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\flushleft \underline{Homework 7 -- MATH 2510 Spring 2019}\\
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\item Prove the following theorem of naive set theory: 
\[ \Big( (x\subseteq y) \wedge \neg ( y \subseteq x)\Big) \rightarrow \exists z(z \in y \wedge \neg(z \in x)) \]

\item Define the intersection notation $x \cap y$: we write $z \in x \cap y$ to mean $z \in x \wedge z \in y$. Prove that $x \cap y \subseteq x$. 

\item Define the union notation $x \cup y$: we write $z \in x \cup y$ to mean $z \in x \vee z \in y$. Prove that $x \subseteq x \cup y$. 

\item Consider a new theory called ``total order theory" which has a two-place predicate ``$\leq$" (meaning it compares two elements) and the following axioms:\\
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\begin{tabular}{|ll|}
\hline 
Axiom T1 & $(\forall x)(\forall y)(\forall z)(x \leq y \wedge y \leq z \rightarrow x \leq z)$ \\
Axiom T2 & $(\forall x)(x \leq x)$ \\
Axiom T3 & $(\forall x)(\forall y)(x \leq y \wedge y \leq x \rightarrow x=y$ \\
Axiom T4 & $(\forall x)(\forall y)(x \leq y \vee y \leq x)$ \\
\hline
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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 $x \leq y \wedge \neg (x \leq z) \rightarrow \neg (y \leq z)$

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