2066 lines
77 KiB
Markdown
2066 lines
77 KiB
Markdown
**Exercise Set 2.1**
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Page 74
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In each of 1-4 represent the common form of each argument using letters to stand
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for component sentences, and fill in the blanks so that the argument in part (b)
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has the same logical form as the argument in part (a).
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1.
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a.
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If all integers are rational, then the number $1$ is rational.
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All integers are rational.
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Therefore, the number $1$ is rational.
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If $p$, then $q$.
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$p$
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Therefore, $q$
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b.
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If all algebraic expressions can be written in prefix notation ,then
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"$(a^2 + 2b)(a^2 - b)$ can be written in prefix notation.".
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"All algebraic expressions can be written in prefix notation".
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Therefore, $(a + 2b)(a^2 - b)$ can be written in prefix notation.
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2.
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a.
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If all computer programs contain errors, then this program contains an error.
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This program does not contain an error.
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Therefore, it is not the case that all computer programs contain errors.
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If $p$, then $\neg q$.
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$\neg q$.
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Therefore $\neg p$.
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b.
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If ______, then ______.
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"2 is odd"
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"all prime numbers are odd."
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2 is not odd.
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Therefore, it is not the case that all prime numbers are odd.
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3.
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a.
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This number is even or this number is odd.
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This number is not even.
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Therefore, this number is odd.
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Either $p$ or $q$.
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$\neg p$
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Therefore $q$
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b.
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______ or logic is confusing.
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"My mind is shot"
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My mind is not shot.
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Therefore, ______.
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"the logic is confusing."
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4.
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a.
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If the program syntax is faulty, then the computer will generate an error
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message.
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If the computer generates an error message, then the program will not run.
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Therefore, if the program syntax is faulty, then the program will not run.
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If $p$, then $q$.
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If $q$ then $\neg r$.
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Therefore, if $p$, then $\neg r$.
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b.
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If this simple graph ______, then it is complete.
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If this graph ______, then any two of its vertices can be joined by a path.
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Therefore, if this simple graph has 4 vertices and 6 edges, then ______.
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"has 4 vertices"
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"has 6 edges"
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"any two of its vertices can be joined by a path."
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5. Indicate which of the following sentences are statements.
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a. 1,024 is the smallest four-digit number that is a perfect square.
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This is a statement (a true one).
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b. She is a mathematics major.
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This is a statement.
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c. $128 = 2^6$
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This is a statement.
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d. $x = 2^6$
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This is not a statement.
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Write the statements in 6-9 in symbolic form using the symbols $\neg$, $\wedge$,
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$\vee$ and the indicated letters to represent component statements.
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6. Let $s = $ "stocks are increasing" and $i = $ "interest rates are steady."
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a. Stocks are increasing but interest rates are steady.
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$$ s \wedge i $$
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b. Neither are stocks increasing nor are interest rates steady.
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$$ \neg s \wedge \neg i $$
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7. Juan is a math major but not a computer science major. ($m = $ "Juan is a
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math major," $c = $ "Juan is a computer science major")
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$$ m \wedge \neg c $$
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8. Let $h = $ "John is healthy," $w = $ "John is wealthy," and $s = $ "John is
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wise."
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a. John is healthy and wealthy but not wise.
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$$ (h \wedge w \wedge) \neg s $$
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b. John is not wealthy but he is healthy and wise.
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$$ \neg w \wedge (h \wedge s) $$
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c. John is neither healthy, wealthy, nor wise.
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$$ (\neg h \wedge \neg w) \wedge \neg s $$
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d. John is neither wealthy nor wise, but he is healthy.
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$$ (\neg w \wedge \neg s) \wedge h $$
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e. John is wealthy, but he is not both healthy and wise.
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$$ h \wedge (\neg h \neg s) $$
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9. Let $p = $ "$x > 5$," $q = $ "$x = 5$," and $r = $ "$10 > x$."
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a. $x \geq 5$
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$$ p \ vee q $$
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b. $10 > x > 5$
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$$ r \wedge p $$
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c. $10 > x \geq 5$
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$$ r \wedge (p \vee q) $$
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10. Let $p$ be the statement "DATAENDFLAG is off," $q$ the statement "ERROR
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equals 0," and $r$ the statement "SUM is less than 1,000." Express the
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following sentences in symbolic notation.
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a. DATAENDFLAG is off, ERROR equals 0, and SUM is less than 1,000.
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$$ p \wedge q \wedge r $$
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b. DATAENDFLAG is off but ERROR is not equal to 0.
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$$ p \wedge \neg q $$
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c. DATAENDFLAG is off; however, ERROR is not 0 or SUM is greater than or equal
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to 1,000.
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$$ p \wedge (\neg q \vee \neg r) $$
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e. Either DATAENDFLAG is on or it is the case that both ERROR equals 0 and SUM
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is less than 1,000.
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$$ \neg p \vee (q \wedge r) $$
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11. In the following sentence, is the word _or_ used in its inclusive or
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exclusive sense? A team wins the playoffs if it wins two games in a row or a
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total of three games.
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This is an inclusive or, as it is possible for the team to win two games in a
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row and a total of three games.
