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\noindent
\parbox{2.25in}{\bf Math 2110}
\hfill {\Large \bf Homework \#1} \hfill
\parbox{2.25in}{\bf \hfill Due: Wed., Aug. $26^\mathrm{th}$, 2020}

\vspace{0.2in}

\begin{enumerate}
\item Consider the following statement: ``Bill is old, but his wife is not.'' Give appropriate definitions for $a$ and $b$ and translate our statement into a formal proposition in terms of $a$ and $b$. Define and translate in such a way that your logical statement involves at least one negation.

\item Let $a$ be the statement ``Tomorrow is Thursday'', $b$ be ``The sky is blue'', and $c$ be ``Becky loves chocolate''.
         Write down a plain English version of the proposition using these translations of $a$, $b$, and $c$. 

\begin{enumerate}[(a)]
   \item Proposition \#1: $a \vee \neg b$
   \item Proposition \#2: $(a \wedge \neg b) \to c$
   \item Proposition \#3: $\neg a \wedge (b \to c)$
\end{enumerate}

\item For each of the following propositions,

\begin{enumerate}[I.]
   \item Write down an abbreviated truth table.
   \item State whether the proposition is a tautology, contradiction, or contingency.
\end{enumerate}

\begin{enumerate}[(a)]
   \item Proposition \#1: $a \to (\neg b \to a)$
   \item Proposition \#2: $(a \wedge b) \to (a \leftrightarrow c)$
   \item Proposition \#3: $\neg (a \vee b) \wedge (a \to \neg b)$
\end{enumerate}

\item The Sheffer stroke\footnote{See for example: \href{https://en.wikipedia.org/wiki/Sheffer_stroke}{https://en.wikipedia.org/wiki/Sheffer\textunderscore{}stroke}} or NAND (= not and) operation is denoted $a \uparrow b$ and has the following truth table:

\begin{center}
\begin{tabular}{c|c|c}
$a$ & $b$ & $a \uparrow b$ \\ \hline
T & T & F \large\strut \\
F & T & T \large\strut \\
T & F & T \large\strut \\
F & F & T \large\strut 
\end{tabular} 
\end{center}

Show that $a \uparrow b$ is logically equivalent to $\neg (a \wedge b)$.

\vspace{0.1in}

{\it Note:} It turns out that the Sheffer stroke is ``sufficient'' for expressing all possible propositions. This follows if our typical operations are already sufficient (they are -- in fact just negation and disjunction are enough) and we know how to translate our usual operations in terms of Sheffer stokes -- which can be done as follows:

$$ \neg a \longleftrightarrow (a \uparrow a) 
\qquad \qquad (a \vee b) \longleftrightarrow ((a \uparrow a) \uparrow (b \uparrow b))
\qquad \qquad (a \wedge b) \longleftrightarrow ((a \uparrow b) \uparrow (a \uparrow b))$$

$$ (a \to b) \longleftrightarrow (a \uparrow (b \uparrow b))
\qquad \qquad (a \leftrightarrow b) \longleftrightarrow ((a \uparrow b) \uparrow ((a \uparrow a) \uparrow (b \uparrow b)))
$$


\end{enumerate}


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