diff --git a/chapter_2/exercises.md b/chapter_2/exercises.md index 36038f7..237a9bc 100644 --- a/chapter_2/exercises.md +++ b/chapter_2/exercises.md @@ -16,13 +16,19 @@ All integers are rational. Therefore, the number $1$ is rational. +If $p$, then $q$. + +$p$ + +Therefore, $q$ + b. If all algebraic expressions can be written in prefix notation ,then -______. +"$(a^2 + 2b)(a^2 - b)$ can be written in prefix notation.". -______. +"All algebraic expressions can be written in prefix notation". Therefore, $(a + 2b)(a^2 - b)$ can be written in prefix notation. @@ -36,10 +42,20 @@ This program does not contain an error. Therefore, it is not the case that all computer programs contain errors. +If $p$, then $\neg q$. + +$\neg q$. + +Therefore $\neg p$. + b. If ______, then ______. +"2 is odd" + +"all prime numbers are odd." + 2 is not odd. Therefore, it is not the case that all prime numbers are odd. @@ -54,14 +70,24 @@ This number is not even. Therefore, this number is odd. +Either $p$ or $q$. + +$\neg p$ + +Therefore $q$ + b. ______ or logic is confusing. +"My mind is shot" + My mind is not shot. Therefore, ______. +"the logic is confusing." + 4. a. @@ -73,6 +99,12 @@ If the computer generates an error message, then the program will not run. Therefore, if the program syntax is faulty, then the program will not run. +If $p$, then $q$. + +If $q$ then $\neg r$. + +Therefore, if $p$, then $\neg r$. + b. If this simple graph ______, then it is complete. @@ -81,16 +113,30 @@ If this graph ______, then any two of its vertices can be joined by a path. Therefore, if this simple graph has 4 vertices and 6 edges, then ______. +"has 4 vertices" + +"has 6 edges" + +"any two of its vertices can be joined by a path." + 5. Indicate which of the following sentences are statements. a. 1,024 is the smallest four-digit number that is a perfect square. +This is a statement (a true one). + b. She is a mathematics major. +This is a statement. + c. $128 = 2^6$ +This is a statement. + d. $x = 2^6$ +This is not a statement. + Write the statements in 6-9 in symbolic form using the symbols $\neg$, $\wedge$, $\vee$ and the indicated letters to represent component statements. @@ -98,60 +144,129 @@ $\vee$ and the indicated letters to represent component statements. a. Stocks are increasing but interest rates are steady. +$$ s \wedge i $$ + b. Neither are stocks increasing nor are interest rates steady. +$$ \neg s \wedge \neg i $$ + 7. Juan is a math major but not a computer science major. ($m = $ "Juan is a math major," $c = $ "Juan is a computer science major") +$$ m \wedge \neg c $$ + 8. Let $h = $ "John is healthy," $w = $ "John is wealthy," and $s = $ "John is wise." a. John is healthy and wealthy but not wise. +$$ (h \wedge w \wedge) \neg s $$ + b. John is not wealthy but he is healthy and wise. +$$ \neg w \wedge (h \wedge s) $$ + c. John is neither healthy, wealthy, nor wise. +$$ (\neg h \wedge \neg w) \wedge \neg s $$ + d. John is neither wealthy nor wise, but he is healthy. +$$ (\neg w \wedge \neg s) \wedge h $$ + e. John is wealthy, but he is not both healthy and wise. +$$ h \wedge (\neg h \neg s) $$ + 9. Let $p = $ "$x > 5$," $q = $ "$x = 5$," and $r = $ "$10 > x$." a. $x \geq 5$ +$$ p \ vee q $$ + b. $10 > x > 5$ +$$ r \wedge p $$ + c. $10 > x \geq 5$ +$$ r \wedge (p \vee q) $$ + 10. Let $p$ be the statement "DATAENDFLAG is off," $q$ the statement "ERROR equals 0," and $r$ the statement "SUM is less than 1,000." Express the following sentences in symbolic notation. a. DATAENDFLAG is off, ERROR equals 0, and SUM is less than 1,000. +$$ p \wedge q \wedge r $$ + b. DATAENDFLAG is off but ERROR is not equal to 0. +$$ p \wedge \neg q $$ + c. DATAENDFLAG is off; however, ERROR is not 0 or SUM is greater than or equal to 1,000. +$$ p \wedge (\neg q \vee \neg r) $$ + e. Either DATAENDFLAG is on or it is the case that both ERROR equals 0 and SUM is less than 1,000. +$$ \neg p \vee (q \wedge r) $$ + 11. In the following sentence, is the word _or_ used in its inclusive or exclusive sense? A team wins the playoffs if it wins two games in a row or a total of three games. +This is an inclusive or, as it is possible for the team to win two games in a +row and a total of three games. + Write truth tables for the statement forms 12-15. 12. $\neg p \wedge q$ +| $p$ | $q$ | $\neg p$ | $\neg p \wedge q$ | +| --- | --- | -------- | ----------------- | +| T | T | F | F | +| T | F | F | F | +| F | T | T | T | +| F | F | T | F | + 13. $\neg (p \wedge q) \vee (p \vee q)$ +| $p$ | $q$ | $(p \wedge q)$ | $\neg (p \wedge q)$ | $(p \vee q)$ | $\neg (p \wedge q) \vee (p \vee q)$ | +| --- | --- | -------------- | ------------------- | ------------ | ----------------------------------- | +| T | T | T | F | T | T | +| T | F | F | T | T | T | +| F | T | F | T | T | T | +| F | F | F | T | F | T | + 14. $p \wedge (q \wedge r)$ +| $p$ | $q$ | $r$ | $(q \wedge r)$ | $p \wedge (q \wedge r)$ | +| --- | --- | --- | -------------- | ----------------------- | +| T | T | T | T | T | +| T | T | F | F | F | +| T | F | T | F | F | +| T | F | F | F | F | +| F | T | T | T | F | +| F | T | F | F | F | +| F | F | T | F | F | +| F | F | F | F | F | + 15. $p \wedge (\neg q \vee r)$ +| $p$ | $q$ | $r$ | $(q \vee r)$ | $\neg (q \vee r)$ | $p \wedge (\neg q \vee r)$ | +| --- | --- | --- | ------------ | ----------------- | -------------------------- | +| T | T | T | T | F | F | +| T | T | F | T | F | F | +| T | F | T | T | F | F | +| T | F | F | F | T | T | +| F | T | T | T | F | F | +| F | T | F | T | F | F | +| F | F | T | T | F | F | +| F | F | F | F | T | F | + Determine whether the statement forms in 16-24 are logically equivalent. In each case, construct a truth table and include a sentence justifying your answer. Your sentence should show that you understand the meaning of logical @@ -159,37 +274,152 @@ equivalence. 16. $p \vee (p \wedge q) \text{ and } p$ +| $p$ | $q$ | $(p \wedge q)$ | $p \vee (p \wedge q)$ | +| --- | --- | -------------- | --------------------- | +| T | T | T | T | +| T | F | F | T | +| F | T | F | F | +| F | F | F | F | + +As the columns for both $p$ and $p \vee (p \wedge q)$ have the same truth +values, they can be said to be equivalent. This proves one of the absorption +laws. + 17. $\neg (p \wedge q) \text{ and } \neg p \wedge \neg q$ +| $p$ | $q$ | $\neg p$ | $\neg q$ | $(p \wedge q)$ | $\neg (p \wedge q)$ | $\neg p \wedge \neg q$ | +| --- | --- | -------- | -------- | -------------- | ------------------- | ---------------------- | +| T | T | F | F | T | F | F | +| T | F | F | T | F | T | F | +| F | T | T | F | F | T | F | +| F | F | T | T | F | T | T | + +No $\neg (p \wedge q) \cancel{\equiv} \neg p \wedge \neg q$, as the columns for +both $\neg (p \wedge q)$ and $\neg p \wedge \neg q$ do not have the same truth +values. + 18. $p \vee \mathbf{t} \text{ and } \mathbf{t}$ -19. $p \wedge \mathbf{t} \text{ and } $ +| $p$ | $\mathbf{t}$ | $p \vee \mathbf{t}$ | +| --- | ------------ | ------------------- | +| T | T | T | +| F | T | T | + +Yes, $p \vee \mathbf{t} \equiv \mathbf{t}$, proving one of the universal bound +laws. + +19. $p \wedge \mathbf{t} \text{ and } p$ + +| $p$ | $\mathbf{t}$ | $p \wedge \mathbf{t}$ | +| --- | ------------ | --------------------- | +| T | T | T | +| F | T | F | + +Yes, $p \wedge \mathbf{t} \equiv p$, proving one of the identity laws. 20. $p \wedge \mathbf{c} \text{ and } p \vee \mathbf{c}$ +| $p$ | $\mathbf{c}$ | $p \wedge \mathbf{c}$ | $p \vee \mathbf{c}$ | +| --- | ------------ | --------------------- | ------------------- | +| T | F | F | T | +| F | F | F | F | + +No, $p \wedge \mathbf{c} \cancel{\equiv} p \vee \mathbf{c}$, as their two +column's truth values are not equal. + 21. $(p \wedge q) \wedge r \text{ and } p \wedge (q \wedge r)$ +| $p$ | $q$ | $r$ | $(p \wedge q)$ | $(q \wedge r)$ | $(p \wedge q) \wedge r$ | $p \wedge (q \wedge r)$ | +| --- | --- | --- | -------------- | -------------- | ----------------------- | ----------------------- | +| T | T | T | T | T | T | T | +| T | T | F | T | F | F | F | +| T | F | T | F | F | F | F | +| T | F | F | F | F | F | F | +| F | T | T | F | T | F | F | +| F | T | F | F | F | F | F | +| F | F | T | F | F | F | F | +| F | F | F | F | F | F | F | + +Yes, $(p \wedge q) \wedge r \equiv p \wedge (q \wedge r)$, proving one of the +associative laws. + 22. $p \wedge (q \vee r) \text{ and } (p \wedge q) \vee (p \wedge r)$ +| $p$ | $q$ | $r$ | $(q \vee r)$ | $(p \wedge q)$ | $(p \wedge r)$ | $p \wedge (q \vee r)$ | $(p \wedge q) \vee (p \wedge r)$ | +| --- | --- | --- | ------------ | -------------- | -------------- | --------------------- | -------------------------------- | +| T | T | T | T | T | T | T | T | +| T | T | F | T | T | F | T | T | +| T | F | T | T | F | T | T | T | +| T | F | F | F | F | F | F | F | +| F | T | T | T | F | F | F | F | +| F | T | F | T | F | F | F | F | +| F | F | T | T | F | F | F | F | +| F | F | F | F | F | F | F | F | + +Yes, $p \wedge (q \vee r) \equiv (p \wedge q) \vee (p \wedge r)$, proving one of +the distributive laws. + 23. $(p \wedge q) \vee r \text{ and } p \wedge (q \vee r)$ +| $p$ | $q$ | $r$ | $(p \wedge q)$ | $(q \vee r)$ | $(p \wedge q) \vee r$ | $p \wedge (q \vee r)$ | +| --- | --- | --- | -------------- | ------------ | --------------------- | --------------------- | +| T | T | T | T | T | T | T | +| T | T | F | T | T | T | T | +| T | F | T | F | T | T | T | +| T | F | F | F | F | F | F | +| F | T | T | F | T | T | F | +| F | T | F | F | T | F | F | +| F | F | T | F | T | T | F | +| F | F | F | F | F | F | F | + +No, $(p \wedge q) \vee r \cancel{\equiv} p \wedge (q \vee r)$, as their columns +in the truth table do not have the same truth values. + 24. $(p \vee q) \vee (p \wedge r) \text{ and } (p \vee q) \wedge r$ +| $p$ | $q$ | $r$ | $(p \vee q)$ | $(p \wedge r)$ | $(p \vee q) \vee (p \wedge r)$ | $(p \vee q) \wedge r$ | +| --- | --- | --- | ------------ | -------------- | ------------------------------ | --------------------- | +| T | T | T | T | T | T | T | +| T | T | F | T | F | T | F | +| T | F | T | T | T | T | T | +| T | F | F | T | F | T | F | +| F | T | T | T | F | T | T | +| F | T | F | T | F | T | F | +| F | F | T | F | F | F | F | +| F | F | F | F | F | F | F | + +No, $(p \vee q) \vee (p \wedge r) \cancel{\equiv} (p \vee q) \wedge r$, as their +two columns in the truth table do not have the same truth values. + Use De Morgan's laws to write negations for the statements in 25-30. 25. Hal is a math major and Hal's sister is a computer science major. +Hal is not a math major or Hal's sister is not a computer science major. + 26. Sam is an orange belt and Kate is a red belt. +Sam is not an orange belt or Kate is not a red belt. + 27. The connector is loose or the machine is unplugged. +The connector is not loose and the machine is not unplugged. + 28. The train is late or my watch is fast. +The train is not late and my watch is not fast. + 29. This computer program has a logical error in the first ten lines or it is being run with an incomplete data set. +This computer program does not have a logical error in the first ten lines and +it is not being run with an incomplete data set. + 30. The dollar is at an all-time high and the stock market is at a record low. +The dollar is not at an all-time high or the stock market is not at a record +low. + 31. Let $s$ be a string of length 2 with characters from $\{0, 1, 2\}$, and define statements $a$, $b$, $c$, and $d$ as follows: @@ -205,61 +435,154 @@ Describe the set of all strings for which each of the following is true. a. $(a \vee b) \wedge (c \vee d)$ +This is the full given set $\{0, 1, 2\}$ as the first letter could be a 0 or 1 +and the second letter could be a 1 or 2. + b. $(\neg(a \vee b)) \wedge (c \vee d)$ +This is the set $\{1, 2\}$, as the first character is neither 0 nor 1, but the +second character is either 1 or 2. + c. $((\neg a) \vee b) \wedge (c \vee (\neg d))$ +This only has the set $\{1\}$, as the first condition says the first character +can either be not 0 or 1, while the second character can be either 1 or not 2. + Assume $x$ is a particular real number and use De Morgan's laws to write negations for the statements in 32-37. 32. $-2 < x < 7$ +$$ -2 \geq x \text{ or } x \geq 7 $$ + 33. $-10 < x < 2$ +$$ -10 \geq x \text{ or } x \geq 2 $$ + 34. $x < 2 \text{ or } x > 5$ +$$ 2 \leq x \leq 5$$ + 35. $x \leq -1 \text{ or } x > 1$ +$$ -1 < x \leq 1 $$ + 36. $1 > x \geq -3$ +$$ 1 \geq x \text{ or } x < -3 $$ + 37. $0 > x \geq -7$ +$$ 0 \leq x \text{ or } x < -7 $$ + In 38 and 39, imagine that _num_orders_ and _num_instock_ are particular values, such as might occur during execution of a computer program. Write negations for the following statements. 