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\flushleft \underline{Homework 6 -- MATH 2510 Spring 2019}\\
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\begin{enumerate}[1.]
\item Prove the following in pure identity theory.
\begin{enumerate}
\item $(x = y) \wedge ((y=z) \wedge (z=w)) \rightarrow x=w$
\item $((x=y) \wedge \neg (y=z)) \rightarrow \neg(x=z)$
\end{enumerate}

\item Prove the following in first order arithmetic.
\begin{enumerate}[a.]
\item $S0 \cdot S0 = S0$
\item $S0 \cdot SS0=S0+S0$
\item $SS0 + SSS0= SSSSS0$
\end{enumerate}

\item Consider a new theory called ``graph theory" which obeys the following two axioms:\\
\textbf{G1} (``Symmetric"): $\forall x \forall y (Rxy \rightarrow Ryx)$ \\
\textbf{G2} (``Anti-reflexive"): $\forall x \neg R(x,x)$ \\
We think of $Rxy$ as standing in for ``$x$ is directly connected to $y$", so \textbf{G1} is saying that ``if $x$ is directly connected to $y$, then $y$ is directly connected to $x$" and \textbf{G2} is saying ``nothing is directly connected to itself" (``no loops").

Assume an extra premise $\forall x R(x,a)$ (where $a$ is a constant). Prove from the axioms of graph theory and this additional premise that $\exists x \exists y (R(x,b) \wedge R(y,c))$

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