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Write truth tables for the statement forms 12-15.
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12. $\neg p \wedge q$
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| $p$ | $q$ | $\neg p$ | $\neg p \wedge q$ |
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| --- | --- | -------- | ----------------- |
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| T | T | F | F |
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| T | F | F | F |
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| F | T | T | T |
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| F | F | T | F |
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13. $\neg (p \wedge q) \vee (p \vee q)$
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| $p$ | $q$ | $(p \wedge q)$ | $\neg (p \wedge q)$ | $(p \vee q)$ | $\neg (p \wedge q) \vee (p \vee q)$ |
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| --- | --- | -------------- | ------------------- | ------------ | ----------------------------------- |
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| T | T | T | F | T | T |
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| T | F | F | T | T | T |
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| F | T | F | T | T | T |
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| F | F | F | T | F | T |
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14. $p \wedge (q \wedge r)$
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| $p$ | $q$ | $r$ | $(q \wedge r)$ | $p \wedge (q \wedge r)$ |
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| --- | --- | --- | -------------- | ----------------------- |
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| T | T | T | T | T |
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| T | T | F | F | F |
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| T | F | T | F | F |
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| T | F | F | F | F |
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| F | T | T | T | F |
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| F | T | F | F | F |
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| F | F | T | F | F |
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| F | F | F | F | F |
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15. $p \wedge (\neg q \vee r)$
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| $p$ | $q$ | $r$ | $(q \vee r)$ | $\neg (q \vee r)$ | $p \wedge (\neg q \vee r)$ |
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| --- | --- | --- | ------------ | ----------------- | -------------------------- |
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| T | T | T | T | F | F |
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| T | T | F | T | F | F |
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| T | F | T | T | F | F |
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| T | F | F | F | T | T |
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| F | T | T | T | F | F |
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| F | T | F | T | F | F |
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| F | F | T | T | F | F |
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| F | F | F | F | T | F |
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Determine whether the statement forms in 16-24 are logically equivalent. In each
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case, construct a truth table and include a sentence justifying your answer.
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Your sentence should show that you understand the meaning of logical
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equivalence.
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16. $p \vee (p \wedge q) \text{ and } p$
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| $p$ | $q$ | $(p \wedge q)$ | $p \vee (p \wedge q)$ |
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| --- | --- | -------------- | --------------------- |
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| T | T | T | T |
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| T | F | F | T |
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| F | T | F | F |
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| F | F | F | F |
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As the columns for both $p$ and $p \vee (p \wedge q)$ have the same truth
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values, they can be said to be equivalent. This proves one of the absorption
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laws.
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17. $\neg (p \wedge q) \text{ and } \neg p \wedge \neg q$
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| $p$ | $q$ | $\neg p$ | $\neg q$ | $(p \wedge q)$ | $\neg (p \wedge q)$ | $\neg p \wedge \neg q$ |
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| --- | --- | -------- | -------- | -------------- | ------------------- | ---------------------- |
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| T | T | F | F | T | F | F |
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| T | F | F | T | F | T | F |
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| F | T | T | F | F | T | F |
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| F | F | T | T | F | T | T |
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No $\neg (p \wedge q) \cancel{\equiv} \neg p \wedge \neg q$, as the columns for
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both $\neg (p \wedge q)$ and $\neg p \wedge \neg q$ do not have the same truth
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values.
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18. $p \vee \mathbf{t} \text{ and } \mathbf{t}$
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| $p$ | $\mathbf{t}$ | $p \vee \mathbf{t}$ |
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| --- | ------------ | ------------------- |
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| T | T | T |
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| F | T | T |
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Yes, $p \vee \mathbf{t} \equiv \mathbf{t}$, proving one of the universal bound
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laws.
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19. $p \wedge \mathbf{t} \text{ and } p$
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| $p$ | $\mathbf{t}$ | $p \wedge \mathbf{t}$ |
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| --- | ------------ | --------------------- |
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| T | T | T |
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| F | T | F |
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Yes, $p \wedge \mathbf{t} \equiv p$, proving one of the identity laws.
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20. $p \wedge \mathbf{c} \text{ and } p \vee \mathbf{c}$
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| $p$ | $\mathbf{c}$ | $p \wedge \mathbf{c}$ | $p \vee \mathbf{c}$ |
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| --- | ------------ | --------------------- | ------------------- |
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| T | F | F | T |
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| F | F | F | F |
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No, $p \wedge \mathbf{c} \cancel{\equiv} p \vee \mathbf{c}$, as their two
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column's truth values are not equal.
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21. $(p \wedge q) \wedge r \text{ and } p \wedge (q \wedge r)$
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| $p$ | $q$ | $r$ | $(p \wedge q)$ | $(q \wedge r)$ | $(p \wedge q) \wedge r$ | $p \wedge (q \wedge r)$ |
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| --- | --- | --- | -------------- | -------------- | ----------------------- | ----------------------- |
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| T | T | T | T | T | T | T |
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| T | T | F | T | F | F | F |
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| T | F | T | F | F | F | F |
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| T | F | F | F | F | F | F |
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| F | T | T | F | T | F | F |
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| F | T | F | F | F | F | F |
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| F | F | T | F | F | F | F |
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| F | F | F | F | F | F | F |
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Yes, $(p \wedge q) \wedge r \equiv p \wedge (q \wedge r)$, proving one of the
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associative laws.
