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chapter_6/exercises.md
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@ -4184,3 +4184,261 @@ Omitted.
both sides with $A$ and deduce the identity.
Omitted.
---
Page 445
**Exercise Set 6.4**
In 1-3 assume that $B$ is a Boolean algebra with operations $+$ and $\cdot$.
Give the reasons needed to fill in the blanks in the proofs using only the
axioms for a Boolean algebra.
1. _Idempotent law for $\cdot$:_ For every $a$ in $B$, $a \cdot a = a$.
**Proof:**
Let $a$ be any element of $B$. Then
$$ a = a \cdot 1 $$
__ (a) __
$$ = a \cdot (a + \overline{a}) $$
__ (b) __
$$ = (a \cdot a) + (a \cdot \overline{a}) $$
__ \(c\) __
$$ = (a \cdot a) + 0 $$
__ (d) __
$$ = a \cdot a $$
__ (e) __
2. _Universal bound law for $+$:_ For every $a$ in $B$, $a + 1 = 1$.
**Proof:**
Let $a$ be any element in $B$. Then
$$ a + 1 = a + (a + \overline{a}) $$
__ (a) __
$$ = (a + a) + \overline{a} $$
__ (b) __
$$ = a + \overline{a} $$
by Example 6.4.2
$$ = 1 $$
__ \(c\) __
3. _Absorption law for $\cdot$ over $+$:_ For all $a$ and $b$ in $B$,
$(a + b) \cdot a = a$.
**Proof:** Let $a$ be any element of $B$. Then
$$ (a + b) \cdot a = a \cdot (a + b) $$
__ (a) __
$$ = a \cdot a + a \cdot b $$
__ (b) __
$$ = a + a \cdot b $$
by exercise 1
$$ = a \cdot 1 + a \cdot b $$
__ \(c\) __
$$ = a \cdot (1 + b) $$
__ (d) __
$$ = a \cdot (b + 1) $$
__ (e) __
$$ = a \cdot 1 $$
by exercise 2
$$ = a $$
__ (f) __
In 4-10 assume that $B$ is a Boolean algebra with operations $+$ and $\cdot$.
Prove each statement using only the axioms for a Boolean algebra and statements
proved in the text or in lower-numbered exercises.
4. _Universal bound for $0$:_ For every $a$ in $B$, $a \cdot 0 = 0$.
5. _Complements of $0$ and $1$:_
a. $\overline{0} = 1$
b. $\overline{1} = 0$
6. _Uniqueness of $0$:_ There is only one element of $B$ that is an identity for
$+$.
7. _Uniqueness of $1$:_ There is only one element of $B$ that 8s an identity for
$\cdot$.
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
$a \cdot b$ has a unique complement.)
9. _De Morgan's law for $+$:_ For all $a$ and $b$ in $B$,
$\overline{a + b} = \overline{a} \cdot \overline{b}$.
10. _Cancellation law:_ For all $x$, $y$, and $z$ in $B$, if $x + y = x + z$ and
$x \cdot y = x \cdot z$, then $y = z$.
11. Let $S = \{0, 1\}$, and define operations $+$ and $\cdot$ on $S$ by the
following tables:
| $+$ | $0$ | $1$ |
| --- | --- | --- |
| $0$ | $0$ | $1$ |
| $1$ | $1$ | $1$ |
| $\cdot$ | $0$ | $1$ |
| ------- | --- | --- |
| $0$ | $0$ | $0$ |
| $1$ | $0$ | $1$ |
a. Show that the elements of $S$ satisfy the following properties:
i. the commutative law for $+$.
ii. the commutative law for $\cdot$.
iii. the associative law for $+$.
iv. the associative law for $\cdot$.
v. the distributive law for $+$ over $\cdot$.
vi. the distributive law for $\cdot$ over $+$.
b. Show that $0$ is an identity element for $+$ and that $1$ is an identity
element 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
$\cdot$.
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
and P. Halmos, Springer, 2009. In order to avoid unneeded parentheses, assume
that $\cdot$ takes precedence over $+$.
12. The universal bound law for $+$ states that for every element $a$ in a
Boolean algebra, $a + 1 = 1$. The proof shown in exercise 2 used the
associative law for $+$. Rederive the law without using the associative law
and using only the other four axioms for a Boolean algebra.
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.
14. _Test for equality law:_ For all elements $a$, $b$, and $c$ in a Boolean
algebra,
If $b \cdot a = c \cdot a$ and $b \cdot \overline{a} = c \cdot \overline{a}$,
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.
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
axioms in the definition and axioms for a Boolean algebra. Then explain how
to use your work to obtain a derivation for the associative law for $\cdot$.
_Hints:_ To prove this theorem, suppose $a$, $b$, and $c$ are any elements in a
Boolean algebra $B$, and divide the proof into three parts. _Part 1:_ Prove that
$(a + (b + c)) \cdot a = ((a + b) + c) \cdot a$. _Part 2:_ Prove that
$(a + (b + c)) \cdot \overline{a} = ((a + b) + c) \cdot \overline{a}$. _Part 3:_
Use the results of parts 1 and 2 to prove that $a + (b + c) = (a + b) + c$. You
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.
