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appendix_b.txt
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chapter_6/exercises.md
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@ -4426,15 +4426,218 @@ _[This is what was to be shown.]_
8. _De Morgan's law for $\cdot$:_ For all $a$ and $b$ in $B$,
$\overline{a \cdot b} = \overline{a} + \overline{b}$. (_Hint:_ Prove that
$(a \cdot b) + (\overline{a} + \overline{b}) = 1$ and that
$(a \cdot b) + (\overline{a} + \overline{b}) = 0$, and use the fact that
$(a \cdot b) \cdot (\overline{a} + \overline{b}) = 0$, and use the fact that
$a \cdot b$ has a unique complement.)
**Proof:**
Suppose $B$ is a Boolean algebra, and that $a$ and $b$ are elements of $B$.
Prove that $(a \cdot b) + (\overline{a} + \overline{b}) = 1$:
$$ (a \cdot b) + (\overline{a} + \overline{b}) $$
$$ = ((a \cdot b) + \overline{a}) + \overline{b} $$
by the associative law for $+$
$$ = ((b \cdot a) + \overline{a}) + \overline{b} $$
by the commutative law for $+$
$$ = ((b + \overline{a}) \cdot (a + \overline{a})) + \overline{b} $$
by the distributive law for $+$ over $\cdot$
$$ = ((b + \overline{a}) \cdot 1) + \overline{b} $$
by the complement law for $+$
$$ = (b + \overline{a}) + \overline{b} $$
by the identity law for $\cdot$
$$ = b + (\overline{b} + \overline{a}) $$
by the commutative law for $+$
$$ = (b + \overline{b}) + \overline{a} $$
by the associative law for $+$
$$ = 1 + \overline{a} $$
by the complement law for $+$
$$ = 1 $$
by the universal bound law for $+$
Prove that $(a \cdot b) \cdot (\overline{a} + \overline{b}) = 0$:
$$ (a \cdot b) \cdot (\overline{a} + \overline{b}) $$
$$ = ((a \cdot b) \cdot \overline{a}) + ((a \cdot b) \cdot \overline{b}) $$
by the distributive law of $\cdot$ over $+$
$$ = ((a \cdot \overline{a}) \cdot b) + (a \cdot (b \cdot \overline{b})) $$
by the commutative and associative laws
$$ = (0 \cdot b) + (a \cdot 0) $$
by the complement laws
$$ = 0 + 0 $$
by the universal bound laws
$$ = 0 $$
by the identity laws
_Conclusion:_
Since $(a \cdot b) + (\overline{a} + \overline{b}) = 1$ and
$(a \cdot b) \cdot (\overline{a} + \overline{b}) = 0$, it can be concluded, by
the uniqueness of complement laws, that
$\overline{a \cdot b} = \overline{a} + \overline{b}$. This is what was to be
shown.
Q.E.D.
9. _De Morgan's law for $+$:_ For all $a$ and $b$ in $B$,
$\overline{a + b} = \overline{a} \cdot \overline{b}$.
**Proof:**
Suppose $B$ is a Boolean algebra, and that $a$ and $b$ are elements of $B$.
Prove that $(a + b) + (\overline{a} \cdot \overline{b}) = 1$:
$$ (a + b) + (\overline{a} \cdot \overline{b}) $$
$$ = ((a + b) + \overline{a}) \cdot ((a + b) + \overline{b}) $$
by the distributive laws for $+$ over $\cdot$
$$ = ((a + \overline{a}) + b) \cdot (a + (b + \overline{b})) $$
by the associative and commutative laws
$$ = (1 + b) \cdot (a + 1) $$
by the complement laws for $+$
$$ = 1 \cdot 1 $$
by the universal bound laws for $+$
$$ = 1 $$
by the identity laws for $\cdot$
Prove that $(a + b) \cdot (\overline{a} \cdot \overline{b}) = 0$:
$$ (a + b) \cdot (\overline{a} \cdot \overline{b}) $$
$$ = (a \cdot \overline{a}) \cdot (b \cdot \overline{b}) $$
by the commutative and associative laws for $\cdot$
$$ = 0 \cdot 0 $$
by the complement laws for $\cdot$
$$ = 0 $$
by the universal bound laws for $\cdot$
_Conclusion:
Since $(a + b) + (\overline{a} \cdot \overline{b}) = 1$ and
$(a + b) \cdot (\overline{a} \cdot \overline{b}) = 0$, it can be concluded, by
the uniqueness of complement laws, that
$\overline{a + b} = \overline{a} \cdot \overline{b}$. This is what was to be
shown.
