%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Beamer Presentation
% LaTeX Template
% Version 1.0 (10/11/12)
%
% This template has been downloaded from:
% http://www.LaTeXTemplates.com
%
% License:
% CC BY-NC-SA 3.0 (http://creativecommons.org/licenses/by-nc-sa/3.0/)
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%----------------------------------------------------------------------------------------
%	PACKAGES AND THEMES
%----------------------------------------------------------------------------------------

\documentclass[9pt]{beamer}
%\setbeamerfont{frametitle}{size=\normalsize}
\usepackage{tikz}
\usepackage{color, xcolor}
\usetikzlibrary{shapes, patterns}
%colorblind-friendly palette
\definecolor{dblue}{RGB}{51,34,136}
\definecolor{lblue}{RGB}{136,204,238}
\definecolor{green}{RGB}{17,119,51}
\definecolor{tan}{RGB}{221,204,119}
\definecolor{mauve}{RGB}{204,102,119}

\newcommand{\ds}{\displaystyle}
\newcommand{\dbox}[1]{\colorbox{tan}{\textbf{#1}}}
\usepackage{amsthm}
\theoremstyle{definition}
\newtheorem*{defn}{Definition}
\newtheorem*{convention}{Convention}
\newtheorem*{recall}{Recall}
\newtheorem*{thm}{Theorem}
\tikzstyle{vertex}=[shape=circle, minimum size=2mm, inner sep=0, fill]
\tikzstyle{opendot}=[shape=circle, minimum size=2mm, inner sep=0, fill=white, draw]
\newcommand{\axis}[2]{\draw[help lines, <->] (-#1,0)--(#1,0) node[right]{$x$};
\draw[help lines, <->] (0,-#2)--(0,#2) node[above]{$y$};}
\newcommand{\aaxis}[4]{\draw[help lines, <->] (-#1,0)--(#2,0) 
node[right]{$x$};
\draw[help lines, <->] (0,-#3)--(0,#4) node[above]{$y$};}

\newcommand{\ct}{\textcolor{mauve}{T}}
\newcommand{\cf}{\textcolor{green}{F}}

\usepackage{multicol}
\setlength{\columnsep}{1cm}

%\usepackage{enumitem, tablists, amsmath,vwcol}
\usepackage{amsmath,vwcol}


\mode<presentation> {

% The Beamer class comes with a number of default slide themes
% which change the colors and layouts of slides. Below this is a list
% of all the themes, uncomment each in turn to see what they look like.

%\usetheme{default}
%\usetheme{AnnArbor}
%\usetheme{Antibes}
%\usetheme{Bergen}
\usetheme{Berkeley}
%\usetheme{Berlin}
%\usetheme{Boadilla}
%\usetheme{CambridgeUS}
%\usetheme{Copenhagen}
%\usetheme{Darmstadt}
%\usetheme{Dresden}
%\usetheme{Frankfurt}
%\usetheme{Goettingen}
%\usetheme{Hannover}
%\usetheme{Ilmenau}
%\usetheme{JuanLesPins}
%\usetheme{Luebeck}
%\usetheme{Madrid}
%\usetheme{Malmoe}
%\usetheme{Marburg}
%\usetheme{Montpellier}
%\usetheme{PaloAlto}
%\usetheme{Pittsburgh}
%\usetheme{Rochester}
%\usetheme{Singapore}
%\usetheme{Szeged}
%\usetheme{Warsaw}

% As well as themes, the Beamer class has a number of color themes
% for any slide theme. Uncomment each of these in turn to see how it
% changes the colors of your current slide theme.

%\usecolortheme{albatross}
%\usecolortheme{beaver}
%\usecolortheme{beetle}
%\usecolortheme{crane}
%\usecolortheme{dolphin}
%\usecolortheme{dove}
%\usecolortheme{fly}
%\usecolortheme{lily}
%\usecolortheme{orchid}
%\usecolortheme{rose}
\usecolortheme{seagull}
%\usecolortheme{seahorse}
%\usecolortheme{whale}
%\usecolortheme{wolverine}
\setbeamercolor{alerted text}{fg=mauve}

\setbeamertemplate{footline} % To remove the footer line in all slides uncomment this line
%\setbeamertemplate{footline}[page number] % To replace the footer line in all slides with a simple slide count uncomment this line

\setbeamertemplate{navigation symbols}{} % To remove the navigation symbols from the bottom of all slides uncomment this line
}

\usepackage{graphicx} % Allows including images
\usepackage{booktabs} % Allows the use of \toprule, \midrule and \bottomrule in tables
\begin{document}

\section{2. Logic}
\begin{frame}{Introduction}
From the Book:
\begin{quote}In proving theorems, we apply logic to information that is considered
obviously true (such as ``Any two points determine exactly one line.") or is
already known to be true (e.g., the Pythagorean theorem). If our logic is
correct, then anything we deduce from such information will also be true
(or at least as true as the ``obviously true" information we began with).\end{quote}\pause
\vfill
Example: \\
If you are the prime minister, then you are human.\\\pause
Justin Trudeau is the prime minster.\\ \pause
Therefore, Justin Trudeau is human.\\ \pause
I am not prime minister.
Am I human?\\\pause

\vfill
Example: \\
No mammal can fly.\\ \pause
A pig is a mammal.\\\pause
Therefore, a pig cannot fly. \textcolor{lblue}{(True)}\\ \pause
A bat is a mammal, therefore a bat cannot fly. \textcolor{lblue}{(False)}
\end{frame}
%----------------------------------------------------------------------------------------
\subsection{2.1 Statements }
%----------------------------------------------------------------------------------------
\begin{frame}{What Are Statements?}
\begin{defn}A \dbox{statement} is a sentence
or a mathematical expression that is either definitely true or definitely
false.\end{defn}\pause

\begin{align*}
&6\times 3 = 18& \onslide<3->{\mbox{\textcolor{lblue}{statement (true)}}}\\
&6\times 3 &\onslide<4->{\mbox{\textcolor{lblue}{not a statement}}}\\
&\mbox{For every finite set $A$, $|\mathcal P(A)|=2^{|A|}$}.&
 \onslide<5->{\mbox{\textcolor{lblue}{statement (true)}}}\\
&\mbox{For every real number $x$, $\sqrt{x^2}=x$.}&
 \onslide<6->{\mbox{\textcolor{lblue}{statement (false)}}}\\
&\mbox{Let $x$ be a real number.}&
 \onslide<7->{\mbox{\textcolor{lblue}{not a statement}}}\\
&\mbox{Multiply by the conjugate.}&
 \onslide<8->{\mbox{\textcolor{lblue}{not a statement}}}\\
\end{align*}
\onslide<9->{We often give statements single-letter names, as we do for variables.

