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\begin{document}
\title{3.3 - 3.4}
\author{Andrew Lounsbury}
\date{February 23, 2020}
\maketitle

\section{3.3, p.93}

\begin{enumerate}
    \item
    $\Sigma:\equiv$ all $\Ell_{NT}$-formulas that are true in $\fN$ \\
    $\phi:\equiv (\fa x)(\ex n)(x = \underbrace{SS\cdots S}_{n S\text{'s}}0)$ is in $\Sigma$ \\
    $\therefore$ There can be no nonstandard elements in any model of $\Sigma$. \\
    Show why this is fallacious. \\

    $\phi:\equiv$``We can repeated apply $S$ to 0 to obtain any $x$ (\underline{in the natural numbers})''

    This argument doesn't take into account any models with a universe other than $\N$. 

    \item[3.]
    If $\fA$ and $\fB$  are $\Ell$-structures such that $\fA \cong \fB$, then $\fA \equiv \fB$. 
    \begin{proof}

    \end{proof}

    \item[5.]
    Every nonstandard model of arithmetic contains an infinite prime number, that is, an infinite number $a$ such that if $a = bc$, then either $b = 1$ or $c = 1$. 
    \begin{proof}

    \end{proof}

    \end{enumerate}

    \section{3.4, p. 101}

    \begin{enumerate}
    \item
    Suppose that $\fB \subseteq \fA$, that $\phi$ is of the form $(\fa x)\psi$, where $\psi$ is quantifier-free, and that $\fA \vDash \phi$. Then, $\fB \vDash \phi$. 
    \begin{proof}
        $\fA \vDash \phi[s]$ with every $s$ \\
        $\fA \vDash (\fa x)\psi[s]$ with every $s$ \\
        $\fA \vDash \psi[s(x|a)]$ with every $s$ and for every $a$ in $A$ \\
        $\fB \vDash \psi[s(x|a)]$ with every $s$ and for every $a$ in $A$ \\
        $\fB \vDash (\fa x) \psi[s]$ with every $s$ \\
        $\fB \vDash \phi[s]$ with every $s$
    \end{proof}

    Suppose that $\fB \subseteq \fA$, that $\phi$ is of the form $(\ex x)\psi$, where $\psi$ is quantifier-free, and that $\fA \vDash \phi$. Then, $\fB \vDash \phi$. 
    \begin{proof}
        $\fA \vDash \phi[s]$ for every $s$ \\
        $\fA (\ex x)\psi[s]$ for every $s$ \\
        $\fA \neg(\fa x)\neq \psi[s]$ for every $s$ \\
        $\fA \not\vDash (\fa x)\neq \psi[s]$ for every $s$ \\
        $\fA \not\vDash\neg \psi[s(x|a)]$ for every $s$ and every $a$ in $A$ \\
        $\fA \vDash \psi[s(x|a)]$ for every $s$ and every $a$ in $A$
    \end{proof}

    \item[8.]

\end{enumerate}
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