diff --git a/appendix_b.txt b/appendix_b.txt index 03b7719..0b8de63 100644 --- a/appendix_b.txt +++ b/appendix_b.txt @@ -1 +1 @@ -964 +966 diff --git a/chapter_6/exercises.md b/chapter_6/exercises.md index 7e12add..c91349f 100644 --- a/chapter_6/exercises.md +++ b/chapter_6/exercises.md @@ -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. diff --git a/leftoff.txt b/leftoff.txt index ce9cd49..5379c47 100644 --- a/leftoff.txt +++ b/leftoff.txt @@ -1 +1 @@ -437 +448