$P$: $6\times 3 = 18$}
\end{frame}
%----------------------------------------------------------------------------------------
\begin{frame}{We don't always know whether a statement is true or false.}
\begin{defn}\dbox{Open sentence}: truth depends on a variable.\\
Example: \textcolor{green}{$x$ is prime}.\\
This is not a statement, at least not until we know what $x$ is.\end{defn}\pause
\vfill

\begin{block}{Goldbach Conjecture}
Every even integer greater than 2 is a sum of two prime numbers.
\end{block}
 $4=2+2$, \hfill $6=3+3$, \hfill $8=3+5$, \hfill $20=7+13$\hfill~
 \pause\\
Holds through $4 \times 10^{18}$ (Wikipedia)\\\pause 
  The Goldbach Conjecture is a statement. (We don't know whether it is true or false, but we know it is one of them!)
\end{frame}
%----------------------------------------------------------------------------------------
\subsection{2.2 And, Or, Not }
%----------------------------------------------------------------------------------------
\begin{frame}{And + Truth Tables}
\begin{defn}
For statements $P$ and $Q$, the statement ''$P$ and $Q$", written \dbox{$P \wedge Q$}, is true if and only if both $P$ and $Q$ are true.
\end{defn}\pause\vfill

$P$: ``I washed the dishes and cleaned the sink."\\\pause\vfill

$Q$: ``I washed the dishes."\\
$R$: ``I cleaned the sink."\\
$P=Q \wedge R$\pause\vfill

If one (or both) of $Q$ or $R$ is false, then $P$ is false.\vfill

\centering
\begin{tabular}{|c|c||c|}
\hline
$\mathbf{Q}$&$\mathbf{R}$&$\mathbf{Q \wedge R}$\\
\hline
\hline
T&T&\onslide<5->{T}\\
\hline
T&F&\onslide<6->{F}\\
\hline
F&T&\onslide<7->{F}\\
\hline
F&F&\onslide<8->{F}\\
\hline
\end{tabular}

\end{frame}
%----------------------------------------------------------------------------------------
\begin{frame}{Or}
\begin{defn}
For statements $P$ and $Q$, the statement ``$P$ or $Q$", written \dbox{$P \vee Q$}, is true if and only if one \emph{or both} of $P$ and $Q$ are true.
\end{defn}\pause\vfill
\emph{This is different from standard English usage.}

If a mathematician says ``take a chocolate cookie or a peanut cookie," you can take one of each, and you may take one cookie, but you must take a cookie.\pause\vfill

\color{mauve}
True or False: $3$ is prime or $5$ is prime. 
\onslide<4->{\hspace{1cm}\textcolor{lblue}{True.}}

True or False: $3$ is prime or $10$ is prime.
\onslide<5->{\hspace{1cm}\textcolor{lblue}{True.}}

True or False: $3$ is prime or some lions have green eyes.
\onslide<6->{\hspace{1cm}\textcolor{lblue}{True.}}

True or False: $10$ is prime or $20$ is prime.
\onslide<7->{\hspace{1cm}\textcolor{lblue}{False.}}
\vfill
\color{black}

\end{frame}
%----------------------------------------------------------------------------------------
\begin{frame}{Or, Not}
\begin{center}
\begin{tabular}{|c|c||c|}
\hline
$\mathbf{P}$&$\mathbf{Q}$&$\mathbf{P\vee Q}$\\
\hline
\hline
T&T&\onslide<2->{T}\\
\hline
T&F&\onslide<3->{T}\\
\hline
F&T&\onslide<4->{T}\\
\hline
F&F&\onslide<5->{F}\\
\hline
\end{tabular}
\end{center}

\onslide<6->{What if we want to express ``$P$ or $Q$, but not both, are true."}\\
\onslide<7->{That is: $\underbrace{P\vee Q}_{\mbox{or}}$ is true, and $\underbrace{P\wedge Q}_{\mbox{and}}$ is false.}

\onslide<8->{\begin{defn}
For a statement $P$, the statement ``not $P$", written \dbox{$\sim P$}, is true when $P$ is false, and false when $P$ is true. (You might also see this written as ``$\lnot P$".)
\end{defn}}

\onslide<9->{\begin{center}
\begin{tabular}{|c|c||c|c|c||c|}
\hline
$\mathbf{P}$&$\mathbf{Q}$&
$\onslide<11->{\color{lblue}\mathbf{P \vee Q}}$&
\color{black}\onslide<13->{$\mathbf{P \wedge Q}$}&
\color{lblue}\onslide<12->{$\mathbf{\sim(P \wedge Q)}$}&
\onslide<10->{\textcolor{mauve}{$\mathbf{(P\vee Q) \wedge \sim(P \wedge Q)}$}}\\
\hline
\hline
T&T&
\onslide<14->{\color{lblue}T}&
\onslide<15->{\color{black}{T}}&
\onslide<16->{\color{lblue}{F}}&
\onslide<17->{\color{mauve}{F}}\\
\hline
T&F&
\onslide<18->{\color{lblue}T}&
\onslide<19->{\color{black}{F}}&
\onslide<20->{\color{lblue}{T}}&
\onslide<21->{\color{mauve}{T}}\\
\hline
F&T&
\onslide<22->{\color{lblue}T}&
\onslide<23->{\color{black}{F}}&
\onslide<24->{\color{lblue}{T}}&
\onslide<25->{\color{mauve}{T}}\\
\hline
F&F&
\onslide<26->{\color{lblue}F}&
\onslide<27->{\color{black}{F}}&
\onslide<28->{\color{lblue}{T}}&
\onslide<29->{\color{mauve}{F}}\\
\hline
\end{tabular}
\end{center}}
\end{frame}

%----------------------------------------------------------------------------------------
%%\begin{frame}
%True or false: the statements 
%\[{(P\vee Q) \wedge \sim(P \wedge Q)}\] and 
%\[(P \vee Q) \wedge (\sim P \vee \sim Q)\] are %equivalent.
%\end{frame}
%----------------------------------------------------------------------------------------
\subsection{2.3 Conditional Statements}
%----------------------------------------------------------------------------------------
\begin{frame}{Conditional Statements (``if-then", implications)}
\begin{defn}
For statements $P$ and $Q$, the statement ``If $P$ then $Q$," written
\dbox{$P \Rightarrow Q$}, is true \emph{unless} $P$ is true and $Q$ is false.\\