In 16-21 determine whether each sentence is a statement. Explain your answers.
16. This sentence is false.
17. If $1 + 1 = 3$, then $1 = 0$.
18. $\boxed{\text{The sentence in this box is a lie.}}$
19. All positive integers with negative squares are prime.
20. This sentence is false or $1 + 1 = 3$.
21. This sentence is false and $1 + 1 = 2$.
22.
a. Assuming that the following sentence is a statement, prove that $1 + 1 = 3$:
If this sentence is true, then $1 + 1 = 3$.
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.)
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.
i. Most of Nixon's assertions about Watergate are false.
ii. Everything Jones says about Watergate is true.
(_Hint:_ Suppose Nixon says (ii) and the only utterance Jones makes about
Watergate is (i).)
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.
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.
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.
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
Bertrand Russell. Explain how it leads to a contradiction. Let $n$ be "the
smallest integer not describable in fewer than 12 English words." (Note that
the total number of strings consisting of 11 or fewer English words is
finite.)
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**.)
29. Use a technique similar to that used to derive Russell's paradox to prove
that for any set $A$, $\mathscr{P}(A) \nsubseteq A$.

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chapter_6/notes.md
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@ -497,3 +497,213 @@ _[This is what was to be shown.]_
_[Since we have proved both the basis step and the inductive step, we conclude
that the theorem is true.]_
---
Page 439
**Definition and Axioms for a Boolean Algebra**
A **Boolean algebra** is a set $B$ together with two operations, generally
denoted $+$ and $\cdot$, such that for all $a$ and $b$ in $B$ both $a + b$ and
$a \cdot b$ are in $B$ and the following axioms are assumed to hold:
1. _Commutative Laws:_ For all $a$ and $b$ in $B$,
$$ \text{(a) } a + b = b + a \quad \text{ and } \quad \text{(b) } a \cdot b = b \cdot a $$
2. _Associative Laws:_ For all $a$, $b$, and $c$ in $B$,
$$ \text{(a) } (a + b) + c = a + (b + c) \quad \text{ and } \quad \text{(b) } (a \cdot b) \cdot c = a \cdot (b \cdot c) $$
3. _Distributive Laws:_ For all $a$, $b$, and $c$ in $B$,
$$ \text{(a) } a + (b \cdot c) = (a + b) \cdot (a + c) \quad \text{ and } \quad \text{(b) } a \cdot (b + c) = (a \cdot b) + (a \cdot c) $$
4. _Identity Laws:_ There exist distinct elements $0$ and $1$ in $B$ such that
for each $a$ in $B$,
$$ \text{(a) } a + 0 = a \quad \text{ and } \quad \text{(b) } a \cdot 1 = a $$
5. _Complement Laws:_ For each $a$ in $B$, there exists an element in $B$,
denoted $\overline{a}$ and called the **complement** or **negation** of $a$,
such that
$$ \text{(a) } a + \overline{a} = 1 \quad \text{ and } \quad \text{(b) } a \cdot \overline{a} = 0 $$
---
Page 439
**Theorem 6.4.1 Properties of a Boolean Algebra**
Let $B$ be any Boolean algebra.
1. _Uniqueness of the Complement Laws:_ For all $a$ and $x$ in $B$, if
$a + x = 1$ and $a \cdot x = 0$ then $x = \overline{a}$.
2. _Uniqueness of $0$ and $1$:_ If there exists $x$ in $B$ such that $a + x = a$
for every $a$ in $B$, then $x = 0$, and if there exists $y$ in $B$ such that
$a \cdot y = a$ for every $a$ in $B$, then $y = 1$.
3. _Double Complement Law:_ For every $a \in B, \overline{(\overline{a})} = a$.
