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\hfill Name: \rule{3in}{.01in}\\[.5in]

\noindent Armstrong Atlantic State University \hfill  September 3, 1998 \\
Department of Computer Science \hfill CSCI 2620, Homework Assignment One\\

Problems: Section 1.1 (21), 1.2 (12, 15), 1.3 (2, 17), 1.4 (11, 15),
and 1.5 (4, 17).\\

{\bf Section 1.1 (21)}

a) bitwise OR 11 11111 , bitwise AND 00 00000, bitwise XOR 11 11111

b) bitwise OR 111 11010, bitwise AND 101 00000, bitwise XOR 010 11010

c) bitwise OR 1011 11001, bitwise AND 00010 00000, bitwise XOR 10001 11001

d) bitwise OR 11111 11111, bitwise AND 00000 00000, bitwise XOR 11111 11111\\

{\bf Section 1.2 (12)}

Let $p$ be false and $q$ be true.

Then $(p \rightarrow q)$ is true.

And so, $\neg p \wedge (p \rightarrow q)$ is true.

But, $(\neg p \wedge (p \rightarrow q)) \rightarrow \neg q$ is false since
true does not imply false.  Thus, the original expression is not a
tautology.\\

{\bf Section 1.2 (15)}

We will find an assignment where the two propositions have different
values.  Let all three variables be false.  Then $(p \rightarrow q)
\rightarrow r$ is false since true does not imply false.  However, $p
\rightarrow (q \rightarrow r)$ is true since false implies true.\\

{\bf Section 1.3 (2)}

P(orange) and P(false) are true since orange and false contain the
letter a.  P(lemon) and P(true) are false since neither of these words
contains the letter a.\\

{\bf Section 1.3 (17)}

a) $\forall x (P(x) \rightarrow \neg Q(x))$

b) $\forall x (Q(x) \rightarrow R(x))$

c) $\forall x (P(x) \rightarrow \neg R(x))$

d) There may be vain people that are not ignorant.  Thus, the conclusion
does not follow.\\

{\bf Section 1.4 (11)}

a) $\{ \{ \}, \{ a \} \}$

b) $\{ \{ \}, \{ a \}, \{ b \}, \{ a, b \} \}$

c) $\{ \{ \}, \{ \{ \} \}, \{ \{ \{ \} \} \}, \{ \{ \}, \{ \{ \} \} \} \}$\\

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{\bf Section 1.4 (15)}

a) $A \times B = \{ (a, y), (b, y), (c, y), (d, y), (a, z), (b, z), (c,
z), (d, z) \}$

b) $B \times A = \{ (y, a), (y, b), (y, c), (y, d), (z, a), (z, b), (z,
c), (z, d) \}$\\

{\bf Section 1.5 (4)}

a)  $A \cup B = \{a, b, c, d, e, f, g, h \}$

b) $A \cap B = \{a, b, c, d, e \}$

c) $A - B = \{ \}$

d) $B - A = \{ f, g, h \}$\\

{\bf Section 1.5 (17)}

a) $A \cap B \cap C = \{ 4, 6 \}$

b) $A \cup B \cup C = \{ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 \}$

c) $(A \cup B) \cap C = \{ 4, 5, 6, 8, 10 \}$

d) $(A \cap B) \cup C = \{0, 2, 4, 5, 6, 7, 8, 9, 10 \}$

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