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Extension Theorem

  1. Aug 25, 2007 #1
    Hi! I'm just reading Wilfrid Hodges's book Logic, chapter 24. Properties of Semantic Entailment. I'm a bit puzzled by the following paragraphs concerning the Extension Theorem:


    What if X contains formulae A, A->B and Y contains just B' (negation of B). Then the resulting set X,Y is semantically inconsistent and therefore the theorem isn't true.

    I'm probably wrong, but I can't see where...
    Last edited: Aug 25, 2007
  2. jcsd
  3. Aug 25, 2007 #2


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    Elaborate upon that "therefore".
  4. Aug 25, 2007 #3
    Well, assume that the first part of the if-clause is true. X implies psi. But then the second part is somewhat tricky - X,Y is semantically inconsistent, so it cannot imply psi. That's what I mean.
  5. Aug 25, 2007 #4
    I'm on the right track now, I think. X |= a (X is a set of formulae, a is a formula) means that there's no structure in which a and all the formulae of X are defined, and all the formulae of X are true while a is false. In the example above, this is satisfied, since there's no structure in which all the formulae of X are true... Is this explanation correct?
  6. Aug 25, 2007 #5


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    That is correct.

    So I think you see, or are at least close to seeing, why you were wrong before. If {X, Y} is inconsistent, then it is clear that X,Y|=P, no matter what P is.
  7. Aug 25, 2007 #6
    thank you.. when i began to elaborate on that "therefore", i was enlighted and saw my mistake:)
  8. Aug 29, 2007 #7
    The property of semantic validity isn't lost by introducing the set Y.
    But the property of soundness will, in general, be lost (assuming the argument was
    sound to begin with). That's important, because (in the final analysis) the question
    of acceptance is usually couched in terms of soundness of argument.

    Unfortunately, I don't have a copy of the book. Maybe the author addresses the issue later.
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