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\tutorial{tutorial1}{Problems 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.}\\


\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.

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

Hint:  First show the following lemma:
\begin{enumerate}
\item $\forall a,b,c \in G, a \cdot b = a \cdot c \Rightarrow b = c$
\item $\forall a,b,c \in G, b \cdot a = c \cdot a \Rightarrow b = c$
\end{enumerate}

\item $\forall a \in G, (a^{-1})^{-1} = a$

\item $\forall a,b \in G, (a \cdot b)^{-1} = b^{-1} \cdot a^{-1}$

\end{enumerate}


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