$P \Rightarrow Q$ can therefore also be written as
$\sim P \vee Q$, or $\sim(P \wedge \sim Q)$.
\end{defn}\pause

$P$: $n$ is divisible by 9\\
$Q$: $n$ is divisible by 3.\pause

\begin{multicols}{2}
{\textcolor{mauve}{$n=27$}}\\
\onslide<7->{\textcolor{green}{$n=10$}}\\
\onslide<11->{\textcolor{dblue}{$n=6$}}

\columnbreak
\begin{tabular}{|c|c||c|c|}
\hline
$\mathbf{P}$&$\mathbf{Q}$&$\mathbf{P \Rightarrow Q}$
&\onslide<15->{$\mathbf{Q \Rightarrow P}$}\\
\hline
\hline
\onslide<4->{\color{mauve}T}&\onslide<5->{\color{mauve}T}&\onslide<6->{\color{mauve}T}&\onslide<17->{\color{mauve}T}\\
\hline
\onslide<8->{\color{green}F}&\onslide<9->{\color{green}F}
&\onslide<10->{\color{green}T}
&\onslide<18->{\color{green}T}\\
\hline
\onslide<12->{\color{dblue} F}
&
\onslide<14->{\color{dblue} T}
&
\onslide<13->{\color{dblue} T}
&
\onslide<16->{\color{dblue} F}\\
\hline

\end{tabular}
\end{multicols}
\end{frame}
%----------------------------------------------------------------------------------------
\begin{frame}{Conditional Example}
$P$: $f(x)$ is a continuous function on $[a,b]$.\\
$Q$: $N$ is a number strictly between $f(a)$ and $f(b)$.\\
$R$: There exists some number $c \in (a,b)$ such that 
$f(c)=N$.\pause

\begin{block}{Intermediate Value Theorem}
 \[(P \wedge Q) \Rightarrow R\]
 %\boxed{If} $f(x)$ is a continuous function on $[a,b]$, and $N$ is a number strictly between $f(a)$ and $f(b)$, \boxed{then} there exists some number $c \in (a,b)$ such that 
%$f(c)=N$.
\end{block}\pause\vfill

\begin{multicols}{2}
\color{mauve}$f(x)=\frac{1}{x^2}$\\
$a=-10$, $b=1$\\
$N=4$\\
\vspace{5mm}
\color{green}
$f(x)=\frac{1}{x^2}$\\
$a=-1$, $b=1$\\
$N=\frac{1}{4}$
\\
\vspace{5mm}
\color{dblue}
$f(x)=\frac{1}{x^2}$\\
$a=1$, $b=10$\\
$N=\frac{1}{4}$

\color{black}
\columnbreak

\begin{tabular}{|c|c|c||c|}
\hline
$\mathbf{P}$&$\mathbf{Q}$&$\mathbf{R}$&$\mathbf{(P \wedge Q)\Rightarrow R}$\\
\hline
\onslide<4->{\textcolor{mauve}{ F}}
&\onslide<6->{\textcolor{mauve}{ F}}
&\onslide<7->{\textcolor{mauve}{T }}
&\onslide<5->{\textcolor{mauve}{ T}}\\
\hline
\onslide<8->{\textcolor{green}{ F}}
&\onslide<9->{\textcolor{green}{F }}
&\onslide<10->{\textcolor{green}{F }}
&\onslide<11->{\textcolor{green}{T }}\\
\hline
\onslide<12->{\textcolor{dblue}{ T}}
&\onslide<13->{\textcolor{dblue}{T }}
&\onslide<14->{\textcolor{dblue}{T }}
&\onslide<15->{\textcolor{dblue}{T}}\\
\hline
\end{tabular}
\end{multicols}
\end{frame}
%----------------------------------------------------------------------------------------
\begin{frame}{Conditional Statements and Truth Tables}
The statement 
\[P \Rightarrow Q\]
is \textbf{true} whenever $P$ is false.\pause

\begin{itemize}
\item A statement must be true or false, and not both.\pause
\item If you don't claim anything, then you didn't lie. \\(So a statement claiming \emph{nothing} is not false.)\pause
\item When mathematicians say a statement is a \emph{theorem}, there should not be any instance in which that statement is false.\\\pause

Compare:
\begin{multicols}{2}
\color{green}{If $\overbrace{n\mbox{ is divisible by 9}}^{A}$,\\
 then $\underbrace{n\mbox{ is divisible by 3}}_{B}$.}
\onslide<12->{\begin{tabular}{|c ||c|c||c|}
\hline
&$A$&$B$& $A \Rightarrow B$\\
\hline
$n=6$& \onslide<13->{F} & \onslide<14->{T} & \only<15>{\textcolor{mauve}{F?}}\only<16->{T} \\
\hline
\end{tabular}
\onslide<17->{\\If $A\Rightarrow B$ were false for some values of $n$, then it would not be a theorem.}}
\columnbreak

\color{dblue}{If $\overbrace{n\mbox{ is divisible by 3}}^{P}$, then\\ 
$\underbrace{n\mbox{ is divisible by 9}}_{Q}$.}
\onslide<6->{\begin{tabular}{|c ||c|c||c|}
\hline
&$P$&$Q$& $P \Rightarrow Q$\\
\hline
\onslide<7->{$n=6}$&\onslide<8->{ T} &\onslide<9->{ F} &\onslide<10->{ F} \\
\hline
\end{tabular}
\onslide<11->{The example $n=6$ shows that this statement is not a theorem, because it is not always true.}}
\end{multicols}
\end{itemize}
\color{black}
\end{frame}
%----------------------------------------------------------------------------------------
\begin{frame}{Further Examples with Conditional Statements}
\color{mauve}
\begin{align*}
(\mbox{the TV is made entirely of cheese}) && \onslide<2->{\textcolor{dblue}{\Rightarrow} }&&&(\mbox{the TV does not turn on})\\
(\mbox{you are in Canada}) &&  \onslide<3->{\textcolor{dblue}{\Leftarrow} } &&&(\mbox{you are in Vancouver, BC})\\
(x>3) &&  \onslide<4->{\textcolor{dblue}{\Rightarrow} } &&&(x \geq 3)\\
(x^2>9) &&  \onslide<5->{\textcolor{dblue}{\Rightarrow} } &&&(|x| \geq 3)\\
\end{align*}
\color{black}
\end{frame}
%----------------------------------------------------------------------------------------
%----------------------------------------------------------------------------------------
\subsection{2.4 Biconditional Statements}
%----------------------------------------------------------------------------------------
\begin{frame}{Converse Statements}
\begin{alertblock}{Careful!}
$P \Rightarrow Q$ does not mean that $Q \Rightarrow P$. \\$Q \Rightarrow P$ is called the \dbox{converse} of $P \Rightarrow Q$.
\end{alertblock}\pause
\textcolor{green}{If you are the prime minister, then you are a human.}\pause\qquad \textcolor{lblue}{True.}\\\pause
If you are a human, then you are the prime minister.\pause\qquad \textcolor{lblue}{False.}\pause\vfill