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22. $p \wedge (q \vee r) \text{ and } (p \wedge q) \vee (p \wedge r)$
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| $p$ | $q$ | $r$ | $(q \vee r)$ | $(p \wedge q)$ | $(p \wedge r)$ | $p \wedge (q \vee r)$ | $(p \wedge q) \vee (p \wedge r)$ |
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| --- | --- | --- | ------------ | -------------- | -------------- | --------------------- | -------------------------------- |
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| T | T | T | T | T | T | T | T |
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| T | T | F | T | T | F | T | T |
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| T | F | T | T | F | T | T | T |
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| T | F | F | F | F | F | F | F |
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| F | T | T | T | F | F | F | F |
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| F | T | F | T | F | F | F | F |
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| F | F | T | T | F | F | F | F |
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| F | F | F | F | F | F | F | F |
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Yes, $p \wedge (q \vee r) \equiv (p \wedge q) \vee (p \wedge r)$, proving one of
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the distributive laws.
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23. $(p \wedge q) \vee r \text{ and } p \wedge (q \vee r)$
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| $p$ | $q$ | $r$ | $(p \wedge q)$ | $(q \vee r)$ | $(p \wedge q) \vee r$ | $p \wedge (q \vee r)$ |
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| --- | --- | --- | -------------- | ------------ | --------------------- | --------------------- |
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| T | T | T | T | T | T | T |
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| T | T | F | T | T | T | T |
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| T | F | T | F | T | T | T |
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| T | F | F | F | F | F | F |
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| F | T | T | F | T | T | F |
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| F | T | F | F | T | F | F |
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| F | F | T | F | T | T | F |
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| F | F | F | F | F | F | F |
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No, $(p \wedge q) \vee r \cancel{\equiv} p \wedge (q \vee r)$, as their columns
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in the truth table do not have the same truth values.
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24. $(p \vee q) \vee (p \wedge r) \text{ and } (p \vee q) \wedge r$
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| $p$ | $q$ | $r$ | $(p \vee q)$ | $(p \wedge r)$ | $(p \vee q) \vee (p \wedge r)$ | $(p \vee q) \wedge r$ |
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| --- | --- | --- | ------------ | -------------- | ------------------------------ | --------------------- |
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| T | T | T | T | T | T | T |
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| T | T | F | T | F | T | F |
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| T | F | T | T | T | T | T |
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| T | F | F | T | F | T | F |
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| F | T | T | T | F | T | T |
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| F | T | F | T | F | T | F |
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| F | F | T | F | F | F | F |
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| F | F | F | F | F | F | F |
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No, $(p \vee q) \vee (p \wedge r) \cancel{\equiv} (p \vee q) \wedge r$, as their
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two columns in the truth table do not have the same truth values.
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Use De Morgan's laws to write negations for the statements in 25-30.
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25. Hal is a math major and Hal's sister is a computer science major.
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Hal is not a math major or Hal's sister is not a computer science major.
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26. Sam is an orange belt and Kate is a red belt.
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Sam is not an orange belt or Kate is not a red belt.
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27. The connector is loose or the machine is unplugged.
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The connector is not loose and the machine is not unplugged.
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28. The train is late or my watch is fast.
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The train is not late and my watch is not fast.
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29. This computer program has a logical error in the first ten lines or it is
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being run with an incomplete data set.
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This computer program does not have a logical error in the first ten lines and
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it is not being run with an incomplete data set.
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30. The dollar is at an all-time high and the stock market is at a record low.
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The dollar is not at an all-time high or the stock market is not at a record
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low.
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31. Let $s$ be a string of length 2 with characters from $\{0, 1, 2\}$, and
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define statements $a$, $b$, $c$, and $d$ as follows:
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$a = $ "the first character of $s$ is 0"
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$b = $ "the first character of $s$ is 1"
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$c = $ "the second character of $s$ is 1"
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$c = $ "the second character of $s$ is 2".
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Describe the set of all strings for which each of the following is true.
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a. $(a \vee b) \wedge (c \vee d)$
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This is the full given set $\{0, 1, 2\}$ as the first letter could be a 0 or 1
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and the second letter could be a 1 or 2.
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b. $(\neg(a \vee b)) \wedge (c \vee d)$
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This is the set $\{1, 2\}$, as the first character is neither 0 nor 1, but the
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second character is either 1 or 2.
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c. $((\neg a) \vee b) \wedge (c \vee (\neg d))$
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This only has the set $\{1\}$, as the first condition says the first character
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can either be not 0 or 1, while the second character can be either 1 or not 2.
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Assume $x$ is a particular real number and use De Morgan's laws to write
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negations for the statements in 32-37.
