Fol 3

**logics** · Nov 7th 2011, 01:11 AM

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.

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(*) 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.

**emakarov** · Nov 7th 2011, 04:51 AM

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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