\textcolor{green}{If you are the murderer, then you were driving a blue van. }\\\pause
Daniel was driving a blue van. \pause Is Daniel the murderer? \pause\\\textcolor{lblue}{Maybe, maybe not.}\vfill\pause

In every day life, we use ``if/then" statements imprecisely.\\
\textcolor{green}{If you are registered for this course, then you will turn in homework on Friday.}
\\ \pause \textcolor{lblue}{Not necessarily true.}\pause\\
If you turn in homework on Friday, then you are registered for this course.\pause\\ \textcolor{lblue}{Not necessarily true.}
\end{frame}
%----------------------------------------------------------------------------------------
\begin{frame}{Biconditional Statements (if and only if)}
\begin{defn}
A statement of the form $P \Leftrightarrow Q$ (meaning $P \Rightarrow Q$ AND $Q \Rightarrow P$) is \dbox{biconditional}. It means that $P$ is true if and only if $Q$ is true. In this case, we think of $P$ and $Q$ as being equivalent.
\end{defn}\pause

\textcolor{green}{A matrix $A$ is invertible if and only if it has nonzero determinant.}\\ \pause
($A$ is an invertible matrix) $\Leftrightarrow$ ($A$ is a matrix with nonzero determinant)\pause
\vfill\vfill
\color{mauve}
Conditional or Biconditional?
\vfill
($n$ is an even integer)\onslide<4>{,}\onslide<5->{\textcolor{lblue}{$\Leftrightarrow$}} ($n$ is an integer divisible by 2)
\vfill
($T$ is teaching our class right now)\onslide<-5>{,}\onslide<6->{\textcolor{lblue}{$\Rightarrow$}} ($T$ is named Elyse)
\vfill
If $x$ is any real number,\\ ($\sqrt{x^2}=x$)
 \onslide<-6>{,}\onslide<7->{\textcolor{lblue}{$\Leftarrow$}}
 ($x \geq 10$)
\vfill
\end{frame}
%----------------------------------------------------------------------------------------
\begin{frame}{Further Examples with (Bi-)Conditional Statements}
\color{mauve}
\begin{align*}
(x=0) && \onslide<2->{\textcolor{dblue}{\Leftrightarrow} }  &&&(4x=0)\\
(x^2-4=0) && \onslide<3->{\textcolor{dblue}{\Leftarrow} }  &&&(x=2)
\\
(x\mbox{ is an integer}) && \onslide<4->{\textcolor{dblue}{\Leftrightarrow} }  &&&(x-17\mbox{ is an integer})\\
(x\mbox{ is odd}) && \onslide<5->{\textcolor{dblue}{\Leftrightarrow} }   &&&(1+x\mbox{ is even})\\
(x\mbox{ is a real number}) &&  \onslide<6->{\textcolor{dblue}{\Rightarrow} } &&&(x^2\mbox{ is a real number})\\
(x^2\geq 9) && \onslide<7->{\textcolor{dblue}{\Leftrightarrow} }   &&&(|x| \geq 3)\\
\end{align*}
\color{black}
\end{frame}
%----------------------------------------------------------------------------------------
%----------------------------------------------------------------------------------------
\subsection{2.5 Truth Tables for Statements}
%----------------------------------------------------------------------------------------
\begin{frame}{Truth Tables}
Suppose you have a statement that involves 5 separate statements, for example:
\[(A \wedge B \wedge C)\vee (\sim A \wedge D) \vee (\sim B \wedge E)\]\pause

\textcolor{dblue}{How many rows will be in the truth table?}

\begin{center}
\begin{tabular}{|c|c|c|c|c||c|}
\hline
$A$&$B$&$C$&$D$&$E$&$(A \wedge B \wedge C)\vee (\sim A \wedge D) \vee (\sim B \wedge E)$\\
\hline 
\only<3->{T}&\onslide<3->{T}&\onslide<3->{T}&\onslide<3->{T}&\onslide<3->{T}&\onslide<3->{T}\\
\hline
\onslide<4->{T}&\onslide<4->{T}&\onslide<4-7,9->{F}&\onslide<4->{T}&\onslide<4-7,9->{F}&\onslide<4->{F}\\
\hline
\onslide<5-7,9->{F}&\onslide<5-7,9->{F}&\onslide<5->{T}&\onslide<5->{T}&\onslide<5-7,9->{F}&\onslide<5->{T}\\
\hline
\onslide<6->{\vdots}&&&&&\\
\hline
\end{tabular}
\end{center}
\onslide<7->{Consider the following model: you choose which statements are true, and the rest are false.}

\end{frame}
%----------------------------------------------------------------------------------------
\subsection{2.6 Logical Equivalence}
%----------------------------------------------------------------------------------------
\begin{frame}{Logical Equivalence and Truth Tables}
\begin{defn}
Two statements are logically equivalent if (and only if) they have exactly the same values in a truth table.\pause

