From 0b378712150a52a2db631298ee1b1957cfd3c138 Mon Sep 17 00:00:00 2001 From: tomit4 Date: Wed, 22 Jul 2026 16:44:17 -0700 Subject: [PATCH] :construction: Setup for 6.4 --- chapter_6/exercises.md | 258 +++++++++++++++++++++++++++++++++++++ chapter_6/notes.md | 210 ++++++++++++++++++++++++++++++ chapter_6/test_yourself.md | 20 +++ 3 files changed, 488 insertions(+) diff --git a/chapter_6/exercises.md b/chapter_6/exercises.md index 70ea5be..e182cc5 100644 --- a/chapter_6/exercises.md +++ b/chapter_6/exercises.md @@ -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 $A$, $\mathscr{P}(A) \nsubseteq A$. diff --git a/chapter_6/notes.md b/chapter_6/notes.md index 88310e1..2b11fce 100644 --- a/chapter_6/notes.md +++ b/chapter_6/notes.md @@ -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.]_ diff --git a/chapter_6/test_yourself.md b/chapter_6/test_yourself.md index 36ecd69..5a8bf1f 100644 --- a/chapter_6/test_yourself.md +++ b/chapter_6/test_yourself.md @@ -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: _____.