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chapter_5/exercises.md
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@ -12305,21 +12305,76 @@ Page 397
a. $\neg p \vee (q \wedge (r \vee \neg s))$
(1) By I, $p$, $q$, $r$, and $s$ are Boolean expressions.
(2) By 1 and II(c), $\neg s$ is a Boolean expression.
(3) By 1, 2, and II(b), $r \vee \neg s$ is a Boolean expression.
(4) By 3 and II(d), $(r \vee \neg s)$ is a Boolean expression.
(5) By 1, 4 and II(a), $q \wedge (r \vee \neg s)$ is a Boolean expression.
(6) By 5 and II(d), $(q \wedge (r \vee \neg s))$ is a Boolean expression.
(7) By 1 and II(c), $\neg p$ is a Boolean expression.
(8) By 6, 7 and II(b), $\neg p \vee (q \wedge (r \vee neg s))$ is a Boolean
expression.
b. $(p \vee q) \vee \neg((p \wedge \neg s) \wedge r)$
2. Consider the set $C$ of parenthesis structures defined in Example 5.9.2. Give
(1) By I, $p$, $q$, $s$, and $r$ are Boolean expressions.
(2) By 1 and II(c), $\neg s$ is a Boolean expression.
(3) By 1, 2, and II(a), $p \wedge \neg s$ is a Boolean expression.
(4) By 3 and II(d), $(p \wedge \neg s)$ is a Boolean expression.
(5) By 1, 4 and II(a), $(p \wedge \neg s) \wedge r$ is a Boolean expression.
(6) By 5 and II(d), $((p \wedge \neg s) \wedge r)$ is a Boolean expression.
(7) By 6 and II(c), $\neg((p \wedge \neg s) \wedge r)$ is a Boolean expression.
(8) By 1 and II(b), $p \vee q$ is a Boolean expression.
(9) By 8 and II(d), $(p \vee q)$ is a Boolean expression.
(10) By 7, 9, and II(b), $(p \vee q) \vee \neg((p \wedge \neg s) \wedge r)$ is a
Boolean expression.
5. Consider the set $C$ of parenthesis structures defined in Example 5.9.2. Give
derivations showing that each of the following is in $C$.
a. $()(())$
(1) By I, $()$ is in $C$.
(2) By 1 and II(a), $(())$ is in $C$.
(3) By 1, 2, and II(b), $()(())$ is in $C$.
b. $(())(())$
(1) By I, $()$ is in $C$.
(2) By 1 and II(a), $(())$ is in $C$.
(3) By 1, 2, and II(b), $(())(())$ is in $C$.
3. Let $S$ be the set of all strings over a finite set $A$ and let $a$, $b$, and
$c$ be any characters in $A$.
a. Using Theorem 5.9.1 but not Theorem 5.9.3 or 5.9.4, show that
$(ab)c = a(bc)$.
(1) By Theorem 5.9.1, $a$ and $b$ are strings because $a$ and $b$ are in $A$.
(2) By (1) and II(c) of the definition of a string, $a(bc) = (ab)c$ because $a$
and $b$ are strings in $S$ and $c$ is in $A$.
b. Show that $ab$ is a string in $S$. Then use the result of part (a) to
conclude that $a(bc)$ is a string in $S$.
@ -12331,8 +12386,38 @@ $abc$.)
a. MIUI
(1) By I, _MI_ is in the _MIU_-system.
(2) By 1 and II(b), _MIU_ is in the _MIU_-system.
(3) By 2 and II(b), _MIIII_ is in the _MIU_-system._
(4) By 3 and II(b), _MIIIIIIII_ is in the _MIU_-system._
(5) By 4 and II(c), _MIUIIII_ is in the _MIU_-system.
(6) By 5 and II(c), _MIUUI_ is in the _MIU_-system._
(7) By 6 and II(d), _MIUI_ is in the _MIU_-system.
b. MUIIU
(1) By I, _MI_ is in the _MIU_-system.
(2) By 1 and II(b), _MII_ is in the _MIU_-system.
(3) By 2 and II(b), _MIIII_ is in the _MIU_-system.
(4) By 3 and II(b), _MIIIIIIII_ is in the _MIU_-system.
(5) By 4 and II(c), _MUIIIII_ is in the _MIU_-system.
