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-\documentclass[a4paper,oneside]{article}
-\usepackage[utf8]{inputenc}
-\usepackage{amsmath}
-\usepackage{amsthm}
-\usepackage{amssymb}
-\usepackage[top=2cm]{geometry}
-\theoremstyle{definition} \newtheorem{exercise}{Exercise}[section]
-\author{Sebastiano Tronto (\texttt{sebastiano.tronto@uni.lu})}
-\title{Elementary Logic exercises (Prep Camp 2020)}
-\begin{document}
-\maketitle
-\section{Logical operations}
-\begin{exercise}
-  Determine if the following statements are \textbf{true} or \textbf{false}:
-  \begin{enumerate}
-    \item ``Today is Tuesday or Germany has more inhabitants than Luxembourg''
-    \item ``$7$ is odd and $2+2=5$''
-    \item Every number of the form $2^{2^n}+1$, for $n=1,2,3...$, is prime.
-  \end{enumerate}
-\end{exercise}
-\begin{proof}[Solution]
-  \begin{enumerate}
-    \item \textbf{True}: regardless of when you solve this exercise, Germany
-          has more inhabitants than Luxembourg.
-    \item \textbf{True}
-    \item \textbf{False}: the number $2^{2^5}+1=4294967297=641\times 6700417$
-          is not prime.
-  \end{enumerate}
-\end{proof}
-\begin{exercise}
-  What is the negation of the sentence ``\emph{I payed attention in class and I
-  did not do my homework}'' ?
-\end{exercise}
-\begin{proof}[Solution]
-  ``\emph{I did not pay attention in class \textbf{or} I did my homework}''
-\end{proof}
-\begin{exercise}
-  Simplify the following logical expressions using the properties of logical
-  operations (where $A,B$ and $C$ are statements):
-  \begin{enumerate}
-    \item $A\land(A\lor B)$
-    \item $A\lor (B\land A)$
-    \item $(A\lor B) \land \neg A$
-    \item $A \lor (\neg A\land B)$
-    \item $(\neg (A\lor \neg B))\land ((A\lor C) \land \neg C)$
-  \end{enumerate}
-\end{exercise}
-\begin{proof}[Solution]
-  They are equivalent to the following (you can check with truth tables):
-  \begin{enumerate}
-    \item $A$
-    \item $A$
-    \item $B\land \neg A$
-    \item $A\lor B$
-    \item Let's do this one in more steps:
-      \begin{align*}
-        (\neg (A\lor \neg B))\land ((A\lor C) \land \neg C)=
-        &(\neg A\land B)\land ((A\lor C)\land \neg C)=\\
-        =&(\neg A\land B)\land ((A\land \neg C)\lor (C\land \neg C))=\\
-        =&(\neg A\land B)\land ((A\land \neg C)\lor \textbf{false})=\\
-        =&(\neg A\land B)\land (A\land \neg C)=\\
-        =&A\land \neg A\land B\land \neg C=\\
-        =&\textbf{false}
-      \end{align*}
-  \end{enumerate}
-\end{proof}
-\section{Implication}
-\begin{exercise}
-  Fill in the following truth table:
-  \begin{align*}
-    \begin{array}{|c|c|c|c|c|}
-      \hline
-      A & B & C & \neg(A\implies B) & (A\implies B) \implies C \\
-      \hline
-      0 & 0 & 0 & 0 & 0 \\
-      \hline
-      0 & 0 & 1 & 0 & 1 \\
-      \hline
-      0 & 1 & 0 & 0 & 0 \\
-      \hline
-      0 & 1 & 1 & 0 & 1 \\
-      \hline
-      1 & 0 & 0 & 1 & 1 \\
-      \hline
-      1 & 0 & 1 & 1 & 1 \\
-      \hline
-      1 & 1 & 0 & 0 & 0 \\
-      \hline
-      1 & 1 & 1 & 0 & 1 \\
-      \hline
-    \end{array}
-  \end{align*}
-\end{exercise}
-\begin{exercise}[Transitivity]
-  Prove that the following statement is true for any statements $A,B$ and $C$:
-  \begin{align*}
-    ((A\implies B)\land (B\implies C))\implies (A\implies C)
-  \end{align*}