($P$ and $Q$ are logically equivalent)$\Leftrightarrow$ ($P \Leftrightarrow Q$)
\end{defn}\pause
Example: $(\sim P \vee \sim Q)$ is equivalent to $\sim(P \wedge Q)$\pause
\centering

\begin{tabular}{|c|c||c|c|c||c|c|}
\hline
$\mathbf{P}$&
$\mathbf{Q}$&
$\mathbf{\sim P}$&
$\mathbf{\sim Q}$&
$\mathbf{P\wedge Q}$&
$\mathbf{(\sim P \vee \sim Q)}$&
$\mathbf{\sim (P\wedge Q)}$\\
\hline
\ct&\ct &\onslide<5->{F}&\onslide<5->{F}&\onslide<5->{T}&\onslide<6->{\cf}&\onslide<7->{\cf}\\
\hline
\ct&\cf&
\onslide<8->{F}&
\onslide<8->{T}&
\onslide<8->{F}&
\onslide<8->{\ct}&
\onslide<8->{\ct}
\\
\hline
\cf&\ct&
\onslide<9->{T}&
\onslide<9->{F}&
\onslide<9->{F}&
\onslide<9->{\ct}&
\onslide<9->{\ct}\\
\hline
\cf&\cf&
\onslide<9->{T}&
\onslide<9->{T}&
\onslide<9->{F}&
\onslide<9->{\ct}&
\onslide<9->{\ct}\\
\hline
\end{tabular}
\end{frame}
%----------------------------------------------------------------------------------------
\begin{frame}{Logical Equivalence}
Which of the following are logically equivalent statements?
\begin{itemize}
\item $P \Rightarrow Q$
\item $\sim P \wedge \sim Q$
\item $\sim P \vee Q$
\item $\sim(P \vee Q)$
\end{itemize}\pause
\begin{center}
\begin{tabular}{|c|c||c|c|c|c|}
\hline
$P$&$Q$&
$P \Rightarrow Q$ &$\sim P \wedge \sim Q$&$\sim P \vee Q$& $\sim(P \vee Q)$\\
\hline
\textcolor{mauve}{T}&\textcolor{mauve}{T}&
\onslide<3->{T}&\onslide<3->{F}&\onslide<3->{T}&
\onslide<3->{F}
\\
\hline
\textcolor{mauve}{T}&\textcolor{green}{F}&
\onslide<3->{F}&\onslide<3->{F}&\onslide<3->{F}&
\onslide<3->{F}\\
\hline
\textcolor{green}{F}&\textcolor{mauve}{T}&
\onslide<3->{T}&\onslide<3->{F}&\onslide<3->{T}&
\onslide<3->{F}\\
\hline
\textcolor{green}{F}&\textcolor{green}{F}&
\onslide<3->{T}&\onslide<3->{T}&\onslide<3->{T}&
\onslide<3->{T}\\
\hline
\end{tabular}\end{center}
\end{frame}
%----------------------------------------------------------------------------------------
\begin{frame}{Common Equivalences}
De Morgan's Laws:
\begin{itemize}
\item $\sim(P \wedge Q) \Leftrightarrow (\sim P \vee \sim Q)$
\item $\sim(P \vee Q) \Leftrightarrow (\sim P \wedge \sim Q)$
\end{itemize}\pause
Associative Laws:
\begin{itemize}
\item $(P \wedge Q) \wedge R\Leftrightarrow P \wedge (Q \wedge R) \Leftrightarrow (P \wedge Q \wedge R)$
\item $(P \vee Q) \vee R\Leftrightarrow P \vee (Q \vee R) \Leftrightarrow (P \vee Q \vee R)$
\end{itemize}
\pause
{Contrapositive:}
\begin{itemize}
\item \textcolor{mauve}{$(P \Rightarrow Q$)} $\Leftrightarrow$ \textcolor{green}{$( (\sim Q) \Rightarrow (\sim P))$}
\end{itemize}\pause
\color{mauve}
\begin{center}
\begin{tikzpicture}
\onslide<4-8>{\draw (0,2) node(p){$P$};
\onslide<6->{\draw (-1,0) node(q){$Q$ is true};}
\onslide<7->{\draw (1,0) node(a){maybe $Q$ true};
\draw (1,-.35) node {maybe $Q$ false};}
\onslide<5->{\draw[thick, ->] (p)--(q) node[midway, left]{$P$ true};}
\onslide<5->{\draw[thick, ->] (p)--(a) node[midway, right]{$P$ false};}
\draw (0,-1) node{\boxed{P \Rightarrow Q}};
\onslide<8->{\draw[green] (0,-1.5) node{$\sim Q \Rightarrow \sim P$};}}
\onslide<9->{\color{green}
\draw (0,2) node(p){$\sim Q$};
\onslide<11->{\draw (-1,0) node(q){$\sim P$ is true};
%\draw (q) node[below]{($P$ is false)};
}
\onslide<12->{\draw (1,0) node(a){maybe $P$ true};
\draw (1,-.35) node[]{maybe $P$ false};}
\onslide<10->{\draw[thick, ->] (p)--(q) node[midway, left]{$\sim Q$ true};}
\onslide<10->{\draw[thick, ->] (p)--(a) node[midway, right]{$\sim Q$ false};}
\draw (0,-1) node{\boxed{\sim Q \Rightarrow \sim P}};
\onslide<13->{\draw[mauve] (0,-1.5) node{$P \Rightarrow Q$};}}
\end{tikzpicture}
\end{center}
\end{frame}%----------------------------------------------------------------------------------------
\begin{frame}{Implication (Conditional
 Statements) and the Contrapositive}
 \[(P \Rightarrow Q) \Leftrightarrow ( (\sim Q) \Rightarrow (\sim P))\]
Let the Venn diagram below represent two sets: the set of all instances in which statement $P$ is true, and the set of all instances in which $Q$ is true. 

If $P \Rightarrow Q$, label $P$ and $Q$ below.