(6) By 5 and II(c), _MUUII_ is in the _MIU_-system.
(7) By 6 and II(d), _MUII_ is in the _MIU_-system.
(8) By 7 and II(a), _MUIIU_ is in the _MIU_-system.
5. The set of arithmetic expressions over the real numbers can be defined
recursively as follows:
@ -12362,8 +12447,35 @@ derivations showing that each of the following is an arithmetic expression.
a. $((2 \cdot (0.3 - 4.2)) + (-7))$
(1) By I, $2$, $0.3$, $4.2$, and $7$ are arithmetic expressions.
(2) By 1 and II(d), $(0.3 - 4.2)$ is an arithmetic expression.
(3) By 2 and II(e) , $(2 \cdot (0.3 - 4.2))$ is an arithmetic expression.
(4) By 1 and II(b), $(-7)$ is an arithmetic expression.
(5) By 3, 4, and II(c), $((2 \cdot (0.3 - 4.2))) + (-7))$ is an arithmetic
expression.
b. $\left(\frac{(9 \cdot(6 \cdot 1 + 2))}{((4 - 7) \cdot 6)}\right)$
(1) By I, $9$, $6$, $1$, $2$, $4$, $7$, and $6$ are arithmetic expressions.
(2) By 1 and II(d), $(4 - 7)$ is an arithmetic expression.
(3) By 1, 2, and II(e), $((4 - 7) \cdot 6)$ is an arithmetic expression.
(4) By 1 and II(e), $(6 \cdot 1)$ is an arithmetic expression.
(5) By 1, 4, and II(c), $(6 \cdot 1 + 2)$ is an arithmetic expression.
(6) By 1, 5, and II(e), $(9 \cdot (6 \cdot 1 + 2))$ is an arithmetic expression.
(7) By 3, 6, and II(f),
$\left(\frac{(9 \cdot (6 \cdot 1 + 2))}{(4 - 7) \cdot 6}\right)$ is an
arithmetic expression.
6. Let $S$ be a set of integers defined recursively as follows:
I. Base: $5$ is in $S$.
@ -12376,6 +12488,68 @@ and II above.
Use structural induction to prove that for every integer $n$ in $S$,
$n \mod 2 = 1$.
**Proof (by structural induction):**
Given any integer $n$ in $S$, let $P(n)$ be the equality:
$$ n \mod 2 = 1 $$
_Basis Step:_
Prove $P(n)$ is true for each integer $n$ in the base for $S$.
The only object in $S$ is $5$, and $P(5)$ is true because $5 \mod 2 = 1$ since
$5 = 2 \cdot 2 + 1$.
Therefore $P(n)$ is true.
_Inductive Step:_
Suppose $n$ is any integer in $S$ such that $P(n)$ is true. That is:
$$ n \mod 2 = 1 $$
This is the inductive hypothesis.
Prove that the recursive definition is true. In other words, we must prove
$P(n + 4)$, that is:
$$ (n + 4) \mod 2 = 1 $$
By the inductive hypothesis:
$$ n \mod 2 = 1 $$
This means that, by the definition of divisibility:
$$ n = 2k + 1 $$
for some integer $k$.
It follows, by substitution:
$$ (n + 4) \mod 2 = [(2k + 1) + 4] \mod 2 $$
$$ = [2k + 4 + 1] mod 2 $$
$$ = [2k + 2(2) + 1] \mod 2 $$
$$ = [2(k + 1) + 1] \mod 2 $$
By the definition of $\mod$:
$$ = 1 $$
Therefore $P(n + 4)$ is true.
_Conclusion:_
Since there are no integers in $S$ other than those obtained from the base and
recursion definitions for $S$, we conclude that every integer $n$ in $S$
satisfies the equation $n \mod 2 = 1$.
Q.E.D.
7. Define a set $S$ of strings over the set $\{0, 1\}$ recursively as follows:
I. Base: $1 \in S$
@ -12391,6 +12565,53 @@ above.
Use structural induction to prove that every string in $S$ ends in a $1$.
**Proof (by structural induction):**
Let $P(n)$ be the sentence: "every string in $S$ ends in a $1$."
_Basis Step:_
Prove the base definition is true. In other words, prove that $P(a)$ is true.
