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\homework{3}{13 September 1995}{Sets, Relations, Functions; Proof Techniques}{}

\problem{1}

Chapter 5 of SEM: 5.1, 5.2, 5.4, 5.6, 5.9, 5.13

\problem{2} 

Chapter 6 of SEM: 6.1, 6.2, 6.3, 6.11

\problem{3}

Simplify this expression using the One-Point Rule.

\begin{verse}
$\exists names: ID \pinj PERSON \bullet \exists enemies: \power ID
\bullet enemies \subseteq {\it dom}~ names \wedge names = \emptyset
\wedge enemies = \emptyset$
\end{verse}

\problem{4}

[GS, Chapter 4, Problem 4.10]

Prove Modus ponens, $p \wedge (p \Rightarrow q) \Rightarrow q$, using
equational reasoning.  Hint: You should also use one of the proof
techniques discussed in Handout 3.

\problem{5}

[GS, Chapter 4, Problem 4.13]

Let $x \downarrow y$ be the minimum of integers $x$ and $y$,
defined by
$x \downarrow y$ = $({\bf if} ~~x~\leq~y~~ {\bf then} ~~x~~ {\bf else} ~~y)$.
Prove that $\downarrow$ is symmetric, i.e., $b \downarrow c = c \downarrow b$.
Use an equational reasoning style of proof.  How many cases do you
have to consider?  You may use the necessary rules of integer
arithmetic,
for example, that $b \leq c \equiv b = c \vee b < c$ and that $b < c
\equiv c > b$.


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