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\begin{document}

\title{Model Theory II Homework 7}
\author{Dakota Blair}
\maketitle

\setcounter{section}{14}

\begin{hw}
%\noindent \begin{itemize}
%\item

Prove
\[
\tp(\tuple{b}/A) \proves \tp(\tuple{b}/A\tuple{c}) \implies
\tp(\tuple{c}/A) \proves \tp(\tuple{c}/A\tuple{b}).
\]
%\item
Deduce if $A \subset C$ and
$\tp(\tuple{b}/A) \proves \tp(\tuple{b}/C)$
then
$\tp(\tuple{c}/A) \proves \tp(\tuple{c}/A\tuple{b})$ for all $\tuple{c} \in C$.
%\end{itemize}
\end{hw}

\begin{proof}
Assume there is a $\tuple{d}$ such that
$\tp(\tuple{d}/A) = \tp(\tuple{c}/A),
\tp(\tuple{d}/A\tuple{b}) = \tp(\tuple{c}/A\tuple{b})$
there is a $\varphi(x,\tuple{a}, \tuple{b}) \in \tp(c/A\tuple{b})$
and $\st{M} \models \symnot \varphi(\tuple{d},\tuple{a},\tuple{b})$.
Then
\[
\{ \exists y \symnot \varphi(x,\tuple{a},y) \symand \psi(y) |
\psi \in \tp(\tuple{b}/A) \} \subset \tp(\tuple{d}/A) = \tp(\tuple{c}/A).
\tag{*}
\label{ast}
\]
Then in a large model we can find $\tuple{b'}$ such that
$\tp(\tuple{b'}/A) = \tp(\tuple{b}/A)$
and $\st{M} \models \symnot \varphi(\tuple{c}, \tuple{a}, \tuple{b'})$
if and only if
$\st{M} \models \symnot \varphi(\tuple{d},\tuple{a},\tuple{d})$.
But $\st{M} \models \varphi(\tuple{c},\tuple{a},\tuple{b}$,
$\tp(\tuple{b}/A) = \tp(\tuple{b'}/A)$
and
$\tp(\tuple{b}/Ac) \neq \tp(\tuple{b'}/A\tuple{c})$
which is a contradiction.

If
$\tp(\tuple{b}/A) \proves \tp(\tuple{b}/C)$
then
$\tp(\tuple{b}/A) \proves \tp(\tuple{b}/A\tuple{c})$ for all $\tuple{c} \in C$
hence from above
$\tp(\tuple{c}/A) \proves \tp(\tuple{c}/A\tuple{b})$ for all $\tuple{c} \in C$.

\end{proof}

\end{document}