\begin{center}
\begin{tikzpicture}
\draw[thick] (0,0) node[shape=circle, minimum size=1cm, draw]{};
\draw[thick] (0,0) node[shape=circle, minimum size=3cm, draw]{};
\onslide<2->{
\draw[thick, mauve] (0,0) node[shape=circle, minimum size=1cm, fill=mauve!20,  pattern=north west lines,  pattern color=mauve!50,  draw]{};
\draw[thick, green] (0,0) node[shape=circle, minimum size=3cm, fill=green!20,  pattern color=green!20, pattern=north east lines, draw]{};
\draw[mauve] (.3,0) node{$P$};
\draw[green] (1.3,0) node{$Q$};}
\end{tikzpicture}
\end{center}
\end{frame}
%----------------------------------------------------------------------------------------
\begin{frame}{Common Equivalences}
Contrapositive:
\begin{itemize}
\item $(P \Rightarrow Q) \Leftrightarrow ( (\sim Q) \Rightarrow (\sim P))$\\\pause\vfill
\textcolor{green}{If you are prime minister, then you are a human.}\\ \pause
If you are not a human, then you are not prime minister.\\ \pause\vfill
\textcolor{green}{If you are the murderer, then you drove a blue van yesterday.}\\ \pause
Jin did not drive a blue van yesterday. \pause Can Jin be the murderer? \pause\\Logically, Jin is definitely not the murderer.\\ \pause\vfill
\textcolor{green}{Let $f(x)$ be a differentiable function. If $f(x)$ has at least two roots, then $f'(x)=0$ somewhere. (Rolle's Theorem)}\\ \pause
 Let $f(x)$ be a differentiable function. If $f'(x)$ is never zero, then $f(x)$ has at most one root.\\
 \pause
 Example: $f(x)=e^x+x-1$
\end{itemize}
\end{frame}
%----------------------------------------------------------------------------------------
%----------------------------------------------------------------------------------------
\begin{frame}{A Statement is Logically Equivalent to its Contrapositive}
For each statement below, choose the statement that is logically equivalent.
\begin{center}
\textcolor{green}{$P \Rightarrow Q$}
\end{center}
\begin{enumerate}[(a)]
\item $Q \Rightarrow P$
\item $\sim P \Rightarrow \sim Q$
\item \alert<2->{ $\sim Q \Rightarrow \sim P$}
\end{enumerate}

\vfill

\onslide<3->{\begin{center}
\textcolor{green}{If it is a square, then it is a rectangle.}
\end{center}
\begin{enumerate}[(a)]
\item If it is a rectangle, then it is a square.
\item If is it not a square, then it is not a rectangle.
\item \alert<4->{If it is not a rectangle, then it is not a square.}
\item It is a square if and only if it is a rectangle.
\item It is a square if and only if it is a square.
\end{enumerate}}

\vfill

\onslide<5->{\begin{center}
\textcolor{green}{If you cook, then you clean.}
\end{center}
\begin{enumerate}[(a)]
\item If you do not cook, then you do not clean.
\item \alert<6->{If you do not clean, then you do not cook.}
\item You cook if and only if you clean.
\item If you do not clean, you cooked.
\end{enumerate}}
\end{frame}
%----------------------------------------------------------------------------------------

%----------------------------------------------------------------------------------------
\begin{frame}{A Statement is Logically Equivalent to its Contrapositive}
For each statement below, choose the statement that is logically equivalent.
\begin{center}
\textcolor{green}{If the year is 1999, then we are partying.}
\end{center}
\begin{enumerate}[(a)]
\item \alert<2->{If we are not partying, then the year is not 1999.}
\item If the year is not 1999, then we are not partying.
\item We are partying if and only if the year is 1999.
\end{enumerate}

\vfill

\begin{center}
\textcolor{green}{If you aren't part of the solution, you're part of the problem.}
\end{center}
\begin{enumerate}[(a)]
\item If you are a part of the problem, then you are not a part of the solution.
\item You are a part of the solution if and only if you are not a part of the problem.
\item \alert<3->{If you are not a part of the problem, then you are a part of the solution.}
\item (none of the above)
\end{enumerate}

\vfill

\begin{center}
\textcolor{green}{If it's Tuesday, then it is raining.}
\end{center}
\begin{enumerate}[(a)]
\item \alert<4->{If it is not raining, then it is not Tuesday.}
\item If it is raining, then it is Tuesday.
\item It is raining if and only if it is Tuesday.
\end{enumerate}
\end{frame}
%----------------------------------------------------------------------------------------
\subsection{2.7 Quantifiers}
%----------------------------------------------------------------------------------------
\begin{frame}{Quantifiers}
\begin{defn}
The \dbox{universal quantifier}, $\forall$, means ``for every" or ``for all."

The \dbox{existential quantifier}, $\exists$, means ``there is" or ``there exists."\pause\vspace{3mm}

We write  $\not\exists$ to mean ``there does not exist."
\end{defn}
\pause
Example: $\forall$ action, $\exists$ equal and opposite reaction.\pause\\
We often omit ``such that," or replace it with a comma or colon.\pause

\color{mauve}\vfill
True or False:
\begin{itemize}
\item \textcolor{mauve}{$\forall x \in \mathbb Q$, $\dfrac{1}{x}\in \mathbb Q$.}\hfill\onslide<6->{\textcolor{lblue}{False: $x=0$}}
\item \textcolor{mauve}{$\forall x \in \mathbb N $, $\forall y \in \mathbb N$, $x-y\in \mathbb Z$.}\hfill
\onslide<7->{\textcolor{lblue}{True}}
\item \textcolor{mauve}{$\forall x \in \mathbb R - \{0\}$, $\exists y \in \mathbb R$ such that $xy=1$.}\hfill
\onslide<8->{\textcolor{lblue}{True: $y=\frac{1}{x}$}}
\item \textcolor{mauve}{$\exists x \in \mathbb R - \{0\}$ such that $\forall y \in \mathbb R$, $xy=1$.}
\hfill\onslide<9->{\textcolor{lblue}{False: $y=0$}}
\item \textcolor{mauve}{$\forall x \in \mathbb Q$, $\forall n \in \mathbb N$, $\exists z \in \mathbb Z$ such that $x=\dfrac{z}{n}$}.\hfill
\onslide<10->{\textcolor{lblue}{False: $x=\frac{1}{3}$, $n=2$}}
\item \textcolor{mauve}{$\forall x \in \mathbb Z$, $\forall n \in \mathbb N$, $\exists z \in \mathbb Z$ such that $x=\dfrac{z}{n}$}.\hfill
\onslide<11->{\textcolor{lblue}{True: $z=xn$}}
\end{itemize}
\end{frame}
%----------------------------------------------------------------------------------------
%----------------------------------------------------------------------------------------
\begin{frame}{Fill in the Blank}
\color{mauve}
Fill in each of the boxes with $\forall$, $\exists$, or $\not\exists$. If multiple answers are possible, choose the best.
\vfill\vfill

\boxed{\onslide<2->{\textcolor{lblue}{\forall}}} integer $n$, $2n$ is even.\hfill\onslide<2->{\textcolor{lblue}{Also $\exists$ is true.}}
\vfill

\boxed{\onslide<3->{\textcolor{lblue}{\exists}}} integer $n$, $3n$ is even.\hfill
\onslide<3->{\textcolor{lblue}{For example, $n=2$.}}\vfill