Since $1$ is the only string in $S$, $P(1)$ is true since $1$ ends in a $1$.
Therefore $P(a)$ is true.
_Inductive Step:_
Suppose $s$ is any string in $S$ such that $P(s)$ is true, that is:
"$s$ ends in a $1$."
This is the inductive hypothesis.
Prove the recursive definition. That is, we must prove both II(a), and II(b):
(a) $0s \in S$
(b) $1s \in S$
_Case (a):_
When rule II(a) is applied to $s$, $0$ is concatenated with $s$. Since $s$ ends
in a $1$ by the inductive hypothesis, it follows that $0s$ also ends with a $1$.
_Case (b):_
When rule II(b) is applied to $s$, $1$ is concatenated with $s$. Since $s$ ends
in a $1$ by the inductive hypothesis, it follows that $1s$ also ends with a $1$.
Therefore in both cases, $0s$ and $1s$ end in a $1$, and the recursive
definition is true, which is what was to be shown.
_Conclusion:_
Since there are no strings in $S$ other than those obtained from the base and
recursion definitions for $s$, we conclude that every string in $S$ ends in a
$1$.
Q.E.D.
8. Define a set $S$ of strings over the set $\{a, b\}$ recursively as follows:
I. Base $a \in S$
@ -12406,6 +12627,52 @@ above.
Use structural induction to prove that every string in $S$ begins with an $a$.
**Proof (by structural induction):**
Let $P(s)$ be the sentence "$s$ begins with an $a$."
_Basis Step:_
Prove the base definition, $P(a)$.
$a$ is the only string in $S$, and $a$ does begin with an $a$.
Therefore $P(a)$ is true.
_Inductive Step:_
Suppose $s$ is a string in $S$ such that $P(s)$ is true, that is:
"$s$ begins with an $a$."
This is the inductive hypothesis.
Prove the recursion definition. That is, we must prove both II(a) and II(b):
(a) $sa \in S$
(b) $sb \in S$
_Case (a):_
$sa$ appends $a$ to $s$. By the inductive hypothesis, $s$ begins with an $a$.
Thus $sa$ begins with an $a$.
_Case (b):_
$sb$ appends $b$ to $s$. By the inductive hypothesis, $s$ begins with an $a$.
Thus $sb$ begins with an $a$.
Therefore in both cases, $sa$ and $sb$ begin with an $a$, therefore the
recursion definition is true, which is what was to be shown.
_Conclusion:_
Since there are no strings in $S$ other than those obtained from the base and
recursion for $S$, we conclude that every string in $S$ begins with an $a$.
Q.E.D.
9. Define a set $S$ of strings over the set $\{a, b\}$ recursively as follows:
I. Base: $\lambda \in S$
@ -12426,6 +12693,109 @@ above.
Use structural induction to prove that every string in $S$ contains an even
number of $a$'s.
**Proof (by structural induction):**
Let $P(s)$ be the sentence "$s$ contains an even number of $a$'s."
_Basis Step:_
Prove the base definition, $P(\lambda)$, that is:
"$\lambda$ contains an even number of $a$'s."
$\lambda$ is the null string, which contains $0$ characters. It follows that
$\lambda$ contains $0$ $a$'s. By the definition of even, $0$ is even.
Therefore $P(\lambda)$ is true.
_Inductive Step:_
Suppose $s$ is a string in $S$ such that $P(s)$ is true, that is:
"$s$ contains an even number of $a$'s"
This is the inductive hypothesis.
We must prove the recursion definition. That is we must prove II(a), II(b),
II(c), and II(d):
(a) $bs \in S$
(b) $sb \in S$
\(c\) $saa \in S$
(d) $aas \in S$
_Case (a):_
$bs$ prepends $b$ to $s$. There are $0$ $a$'s in the string $b$, and $s$ has an
even number of $a$'s by the inductive hypothesis. Therefore, $bs$ contains:
$$ 2k + 0 $$
$$ = 2k $$
$a$'s for some integer $k$.
Thus, by the product of integers and the definition of even, $bs$ has an even
number of $a$'s.
_Case (b):_
$sb$ appends $b$ to $s$. There are $0$ $a$'s in the string $b$, and $s$ has an
even number of $a$'s by the inductive hypothesis. Therefore, $sb$ contains:
$$ 0 + 2k $$
$$ = 2k $$
$a$'s for some integer $k$.