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32. $-2 < x < 7$
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$$ -2 \geq x \text{ or } x \geq 7 $$
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33. $-10 < x < 2$
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$$ -10 \geq x \text{ or } x \geq 2 $$
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34. $x < 2 \text{ or } x > 5$
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$$ 2 \leq x \leq 5$$
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35. $x \leq -1 \text{ or } x > 1$
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$$ -1 < x \leq 1 $$
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36. $1 > x \geq -3$
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$$ 1 \geq x \text{ or } x < -3 $$
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37. $0 > x \geq -7$
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$$ 0 \leq x \text{ or } x < -7 $$
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In 38 and 39, imagine that _num_orders_ and _num_instock_ are particular values,
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such as might occur during execution of a computer program. Write negations for
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the following statements.
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38. $(\text{num_orders } > 100 \text{ and } \text{num_instock } \leq 500) \text{ or } \text{num_instock } < 200$
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$$ (\text{num_orders } \leq 100 \text{ or } \text{num_instock } > 500) \text{ and } \text{num_instock } \geq 200 $$
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39. $(\text{num_orders } < 50 \text{ and } \text{num_instock } > 300) \text{ or } (50 \leq \text{ num_orders } < 75 \text{ and } \text{num_instock} > 500)$
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$$ (\text{num_orders } \geq 50 \text{ or } \text{num_instock } \leq 300) \text{ and } (50 > \text{ num_orders } \geq 75 \text{ or } \text{num_instock} \leq 500) $$
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Use truth tables to establish which of the statement forms in 40-43 are
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tautologies and which are contradictions.
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40. $(p \wedge q) \vee (\neg p \vee (p \wedge \neg q))$
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| $p$ | $q$ | $\neg p$ | $\neg q$ | $(p \wedge q)$ | $(p \wedge \neg q)$ | $(\neg p \vee (p \wedge \neg q))$ | $(p \wedge q) \vee (\neg p \vee (p \wedge \neg q))$ |
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| --- | --- | -------- | -------- | -------------- | ------------------- | --------------------------------- | --------------------------------------------------- |
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| T | T | F | F | T | F | F | T |
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| T | F | F | T | F | T | T | T |
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| 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:
|
|
|
|
| $p$ | $q$ | $p \plus q$ |
|
|
| --- | --- | ----------- |
|
|
| T | T | F |
|
|
| T | F | T |
|
|
| F | T | T |
|
|
| F | F | F |
|
|
|
|
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.
|
|
|
|
48.
|
|
|
|
$$ p \vee \neg q \vee (p \wedge q) \equiv p \wedge (\neg q \vee q) \text{ by (a)} $$
|
|
|
|
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.
|
|
|
|
$$ (p \vee \neg q) \wedge (\neg p \vee \neg q) $$
|
|
|
|
$$ \quad \equiv (\neg q \vee p) \wedge (\neg q \vee \neg p) \text{ by (a)} $$
|
|
|
|
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
|
|
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. $(p \wedge (\neg(\neg p \vee q))) \vee (p \wedge q) \equiv p$
|
|
|
|
$$ (p \wedge (\neg(\neg p \vee q))) \vee (p \wedge q) \equiv p $$
|
|
|
|
$$ (p \wedge (p \wedge \neg q)) \vee (p \wedge q) \equiv p \text{ by De Morgan's law for } \vee $$
|
|
|
|
$$ ((p \wedge p) \wedge \neg q) \vee (p \wedge q) \equiv p \text{ by associative law for } \wedge $$
|
|
|
|
$$ (p \wedge \neg q) \vee (p \wedge q) \equiv p \text{ by idempotent law for } \wedge $$
|
|
|
|
$$ p \wedge (\neg q \vee q) \equiv p \text{ by distributive law for } \wedge $$
|
|
|
|
$$ p \wedge \mathbf{t} \equiv p \text{ by negation law for } \vee $$
|
|
|
|
$$ p \equiv p \text{ by identity law for } \wedge $$
|
|
|
|
---
|
|
|
|
**Exercise Set 2.2**
|
|
|
|
Page 86
|
|
|
|
Rewrite the statements in 1-4 in if-then form.
|
|
|
|
1. This loop will repeat exactly $n$ times if it does not contain a **stop** or
|
|
a **go to**.
|
|
|
|
"If this lop does not contain a **stop** or a **go to**, then it will repeat
|
|
exactly $N$ times."
|
|
|
|
2. I am on time for work if I catch the 8:05 bus.
|
|
|
|
"If I catch the 8:05 bus, then I am on time for work."
|
|
|
|
3. Freeze or I'll shoot.
|
|
|
|
"If you do not freeze, then I'll shoot."
|
|
|
|
4. Fix my ceiling or I won't pay my rent.
|
|
|
|
"If you do not fix my ceiling, then I won't pay my rent."
|
|
|
|
Construct truth tables for the statements forms in 5-11.