\boxed{\onslide<4->{\textcolor{lblue}{\forall}}}  real $x \neq 0$,
 \boxed{\onslide<4->{\textcolor{lblue}{\exists}}} real $y$ such that $xy=1$.\hfill
 \onslide<4->{\textcolor{lblue}{Since $x$ is nonzero, choose $y=\frac{1}{x}$.}}\vfill

\boxed{\onslide<5->{\textcolor{lblue}{\exists}}} real $x$ such that $\forall y \in \mathbb R$, $xy=0$. \hfill
\onslide<5->{\textcolor{lblue}{That $x$ is $x=0$.}}\vfill

\boxed{\onslide<6->{\textcolor{lblue}{\not\exists}}} $x\in \mathbb R$ such that $\forall y\in\mathbb R$, $xy\neq 0$.
\hfill\onslide<6->{\textcolor{lblue}{No $x$ has $xy \neq 0$ when $y=0$.}}\vfill 
\end{frame}
%----------------------------------------------------------------------------------------
\subsection{2.8 More on Conditional Statements}
%----------------------------------------------------------------------------------------
\begin{frame}{Conditional Statements with Variables}
Every universally qualified statement (that is, a statement containing ``$\forall$")
 can be expressed as a conditional statement with a variable.\pause\vfill
 
\textcolor{green}{ $\forall  x \in \mathbb N$, $x>0$}\\ \pause
\textcolor{dblue}{$x \in \mathbb N \Rightarrow x>0$} \pause \vfill

\textcolor{green}{$\forall x,\,y \in \mathbb Q,$ $xy \in \mathbb Q$}\\ \pause
\textcolor{dblue}{$x,y \in \mathbb Q \Rightarrow xy \in \mathbb Q$}
 \vfill
\end{frame}

%----------------------------------------------------------------------------------------
\subsection{2.9 Translating English to Symbolic Logic}
%----------------------------------------------------------------------------------------
\begin{frame}{Translating English to Symbolic Logic}
\textcolor{green}{Every rational number is also a real number.}\\
\onslide<2->{$\forall r \in \mathbb Q$, $r \in \mathbb R$\\
$\mathbb Q \subseteq \mathbb R$}\vfill

\textcolor{green}{If $x$ is a rational number, then $x^2$ is rational as well.}\\
\onslide<3->{$x \in \mathbb R \Rightarrow x^2 \in \mathbb R$\\
$\forall x \in \mathbb R$, $x^2 \in \mathbb R$}.\vfill
\textcolor{green}{Every integer is even or odd.}\\
\onslide<4->{$\forall z \in \mathbb Z$, ($z$ even) $\vee$ ($z$ odd)}
\vfill
\textcolor{green}{No real number is both even and irrational.}\\
\onslide<5->{
$\forall x\in\mathbb R$, $\sim(x \mbox{ even }\wedge x \not\in \mathbb Q)$}
\vfill
\textcolor{green}{Every quadratic function with real coefficients has at least one (possibly complex) root.}\\
( $\mathbb C$ is the set of complex numbers.)\pause\\
\onslide<6->{$\forall a,b,c \in \mathbb R$, $\exists x \in \mathbb C$, $ax^2+bx+c=0$.}
\end{frame}
%----------------------------------------------------------------------------------------
\begin{frame}{Translating}
\textcolor{green}{At least one of the numbers $x$ and $y$ is even.}\\ \pause
\textcolor{dblue}{$(x \mbox{ even}) \vee (y \mbox{ even})$}\\ \pause\vfill

\textcolor{green}{$x$ is even, but $y$ is odd}\\
\onslide<4->{\textcolor{dblue}{($x$ even) $\wedge$ ($y$ odd)}}\vfill

\textcolor{green}{Whenever $x$ is a perfect square, $x^3$ is as well.}\\
\onslide<5->{\textcolor{dblue}{$\sqrt{x}\in\mathbb Z \Rightarrow \sqrt{x^3} \in \mathbb Z$}}\vfill

\textcolor{green}{$P$ is true only if $Q$ is true.}\\
\textcolor{green}{(For example: ''I sing only if I'm in the shower.")}\\
\onslide<6->{\textcolor{dblue}{$P \Rightarrow Q$}}\vfill

\textcolor{green}{$P$ is true if $Q$ is true.}\\
\textcolor{green}{(For example: ''I sing if I'm in the shower.")}\\
\onslide<7->{\textcolor{dblue}{$Q \Rightarrow P$\\$P \Leftarrow Q$}}\vfill
\end{frame}
%----------------------------------------------------------------------------------------
\subsection{2.10 Negating Statements}
%----------------------------------------------------------------------------------------
\begin{frame}{Negating Statements using DeMorgan's Laws}
\begin{block}{DeMorgan's Laws}
\[\sim(P \wedge Q) \qquad\Leftrightarrow\qquad (\sim P) \vee (\sim Q)\]
\[\sim(P \vee Q) \qquad\Leftrightarrow\qquad (\sim P) \wedge (\sim Q)\]
\end{block}\pause
\vfill

\textcolor{green}{A number can't be both even and odd.}\\
\onslide<3->{\textcolor{dblue}{ $\forall x \,\sim\left[(x\mbox{ even})\wedge(x\mbox{ odd})\right]=$\\ $\forall x \, \left(\sim(x \mbox{ even})\right) \vee \left(\sim(x\mbox{ odd})\right)$}}
\vfill

\textcolor{green}{Ted isn't smart or handsome.}\\
\onslide<4->{\textcolor{dblue}{
$\sim\left[(\mbox{Ted is smart})\vee(\mbox{Ted is handsome})\right]=$\\
$\left(\sim(\mbox{Ted is smart})\right)\wedge \left(\sim(\mbox{Ted is handsome})\right)$}}
\end{frame}
%----------------------------------------------------------------------------------------
\begin{frame}{Negating Quantified Statements}
\textcolor{green}{$P$: Every swan is white.}\\ \pause
\textcolor{dblue}{$\sim P$: At least one swan is not white.}\pause\vfill

\textcolor{green}{$Q$: No student gets an A.}\\ \pause
\textcolor{dblue}{$\sim Q$: At least one student gets an A.}\\ \pause \vfill

\textcolor{green}{$R$: Everything in the library is also on the internet.}\\ \pause
\textcolor{dblue}{$\sim R$: At least one thing in the library is not also on the internet.}\pause\vfill

\textcolor{green}{$S$: There exists a human who has walked on the moon.}\\ \pause
\textcolor{dblue}{$\sim S$: No human has walked on the moon. \\ $\sim S$: For every human $H$, $H$ has not walked on the moon.}\\ \pause \vfill

