Started definable sets handout

Advanced/Definable Sets/parts/0 logic.tex

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+\section{Logical Algebra}
+
+\definition{}
+Odds are, you are familiar with \textit{logical symbols}. \par
+In this handout, we'll use the following:
+\begin{itemize}
+	\item $\lnot$: not
+	\item $\land$: and
+	\item $\lor$: or
+	\item $\rightarrow$: implies
+	\item $()$, parenthesis.
+\end{itemize}
+
+The function of these is defined by \textit{truth tables}:
+\begin{center}
+\begin{tabular}{ c | c | c }
+	\multicolumn{3}{ c }{and} \\
+	\hline
+		$A$ & $B$ & $A \land B$ \\
+	\hline
+	F & F & F \\
+	F & T & F \\
+	T & F & F \\
+	T & T & T
+\end{tabular}
+\hfill
+\begin{tabular}{ c | c | c }
+	\multicolumn{3}{ c }{or} \\
+	\hline
+		$A$ & $B$ & $A \lor B$ \\
+	\hline
+	F & F & F \\
+	F & T & T \\
+	T & F & T \\
+	T & T & T
+\end{tabular}
+\hfill
+\begin{tabular}{ c | c | c }
+	\multicolumn{3}{ c }{implies} \\
+	\hline
+		$A$ & $B$ & $A \rightarrow B$ \\
+	\hline
+	F & F & T \\
+	F & T & T \\
+	T & F & F \\
+	T & T & T
+\end{tabular}
+\hfill
+\begin{tabular}{ c | c }
+	\multicolumn{2}{ c }{not} \\
+	\hline
+		$A$ & $\lnot A$ \\
+	\hline
+	T & F \\
+	F & T \\
+	~ & ~ \\
+	~ & ~ \\
+\end{tabular}
+\end{center}
+
+\vspace{2mm}
+
+$A \land B$ is only true if both $A$ and $B$ are true. $A \lor B$ is only true if $A$ or $B$ (or both) are true. \par
+$\lnot A$ is the opposite of $A$, which is why it looks like a \say{negative} sign. \par
+
+\vspace{2mm}
+
+$A \rightarrow B$ is a bit harder to understand. Read aloud, this is \say{$A$ implies $B$.} \par
+The only time $\rightarrow$ is false is when $T \rightarrow F$. Think about it: why does this make sense? \par
+
+\problem{}
+Evaluate the following.
+\begin{itemize}
+	\item $(T \land F) \lor T$
+	\item $(\lnot (F \lor \lnot T) ) \rightarrow T$
+	\item $A \rightarrow T$ for any $A$
+	\item $(\lnot (A \rightarrow B)) \rightarrow A$ for any $A,B$
+\end{itemize}
+
+\vfill
+\pagebreak
+
+\problem{}
+Show that $\lnot (A \rightarrow \lnot B)$ is equivalent to $A \land B$. \par
+\hint{Use a truth table}
+
+\vfill
+
+\problem{}
+Can you express $A \lor B$ using only $\lnot$, $\rightarrow$, and $()$?
+
+\begin{solution}
+	$((\lnot A) \rightarrow B)$
+\end{solution}
+
+\vfill
+
+Note that both $\land$ and $\lor$ can be defined using the other logical symbols. \par
+The only logical symbols we \textit{need} are $\lnot$, $\rightarrow$, and $()$. \par
+We include $\land$ and $\lor$ to simplify our logical expressions.
+
+\pagebreak