|
|
|
|
5. $\neg p \vee q \to \neg q$
|
|
|
|
| $p$ | $q$ | $\neg p$ | $\neg q$ | $\neg p \vee q$ | $\neg p \vee q \to \neg q$ |
|
|
| --- | --- | -------- | -------- | --------------- | -------------------------- |
|
|
| T | T | F | F | T | F |
|
|
| T | F | F | T | F | T |
|
|
| F | T | T | F | T | F |
|
|
| F | F | T | T | T | T |
|
|
|
|
6. $(p \vee q) \vee (\neg p \wedge q) \to q$
|
|
|
|
| $p$ | $q$ | $\neg p$ | $(p \vee q)$ | $(\neg p \wedge q)$ | $(p \vee q) \vee (\neg p \wedge q)$ | $(p \vee q) \vee (\neg p \wedge q) \to q$ |
|
|
| --- | --- | -------- | ------------ | ------------------- | ----------------------------------- | ----------------------------------------- |
|
|
| T | T | F | T | F | T | T |
|
|
| T | F | F | T | F | T | F |
|
|
| F | T | T | T | T | T | T |
|
|
| F | F | T | F | F | F | T |
|
|
|
|
7. $p \wedge \neg q \to r$
|
|
|
|
| $p$ | $q$ | $r$ | $\neg q$ | $p \wedge \neg q$ | $p \wedge \neg q \to r$ |
|
|
| --- | --- | --- | -------- | ----------------- | ----------------------- |
|
|
| T | T | T | F | F | T |
|
|
| T | T | F | F | F | T |
|
|
| T | F | T | T | T | T |
|
|
| T | F | F | T | T | F |
|
|
| F | T | T | F | F | T |
|
|
| F | T | F | F | F | T |
|
|
| F | F | T | T | F | T |
|
|
| F | F | F | T | F | T |
|
|
|
|
8. $\neg p \vee q \to r$
|
|
|
|
| $p$ | $q$ | $r$ | $\neg p$ | $\neg p \vee q$ | $\neg p \vee q \to r$ |
|
|
| --- | --- | --- | -------- | --------------- | --------------------- |
|
|
| T | T | T | F | T | T |
|
|
| T | T | F | F | T | F |
|
|
| T | F | T | F | F | T |
|
|
| T | F | F | F | F | T |
|
|
| F | T | T | T | T | T |
|
|
| F | T | F | T | T | F |
|
|
| F | F | T | T | T | T |
|
|
| F | F | F | T | T | F |
|
|
|
|
9. $p \wedge \neg r \leftrightarrow q \vee r$
|
|
|
|
| $p$ | $q$ | $r$ | $\neg r$ | $p \wedge \neg r$ | $q \vee r$ | $p \wedge \neg r \leftrightarrow q \vee r$ |
|
|
| --- | --- | --- | -------- | ----------------- | ---------- | ------------------------------------------ |
|
|
| T | T | T | F | F | T | F |
|
|
| T | T | F | T | T | T | T |
|
|
| T | F | T | F | F | T | F |
|
|
| T | F | F | T | T | F | F |
|
|
| F | T | T | F | F | T | F |
|
|
| F | T | F | T | F | T | F |
|
|
| F | F | T | F | F | T | F |
|
|
| F | F | F | T | F | F | T |
|
|
|
|
10. $(p \to r) \leftrightarrow (q \to r)$
|
|
|
|
| $p$ | $q$ | $r$ | $(p \to r)$ | $(q \to r)$ | $(p \to r) \leftrightarrow (q \to r)$ |
|
|
| --- | --- | --- | ----------- | ----------- | ------------------------------------- |
|
|
| T | T | T | T | T | T |
|
|
| T | T | F | F | F | T |
|
|
| T | F | T | T | T | T |
|
|
| T | F | F | F | T | F |
|
|
| F | T | T | T | T | T |
|
|
| F | T | F | T | F | F |
|
|
| F | F | T | T | T | T |
|
|
| F | F | F | T | T | T |
|
|
|
|
11. $(p \to (q \to r)) \leftrightarrow ((p \wedge q) \to r)$
|
|
|
|
| $p$ | $q$ | $r$ | $(q \to r)$ | $(p \wedge q)$ | $(p \to (q \to r))$ | $((p \wedge q) \to r)$ | $(p \to (q \to r)) \leftrightarrow ((p \wedge q) \to r)$ |
|
|
| --- | --- | --- | ----------- | -------------- | ------------------- | ---------------------- | -------------------------------------------------------- |
|
|
| T | T | T | T | T | T | T | T |
|
|
| T | T | F | F | T | F | F | T |
|
|
| T | F | T | T | F | T | T | T |
|
|
| T | F | F | T | F | T | T | T |
|
|
| F | T | T | T | F | T | T | T |
|
|
| F | T | F | F | F | T | T | T |
|
|
| F | F | T | T | F | T | T | T |
|
|
| F | F | F | T | F | T | T | T |
|
|
|
|
12. Use the logical equivalence established in Example 2.2.3,
|
|
$p \vee q \to r \equiv (p \to r) \wedge (q \to r)$, to rewrite the following
|
|
statement. (Assume that $x$ represents a fixed real number.)
|
|
|
|
If $x > 2 $ or $x < -2$, then $x^2 > 4$.
|
|
|
|
If $x > 2$, then $x^2 > 4$ and if $x < -2$, then $x^2 > 4$.
|
|
|
|
13. Use truth tables to verify the following logical equivalences. Include a few
|
|
words of explanation with your answers.
