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\begin{document}
\title{3.2}
\author{Andrew Lounsbury}
\date{February 19, 2020}
\maketitle

\section{3.2, p.86}

\begin{enumerate}
    \item
    If $\Sigma$ is inconsistent and $\phi$ is an $\Ell$-formula, then $\Sigma \vdash \phi$. 
    \begin{proof}
        $\Sigma$ is inconsistent \\
        $\Sigma \vdash \bot$ \\
        No truth assignment makes $\bot_P$ true \\
        $\phi_P$ is true for every truth assignment that $\bot_P$ is true \\
        $\phi_P$ is a propositional consequence of $\{\bot\}_P$ \\ 
        $(\{\bot\},\phi)$ is a rule of inference of type (PC) \\
        $\therefore \Sigma \vdash \phi$
    \end{proof}

    \item
    Let $\Sigma_0 \subseteq \Sigma_1 \subseteq \Sigma_2 \subseteq \cdots$ be such that each $\Sigma_i$ is a consistent set of sentences in a language $\Ell$. Then $\bigcup \Sigma_i$ is consistent. 
    \begin{proof}
        Deny \\
        $\bigcup \Sigma_i$ is inconsistent \\
        $\bigcup \vdash \bot$ \\
        There is a deduction $D = (\Gamma, \bot)$, where $\Gamma \subseteq \bigcup \Sigma_i$ \\
        $\Gamma$ is finite $\implies \Gamma \subset \Sigma_j$, for some $j$. \\
        $\Sigma_j,\dots \vdash \bot$ \\
        $\Sigma_j,\dots$ are inconsistent, a contradiction
    \end{proof}

    \item[7.]
    The equivalence relation $\sim$ on the set $T$ of variable-free terms of the language $\Ell'$ defined by 
    \[
        t_1 \sim t_2 \text{ if and only if } (t_1 = t_2) \in \Sigma'
    \]
    is an equivalence relation. 
    \begin{proof}
        Note that $(t_1 = t_1) \in \Sigma'$ since this is the logical axiom E1. So, $\sim$ is reflexive. \\
        Let $\Gamma = \{t_1 = t_2, t_2 = t_3\}$ and let $\phi :\equiv t_1 = t_3$. If $\Gamma_P = \{A, B\}$, then when $A \land B$ is true, $\phi_P$ is true. So, $\phi$ is a propositional consequence of $\Gamma$. Hence, $(\Gamma, \phi)$ is a rule of inference of type PC, and $\Sigma' \vdash \phi$. 

        Or $\Sigma' \vdash (t_1 = t_3)$ by Thm 2.7.1. 
    \end{proof}
\end{enumerate}
\end{document}