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\markboth{}{Foundations of Mathematics \--- Dr. Russell E. Goodman}
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\bf\emph{\Large{Scrambled Proof \#1}}
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\noindent Consider the following theorem and the ``pieces" of the proof you've been given. Please put the pieces in logical order to form an understandable proof.

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\noindent \textbf{Theorem:} Let $x,y\in\mathbb{Z}$ and assume $x$ is even and $y$ is odd.  Then $x+y$ is odd and $x\cdot y$ is even.

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\noindent\textbf{Proof:}

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\begin{enumerate}
\item $2(c(2c'+1))=2(2cc'+c)$
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\item Finally, we conclude that $x$ even and $y$ odd implies $x+y$ is odd and $x\cdot y$ is even, as desired.
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\item $x+y=(2c)+(2c'+1)$
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\item Thus, by definition of \emph{odd}, we know $x+y=2(c+c')+1$ is an odd integer.
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\item Thus, by definition of \emph{even}, we know that $x\cdot y=2(2cc'+c)$ is an even integer.
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\item $(2c)(2c'+1)=2(c(2c'+1))$
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\item This is because there exists an integer, in this case $k=(c+c')$, such that $x+y=2k+1$.
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\item There exists $c\in\mathbb{Z}$ such that $x=2c$.
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\item Next, we consider $x\cdot y=(2c)(2c'+1)$
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\item Assume $x,y\in\mathbb{Z}$.
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\item $(2c+2c')+1=2(c+c')+1$
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\item $(2c)+(2c'+1)=(2c+2c')+1$
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\item This is because there exists an integer, in this case $j=(2cc'+c)$, such that $x\cdot y=2j$.
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\item Assume $x$ is even and $y$ is odd.
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\item There exists $c'\in\mathbb{Z}$ such that $y=2c'+1$.
\end{enumerate}

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\noindent\textbf{Key:} 10, 14, 8, 15, 3, 12, 11, 4, 7, 9, 6, 1, 5, 13, 2\\

\noindent\textbf{Note \#1:} Statements 8 and 15 are interchangeable without altering the logic of the proof.\\

\noindent\textbf{Note \#2:} Two portions of the proof are interchangeable without altering the logic of the proof. One could prove the statement about $x\cdot y$ before proving the statement about $x+y$.
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