|
|
|
|
a. $p \to q \equiv \neg p \vee q$
|
|
|
|
| $p$ | $q$ | $\neg p$ | $p \to q$ | $\neg p \vee q$ |
|
|
| --- | --- | -------- | --------- | --------------- |
|
|
| T | T | F | T | T |
|
|
| T | F | F | F | F |
|
|
| F | T | T | T | T |
|
|
| F | F | T | T | T |
|
|
|
|
Both columns for $p \to q$ and $\neg p \vee q$ have the same truth values, so
|
|
they are logically equivalent.
|
|
|
|
b. $\neg(p \to q) \equiv p \wedge \neg q$
|
|
|
|
| $p$ | $q$ | $\neg q$ | $(p \to q)$ | $\neg(p \to q)$ | $p \wedge \neg q$ |
|
|
| --- | --- | -------- | ----------- | --------------- | ----------------- |
|
|
| T | T | F | T | F | F |
|
|
| T | F | T | F | T | T |
|
|
| F | T | F | T | F | F |
|
|
| F | F | T | T | F | F |
|
|
|
|
Both columns for $\neg(p \to q)$ and $p \wedge \neg q$ have the same truth
|
|
values, so they are logically equivalent.
|
|
|
|
14.
|
|
|
|
a. Show that the following statement forms are all logically equivalent:
|
|
|
|
$p \to (q \vee r)$, $(p \wedge \neg q) \to r$, and $(p \wedge \neg r) \to q$
|
|
|
|
| $p$ | $q$ | $r$ | $\neg q$ | $\neg r$ | $(q \vee r)$ | $(p \wedge \neg q)$ | $(p \wedge \neg r)$ | $p \to (q \vee r)$ | $(p \wedge \neg q) \to r$ | $(p \wedge \neg r) \to q$ |
|
|
| --- | --- | --- | -------- | -------- | ------------ | ------------------- | ------------------- | ------------------ | ------------------------- | ------------------------- |
|
|
| T | T | T | F | F | T | F | F | T | T | T |
|
|
| T | T | F | F | T | T | F | T | T | T | T |
|
|
| T | F | T | T | F | T | T | F | T | T | T |
|
|
| T | F | F | T | T | F | T | T | F | F | F |
|
|
| F | T | T | F | F | T | F | F | T | T | T |
|
|
| F | T | F | F | T | T | F | F | T | T | T |
|
|
| F | F | T | T | F | T | F | F | T | T | T |
|
|
| F | F | F | T | T | F | F | F | T | T | T |
|
|
|
|
b. Use the logical equivalences established in part (a) to rewrite the following
|
|
sentence in two different ways. (Assume that $n$ represents a fixed integer.)
|
|
|
|
If $n$ is prime, then $n$ is odd or $n$ is $2$.
|
|
|
|
"If $n$ is prime and $n$ is not odd then $n$ is $2$"
|
|
|
|
"If $n$ is prime and $n$ is not $2$, then $n$ is odd."
|
|
|
|
15. Determine whether the following statement forms are logically equivalent:
|
|
|
|
$p \to (q \to r)$ and $(p \to q) \to r$
|
|
|
|
| $p$ | $q$ | $r$ | $(q \to r)$ | $(p \to q)$ | $p \to (q \to r)$ | $(p \to q) \to r$ |
|
|
| --- | --- | --- | ----------- | ----------- | ----------------- | ----------------- |
|
|
| T | T | T | T | T | T | T |
|
|
| T | T | F | F | F | F | T |
|
|
| T | F | T | T | T | T | T |
|
|
| T | F | F | T | F | T | T |
|
|
| F | T | T | T | T | T | T |
|
|
| F | T | F | F | T | T | F |
|
|
| F | F | T | T | T | T | T |
|
|
| F | F | F | T | T | T | F |
|
|
|
|
No, they are not equivalent, $p \to (q \to r) \cancel{\equiv} (p \to q) \to r$
|
|
as their truth values are not the same.
|
|
|
|
In 16 and 17, write each of the two statements in symbolic form and determine
|
|
whether they are logically equivalent. Include a truth table and a few words of
|
|
explanation to show that you understand what it means for statements to be
|
|
logically equivalent.
|
|
|
|
16. If you paid full price, you didn't buy it at Crown Books. You didn't buy it
|
|
at Crown Books or you paid full price.
|
|
|
|
$p$ is "You paid full price"
|
|
|
|
$q$ is "You bought it at Crown Books"
|
|
|
|
$$ p \to \neg q \text{ and } \neg q \vee p $$
|
|
|
|
| $p$ | $q$ | $\neg q$ | $p \to \neg q$ | $\neg q \vee p$ |
|
|
| --- | --- | -------- | -------------- | --------------- |
|
|
| T | T | F | F | T |
|
|
| T | F | T | T | T |
|
|
| F | T | F | T | F |
|
|
| F | F | T | T | T |
|
|
|
|
No, these two statements are not logically equivalent as their truth tables do
|
|
not have the same truth values.
|
|
|
|
$$ p \to \neg q \cancel{\equiv} \neg q \vee p $$
|
|
|
|
One could not buy it at Crown Books, and not paid full price, and also one could
|
|
have paid full price and have also bought it at Crown books.
|
|
|
|
17. If $2$ is a factor of $n$ and $3$ is a factor of $n$, then $6$ is a factor
|
|
of $n$. $2$ is not a factor of $n$ or $3$ is not a factor of $n$ or $6$ is a
|
|
factor of $n$.
