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Math Help - inconsistent set

  1. #1
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    inconsistent set

    Hey everyone,
    I am stuck on this proof and don't know what direction to take. I know what inconsistent means, but I don't know how to show  G_{\sigma} \vdash \lnot \sigma .

    Here is the problem...

    Suppose that  G is an inconsistent set of sentences. For each \sigma \in G , let  G_{\sigma} be the set obtained by removing  \sigma from  G . Prove
    for any \sigma \in G, G_{\sigma} \vdash \lnot \sigma

    Thanks!
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  2. #2
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    Hi

    Does inconsistent mean "no model" for you? If so, what can you say of a model of G_{\sigma}, can it be model of \sigma? Conclude. ( \sigma is a sentence: in any structure, either it is true or false)
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  3. #3
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    Inconsistent to me means there is a  \sigma such that  G \vdash \sigma and  G \vdash \lnot \sigma . How do you know that by removing  \sigma , you can still derive  \lnot \sigma ? Why couldn't you derive  \sigma ?
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  4. #4
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    Quote Originally Posted by Kiwili49 View Post
    Hey everyone,
    I am stuck on this proof and don't know what direction to take. I know what inconsistent means, but I don't know how to show  G_{\sigma} \vdash \lnot \sigma .

    Here is the problem...

    Suppose that  G is an inconsistent set of sentences. For each \sigma \in G , let  G_{\sigma} be the set obtained by removing  \sigma from  G . Prove
    for any \sigma \in G, G_{\sigma} \vdash \lnot \sigma

    Thanks!
    You're given that G is an inconsistent set of formulas.
    This means that from G a contradiction can be derived.

    Suppose you pull any one of the formulas, say P, out of G to form G-{P}. Call it G'.

    Now I think there are two cases to consider:

    Case 1: G' is still an inconsistent set. But then G' |- "any formula you like". So surely, G' |- ~P.

    Case 2: G' is now a consistent set. But clearly, G'U{P} is simply G (our inconsistent set). All we've done is to put P back into G'.
    Now all we need to show is that if G'U{P} is inconsistent (which we know it is), then G' |- ~P. We can work RAA on this one.
    Suppose G'U{P} is inconsistent. Then for some formula say Q we can derive both G',P |- Q and G',P |- ~Q
    But by RAA, we can then conclude that G' |- ~P.
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  5. #5
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    Thank you so much. I really appreciate all of your help!

    I am still confused though on case two. Does it work because you can assume anything from the contradiction?
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  6. #6
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    Quote Originally Posted by Kiwili49 View Post
    Thank you so much. I really appreciate all of your help!

    I am still confused though on case two. Does it work because you can assume anything from the contradiction?
    The wording of your question is a bit of a problem. We don't assume anything from a contradition. Replace the word assume with conclude.

    Case 1 should be immediate.

    I thought case 2 had enough detail to make it clear what was done. So maybe it wasn't enough.

    Here's case 2 in more detail. (Suppose we're in some NDS.)
    We know by pulling P out of G that we've made the result, G', consistent (this is case 2).
    Now take G' as our premiss-set and see if we can reach ~P from G'.
    So our derivation initially is nothing more than a listing of the formulas in G'.
    At that point we can introduce an assumption, namely, P.
    Essentially, we've now opened a subderivation within the our main derivation.
    Why choose P? It's because we know that G' plus P is an inconsistent set.
    Let's get all our premisses (i.e., formulas in G') into the subderivation by reiteration.
    So now, in this subderivation we have the means for deriving a contradiction (say Q and ~Q).
    Once we obtain the contradiction, what can we say about our assumption, P? Well, I'd say it doesn't look to good.
    So RAA (or typically ~I in the NDS) allows us to conclude, ~P.
    At that point, because ~I is a discharge rule, we have in effect abandoned the assumption P and closed (or sealed) the subderivation.
    What's left? We're left with G' taking us to ~P, i.e., G' |- ~P.
    Last edited by PiperAlpha167; May 14th 2010 at 04:50 AM. Reason: clean-up
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