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Thread: Fol 3

  1. #1
    Junior Member
    Nov 2011

    Fol 3

    Show that the formula x=y \rightarrow Pzfx \rightarrow Pzfy (where f is a one-place function symbol and P is a two-place predicate symbol) is valid.


    (*) If \gamma;\alpha \models \phi, then \gamma \models (\alpha \rightarrow \phi).

    We show \models x=y \rightarrow Pzfx \rightarrow Pzfy.
    By (*), it suffices to show that \{x=y, Pzfx\} \models Pzfy. Therefore, we need to show every A that satisfies x=y and Pzfx with every function s:V \rightarrow |A| satisfies Pzfy with s. That is, if \models_Ax=y[s] and \models_A Pzfx[s], then \models_A Pzfy[s].

    I am stuck here. Any help will be appreciated.
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  2. #2
    MHF Contributor
    Oct 2009

    Re: Fol 3

    As you know, given A, the function s:V\to|A| can be extended to a function s_A from the set of all terms to |A|. In particular, A associates some function \mathbf{f}:|A|\to|A| to a unary functional symbol f and some function \mathbf{P}:|A|\times|A|\to\{T,F\} to a binary predicate symbol P. Then s_A(x)=s(x), s_A(fx)=\mathbf{f}(s_A(x)) and \models_A Pz(fx) iff \mathbf{P}(s_A(z),s_A(fx))=T.

    By definition of \models, if \models_Ax=y[s], then s(x) = s(y). Therefore,

    \models_A Pz(fx) iff
    \mathbf{P}(s_A(z),s_A(fx))=T iff
    \mathbf{P}(s_A(z),\mathbf{f}(s(x)))=T iff
    \mathbf{P}(s_A(z),\mathbf{f}(s(y)))=T iff
    \mathbf{P}(s_A(z),s_A(fy))=T iff
    \models_A Pz(fy)
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