-\end{exercise}
-\begin{proof}[Solution]
-  Let's rewrite the first part in terms of basic logical operations:
-  \begin{align*}
-    (A\implies B)\land (B\implies C)=&(B\lor \neg A)\land(C\lor \neg B)
-  \end{align*}
-  now we can write a truth table for the two parts
-  \begin{align*}
-    \begin{array}{|c|c|c|c|c|}
-      \hline
-      A & B & C & (B\lor\neg A)\land(C\lor\neg B) & A\implies C \\
-      \hline
-      0 & 0 & 0 & 1 & 1 \\
-      \hline
-      0 & 0 & 1 & 1 & 1 \\
-      \hline
-      0 & 1 & 0 & 0 & 1 \\
-      \hline
-      0 & 1 & 1 & 1 & 1 \\
-      \hline
-      1 & 0 & 0 & 0 & 0 \\
-      \hline
-      1 & 0 & 1 & 0 & 1 \\
-      \hline
-      1 & 1 & 0 & 0 & 0 \\
-      \hline
-      1 & 1 & 1 & 1 & 1 \\
-      \hline
-    \end{array}
-  \end{align*}
-  With the help of the truth table we see that whenever the first part
-  ``$(\neg (A\lor \neg B))\land ((A\lor C) \land \neg C)$'' is true, also the
-  implication ``$A\implies C$'' is true. This shows that the ``big
-  implication'' is true. (If you are not convinced, you can add more details to
-  this proof, for example by writing more truth tables.)
-\end{proof}
-\begin{exercise}
-What is the contrapositive of ``\emph{If this table is not reserved, we sit
-here}'' ?
-\end{exercise}
-\begin{proof}[Solution]
-  ``\emph{If we do not sit here, this table is reserved}''. One could also say
-  this in another way, for example ``\emph{We do not sit here because this
-  table is reserved}''.
-\end{proof}
-\section{Quantifiers}
-\begin{exercise}
-  Write the negation of the following statements:
-  \begin{enumerate}
-    \item $\exists x\in \mathbb N,\, x^2-2=0$
-    \item ``Every prime number is odd''
-    \item ``Every person I have met likes pizza''
-    \item ``There is at least one number greater than $7$''
-    \item $\forall x\in \mathbb N,\,x\geq 0$
-    \item $\forall x\in \mathbb Z,\,(\exists y\in\mathbb Z,\,x+y=0)$
-  \end{enumerate}
-\end{exercise}
-\begin{proof}[Solution]
-  \begin{enumerate}
-    \item $\forall x\in\mathbb N,\, x^2-2\neq 0$
-    \item ``There is at least one prime number which is even''
-    \item ``I have met at least one person that does not like pizza''
-    \item ``Every number is less or equal than $7$''
-    \item $\exists x\in\mathbb N,\, x<0$
-    \item $\exists x\in \mathbb Z,\, (\forall y\in \mathbb Z,\, x+y\neq 0)$
-  \end{enumerate}
-\end{proof}
-\begin{exercise}
-  There is another quantifier that we did not cover in the lecture, namely
-  $\exists!$ (read ``there exists exactly one''). For example, the sentence
-  ``\emph{there exists exactly one natural number x such that x+2=5}'' can be
-  written in symbols as ``$\exists!x\in \mathbb N,\,x+2=5$''.
-  In this exercise, your task is to give a formal definition of this quantifier
-  using the logical symbols that we have defined in class. In particular, you
-  will need the following:
-  \begin{itemize}
-    \item the universal ($\forall$) and existential ($\exists$) quantifiers
-    \item the conjunction $\land$
-    \item the implication $\implies$
-  \end{itemize}
-  Moreover, you will need the equality symbol $=$ between two elements of a set
-  (if $a$ and $b$ are two elements of the same set, ``$a=b$'' is a mathematical
-  statement and it is \textbf{true} if and only if $a$ and $b$ are the same
-  element).