\centering
$\sim(\forall x \in A, P(x)) = \exists x \in A, \sim P(x)$
\\$\sim(\exists x \in A, P(x)) = \forall x \in A, \sim P(x)$
\end{frame}
%----------------------------------------------------------------------------------------
\begin{frame}{Negating  Statements with Multiple Quantifiers}
\begin{center}\color{mauve}
$\sim(\forall x \in A, P(x)) = \exists x \in A, \sim P(x)$
\\$\sim(\exists x \in A, P(x)) = \forall x \in A, \sim P(x)$
\end{center}\color{black}
\textcolor{green}{$P$: $\forall x \in \mathbb R$, $\exists y \in \mathbb R$, $y^3=x$. \\
That is: $P$ says that every real number has a real cube root.}\\
\onslide<2->{\textcolor{dblue}{$\sim P$: $\exists x \in \mathbb R$, $\forall y \in \mathbb R$, $y^3 \neq x$.\\
That is: $\sim P$ says that there exists some real number such that every real number is not its cube root.}}
\vfill

\textcolor{green}{$Q$: $\exists x\in \mathbb Z$, $\forall y\in \mathbb R$, $xy=0$.\\
That is: there is some integer $x$ such that $xy=0$ for every real number $y$.}\\
\onslide<3->{\textcolor{dblue}{$\sim Q$: $\forall x \in \mathbb Z$, $\exists y \in \mathbb R$, $xy \neq 0$.\\
That is: for every integer $x$, there is some real number $y$ such that $xy \neq 0$.}}\vfill

\textcolor{green}{$\forall A \subseteq \mathbb N$, $\exists a \in A$, $\forall b \in A$ $a \leq b$\\ Every subset of the natural numbers has a smallest element.}
\onslide<4->{\textcolor{dblue}{$\exists A \subseteq \mathbb N, \forall a \in A, \exists b \in A, a>b$\\
There is some subset of the natural numbers so that, no matter what element we choose, there is some element that is smaller.}}
\vfill

\textcolor{green}{$\forall x \in \mathbb R, \forall y \in \mathbb Z, \exists z \in \mathbb \{-5,3.8\},
xyz \geq 0$}\\
\onslide<5->{\textcolor{dblue}{$\exists x \in \mathbb R, \exists y \in \mathbb Z, \forall z \in \{-5,3.8\}, xyz<0$\\
$\exists x \in \mathbb R, \exists y \in \mathbb Z, \left[(-5xy<0)\wedge(3.8xy<0)\right]$}}
\end{frame}
%----------------------------------------------------------------------------------------
\begin{frame}{Negating Conditional Statements}
\textcolor{green}{$\forall x \in \mathbb R,\left[ \frac{x}{9} \in \mathbb Z \Rightarrow \frac{x}{3}\in\mathbb Z\right]$}\\ 
\onslide<2->{\textcolor{dblue}{$\exists x \in \mathbb R, \left[\frac{x}{9}\in \mathbb Z\wedge \frac{x}{3}\not\in\mathbb{Z}\right]$}}\vfill

\textcolor{green}{$\forall x \in \mathbb Z$, $\left[x<a \Rightarrow x^2<a^2\right]$}\\ 
\onslide<3->{\textcolor{dblue}{$\exists x \in \mathbb Z, \left[(x<a)\wedge(x^2\ge a^2)\right]$}}\vfill


\textcolor{green}{$ \forall x,y,z,n \in \mathbb N, \,\left[ (x^n+y^n = z^n)\Rightarrow n \in \{1,2\}\right]$.\\Fermat's Last Theorem}\\
\onslide<4->{\textcolor{dblue}{$\exists x,y,z,n \in \mathbb N$, $\left[(x^n+y^n = z^n)\wedge(n>2)\right] $}}
\vfill

\textcolor{green}{$\forall $ function $f(x)$, $\forall a<b \in \mathbb R$,
$\left[f(x)\mbox{ continuous on }[a,b]\right]\Rightarrow\left[\forall N \mbox{ between }f(a)\mbox{ and }f(b),\exists c \in (a,b),f(c)=N\right]$\\
Intermediate Value Theorem}\\
\onslide<5->{\textcolor{dblue}{$\exists $ function $f(x)$, $\exists a<b \in \mathbb R$,
$\left[f(x)\mbox{ continuous on }[a,b]\right]\wedge\left[\exists N \mbox{ between }f(a)\mbox{ and }f(b),\forall c \in (a,b),f(c)\neq N\right]$}}
\end{frame}
%----------------------------------------------------------------------------------------
\subsection{2.11 Logical Inference}
%-----------------------------------------
\begin{frame}{Logical Inference}
\begin{defn}
Given two true statements, we can conclude using pure logic that a third is true. This process is called \dbox{logical inference}.
\end{defn}\pause\vfill
\begin{multicols}{2}
\begin{itemize}
\item $P \Rightarrow Q$ \pause
\item $P$ \pause
\item \textcolor{mauve}{Therefore, $Q$.}
\end{itemize}\pause\vfill


\begin{itemize}
\item $P \Rightarrow Q$ 
\item $\sim Q$\pause 
\item \textcolor{mauve}{Therefore, $\sim P$.}
\end{itemize}\pause\vfill


\begin{itemize}
\item $P \vee Q$ 
\item $\sim P$ \pause
\item \textcolor{mauve}{Therefore, $Q$.}
\end{itemize}\pause

\columnbreak

\begin{itemize}
\item $P \wedge Q$\pause 
\item \textcolor{mauve}{Therefore, $Q$.}
\item \textcolor{mauve}{Also therefore, $P$.}
\end{itemize}\pause\vfill


\begin{itemize} 
\item $P$\pause 
\item \textcolor{mauve}{Therefore, $P\vee Q$.}
\end{itemize}\pause\vfill


\begin{itemize}
\item $\sim(P \wedge Q)$ 
\item $P$ \pause
\item \textcolor{mauve}{Therefore, $\sim Q$.}
\end{itemize}
\end{multicols}
\end{frame}
%----------------------------------------------------------------------------------------
\subsection*{}
%----------------------------------------------------------------------------------------
\begin{frame}{Transition to Proofs}
We won't usually write out these formal symbols while we're proving something, but the logic behind them is always in the back of our minds.
\end{frame}
%----------------------------------------------------------------------------------------
\end{document} 