Thus, by the product of integers and the definition of even, $sb$ has an even
number of $a$'s.
_Case \(c\):_
$saa$ appends $aa$ to $s$. There are $2$ $a$'s in the string $aa$, and $s$ has
an even number of $a$'s by the inductive hypothesis. Therefore, $saa$ contains:
$$ 2 + 2k $$
$$ = 2(k + 1) $$
$a$'s for some integer $k$.
Thus, by the sum of integers and the definition of even, $saa$ has an even
number of $a$'s.
_Case (d):_
(d) $aas \in S$
$aas$ prepends $aa$ to $s$. There are $2$ $a$'s in the string $aa$, and $s$ has
an even number of $a$'s by the inductive hypothesis. Therefore, $aas$ contains:
$$ 2k + 2 $$
$$ = 2(k + 1) $$
$a$'s for some integer $k$.
Thus, by the sum of integers and the definition of even, $aas$ has an even
number of $a$'s.
Therefore, in all four cases, each string has an even number of $a$'s. Therefore
the recursive definition is true.
_Conclusion:_
Since there are no strings in $S$ other than those obtained from the base and
recursion definitions for $S$, we conclude that every string in $S$ has an even
number of $a$'s.
Q.E.D.
10. Define a set $S$ of strings over the set of all integers recursively as
follows:
@ -12444,6 +12814,60 @@ above.
Use structural induction to prove that no string in $S$ represents an integer
with a leading zero.
**Proof (by structural induction):**
Let $P(s)$ be the sentence "$s$ does not represents an integer with a leading
zero."
_Basis Step:_
Prove the base definition. In other words prove $P(a)$, where $a$ is a string
defined in the base:
The base defines the following strings as being in $S$:
$$ 1, 2, 3, 4, 5, 6, 7, 8, 9 $$
None of these strings represent an integer with a leading zero. Therefore $P(a)$
is true.
_Inductive Step:_
Let $s$ be a string in $S$ such that P(s) is true, that is:
"$s$ does not represents an integer with a leading zero."
This is the inductive hypothesis.
We must prove the recursion definition. That is we must prove II(a) and II(b):
(a) $s0 \in S$
(b) $st \in S$
_Case (a):_
$s0$ appends a $0$ to $s$. By the inductive hypothesis, $s$ does not represent
an integer with a leading zero. Appending $0$ to $s$ does not change this fact.
Thus $s0$ does not represent an integer with a leading zero.
_Case (b):_
$st$ appends a $t$ to $s$. By the inductive hypothesis, $s$ does not represent
an integer with a leading zero. Appending $t$ to $s$ does not change this fact
($t$ is also in $S$, and thus does not represent an integer with a leading zero,
but appending $t$ to $s$ still does not change this fact). Thus $st$ does not
represent an integer with a leading zero.
In both cases, the strings do not represent an integer with a leading zero.
Therefore the recursion definition is true, which is what was to be shown.
_Conclusion:_
Since there are no strings in $S$ other than those obtained from the base and
recursion definitions for $S$, we conclude that every string in $S$ do not
represent an integer with a leading zero.
11. Define a set $S$ of strings over the set of all integers recursively as
follows:
@ -12467,6 +12891,135 @@ above.
Use structural induction to prove that every string in $S$ represents an odd
integer when written in decimal notation.
**Proof (by structural induction):**
Let $P(s)$ be the sentence:
"$s$ represents an odd integer when written in decimal notation."
_Basis Step:_
Prove the base definition. In other words, prove $P(a)$.
The base definition defines $S$ as being:
$$ \{1, 3, 5, 7, 9\} $$
$1$ represents an odd integer since $1 = 2(0) + 1$.
$3$ represents an odd integer since $3 = 2(1) + 1$.
$5$ represents an odd integer since $5 = 2(2) + 1$.
$7$ represents an odd integer since $7 = 2(3) + 1$.
$9$ represents an odd integer since $9 = 2(4) + 1$.
Therefore for all strings in $S$ obtained from the base definition, $a$
represents an odd integer when written in decimal notation, and $P(a)$ is true.