38. $(\text{num_orders } > 100 \text{ and } \text{num_instock } \leq 500) \text{ or } \text{num_instock } < 200$ +$$ (\text{num_orders } \leq 100 \text{ or } \text{num_instock } > 500) \text{ and } \text{num_instock } \geq 200 $$ + 39. $(\text{num_orders } < 50 \text{ and } \text{num_instock } > 300) \text{ or } (50 \leq \text{ num_orders } < 75 \text{ and } \text{num_instock} > 500)$ +$$ (\text{num_orders } \geq 50 \text{ or } \text{num_instock } \leq 300) \text{ and } (50 > \text{ num_orders } \geq 75 \text{ or } \text{num_instock} \leq 500) $$ + Use truth tables to establish which of the statement forms in 40-43 are tautologies and which are contradictions. 40. $(p \wedge q) \vee (\neg p \vee (p \wedge \neg q))$ +| $p$ | $q$ | $\neg p$ | $\neg q$ | $(p \wedge q)$ | $(p \wedge \neg q)$ | $(\neg p \vee (p \wedge \neg q))$ | $(p \wedge q) \vee (\neg p \vee (p \wedge \neg q))$ | +| --- | --- | -------- | -------- | -------------- | ------------------- | --------------------------------- | --------------------------------------------------- | +| T | T | F | F | T | F | F | T | +| T | F | F | T | F | T | T | T | +| F | T | T | F | F | F | T | T | +| F | F | T | T | F | F | T | T | + +So the statement $(p \wedge q) \vee (\neg p \vee (p \wedge \neg q))$ is a +tautology as all of its truth values are true. + 41. $(p \wedge \neg q) \wedge (\neg p \vee q)$ +| $p$ | $q$ | $\neg p$ | $\neg q$ | $(p \wedge neg q)$ | $(\neg p \vee q)$ | $(p \wedge \neg q) \wedge (\neg p \vee q)$ | +| --- | --- | -------- | -------- | ------------------ | ----------------- | ------------------------------------------ | +| T | T | F | F | F | T | F | +| T | F | F | T | T | F | F | +| F | T | T | F | F | T | F | +| F | F | T | T | F | T | F | + +So the statement $(p \wedge \neg q) \wedge (\neg p \vee q)$ is a contradiction, +as all of its truth values are false. + 42. $((\neg p \wedge q) \wedge (q \wedge r)) \wedge \neg q$ +| $p$ | $q$ | $r$ | $\neg p$ | $\neg q$ | (\neg p \wedge q) | $(q \wedge r)$ | $((\neg p \wedge q) \wedge (q \wedge r))$ | $((\neg p \wedge q) \wedge (q \wedge r)) \wedge \neg q$ | +| --- | --- | --- | -------- | -------- | ----------------- | -------------- | ----------------------------------------- | ------------------------------------------------------- | +| T | T | T | F | F | F | T | F | F | +| T | T | F | F | F | F | F | F | F | +| T | F | T | F | T | F | F | F | F | +| T | F | F | F | T | F | F | F | F | +| F | T | T | T | F | T | T | T | F | +| F | T | F | T | F | T | F | F | F | +| F | F | T | T | T | F | F | F | F | +| F | F | F | T | T | F | F | F | F | + +So the statement $((\neg p \wedge q) \wedge (q \wedge r)) \wedge \neg q$ is a +contradiction as all of its truth values are false. + 43. $(\neg p \vee q) \vee (p \wedge \neg q)$ +| $p$ | $q$ | $\neg p$ | $\neg q$ | $(\neg p \vee q)$ | $(p \wedge \neg q)$ | $(\neg p \vee q) \vee (p \wedge \neg q)$ | +| --- | --- | -------- | -------- | ----------------- | ------------------- | ---------------------------------------- | +| T | T | F | F | T | F | T | +| T | F | F | T | F | T | T | +| F | T | T | F | T | F | T | +| F | F | T | T | T | F | T | + +So the statement $(\neg p \vee q) \vee (p \wedge \neg q)$ is a tautology, as all +of the truth values are true. + 44. Recall that $a < x < b$ means that $a < x$ and $x < b$. Also $a \leq b$ means that $a < b$ or $a = b$. Find all real numbers that satisfy the following inequalities. a. $2 < x \leq 0$ +No real numbers satisfy this inequality ($x$ cannot both be greater than $2$ and +less than or equal to $0$). + b. $1 \leq x < -1$ +No real numbers satisfy this inequality ($x$ cannot both be greater than or +equal to $1$ and also less than $-1$). + 45. Determine whether the statements in (a) and (b) are logically equivalent. a. Bob is both a math and computer science major and Ann is a math major, but Ann is not both a math and computer science major. +$$ (p \wedge q \wedge r) \wedge \neg(r \wedge s) $$ + +$$ (p \wedge q \wedge r) \wedge (\neg r \vee \neg s) $$ + +$$ p \wedge q \wedge r \wedge \neg s $$ + b. It is not the case that both Bob and Ann are both math and computer science majors, but it is the case that Ann is a math major and Bob is both a math and computer science major. +$$ \neg((p \wedge r) \wedge (q \wedge s)) \wedge (r \wedge (p \wedge q)) $$ + +$$ (\neg(p \wedge r) \vee \neg(q \wedge s)) \wedge (r \wedge (p \wedge q)) $$ + +$$ (\neg p \vee \neg r) \vee (\neg q \vee \neg s) \wedge (r \wedge p \wedge q) $$ + +$$ \neg s \wedge r \wedge p \wedge q $$ + +$$ p \wedge q \wedge r \neg s $$ + +Both parts (a) and (b) are logically equivalent. + 46. Let the symbol $\oplus$ denote _exclusive or_; so $p \plus q \equiv (p \vee q) \wedge \neg(p \wedge q)$. Hence the truth table for $p \plus q$ is as follows: @@ -271,18 +594,58 @@ computer science major. | F | T | T | | F | F | F | -a. Find simpler statement forms that are logically equivalent to $p \plus p$ and -$(p \oplus p) \oplus p$. +a. Find simpler statement forms that are logically equivalent to $p \oplus p$ +and $(p \oplus p) \oplus p$. + +| $p$ | $p$ | $p \oplus p$ | $(p \oplus p) \oplus p$ | +| --- | --- | ------------ | ----------------------- | +| T | T | F | T | +| F | F | F | F | + +$$ p \oplus p \equiv \mathbf{c} $$ + +$$ (p \oplus p) \oplus p \equiv p $$ b. Is $(p \oplus q) \oplus r \equiv p \oplus (q \oplus r)$? Justify your answer. +| $p$ | $q$ | $r$ | $(p \oplus q)$ | $(q \oplus r)$ | $(p \oplus q) \oplus r$ | $p \oplus (q \oplus r)$ | +| --- | --- | --- | -------------- | -------------- | ----------------------- | ----------------------- | +| T | T | T | F | F | T | T | +| T | T | F | F | T | F | F | +| T | F | T | T | T | F | F | +| T | F | F | T | F | T | T | +| F | T | T | T | F | F | F | +| F | T | F | T | T | T | T | +| F | F | T | F | T | T | T | +| F | F | F | F | F | F | F | + +They are equivalent, $(p \oplus q) \oplus r \equiv p \oplus (q \oplus r)$, as +their columns in the truth table show they have the same truth values. + c. Is $(p \oplus q) \wedge r \equiv (p \wedge r) \oplus (q \wedge r)$? Justify your answer. +| $p$ | $q$ | $r$ | $(p \oplus q)$ | $(p \wedge r)$ | $(q \wedge r)$ | $(p \oplus q) \wedge r$ | $(p \wedge r) \oplus (q \wedge r)$ | +| --- | --- | --- | -------------- | -------------- | -------------- | ----------------------- | ---------------------------------- | +| T | T | T | F | T | T | F | F | +| T | T | F | F | F | F | F | F | +| T | F | T | T | T | F | T | T | +| T | F | F | T | F | F | F | F | +| F | T | T | T | F | T | T | T | +| F | T | F | T | F | F | F | F | +| F | F | T | F | F | F | F | F | +| F | F | F | F | F | F | F | F | + +They are equivalent, +$(p \oplus q) \wedge r \equiv (p \wedge r) \oplus (q \wedge r)$, as their +columns in the truth table show they have the same truth values. + 47. In logic and in standard English, a double negative is equivalent to a positive. There is one fairly common English usage in which a "double positive" is equivalent to a negative. What is it? Can you think of others? +"Yeah, yeah" (see page 902) + In 48 and 49 below, a logical equivalence is derived from Theorem 2.1.1. Supply a reason for each step. @@ -290,12 +653,20 @@ a reason for each step. $$ p \vee \neg q \vee (p \wedge q) \equiv p \wedge (\neg q \vee q) \text{ by (a)} $$ +by distributive law + $$ \quad \equiv p \wedge (q \vee \neg q) \text{ by (b)} $$ +by commutative law + $$ \quad \equiv p \wedge \mathbf{t} \text{ by (c)} $$ +by universal bound law + $$ \quad \equiv p \text{ by (d)} $$ +by identity law + Therefore, $(p \wedge \neg q) \vee (p \wedge q) \equiv p$. 49. @@ -304,12 +675,20 @@ $$ (p \vee \neg q) \wedge (\neg p \vee \neg q) $$ $$ \quad \equiv (\neg q \vee p) \wedge (\neg q \vee \neg p) \text{ by (a)} $$ +by commutative law + $$ \quad \equiv \neg q \vee (p \wedge \neg p) \text{ by (b)} $$ +by distributive law + $$ \quad \equiv q \vee \mathbf{c} \text{ by (c)} $$ +by universal bound law + $$ \quad \equiv \neg q \text{ by (d)} $$ +by negation law + Therefore, $(p \vee \neg q) \wedge (\neg p \vee \neg q) \equiv \neg q$. Use Theorem 2.1.1 to verify the logical equivalences in 50-54. Supply a reason @@ -317,10 +696,64 @@ for each step. 50. $(p \wedge \neg q) \vee p \equiv p$ +$$ (p \wedge \neg q) \vee p \equiv p $$ + +$$ p \vee (p \wedge \neg q) \equiv p \text{ by communative law for } \vee $$ + +$$ \equiv p \text{ by the absorption law for } \vee \text{ with } \neg q \text{ replacing } q $$ + 51. $p \wedge (\neg q \vee p) \equiv p$ +$$ p \wedge (\neg q \vee p) \equiv p $$ + +$$ (p \wedge \neg q) \vee (p \wedge p) \equiv p \text{ by distributive law for } \wedge $$ + +$$ (p \wedge \neg q) \vee p \equiv p \text{ by idempotent law for } \wedge $$ + +$$ p \vee (p \wedge \neg q) \equiv p \text{ by commutative law for } \vee $$ + +$$ \equiv p \text{ by absorption law for } \vee \text{ with } \neg q \text{ replacing } q $$ + 52. $\neg(p \vee \neg q) \vee (\neg p \wedge \neg q) \equiv \neg p$ +$$ \neg(p \vee \neg q) \vee (\neg p \wedge \neg q) \equiv \neg p $$ + +$$ (\neg p \wedge q) \vee (\neg p \wedge \neg q) \equiv \neg p \text{ by De Morgan's law for } \vee $$ + +$$ \neg p \wedge (q \vee \neg q) \equiv \neg p \text{ by distributive law for } \wedge $$ + +$$ \neg p \wedge \mathbf{t} \equiv \neg p \text{ by negation law for } \vee $$ + +$$ \neg p \equiv \neg p \text{ by identity law for } \wedge $$ + 53. $\neg((\neg p \wedge q) \vee (\neg p \wedge \neg q)) \vee (p \wedge q) \equiv p$ +$$ \neg((\neg p \wedge q) \vee (\neg p \wedge \neg q)) \vee (p \wedge q) \equiv p $$ + +$$ (\neg(\neg p \wedge q) \wedge \neg(\neg p \wedge \neg q)) \vee (p \wedge q) \equiv p \text{ by De Morgan's law for } \vee $$ + +$$ ((p \vee \neg q) \wedge (p \vee q)) \vee (p \wedge q) \equiv p \text{ by De Morgan's law for } \wedge $$ + +$$ (p \vee (\neg q \wedge q)) \vee (p \wedge q) \equiv p \text{ by distributive law for } \vee $$ + +$$ (p \vee \mathbf{c}) \vee (p \wedge q) \equiv p \text{ by negation law for } \wedge $$ + +$$ p \vee (p \wedge q) \equiv p \text{ by identity law for } \vee $$ + +$$ p \equiv p \text{ by absorption law for } \vee $$ + 54. $(p \wedge (\neg(\neg p \vee q))) \vee (p \wedge q) \equiv p$ + +$$ (p \wedge (\neg(\neg p \vee q))) \vee (p \wedge q) \equiv p $$ + +$$ (p \wedge (p \wedge \neg q)) \vee (p \wedge q) \equiv p \text{ by De Morgan's law for } \vee $$ + +$$ ((p \wedge p) \wedge \neg q) \vee (p \wedge q) \equiv p \text{ by associative law for } \wedge $$ + +$$ (p \wedge \neg q) \vee (p \wedge q) \equiv p \text{ by idempotent law for } \wedge $$ + +$$ p \wedge (\neg q \vee q) \equiv p \text{ by distributive law for } \wedge $$ + +$$ p \wedge \mathbf{t} \equiv p \text{ by negation law for } \vee $$ + +$$ p \equiv p \text{ by identity law for } \wedge $$ diff --git a/chapter_2/notes.md b/chapter_2/notes.md index bb8eefd..a151d5f 100644 --- a/chapter_2/notes.md +++ b/chapter_2/notes.md @@ -91,3 +91,76 @@ A **contradiction** is a statement form that is always false regardless of the truth values of the individual statements substituted for its statement variables. A statement whose form is a contradiction is a **contradictory statement**. + +--- + +Page 72 + +**Theorem 2.1.1 Logical Equivalences** + +Given any statement variables $p$, $q$, and $r$, a tautology $\mathbf{t}$ and a +contradiction $\mathbf{c}$, the following logical equivalences hold. + +1. _Communitative laws:_ + +$$ p \wedge q \equiv q \wedge p $$ + +$$ p \vee q \equiv q \vee p $$ + +2. _Associative laws:_ + +$$ (p \wedge q) \wedge r \equiv p \wedge (q \wedge r) $$ + +$$ (p \vee q) \vee r \equiv p \vee (q \vee r) $$ + +3. _Distributive laws:_ + +$$ p \wedge (q \vee r) \equiv (p \wedge q) \vee (p \wedge r) $$ + +$$ p \vee (q \wedge r) \equiv (p \vee q) \wedge (p \vee r) $$ + +4. _Identity laws:_ + +$$ p \wedge \mathbf{t} \equiv p $$ + +$$ p \vee \mathbf{c} \equiv p $$ + +5. _Negation laws:_ + +$$ p \vee \neg p \equiv \mathbf{t} $$ + +$$ p \wedge \neg p \equiv \mathbf{c} $$ + +6. _Double negative law:_ + +$$ \neg(\neg p) \equiv p $$ + +6. _Idempotent laws:_ + +$$ p \wedge p \equiv p $$ + +$$ p \vee p \equiv p $$ + +8. _Universal bound laws:_Double + +$$ p \vee \mathbf{t} \equiv \mathbf{t} $$ + +$$ p \wedge \mathbf{c} \equiv \mathbf{c} $$ + +9. _De Morgan's laws:_ + +$$ \neg (p \wedge q) \equiv \neg p \vee \neg q $$ + +$$ \neg (p \vee q) \equiv \neg p \wedge \neg q $$ + +10 _Absorption laws:_ + +$$ p \vee (p \wedge q) \equiv p $$ + +$$ p \wedge (p \vee q) \equiv p $$ + +11. _Negations of $\mathbf{t}$ and $\mathbf{c}$:_ + +$$ \neg \mathbf{t} \equiv \mathbf{c} $$ + +$$ \neg \mathbf{c} \equiv \mathbf{t} $$