4. _Idempotent Laws:_ For every $a \in B$ ,
$$ \text{(a) } a + a = a \quad \text{ and } \quad \text{(b) } a \cdot a = a $$
5. _Universal Bound Laws:_ For every $a \in B$,
$$ \text{(a) } a + 1 = 1 \quad \text{ and } \quad \text{(b) } a \cdot 0 = 0 $$
6. _De Morgan's Laws:_ For all $a$ and $b \in B$,
$$ \text{(a) } \overline{a + b} = \oveline{a} \cdot \overline{b} \quad \text{ and } \quad \text{(b) } \overline{a \cdot b} = \overline{a} + \overline{b} $$
7. _Absorption Laws:_ For all $a$ and $b \in B$,
$$ \text{(a) } (a + b) \cdot a = a \quad \text{ and } \quad \text{(b) } (a \cdot b) + a = a $$
8. _Complements of $0$ and $1$:_
$$ \text{(a) } \overline{0} = 1 \quad \text{ and } \quad \text{(b) } \overline{1} = 0 $$
**Proof:**
_Part 1: Uniqueness of the Complement Law_
Suppose $a$ and $x$ are particular, but arbitrarily chosen, elements of $B$ that
satisfy the following hypothesis: $a + x = 1$ and $a \cdot x = 0$. Then
$$ x = x \cdot 1 $$
because $1$ is an identity for $\cdot$
$$ = x \cdot (a + \overline{a}) $$
by the complement law for $+$
$$ = x \cdot a + x \cdot \overline{a} $$
by the distributive law for $\cdot$ over $+$
$$ = a \cdot x + x \cdot \overline{a} $$
by the commutative law for $\cdot$
$$ = 0 + x \cdot \overline{a} $$
by hypothesis
$$ = a \cdot \overline{a} + x \cdot \overline{a} $$
by the complement law for $\cdot$
$$ = (\overline{a} \cdot a) + (\overline{a} \cdot x) $$
by the commutative law for $\cdot$
$$ = \overline{a} \cdot (a + x) $$
by the distributive law for $\cdot$ over $+$
$$ = \overline{a} \cdot 1 $$
by hypothesis
$$ = \overline{a} $$
because $1$ is an identity for $\cdot$.
Proofs of the other parts of the theorem are discussed in the examples that
follow and in the exercises.
---
Page 441
**Theorem 6.4.1(3) Double Complement Law**
For every element $a$ in a Boolean algebra $B$, $\overline{(\overline{a})} = a$.
**Proof:**
Suppose $B$ is a Boolean algebra and $a$ is any element of $B$. Then
$$ \overline{a} + a = a + \overline{a} $$
by the commutative law for $+$
$$ = 1 $$
by the complement law for $1$
and
$$ \overline{a} \cdot a = a \cdot \overline{a} $$
by the commutative law for $\cdot$
$$ = 0 $$
by the complement law for $0$
Thus $a$ satisfies the two equations with respect to $\overline{a}$ that are
satisfied by the complement of $\overline{a}$. From the fact that the complement
of $a$ is unique, we conclude that $\overline{(\overline{a})} = a$.
---
Page 444
**Theorem 6.4.2**
There is no computer algorithm that will accept any algorithm $X$ and data set
$D$ as input and then will output "halts" or "loops forever" to indicate whether
or not $X$ terminates in a finite number of steps when $X$ is run with data set
$D$.
**Proof (by contradiction):**
Suppose there is an algorithm, CheckHalt, such that if an algorithm $X$ and a
data set $D$ are input, then
$\text{CheckHalt}(X, D)$ prints
"halts" if $X$ terminates in a finite number of steps when run with data set $D$
or
"loops forever" if $X$ does not terminate in a finite number of steps when run
with data set $D$.
_[To show that no algorithm such as CheckHalt can exist, we will deduce a
contradiction.]_
Observe that the sequence of characters making up an algorithm $X$ can be
regarded as a data set itself. Thus it is possible to consider running CheckHalt
with input $(X, X)$. Define a new algorithm, Test, as follows: For any input
algorithm $X$,
$\text{Test}(X)$
loops forever if $\text{CheckHalt}(X, X)$ prints "halts"
or
stops if $\text{CheckHalt}(X, X)$ prints "loops forever".
Now run algorithm Test with input Test. If $\text{Test}(\text{Test})$ terminates
after a finite number of steps, then the value of
$\text{Checkhalt}(\text{Test}, \text{Test})$ is "halts" and so
$\text{Test}(\text{Test})$ loops forever.
On the other hand, if $\text{Test}(\text{Test})$ does not terminate after a
finite number of steps, then $\text{CheckHalt}(\text{Test}, \text{Test})$ prints
"loops forever" and so $\text{Test}(\text{Test})$ terminates.
The two paragraphs above show that $\text{Test}(\text{Test})$ loops forever and
also that it terminates. This is a contradiction. But the existence of Test
follows logically from the supposition of the existence of an algorithm
CheckHalt that can check any algorithm and data set for termination. _[Hence the
supposition must be false, and there is no such algorithm.]_

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@ -114,3 +114,23 @@ cite the property from 6.2.2 used
stated.
exactly
---
Page 445
**Test Yourself**
1. In the comparison between the structure of the set of statement forms and the
set of subsets of a universal set, the _or_ operation $\vee$ corresponds to
_____, the _and_ operation $\wedge$ corresponds to _____, a tautology
$\mathbf{t}$ corresponds to _____, a contradiction $\mathbf{c}$ corresponds
to _____, and the negation operation, denoted $\neg$, corresponds to _____.
2. The operations of $+$ and $\cdot$ in a Boolean algebra are generalizations of
the operations of _____ and _____ in the set of all statement forms in a
given finite number of variables and the operations of _____ and _____ in the
set of all subsets of a given set.
3. Russell showed that the following proposed "set definition" could not
actually define a set: _____.