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\noindent \underline{Homework 10} \\
\noindent 1. Prove the following in pure identity theory or give an interpretation to show it is not true:
\begin{enumerate}[a.)]
\item $(x=y \wedge (y=z \wedge z=w)) \rightarrow x=w$ 
\item $\neg (x=y) \wedge \neg (y=z) \rightarrow \neg (x=z)$
\end{enumerate}
\noindent 2. Prove the following in first order arithmetic or give an interpretation to show it is not true: 
\begin{enumerate}[a.)]
\item $S0 \cdot S0=S0$ 
\item $S0 \cdot SS0 = S0 + S0$
\item $SS0 \cdot SSS0 = SSSS0$
\end{enumerate}
\noindent 3. Consider a new theory called ``total order theory" which has a two-place predicate ``$\leq$" and the following axioms: 
\begin{tabular}{|ll|}
\hline 
Axiom 1 & $(\forall x)(\forall y)(\forall z)(x \leq y \wedge y \leq z \rightarrow x \leq z)$ \\
Axiom 2 & $(\forall x)(x \leq x)$ \\
Axiom 3 & $(\forall x)(\forall y)(x \leq y \wedge y \leq x \rightarrow x=y$ \\
Axiom 4 & $(\forall x)(\forall y)(x \leq y \vee y \leq x)$ \\
\hline
\end{tabular} \\
Prove the following in total order theory or give an interpretation to show it is not true: 
\begin{enumerate}[a.)]
\item $(\forall x)(\forall y)(\neg (x \leq y) \rightarrow y \leq x)$
\item $x \leq y \wedge \neg (x \leq z) \rightarrow \neg (y \leq z)$
\item $\neg (x \leq y) \wedge \neg (y \leq z) \rightarrow \neg (x \leq z)$
\end{enumerate}
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