Q.E.D.
10. _Cancellation law:_ For all $x$, $y$, and $z$ in $B$, if $x + y = x + z$ and
$x \cdot y = x \cdot z$, then $y = z$.
**Proof:**
Suppose $B$ is a Boolean Algebra, and that $x$, $y$, and $z$ are elements in $B$
such that $x + y = x + z$ and $x \cdot y = x \cdot z$.
$$ y = (y + x) \cdot y $$
by exercise 3
$$ = y \cdot (y + x) $$
by the commutative laws
$$ = y \cdot (x + y) $$
by the commutative laws
$$ = y \cdot (x + z) $$
by the supposition
$$ = (y \cdot x) + (y \cdot z) $$
by the distributive laws for $\cdot$ over $+$
$$ = (x \cdot y) + (y \cdot z) $$
by the commutative laws
$$ = (x \cdot z) + (y \cdot z) $$
by the supposition
$$ = (z \cdot x) + (z \cdot y) $$
by the commutative laws
$$ = z \cdot (x + y) $$
by the distributive laws of $\cdot$ over $+$
$$ = z \cdot (x + z) $$
by the supposition
$$ = (z \cdot x) + (z \cdot z) $$
by the distributive laws of $\cdot$ over $+$
$$ = (z \cdot x) + z $$
by the idempotent laws
$$ = (z \cdot x) + (z \cdot 1) $$
by the identity laws
$$ = z \cdot (x + 1) $$
by the distributive laws of $\cdot$ over $+$
$$ = z \cdot 1 $$
by the universal bound laws
$$ = z $$
by the identity laws
Q.E.D.
11. Let $S = \{0, 1\}$, and define operations $+$ and $\cdot$ on $S$ by the
following tables:
@ -4457,14 +4660,77 @@ a. Show that the elements of $S$ satisfy the following properties:
v. the distributive law for $+$ over $\cdot$.
vi. the distributive law for $\cdot$ over $+$.
i.
$$ 0 + 1 = 1 + 0 $$
$$ 1 = 1 $$
ii.
$$ 0 \cdot 1 = 1 \cdot 0 $$
$$ 0 = 0 $$
iii.
$$ (0 + 0) + 1 = 0 + (0 + 1) $$
$$ 1 = 1 $$
iv.
$$ (0 \cdot 0) \cdot 1 = 0 \cdot (0 \cdot 1) $$
$$ 0 \cdot 1 = 0 \cdot 0 $$
$$ 0 = 0 $$
v.
$$ 0 + (0 \cdot 1) = (0 + 0) \cdot (0 + 1) $$
$$ 0 + 0 = 0 \cdot 1 $$
$$ 0 = 0 $$
vi.
$$ 0 \cdot (0 + 1) = (0 \cdot 0) + (0 \cdot 1) $$
$$ 0 \cdot 1 = 0 + 0 $$
$$ 0 = 0 $$
NOTE: part a many cases are omitted as you have to explore each case (2 for both
commutative and associative, 8 for distributive).
b. Show that $0$ is an identity element for $+$ and that $1$ is an identity
element for $\cdot$.
_Hint:_ Verify that $0 + x = x$ and that $1 \cdot x = x$ for every $x \in S$.
