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Thread: Formula with parameters

  1. #1
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    Formula with parameters

    Hi everybody. Can someone help me finding a solution for this problem:

    "Let M be a model and phi(x) be a formula with parameters in M, x being a single variable. Prove that if phi(M) contains the parameters of phi(x), then phi(M) is not an elementary substructure of M."

    I'm not even fully understanding the problem, since I can't figure it out what the "parameters of phi(x)" are.

    Thank you so much
    bye!
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  2. #2
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    Hi, and welcome to the forum.

    Let $\displaystyle M=\langle D,\Sigma,I\rangle$ where $\displaystyle D$ is a domain, $\displaystyle \Sigma$ is a signature, and $\displaystyle I$ is an interpretation of $\displaystyle \Sigma$ on $\displaystyle D$.

    I can't figure it out what the "parameters of phi(x)" are.
    This probably means elements of D that occur in phi.

    Prove that if phi(M) contains the parameters of phi(x), then phi(M) is not an elementary substructure of M.
    What is phi(M)? I'll assume it is $\displaystyle \langle D',\Sigma,I'\rangle$ where $\displaystyle D'=\{a\in D\mid M\models\varphi(a)\}$, and $\displaystyle I'$ is the restriction of $\displaystyle I$ on $\displaystyle D'$. However, it is not clear that this is a structure, namely, that $\displaystyle D'$ is closed with respect to functions. (It may happen that for a functional symbol $\displaystyle f\in\Sigma$ and $\displaystyle a\in D'$, $\displaystyle I(f)(a)\notin D'$.) So, I'll assume that there are no functional symbols in $\displaystyle \Sigma$.

    Next, what happens if $\displaystyle D'=D$, e.g., when $\displaystyle \varphi(x)$ is $\displaystyle x=x$? Then phi(M) = M and so phi(M) is an elementary substructure of M. So, I'll assume that $\displaystyle D'\subset D$, but $\displaystyle D'\ne D$, i.e., there exists an $\displaystyle a\in D$ such that $\displaystyle M\not\models\varphi(a)$.

    In this case, what about $\displaystyle \exists x\,\neg\varphi(x)$?

    Maybe I made too many assumptions here...
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  3. #3
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    Hi and thank you for the quick reply! You made the right assumptions, I had a reply from a classmate of mine that clarified me the meaning of "parameters"... Apparently (I can't follow the lectures) our professor has a slightly uncommon approach to the subject, so the "parameters" are elements of a set that in this case, but not necessarily, coincide with D.

    However, this:

    Quote Originally Posted by emakarov View Post
    Next, what happens if $\displaystyle D'=D$, e.g., when $\displaystyle \varphi(x)$ is $\displaystyle x=x$? Then phi(M) = M and so phi(M) is an elementary substructure of M.
    is the exact same thought I had, that made me submit the problem to the community, since the definition of elementary substructure the professor gave to us implies $\displaystyle D'\subseteq D$.

    Only a bit of a correction to your statement... in the hypotesis of the exercise it says "a formula with parameters" and $\displaystyle x=x$ doesn't have any. But the core of the reasoning is correct, I think it is sufficient putting $\displaystyle D=\{0,1,2\}, \Sigma=\{<\}$ with its normal interpretation, and $\displaystyle \varphi(x)=x<2 \vee x=2$. Am I right?
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  4. #4
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    I believe so.

    I tend to understand "a formula with parameters" as "a formula that is allowed to contain parameters". But to change my example, you could take phi to be x = x /\ a = a for some a in D, or x = a -> x = a.

    Maybe your instructor wanted to claim that phi(M) is not a proper elementary substructure of M.
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  5. #5
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    Quote Originally Posted by emakarov View Post
    Maybe your instructor wanted to claim that phi(M) is not a proper elementary substructure of M.
    This could be, I'll inspect this possibility!

    Thank you!!
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