|
|
|
|
$p$ is "$2$ is a factor of $n$"
|
|
|
|
$q$ is "$3$ is a factor of $n$"
|
|
|
|
$r$ is "$6$ is a factor of $n$"
|
|
|
|
$$ p \wedge q \to r\text{ and } \neg p \vee \neg q \vee r $$
|
|
|
|
| $p$ | $q$ | $r$ | $\neg p$ | $\neg q$ | $p \wedge q$ | $\neg p \vee \neg q$ | $p \wedge q \to r$ | $\neg p \vee \neg q \vee r$ |
|
|
| --- | --- | --- | -------- | -------- | ------------ | -------------------- | ------------------ | --------------------------- |
|
|
| T | T | T | F | F | T | F | T | T |
|
|
| T | T | F | F | F | T | F | F | F |
|
|
| T | F | T | F | T | F | T | T | T |
|
|
| T | F | F | F | T | F | T | T | T |
|
|
| F | T | T | T | F | F | T | T | T |
|
|
| F | T | F | T | F | F | T | T | T |
|
|
| F | F | T | T | T | F | T | T | T |
|
|
| F | F | F | T | T | F | T | T | T |
|
|
|
|
Yes, they are equivalent, as their truth tables show the same truth values.
|
|
|
|
$$ p \wedge q \to r \equiv \neg p \vee \neg q \vee r $$
|
|
|
|
18. Write each of the following three statements in symbolic form and determine
|
|
which pairs are logically equivalent. Include truth tables and a few words
|
|
of explanation.
|
|
|
|
$p$ = "It walks like a duck"
|
|
|
|
$q$ = "It talks like a duck"
|
|
|
|
$r$ = "It is a duck"
|
|
|
|
If it walks like a duck and it talks like a duck, then it is a duck.
|
|
|
|
$$ p \wedge q \to r $$
|
|
|
|
Either it does not walk like a duck or it does not talk like a duck, or it is a
|
|
duck.
|
|
|
|
$$ p \vee \neg q \vee r $$
|
|
|
|
If it does not walk like a duck and it does not talk like a duck, then it is not
|
|
a duck.
|
|
|
|
$$ \neg p \wedge \neg q \to \neg r $$
|
|
|
|
| $p$ | $q$ | $r$ | $\neg p$ | $\neg q$ | $\neg r$ | $p \wedge q$ | $p \vee \neg q$ | $\neg p \wedge \neg q$ | $p \wedge q \to r$ | $p \vee \neg q \vee r$ | $\neg p \wedge \neg q \to \neg r$ |
|
|
| --- | --- | --- | -------- | -------- | -------- | ------------ | --------------- | ---------------------- | ------------------ | ---------------------- | --------------------------------- |
|
|
| T | T | T | F | F | F | T | T | F | T | T | T |
|
|
| T | T | F | F | F | T | T | T | F | F | T | T |
|
|
| T | F | T | F | T | F | F | T | F | T | T | T |
|
|
| T | F | F | F | T | T | F | T | F | T | T | T |
|
|
| F | T | T | T | F | F | F | F | F | T | T | T |
|
|
| F | T | F | T | F | T | F | F | F | T | F | T |
|
|
| F | F | T | T | T | F | F | T | T | T | T | F |
|
|
| F | F | F | T | T | T | F | T | T | T | T | T |
|
|
|
|
No, none of these statements are equivalent, as you can see in their
|
|
corresponding truth table's truth values.
|
|
|
|
$$ p \wedge q \to r \cancel{\equiv} p \vee \neg q \vee r \cancel{\equiv} \neg p \wedge \neg q \to \neg r $$
|
|
|
|
19. True or false? The negation of "If Sue is Luiz's mother, then Ali is his
|
|
cousin" is "If Sue is Luiz's mother, then Ali is not his cousin."
|
|
|
|
$p$ = "Sue is Luiz's mother"
|
|
|
|
$q$ = "Ali is his cousin"
|
|
|
|
Firs statement is:
|
|
|
|
$$ p \to q $$
|
|
|
|
Negation is:
|
|
|
|
$$ p \wedge \neg q $$
|
|
|
|
"Sue is Luiz's mother and Ali is not his cousin."
|
|
|
|
Second statement is:
|
|
|
|
$$ p \to \neg q $$
|
|
|
|
Negation is:
|
|
|
|
$$ p \to q $$
|
|
|
|
"If Sue is Luiz's mother, then Ali is his cousin."
|
|
|
|
So no, they are not equivalent:
|
|
|
|
$$ p \wedge \neg q \cancel{\equiv} p \wedge q $$
|
|
|
|
20. Write negations for each of the following statements. (Assume that all
|
|
variables represent fixed quantities or entities, as appropriate.)
|
|
|
|
a. If $P$ is a square, then $P$ is a rectangle.
|
|
|
|
$$ p \to q $$
|
|
|
|
Negation:
|
|
|
|
$$ p \wedge \neg q $$
|
|
|
|
"P is a square and P is not a rectangle."
|
|
|
|
b. If today is New Year's Eve, then tomorrow is January.
