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Math Help - Enderton 3.7 Problem 1

  1. #1
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    Enderton 3.7 Problem 1

    Let \sigma be a sentence such that

    \text{PA} \vdash (\sigma \leftrightarrow \text{Prb}_{PA}\sigma).

    (Thus \sigma says "I am provable," in contrast to the sentence "I am unprovable" that has been found to have such interesting proberties.)

    Does \text{PA} \vdash \sigma?

    (Notation) PA stands for "Peano arithmetic", and Prb (textbook page 266) abbreviates "Provable".

    ================================================== ===

    Löb's Theorem: Assume that T is sufficiently strong recursively axiomatizable theory. If \tau is any sentence for which T \vdash (\text{Prb}_T \tau \rightarrow \tau), then T \vdash \tau.

    Textbook says that PA is sufficiently strong theory (page 269), so I think I can apply
    Löb's Theorem directly.

    By hypothesis,  \textbook{PA} \vdash (\text{Prb}_\TEXT{PA} \sigma \leftrightarrow \sigma).

    We see that if  \textbook{PA} \vdash (\text{Prb}_\TEXT{PA} \sigma \rightarrow \sigma), then, \text{PA} \vdash \sigma by Löb's Theorem.

    Further, if  \textbook{PA} \vdash (\sigma \rightarrow \text{Prb}_\TEXT{PA} \sigma)}, then we claim that \text{PA} \vdash \sigma. Suppose to the contrary that \text{PA} \not\vdash \sigma. Then  \textbook{PA} \not\vdash \text{Prb}_\text{PA} \sigma} by the choice of \sigma, which follows that  \textbook{PA} \not\vdash (\sigma \rightarrow \text{Prb}_\TEXT{PA} \sigma)}.

    Thus \text{PA} \vdash \sigma.

    Could you check the above?

    Thanks.
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  2. #2
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    Re: Enderton 3.7 Problem 1

    Quote Originally Posted by logics View Post
    By hypothesis,  \textbook{PA} \vdash (\text{Prb}_\TEXT{PA} \sigma \leftrightarrow \sigma).

    We see that if  \textbook{PA} \vdash (\text{Prb}_\TEXT{PA} \sigma \rightarrow \sigma), then, \text{PA} \vdash \sigma by Löb's Theorem.
    I agree.

    Quote Originally Posted by logics View Post
    Further, if  \textbook{PA} \vdash (\sigma \rightarrow \text{Prb}_\TEXT{PA} \sigma)}, then we claim that \text{PA} \vdash \sigma. Suppose to the contrary that \text{PA} \not\vdash \sigma. Then  \textbook{PA} \not\vdash \text{Prb}_\text{PA} \sigma} by the choice of \sigma
    To conclude this, you need \textbook{PA} \vdash (\text{Prb}_\TEXT{PA} \sigma \rightarrow \sigma).

    Quote Originally Posted by logics View Post
    which follows that  \textbook{PA} \not\vdash (\sigma \rightarrow \text{Prb}_\TEXT{PA} \sigma)}.
    It is important that T\not\vdash B does not imply T\not\vdash A\to B. For example, if T is consistent, then T\not\vdash\bot, but T\vdash\bot\to\bot.
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