don't know if that's good enough either...
    Right, but you're trying to prove (E/c) is valid, aren't you?
     
  10. Mar 18, 2005 #9

    honestrosewater

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    The proof was right below.
    It's actually not quite right. The only way (X, ~p) could be inconsistent is if X contains ~r, where r is some subformula of p. So they still must have at least one subformula in common.
    Right, that's what I was acknowledging...
     
  11. Mar 19, 2005 #10

    honestrosewater

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    Okay, so (X l= p) and E is consistent and c is self-contradictory. Then (E, ~c) is consistent, since ~c is tautologous. From a previous post, if (E, ~c) is consistent, then (X, ~p) is consistent. But (X l= p), so (X, ~p) is inconsistent. What does that get me- that (X l= p) and E is consistent and c is self-contradictory is impossible? Should I just let (X l= p) and E be consistent, show that these conditions are consistent, and then add that c is self-contradictory so that I can pinpoint the last addition as the source of the contradiction?
     
  12. Mar 19, 2005 #11

    AKG

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    That does it, because you're assuming (X |= p) and X and E are consistent, so this shows that the case that c is self-contradictory is impossible.
     
  13. Mar 21, 2005 #12

    honestrosewater

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    Okay, so I just need to find the reason for I;3;b. It'll come to me eventually. Thanks for all the help.
     
  14. Mar 21, 2005 #13

    AKG

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    X is consistent, and (X |= p) is valid, so (X, p) is consistent, that is, on some assignment of truth values to atomic sentences, all of (X, p) (i.e. X U {p}) are true. In X, replace each instance of (p v q) with T, (p & q) with Q, (p --> q) with q, (p = q) with q, (~p) with F, and (p) with T. If we make these replacements and remove p from (X, p), we are left with X' which must also be consistent. X' is true on the truth value assignment where p is true and all the rest are true, i.e. there is an assignment when p is true and so is the rest. If we make a similar replacement and get E', then there is some truth-value assignment where c is true (since it's not self-contradictory) and if E' were inconsistent, then so would X' be, but it's not. To be honest, I don't think this is correct, but maybe this will give you an idea to work with...
     
  15. Mar 22, 2005 #14

    honestrosewater

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    In I;3;b, E is consistent. If E were inconsistent, (E/c) would still be valid. If E is inconsistent, X can be either consistent or inconsistent. It's only when E is consistent that X must also be consistent.
     
    Last edited: Mar 22, 2005
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