$0 + 0 = 0$ and $0 + 1 = 1$, so $0$ is an identity for $+$.
$1 \cdot 0 k 0$ and $1 \cdot 1 = 1$, so $1$ is an identity for $\cdot$.
c. Define $\overline{0} = 1$ and $\overline{1} = 0$. Show that for every $a$ in
$S$, $a + \overline{a} = 1$ and $a \cdot \overline{a} = 0$. It follows from
parts (a)-\(c\) that $S$ is a Boolean algebra witgh the operations $+$ and
parts (a)-\(c\) that $S$ is a Boolean algebra with the operations $+$ and
$\cdot$.
$a = 0$:
$$ 0 + \overline{0} = 0 + 1 = 1 $$
$$ 0 \cdot \overline{0} = 0 \cdot 1 = 0 $$
$a = 1$:
$$ 1 + \overline{1} = 1 + 0 = 1 $$
$$ 1 \cdot \overline{1} = 1 \cdot 0 = 0 $$
Exercises 12-15 provide an outline for a proof that the associative laws, which
were included as an axiom for a Boolean algebra, can be derived from the other
four axioms. The outline is from _Introduction to Boolean Algebra_ by S. Givant
@ -4476,11 +4742,15 @@ that $\cdot$ takes precedence over $+$.
associative law for $+$. Rederive the law without using the associative law
and using only the other four axioms for a Boolean algebra.
Omitted.
13. The absorption law for $+$ states that for all elements $a$ and $b$ in a
Boolean algebra, $a \cdot b + a = a$. Prove this law without using the
associative law and using only the other four axioms for a Boolean algebra
plus the result of exercise 12.
Omitted.
14. _Test for equality law:_ For all elements $a$, $b$, and $c$ in a Boolean
algebra,
@ -4490,6 +4760,8 @@ then $b = c$.
Without using the associative law, derive this law from the other four laws in
the axioms for a Boolean algebra plus the result of exercise 12.
Omitted.
15. The associative law for $+$ states that for all elements $a$, $b$, and $c$
in a Boolean algebra, $a + (b + c) = (a + b) + c$. Show that this law, as
well as the associative law for $\cdot$, can be derived from the other four
@ -4505,29 +4777,82 @@ may use the universal bound law for $+$, the absorption law for $+$, and the
test for equality law from exercises 12, 13, and 14 because the associative laws
were not used to derive these properties.
Omitted.
In 16-21 determine whether each sentence is a statement. Explain your answers.
16. This sentence is false.
In order for a sentence to be a statement, it must be either true or false.
The sentence, "This sentence is false.", is not a statement. If the sentence is
false, then "This sentence is false", is false and therefore the sentence is
true. On the other hand, if the sentence is true, then "This sentence is false."
is true and therefore the sentence is false. Consequently, the sentence is both
true and false and not a statement.
17. If $1 + 1 = 3$, then $1 = 0$.
This sentence is a statement. By logical deduction, if $1 + 1 = 3$, which is a
false hypothesis, then $1 = 0$, which is a false conclusion. Thus the sentence
is vacuously true.
18. $\boxed{\text{The sentence in this box is a lie.}}$
This sentence is not a statement. Since the sentence is in the box, the
hypothesis is true. The conclusion however can be either true or false for much
the same reasons as given in problem 16.
19. All positive integers with negative squares are prime.
This sentence is a statement. The hypothesis is that for all positive integers
with negative squares, but there are no such integers. This hypothesis is false,
therefore the conclusion that they are all prime is vacuously true.
20. This sentence is false or $1 + 1 = 3$.
This is not a statement. Since the conditional starts with the paradoxical
statement from problem 16, the addition of an "or" conditional does not change
the fact that this is not a statement.
21. This sentence is false and $1 + 1 = 2$.
This is not a statement. For reasons similar to 20. Think on the wording "true
and false" and "true." This is what this sentence is saying. It is not a
statement.
