\section{Equivalence (Bonus)}

\generic{Notation:}
Let $S$ be a structure and $\varphi$ a formula. \par
If $\varphi$ is true in $S$, we write $S \models \varphi$.

\definition{}
Let $S$ and $T$ be structures. \par
We say $S$ and $T$ are \textit{equivalent} and write $S \equiv T$ if for any formula $\varphi$, $S \models \varphi \Longleftrightarrow T \models \varphi$.

\problem{}
Show that $
	\Bigl(\mathbb{Z} ~\big|~ \{ +, 0 \}\Bigr)
	\not\equiv
	\Bigl(\mathbb{R} ~\big|~ \{ +, 0 \}\Bigr)
$
\vfill

\problem{}
Show that $
	\Bigl(\mathbb{Z} ~\big|~ \{ +, 0 \}\Bigr)
	\not\equiv
	\Bigl(\mathbb{N} ~\big|~ \{ +, 0 \}\Bigr)
$
\vfill

\problem{}
Show that $
	\Bigl(\mathbb{R} ~\big|~ \{ +, 0 \}\Bigr)
	\not\equiv
	\Bigl(\mathbb{N} ~\big|~ \{ +, 0 \}\Bigr)
$
\vfill

\problem{}
Show that $
	\Bigl(\mathbb{R} ~\big|~ \{ +, 0 \}\Bigr)
	\not\equiv
	\Bigl(\mathbb{Z}^2 ~\big|~ \{ +, 0 \}\Bigr)
$
\vfill

\problem{}
Show that $
	\Bigl(\mathbb{Z} ~\big|~ \{ +, 0 \}\Bigr)
	\not\equiv
	\Bigl(\mathbb{Z}^2 ~\big|~ \{ +, 0 \}\Bigr)
$

\begin{solution}
	All of the above are easy, but the last one can take a while. \par
	The trick is to notice that $\mathbb{Z}$ has two equivalence classes mod 2, while $\mathbb{Z}^2$ has four.
\end{solution}

\vfill