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\noindent
\parbox{2in}{\bf Math 2510} 
\hfill {\Large \bf Test \#1} \hfill
\parbox{2in}{\bf \hfill February $23^{\mathrm{rd}}$, 2010}

\vspace{0.3in}

\noindent {\large\bf Name:} \underline{\hskip 3.0 truein}\\
Don't merely state answers, prove your statements. {\bf Be sure to show your work!}

\begin{description}
   \item[\large 1. (\underline{\hskip 0.35 truein}/12 points)] Consider theorem L10: $\vdash (\neg B \to \neg A) \to (A \to B)$
\begin{enumerate}[(a)]

   \item Show L10 is a tautology using by filling out an abbreviated truth table.

\vspace{4in}

   \item Prove L10 in system L. [You may use the deduction theorem and theorems L1 -- L9.]

\end{enumerate}

\newpage
   \item[\large 2. (\underline{\hskip 0.35 truein}/12 points)] Flawed proof\\

I am in the middle of proving ``$A \to B, \neg A \to B \vdash B$''. First, show (using an abbreviated truth table) that
            this is indeed a theorem of L. Then explain why I know I've made a mistake if I have the following lines in my proof:
            
            \begin{tabular}{llcc}
            \vdots & \vdots & & \vdots \\
            10 & $A$         &  & M.P. (lines 5 and 9) \\
            11 & $\neg \neg B \to \neg A$ & & M.P. (lines 2 and 7) \\
            12 & $(\neg \neg B \to \neg A) \to (A \to \neg B)$ & & L10 \\
            \vdots & \vdots & & \vdots 
            \end{tabular}

\newpage
   \item[\large 3. (\underline{\hskip 0.35 truein}/10 points)] Construct models whose objects are $\mathbb{Z}$ (integers) to show that 
    $$\forall x \forall y \, (P(x,y) \to P(f(y),f(x)))$$ is satisfiable but not logically valid. [Note: $P(x,y)$ is a predicate and $f(x)$ is a function.]

\newpage
   \item[\large 4. (\underline{\hskip 0.35 truein}/14 points)] Proofs in K.
   
   \begin{enumerate}[(a)]
      \item Prove theorem K13: $\vdash \forall x A(x) \to \exists x A(x)$
      
      \vspace{4.5in}
      
      \item Prove theorem K31: $\vdash \exists x (A(x) \wedge B(x)) \to (\exists x A(x) \wedge \exists x B(x))$
   \end{enumerate}   

\newpage
   \item[\large 5. (\underline{\hskip 0.35 truein}/14 points)] How about...more proofs?
   
   \begin{enumerate}[(a)]
      \item Use induction to show that $1+4+7+\cdots+(3n-2) = \frac{1}{2}n(3n-1)$ for all positive integers $n$. 
      
      \vspace{4.5in}
      
      \item Prove that $\sqrt{2}$ is irrational. [Recall: $x$ is rational means $\exists$ integers $p$, $q \not= 0$ such that $x=p/q$.]
      
      {\it Hint:} Proof by contradiction.
   \end{enumerate}   

\end{description}

\newpage
    
\noindent
\parbox{1.5in}{\bf Math 2510} 
\hfill {\Large \bf Test \#1 -- TAKE HOME} \hfill
\parbox{1.5in}{\bf \hfill Due: Feb. $24^{\mathrm{th}}$, 2010}

\vspace{0.3in}

\noindent {\large\bf Name:} \underline{\hskip 3.0 truein}\\
You may use notes and your textbook, but no help from other people especially your classmates.
 
\vspace{0.2in} 
   
\begin{description}
   \item[\large 6. (\underline{\hskip 0.35 truein}/12 points)]  A few more proofs in L
   \begin{enumerate}[(a)]
      \item Prove theorem L7: $A \to (B \to C) \vdash B \to (A \to C)$. You may use the deduction theorem and theorems L1 -- L6.

\vspace{0.2in}

      \item Prove $\vdash (A \wedge \neg A) \to B$. You may use the deduction theorem and theorems L1 -- L15.

\vspace{0.2in}

      \item Prove $A \vee B \vdash B \vee A$. You may use the deduction theorem and theorems L1 -- L12.

\vspace{0.5in}

   \end{enumerate}

   \item[\large 7. (\underline{\hskip 0.35 truein}/15 points)]  A few more proofs in K.  Prove (at least) 5 of the theorems K27 -- K37.

\vspace{0.5in}

   \item[\large 8. (\underline{\hskip 0.35 truein}/12 points)]  More proofs !?!
   \begin{enumerate}[(a)]
      \item Prove that $1 \cdot 1! + 2 \cdot 2! + \cdots n \cdot n! = (n+1)! - 1$ for all positive integers $n$.

\vspace{0.2in}

      \item Let $n$ be an odd integer. Prove that $n^2-1$ is a multiple of 4.

\vspace{0.2in}

      \item Let $r$ be an irrational number and $m$ be an integer. Prove that $mr$ is irrational.

\vspace{0.2in}

      \item Let $a$, $b$, and $c$ be integers. Show that if $a \,\Big|\, b$ and $b \,\Big|\, c$, then $a \,\Big|\, c$.
   \end{enumerate}


\end{description}
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