22.
a. Assuming that the following sentence is a statement, prove that $1 + 1 = 3$:
If this sentence is true, then $1 + 1 = 3$.
**Proof (by contradiction):**
Suppose that the sentence "If this sentence is true, then $1 + 1 = 3$" is false.
Since the sentence is false, then the hypothesis "If this sentence is true,"
must be true, and the conclusion, "$1 + 1 = 3$", must be false.
So the sentence is true (by the hypothesis), and false (by the conclusion). This
is a contradiction.
Therefore the supposition that the sentence is false is false. Therefore, the
supposition must be true, and its conclusion that $1 + 1 = 3$ must follow.
Q.E.D.
b. What can you deduce from part (a) about the status of "This sentence is
true"? Why? (This example is known as Lob's paradox.)
We can deduce that "This sentence is true" is paradoxical, _i.e._ it is both
true and not true. As such, "This sentence is true" is a sentence, but not a
statement.
It is worth noting that any conclusion that follows it is true by logical
deduction. This makes any conclusion following the hypothesis true, and thereby
any sentence true.
23. The following two sentences were devised by the logician Saul Kripke. While
not intrinsically paradoxical, they could be paradoxical under certain
circumstances. Describe such circumstances.
@ -4539,19 +4864,48 @@ true"? Why? (This example is known as Lob's paradox.)
(_Hint:_ Suppose Nixon says (ii) and the only utterance Jones makes about
Watergate is (i).)
Omitted.
24. Can there exist a computer program that has as output a list of all the
computer programs that do not list themselves in their output? Explain your
answer.
No. Suppose there exists a computer program $P$ that has as output a list of all
computer programs that do not list themselves in their output. If $P$ lists
itself as output, then it would be on the output list of $P$, which consists of
all computer programs that do not list themselves in their output. Hence $P$
would not list itself as output. But if $P$ does not list itself as output, then
$P$ would be a member of the list of all computer programs that do not list
themselves in their output, and this list is exactly the output of $P$. Hence
$P$ would list itself as output. This analysis shows that the assumption of the
existence of such a program $P$ is contradictory, and so no such program exists.
25. Can there exist a book that refers to all those books and only those books
that do not refer to themselves? Explain your answer.
This is the same as number 24.
Say there is a book $B$ that refers to all books that do not refer to
themselves. If $B$ refers to itself, then by the definition of $B$, $B$ would
not refer to itself. On the other hand, if $B$ does not refer to itself, then by
definition of $B$, $B$ would refer to itself. This is a paradox and therefore no
such book can exist.
26. Some English adjectives are descriptive of themselves (for instance, the
word _polysyllabic_ is polysyllabic) whereas others are not (for instance,
the word _monosyllabic_ is not monosyllabic). The word _heterological_
refers to an adjective that does not describe itself. Is _heterological_
heterological? Explain your answer.
If _heterological_ is heterological, then _heterological_ does not describe
itself. But since _heterological_ is heterological, it is describing itself by
the supposition. This is a contradiction.
If _heterological_ is not heterological, then _heterological_ does describe
itself, which contradicts its own definition.
It is paradoxical, _heterological_ is both heterological and not heterological.
27. As strange as it may seem, it is possible to give a precise-looking verbal
definition of an integer that, in fact, is not a definition at all. The
following was devised by an English librarian, G.G. Berry, and reported by
@ -4560,10 +4914,16 @@ Watergate is (i).)
the total number of strings consisting of 11 or fewer English words is
finite.)
Omitted.
28. Is there an algorithm which, for a fixed quantity $a$ and any input
algorithm $X$ and data set $D$, can determine whether $X$ prints $a$ when
run with data set $D$? Explain. (This problem is called the **printing
problem**.)
Omitted.
29. Use a technique similar to that used to derive Russell's paradox to prove
that for any set $A$, $\mathscr{P}(A) \nsubseteq A$.
Omitted.

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