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week1/13-01-2022/13-01-2022.tex

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+%        File: 13-01-2022.tex
+%     Created: 12:22:44 Thu, 13 Jan 2022 EST
+% Last Change: 12:22:44 Thu, 13 Jan 2022 EST
+%
+\documentclass[letterpaper]{article}
+\usepackage{amsmath}
+\usepackage{graphicx}
+\usepackage{cancel}
+\usepackage{amssymb}
+\usepackage{listings}
+\usepackage[shortlabels]{enumitem}
+\usepackage{soul}
+%\usepackage[smartEllipses,hashEnumerators,hybrid]{markdown}
+%\usepackage{minted}
+\usepackage{geometry}
+\usepackage{dirtytalk}
+\usepackage{lplfitch}
+
+\geometry{portrait, margin=1in}
+
+%\begin{minted}[linenos,bgcolor=LightGray]{[language]}
+
+\date{01/13/2022}
+\title{%
+       Lecture \\
+        \large MATH--190--04: Discrete Mathematics for Computing}
+\author{Blizzard MacDougall}
+\begin{document}
+\maketitle
+\pagenumbering{arabic}
+
+\section{Beginning Practice Problems}
+Make a truth table for the following:
+\begin{equation}
+    (p\land q) \leftrightarrow (p\lor q)
+\end{equation}
+
+\begin{center}
+\begin{tabular}{c|c||c|c|c}
+    $p$ & $q$ & $p\land q$ & $p\lor q$ & $(p\land q) \leftrightarrow (p \lor q)$\\
+    \hline
+    1 & 1 & 1 & 1 & 1\\
+    1 & 0 & 0 & 1 & 0\\
+    0 & 1 & 0 & 1 & 0\\
+    0 & 0 & 0 & 0 & 1
+\end{tabular}
+\end{center}
+
+Note that this is functionally equivalent to $p\leftrightarrow q$.
+
+
+\begin{equation}
+    [p\land (q \lor r)]\to(\lnot r)
+\end{equation}
+
+\begin{center}
+\begin{tabular}{c|c|c||c|c|c|c}
+    $p$ & $q$ & $r$ & $q\lor r$ & $p\land (q\lor r)$ & $\lnot r$ & $[p\land (q\lor r)]\to(\lnot r)$\\
+    \hline
+\end{tabular}
+\end{center}
+
+\section{New Notes}
+Remember from PHIL-205:
+\begin{itemize}
+    \item Tautology: A statement which is \emph{always} true.
+    \item Contradiction: A statement which is \emph{always} false.
+    \item Commutative Property: $p\land q \equiv q\land p$ (also works for \verb|OR|)
+    \item Associative Property: $(p\land q) \land r \equiv p\land (q\land r)$ (also works for \verb|OR|)
+    Note that this does \emph{not} work with mixed symbols.
+    \item Distributive Property: $p\land (q\land r) \equiv (p\land q)\land (p\land r)$ (also works for \verb|OR|)
+    Note that this \emph{does} work with mixed symbols.
+    \item Identity: $p\land T \equiv p$; $p\lor F \equiv p$
+    \item Negation: $p\land \lnot p \equiv F$; $p\lor \lnot p \equiv T$
+    \item Double Negation (obvious)
+    \item Idempotent $p\land p\equiv p$; $p\lor p\equiv p$
+    \item Domination Laws: True dominates \verb|OR|, false dominates \verb|AND|
+    \item DeMorgan's: Split the line, change the sign
+    \item Absorption: $p\lor (p\land q)\equiv p$;\ $p\land (\lor q)\equiv p$
+\end{itemize}
+
+\subsection{Conditionals}
+Now, how do we negate a conditional?
+\begin{equation}
+    \lnot(p\to q)\equiv \lnot(\lnot p\lor q)\equiv p\land \lnot q
+\end{equation}
+
+There is also a \emph{contrapositive}. Given $p\to q$, the contrapositive is
+$\lnot q\to\lnot p$. Note that these two statements have the same truth value.
+
+There is also the converse, and the inverse.
+The converse swaps the two terms of a conditional. The inverse negates the terms.
+Note that these "transforms" don't preserve the truth value of the statement.
+However, the converse and the inverse \emph{do} preserve truth.
+
+
+
+\end{document}