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\noindent
{\bf Solutions to HW 8.}

\bigskip

  
\begin{enumerate}
\item[1.]
  Show that ``$A\subseteq B$ and $B\subseteq A$ implies $A=B$''
  in each of the following two ways.
\end{enumerate}
\bigskip


\begin{enumerate}
\item[(a)] With a direct proof.
\end{enumerate}
\bigskip
\color{blue}
We assume that $(A\subseteq B)\wedge (B\subseteq A)$,
and argue that $A=B$.

Choose any $a\in A$.  Since $A\subseteq B$, it follows that $a\in B$.
Now choose any $b\in B$.  Since $B\subseteq A$, it follows that $b\in A$.
We have shown that $A$ and $B$ have the same elements, so $A=B$
by the Axiom of Extensionality.
\color{black}

\bigskip

\begin{enumerate}
\item[(b)]
  With a proof of the contrapositive.
\end{enumerate}
\bigskip

  \color{blue}
  Assume that $A\neq B$. The two sets do not have the same elements,
  so either (i) there is some $x$ in $A$
  that is not in $B$ ($x\in A-B$) or (ii) there is some
  $y$ in $B$ that is not in $A$ ($y\in B-A$).
  In Case (i), $A\not\subseteq B$, while in Case (ii),
  $B\not\subseteq A$.
  This shows that
  \[
  (A\neq B) \to (A\not\subseteq B)\vee (B\not\subseteq A),
  \]
which is the contrapositive statement.  
 \color{black}

 \vfill

\begin{enumerate}
\item[2.]
  Prove the statement ``If $A\cap B=\emptyset$ and $A\cup B=B$,
  then $A=\emptyset$'' in each of the following two ways.
\end{enumerate}
\bigskip

\begin{enumerate}
\item[(a)] With a direct proof.
\end{enumerate}
\bigskip

\color{blue}
Assume that $A\cap B=\emptyset$ and $A\cup B=B$.
We have
\[
\begin{array}{rll}
\emptyset&=A\cap \underline{B} &\textrm{Assumption}\\
&=A\cap(\underline{A\cup B})&\textrm{Assumption}\\
&=A&{\textrm{Absorption Law}, (\textrm{true of any sets} \;A, B)}\\
\end{array}
\]
\medskip

\color{purple}
A second direct proof:
\[
\begin{array}{rll}
\emptyset &= A\cap \underline{B}&{\textrm{Assumption}}\\
  &=A\cap (\underline{A\cup B})& \textrm{Assumption}\\
  &=(A\cap A)\cup (A\cap B)& \textrm{Distributive Law}\\
  &=(A\cap A)\cup \emptyset& \textrm{Assumption}\\
  &=A\cap A& \textrm{$\emptyset$ is a unit element for $\cup$}\\
  &=A& \textrm{Idempotence of $\cap$}\\
\end{array}
\]

\color{black}

\bigskip

\begin{enumerate}
\item[(b)] With a proof by contradiction.
\end{enumerate}
\bigskip

  \color{blue}
  Assume that $A\cap B=\emptyset$, $A\cup B=B$, but $A\neq \emptyset$.
  Since $A\neq \emptyset$, there exists some $x\in A$.
  From the definition of union, $x\in A\cup B\;(=B)$.
  Therefore $x\in A$ and $x\in B$, so $x\in A\cap B$,
  contradicting $A\cap B=\emptyset$.
 \color{black}
 \bigskip
 
\vfill
  
\begin{enumerate}
\item[3.]
  The goal of this problem is to prove that the
  composition of two injective functions is injective.
  The type of structure involved looks like
  $\mathbb X=\langle A,B,C;f,g\rangle$
  where $f:A\to B$ and $g:B\to C$ are functions.
  Let the variables $a, a'$ range over the set $A$ and
  and the variables $b, b'$ range over the set $B$.

  The functions (i) $f$, (ii) $g$, (iii) $g\circ f$
  are injective if the following sentences hold in $\mathbb X$:
  \begin{enumerate}
\item[(i)] $(\forall a)(\forall a')((f(a)=f(a'))\to (a=a'))$.
\item[(ii)] $(\forall b)(\forall b')((g(b)=g(b')\to (b=b'))$.
\item[(iii)] $(\forall a)(\forall a')((g\circ f(a)=g\circ f(a'))\to (a=a'))$.
\end{enumerate}

  To prove that the composition of injective functions
  is injective, you must give a winning strategy for
  $\exists$ in the sentence in (iii).
  YOU ARE ALLOWED TO USE the fact that there exist winning
  strategies for $\exists$ in the sentences in (i) and (ii).
  Write a proof that indicates the winning strategy for $\exists$
  in (iii), which accesses the information of the strategies
  for $\exists$ in (i) and (ii). 
\end{enumerate}
\bigskip

\color{blue}
We must provide a winning strategy for $\exists$ for Game (iii). We
are allowed to use that there exist winning strategies for
$\exists$ in Games (i) and (ii).

Since the existential quantifier $\exists$ does not appear
in any of the sentences, the only possible strategy for $\exists$
in any of these games 
is ``Do nothing''. That is, the strategy
for $\exists$ is ``just watch while $\forall$ plays''.
We need to explain why this is a winning strategy
for $\exists$ in Game (iii).

Each of the sentences has the form
\begin{center}
  (Quantifiers)(Premise $\to$ Conclusion).
\end{center}
Recall that $P\to C$ fails only when the premise $P$ is true
and the conclusion $C$ is false. We will use this fact.

We argue that ``Do nothing'' is a winning strategy for
$\exists$ for Game (iii). Suppose 

\begin{itemize}
\item (Play 1): $\forall$ chooses some $a\in A$.
\item (Play 2): $\forall$ chooses some $a'\in A$.
\end{itemize}
We must show that $(g\circ f(a)=g\circ f(a'))\to (a=a')$ holds.

\medskip

{\bf Case 1.} (The premise
$P$: ``$g\circ f(a)=g\circ f(a')$''
of sentence (iii) does not hold.)

In this case, $\exists$ has won, since 
$(P\to C)$ is true when the premise does not hold.

\medskip

{\bf Case 2.} (The premise
$P$: ``$g\circ f(a)=g\circ f(a')$''
of sentence (iii) DOES hold.)

In this case, let $b=f(a)$ and $b'=f(a')$.
(Here $\exists$ is imagining an instance
of Game (ii) where 
$\forall$ chooses $b=f(a)$ and $b'=f(a')$.)
Since the premise of sentence (iii) holds, we obtain
$g(b)=g(f(a))=g(f(a'))=g(b')$, establishing that the premise 
of sentence (ii) holds. Since we are assuming
that the sentence in (ii) is true, the
conclusion $b=b'$ of Game (ii) holds.
Now, $f(a)=b=b'=f(a')$, establishing that the premise of statement (i)
holds. Since we are assuming
that the sentence in (i) is true, the
conclusion $a=a'$ holds.
This establishes that $\exists$ wins in Case 2.
\medskip

These are the only cases, so ``Do nothing'' is a winning strategy for $\exists$.
\color{black}

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