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Completeness notions in logic

  1. Feb 23, 2013 #1
    Is it true that the following definitions of completeness are equivalent?
    [itex]\mbox{For theory } \Sigma \mbox{ and for any sentence } A[/itex].

    [itex]\mbox{ Either } \Sigma \vdash A \mbox{ or } \Sigma \vdash \lnot A [/itex]
    and
    [itex]\mbox{ Either } A \in \Sigma \mbox{ or } (\lnot A) \in \Sigma[/itex].

    (The second clearly implies the first.)
     
  2. jcsd
  3. Feb 27, 2013 #2

    MLP

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    Did you mean to include that Ʃ was maximally consistent? If not then let Ʃ be the unit set that contains only the sentence letter p. If it is granted that we can then derive q→p, we have a situation where
    (1) Either q→p is derivable from Ʃ or ~(q→p) is derivable from Ʃ (because q→p is derivable)
    but
    (2) neither q→p nor ~(q→p) is a member of Ʃ

    No?
     
  4. Feb 27, 2013 #3
    No consistency is assumed.

    I'm a bit unsure here.


    We assume {P} derives/turnstile Q→P and show that (1) holds while (2) fails.

    (Using the rules of inference my class uses at least)

    {P} derives P, and
    {~Q v P} derives Q → P.

    So, (while skipping some steps)

    either {P} derives ~Q or {P} derives P.
    hence, {P} derives ~Q v P.
    hence, {P} derives Q → P.

    So, either {P} derives ~(Q → P) or {P} derives Q → P.
    Neither of which are in {P}.

    So they are not equivalent.

    Is this what you mean? (Thanks for the help)
     
  5. Feb 27, 2013 #4

    MLP

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    I think so. I was assuming that there was some way of deriving q→p from {p} without really saying what it was because systems can differ. I think what you did was show how you could get there in the system you are using.
     
  6. Feb 27, 2013 #5
    There's a flaw though.

    {P} does not satisfy (1). For example, neither {P} derives Q nor {P} derives ~Q. And we show that {P} doesn't satisfy (2) by example. I.e. {P} is incomplete.
     
  7. Feb 27, 2013 #6
    A theory is a set of formulas closed under [itex]\vdash[/itex]. So, trivially, [itex]\Sigma \vdash A[/itex] and [itex]A \in \Sigma[/itex] mean the same thing for any theory [itex]\Sigma[/itex].
     
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