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\tutorial{tutorial1}{Solutions for Tutorial 1}{February 10, 1997}

\section {The Superman Problem}

\noindent Cast the following into propositional symbols, and verify or
refute:\\

{\em \noindent If Superman were able and willing to prevent evil, he would do
  so. If Superman were unable to prevent evil, he would be impotent.
  If he were unwilling to prevent evil, he would be malevolent.
  Superman does not prevent evil. If Superman exists, he is neither
  impotent nor malevolent. Therefore, Superman does not exist.}\\

A: Superman were able to prevent evil. \\
W: Superman were willing to prevent evil. \\
I: Superman is impotent. \\
M: Superman is malevolent. \\
P: Superman prevents evil. \\
E: Superman exists. \\

$A \wedge W \rightarrow P$
If Superman were able and willing to prevent evil, he would do so.  \\
$\neg A \rightarrow I$
If Superman were unable to prevent evil, he would be impotent. \\
$\neg W \rightarrow M$
If he were unwilling to prevent evil, he would be malevolent. \\
$\neg P$
Superman does not prevent evil.  \\
$E \rightarrow \neg (M \vee I)$
If Superman exists, he is neither impotent nor malevolent.  \\
$\neg E$
Therefore, Superman does not exist. \\



\section {A Touch of Group Theory}

A {\em group} is a set $G$ together with a binary operation $\cdot: GxG \rightarrow G$ that satisfies the following axioms.

\bparts

\ppart $\forall a,b,c \in G,(a \cdot b) \cdot c = a \cdot (b \cdot c)$.

\ppart $\exists e \in G$ (called the {\em identity}) 
such that $\forall a \in G, a \cdot e = e \cdot a = a$.

\ppart $\forall a \in G, \exists a^{-1} \in G$ (called the {\em inverse} 
of $a$), such that $a \cdot a^{-1} = a^{-1} \cdot a = e$.

\eparts

Using the three group axioms prove the following propositions.

\begin{enumerate}

\item Show, by contradiction, that the identity element is unique.

Assume that there are two identity elements, $e_1$ and $e_2$.

So, $ \forall a (\exists e_1, e_2) ((a \cdot e_1 = e_1 \cdot a = a)
\wedge (a \cdot e_2 = e_2 \cdot a = a) \wedge (e_1 \neq e_2))$.
\begin {eqnarray*}
a &=& a\\
a \cdot e_1 &=& a \cdot e_2\\
a^{-1} \cdot a \cdot e_1 &=& a^{-1} \cdot a \cdot e_2\\
e_1 \cdot e_1 &=& e_2 \cdot e_2\\
e_1 &=& e_2 \\
\end{eqnarray*}
But that contradicts $e_1 = e_2$, so there must be a unique inverse.


\item Show that for every $a \in G$, the inverse $a^{-1}$ is unique.

Assume that there are two inverses, $b$ and $c$, which satisfy
\begin{eqnarray*}
e = e
a^{-1} a = e = a^{-1}a
b \cdot a = c \cdot a
b \cdot a = c \cdot a

\end{eqnarray*}


\item $\forall a \in G, (a^{-1})^{-1} = a$
\begin {eqnarray*}
  (a^{-1})^{-1} &=& (a^{-1})^{-1}e \\
  &=& (a^{-1})^{-1}a^{-1}a \\
  &=& ea \\
  &=& a \\
\end{eqnarray*}

\item $\forall a,b \in G, (a \cdot b)^{-1} = b^{-1} \cdot a^{-1}$
\begin {eqnarray*}
(a \cdot b)^{-1} &=& (a\cdot b)^{-1} \\
&=& (a\cdot b)^{-1} e \\
&=& (a\cdot b)^{-1} a \cdot a^{-1} \\
&=& (a\cdot b)^{-1} a e a^{-1} \\
&=& (a\cdot b)^{-1} a  b \cdot b^{-1}  a^{-1} \\
&=& (a\cdot b)^{-1} (a  b) \cdot b^{-1}  a^{-1} \\
&=& e b^{-1}  a^{-1} \\
&=& b^{-1}  a^{-1} \\
\end{eqnarray*}


\end{enumerate}


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