-  
-  \emph{Warning: your definition must depend on a set $S$ and on a ``variable
-  statement'' $A(x)$, as the existential and universal quantifiers.}
-\end{exercise}
-\begin{proof}[Solution]
-  The idea is that we want to write ``\emph{there is $x\in S$ such that $A(x)$
-  is true, \textbf{and}, for any other $y\in S$, $A(y)$ is false}. From this 
-  we see that the structure of the statement is
-  \begin{align*}
-    \exists x\in S,\,(\text{``something''}\land\text{``something else''})
-  \end{align*}
-  The ``something'' part is just $A(x)$. The ``something else'' part can be
-  written in different ways, for example
-  \begin{align*}
-    \forall y\in S,\,(y\neq x\implies \neg A(y)) \quad \text{or}
-    \quad \forall y\in S,\,(A(y)\implies y=x)
-  \end{align*}
-  (notice that the two implications above are one the contrapositive of the
-  other); or also
-  \begin{align*}
-    \forall y\in S\setminus \{x\},\, \neg A(y)
-  \end{align*}
-  So in conclusion, one way to define ``$\exists!$'' is the following:
-  \begin{align*}
-    \exists!x\in S,\,A(x)\quad:=\quad\exists x\in S,\,(A(x)\land
-    (\forall y\in S,\,(A(y)\implies y=x)))
-  \end{align*}
-    
-\end{proof}
-\section{Proofs}
-\begin{exercise}
-  Prove by induction that
-  \begin{align*}
-    \forall n\in\mathbb N,\quad \sum_{k=1}^n(2k-1)=n^2
-  \end{align*}
-    (here $\sum_{k=1}^n(2k-1)$ means $1+3+5+\cdots+ (2n-1)$).
-\end{exercise}
-\begin{proof}[Solution]
-  \textbf{Base case:} for $n=0$ the sum is empty, so we have $0=0$ which is
-  true. (If you do not think that the formula makes sense for $n=0$, you can do
-  prove it for $n\geq 1$ and fo $n=1$ as a base case.)
-  \textbf{Inductive step:} we can assume that the formula works for a generic
-  (but fixed) $n\in\mathbb N$ and prove that then it also works for $n+1$. So:
-  \begin{align*}
-    \sum_{k=1}^{n+1}(2k-1)&=\left(\sum_{k=1}^{n}(2k-1)\right)+2(n+1)-1=\\
-                          &=n^2 +2(n+1)-1=\\
-                          &=n^2+2n+1=\\
-                          &=(n+1)^2
-  \end{align*}
-  which is the formula for $n+1$.
-\end{proof}
-\begin{exercise}
-  If $n\in \mathbb N$ the \emph{factorial} of $n$, denoted by $n!$ is defined
-  as follows:
-  \begin{align*}
-    n!=\begin{cases}
-      1&\text{if } n=0,\\
-      n\times (n-1)! & \text{if } n> 0.
-    \end{cases}
-  \end{align*}
-  Prove by induction that if $n\geq 4$ then $n!\geq 2^n$.
-\end{exercise}
-\begin{proof}[Solution]
-  \textbf{Base case:} here the base case is $n=4$, and we have
-  $4!=24\geq 16=2^4$.
-  \textbf{Inductive step:} we can assume that the inequality is true for $n$,
-  and prove that it is also true for $n+1$. We have:
-  \begin{align*}
-    (n+1)!=(n+1)\times n!\geq (n+1)\times 2^n\geq 2^{n+1}
-  \end{align*}
-\end{proof}
-\begin{exercise}
-  Is the following statement true or false? Give a proof of your answer.
-  \begin{align*}
-    \forall n\in \mathbb N,\, n^2 -4n +5>n
-  \end{align*}
-\end{exercise}
-\begin{proof}[Solution]
-  The statement is \textbf{false}. Since it starts with a universal
-  quantifier, in order to prove that it is false we just need to provide one
-  example of $n\in \mathbb N$ which makes it false. In other words, we need to
-  prove that
-  \begin{align*}
-    \exists n\in \mathbb N,\, n^2-4n+5\leq n
-  \end{align*}
-  and a proof of this fact is very simple: for $n=2$ we have
-  $2^2-4\times 2+5=1\leq 2$.
-\end{proof}
-\begin{exercise}
-  Do the last point of Exercise 1.1 again, but this time give a proof of your
-  answer.
-\end{exercise}
-\begin{proof}[Solution]
-  Again, since we have to prove that the statement is false, we just need to
-  show one counterexample. for example, the number
-  $2^{2^5}+1=4294967297=641\times 6700417$ is not prime.
-\end{proof}
-\end{document}