_Inductive Step:_
Suppose $s$ and $t$ are strings in $S$ such that $P(s)$ and $P(t)$ are true,
that is:
"$s$ represents an odd integer when written in decimal notation."
"$t$ represents an odd integer when written in decimal notation."
This is the inductive hypothesis.
We must prove the recursion definition. That is we must prove II(a), II(b),
II\(c\), II(d), and II(e):
(a) $st \in S$
(b) $2s \in S$
\(c\) $4s \in S$
(d) $6s \in S$
(e) $8s \in S$
_Case (a):_
By the inductive hypothesis, $s$ represents an odd integer when written in
decimal notation, and $t$ represents and odd integer when written in decimal
notation.
By the definition of odd, odd integers always end with an odd integer in the
$1$'s place.
Since $t$ appends $s$, and since $t$ is an odd integer when written in decimal
notation, it follows that $t$'s $1$'s place contains a representation of an odd
integer when written in decimal notation. Thus $st$ also contains an odd integer
in it's $1$'s place and $st$ is an odd integer when written in decimal notation.
_Case (b):_
By the inductive hypothesis, $s$ represents an odd integer when written in
decimal notation.
By the definition of odd, odd integers always end with an odd integer in the
$1$'s place.
Since $s$ is an odd integer when written in decimal notation, it follows that
$s$'s $1$'s place contains a representation of an odd integer when written in
decimal notation. Thus $2s$ also contains an odd integer in it's $1$'s place and
$2s$ is an odd integer when written in decimal notation.
_Case \(c\):_
By the inductive hypothesis, $s$ represents an odd integer when written in
decimal notation.
By the definition of odd, odd integers always end with an odd integer in the
$1$'s place.
Since $s$ is an odd integer when written in decimal notation, it follows that
$s$'s $1$'s place contains a representation of an odd integer when written in
decimal notation. Thus $4s$ also contains an odd integer in it's $1$'s place and
$4s$ is an odd integer when written in decimal notation.
_Case (d):_
By the inductive hypothesis, $s$ represents an odd integer when written in
decimal notation.
By the definition of odd, odd integers always end with an odd integer in the
$1$'s place.
Since $s$ is an odd integer when written in decimal notation, it follows that
$s$'s $1$'s place contains a representation of an odd integer when written in
decimal notation. Thus $6s$ also contains an odd integer in it's $1$'s place and
$6s$ is an odd integer when written in decimal notation.
_Case (e):_
By the inductive hypothesis, $s$ represents an odd integer when written in
decimal notation.
By the definition of odd, odd integers always end with an odd integer in the
$1$'s place.
Since $s$ is an odd integer when written in decimal notation, it follows that
$s$'s $1$'s place contains a representation of an odd integer when written in
decimal notation. Thus $8s$ also contains an odd integer in it's $1$'s place and
$8s$ is an odd integer when written in decimal notation.
In all cases, the obtained string is an odd integer when written in decimal
notation. Therefore the recursion definition is true, which is what was to be
shown.
_Conclusion:_
Since there are no strings in $S$ other than those obtained from the base and
recursion definitions for $S$, we conclude that every string in $S$ represents
an odd integer when written in decimal notation.
Q.E.D.
12. Define a set $S$ of integers recursively as follows:
I. Base: $0 \in S, 5 \in S$
@ -12482,6 +13035,89 @@ above.
Use structural induction to prove that every integer in $S$ is divisible by $5$.
**Proof (by structural induction):**
Let $P(n)$ be the sentence:
"$n$ is divisible by $5$."
_Basis Step:_
Prove the base definition. In other words, prove $P(a)$.
By the base definition, $S$ is defined as only containing $0$ and $5$.
Both $0$ and $5$ are divisible by $5$ by the definition of divisibility.
Therefore $P(a)$ is true.
_Inductive Step:_
Let $k$ and $p$ be any integers such that $P(k)$ and $P(p)$ is true, that is:
"$k$ is divisible by $5$."
and
"$p$ is divisible by $5$."
This is the inductive hypothesis.
We must prove the recursion definition. That is we must prove II(a) and II(b):
(a) $k + p \in S$
(b) $k - p \in S$
_Case (a):_
By the inductive hypothesis, $k$ and $p$ are both divisible by $5$. It follows
that:
$$ k = 5m $$
$$ p = 5q $$
for some integers $m$ and $q$.
By substitution:
$$ k + p = 5m + 5q $$
$$ = 5(m + q) $$
Thus, by the sum of integers and the definition of divisibility, $k + p$ is
divisible by $5$.
_Case (b):_
By the inductive hypothesis, $k$ and $p$ are both divisible by $5$. It follows
that:
$$ k = 5m $$
$$ p = 5q $$
for some integers $m$ and $q$.
By substitution:
$$ k + p = 5m - 5q $$
$$ = 5(m - q) $$
Thus, by the difference of integers and the definition of divisibility, $k - p$
is divisible by $5$.
In both cases, the obtained integers are divisible by $5$. Therefore the
recursion definition is true, which is what was to be shown.
_Conclusion:_
Since there are no integers in $S$ other than those obtained from the base and
recursion definitions for $S$, we conclude that every integer in $S$ is
divisible by $5$.
13. Define a set $S$ of integers recursively as follows:
I. Base: $0 \in S$
@ -12497,17 +13133,98 @@ above.
Use structural induction to prove that every integer in $S$ is divisible by $3$.
**Proof (by structural induction):**
Let $P(n)$ be the sentence:
"$n$ is divisible by $3$."
_Basis Step:_
Prove the base definition. In other words prove $P(a)$.
$0$ is the only integer in $S$ by the base definition.
$0$ is divisible by $3$ by the definition of divisibility.
Therefore $P(a)$ is true.
_Inductive Step:_
Let $k$ be any integer such that $P(k)$ is true, that is:
"$k$ is divisible by $3$."
This is the inductive hypothesis.
We must prove the recursion definition. That is we must prove II(a) and II(b):
(a) $k + 3 \in S$
(b) $k - 3 \in S$
_Case (a):_
By the inductive hypothesis, $k$ is divisible by $3$. By the definition of
divisibility, it follows that:
$$ k = 3p $$
for some integer $p$.
By substitution:
$$ k + 3 = 3p + 3 $$
$$ = 3(p + 1) $$
Thus, by the sum of integers and by the definition of divisibility, $k + 3$ is
divisible by $3$.
_Case (b):_
By the inductive hypothesis, $k$ is divisible by $3$. By the definition of
divisibility, it follows that:
$$ k = 3p $$
for some integer $p$.
By substitution:
$$ k - 3 = 3p - 3 $$
$$ = 3(p - 1) $$
Thus, by the sum of integers and by the definition of divisibility, $k - 3$ is
divisible by $3$.
In both cases, the integers obtained are divisible by $3$. Therefore, the
recursion definition is true, which is what was to be shown.
_Conclusion:_
Since there are no integers in $S$ other than those obtained from the base and
recursion definitions for $S$, we conclude that every integer in $S$ is
divisible by $3$.
14. Is the string _MU_ in the _MIU_-system? Use structural induction to prove
your answer.
Omitted.
15. Determine whether either of the following parenthesis configuration is in
the set $c$ defined in Example 5.9.2. Use structural induction to prove your
answers.
a. $()(()$
Omitted.
b. $(()()))(()$
Omitted.
16. Give a recursive definition for the set of all strings of $0$'s and $1$'s
that have the same number of $0$'s and $1$'s.

15
chapter_5/test_yourself.md
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@ -262,16 +262,31 @@ Page 397
1. The base for a recursive definition of a set is _____.
a statement that certain objects belong to the set
2. The recursion for a recursive definition of a set is _____.
a collection of rules indicating how to form new set objects from those already
known to be in the set
3. The restriction for a recursive definition of a set is _____.
a statement that no objects belong to the set other than those coming from the
base and the recursion
4. One way to show that a given element is in a recursively defined set is to
start with an element or elements in the _____ and apply the rules from the
_____ until you obtain the given element.
base; recursion
5. To use structural induction to prove that every element in a recursively
defined set $S$ satisfies a certain property, you show that _____ and that,
for each rule in the recursion, if _____ then _____.
each object in the base satisfies the property; the rule is applied to the
objects in the base; the objects defined by the rule also satisfy the property
6. A function is said to be defined recursively if, and only if, _____.
its rule of definition refers to itself

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leftoff.txt
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@ -1 +1 @@
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