|
|
|
|
$$ p \to q $$
|
|
|
|
$$ p \wedge \neg q $$
|
|
|
|
"Today is New Year's Eve and tomorrow is not January."
|
|
|
|
c. If the decimal expansion of $r$ is terminating, then $r$ is rational.
|
|
|
|
"The decimal expansion of $r$ is terminating and $r$ is not rational."
|
|
|
|
d. If $n$ is prime, then $n$ is odd or $n$ is $2$.
|
|
|
|
$n$ is prime and $n$ is not odd and $n$ is not $2$.
|
|
|
|
e. If $x$ is nonnegative, then $x$ is positive or $x$ is $0$.
|
|
|
|
$x$ is nonnegative and $x$ is both negative and not $0$.
|
|
|
|
f. If Tom is Ann's father, then Jim is her uncle and Sue is her aunt.
|
|
|
|
Tom is Ann's father and either Jim is not her uncle or Sue is not her aunt.
|
|
|
|
g. If $n$ is divisible by $6$, then $n$ is divisible by $2$ and $n$ is divisible
|
|
by $3$.
|
|
|
|
$n$ is divisible by $6$ and either $n$ is not divisible by $2$ or $n$ is not
|
|
divisible by $3$.
|
|
|
|
21. Suppose that $p$ and $q$ are statements so that $p \to q$ is false. Find the
|
|
truth values of each of the following.
|
|
|
|
| $p$ | $q$ | $p \to q$ |
|
|
| --- | --- | --------- |
|
|
| T | T | T |
|
|
| T | F | F |
|
|
| F | T | T |
|
|
| F | F | T |
|
|
|
|
This means that $p = T$ and $q = F$.
|
|
|
|
a. $\neg p \to q$
|
|
|
|
$$ F \to F = T $$
|
|
|
|
b. $p \vee q$
|
|
|
|
$$ T \vee F = T $$
|
|
|
|
c. $q \to p$
|
|
|
|
$$ F \to T = T $$
|
|
|
|
22. Write contrapositives for the statements of exercise 20.
|
|
|
|
a. $\neg p \to q$
|
|
|
|
$$ \neg q \to p $$
|
|
|
|
b. $p \vee q$
|
|
|
|
$$ \neg q \to p $$
|
|
|
|
c. $q \to p$
|
|
|
|
$$ \neg p \to \neg q $$
|
|
|
|
23. Write the converse and inverse for each statement of exercise 20.
|
|
|
|
a. $\neg p \to q$
|
|
|
|
Converse:
|
|
|
|
$$ q \to \neg p $$
|
|
|
|
Inverse:
|
|
|
|
$$ p \to \neg q $$
|
|
|
|
b. $p \vee q$
|
|
|
|
Because the contrapositive is equivalent, we can take:
|
|
|
|
$$ \neg q \to p $$
|
|
|
|
Converse:
|
|
|
|
$$ p \to \neg q $$
|
|
|
|
Inverse:
|
|
|
|
$$ q \to \neg p $$
|
|
|
|
c. $q \to p$
|
|
|
|
Converse:
|
|
|
|
$$ p \to q $$
|
|
|
|
Inverse:
|
|
|
|
$$ \neg q \to \neg p $$
|
|
|
|
Use truth tables to establish the truth of each statement in 24-27.
|
|
|
|
24. A conditional statement is not logically equivalent to its converse.
|
|
|
|
| $p$ | $q$ | $p \to q$ | $q \to p$ |
|
|
| --- | --- | --------- | --------- |
|
|
| T | T | T | T |
|
|
| T | F | F | T |
|
|
| F | T | T | F |
|
|
| F | F | T | T |
|
|
|
|
As you can see $p \to q \cancel{\equiv} q \to p$, so a conditional statement is
|
|
not equivalent to its converse.
|
|
|
|
25. A conditional statement is not logically equivalent to its inverse.
|
|
|
|
| $p$ | $q$ | $\neg p$ | $\neg q$ | $p \to q$ | $\neg p \to \neg q$ |
|
|
| --- | --- | -------- | -------- | --------- | ------------------- |
|
|
| T | T | F | F | T | T |
|
|
| T | F | F | T | F | T |
|
|
| F | T | T | F | T | F |
|
|
| F | F | T | T | T | T |
|
|
|
|
As you can see $p \to q \cancel{\equiv} \neg p \to \neg q$, so a conditional
|
|
statement is not equivalent to its inverse.
|
|
|
|
26. A conditional statement and its contrapositive are logically equivalent to
|
|
each other.
|
|
|
|
| $p$ | $q$ | $\neg q$ | $\neg p$ | $p \to q$ | $\neg q \to \neg p$ |
|
|
| --- | --- | -------- | -------- | --------- | ------------------- |
|
|
| T | T | F | F | T | T |
|
|
| T | F | T | F | F | F |
|
|
| F | T | F | T | T | T |
|
|
| F | F | T | T | T | T |
|
|
|
|
As you can see $p \to q \equiv \neg q \to \neg p$, so a conditional statement is
|
|
logically equivalent to it's contrapositive.
|
|
|
|
27. The converse and inverse of a conditional statement are logically equivalent
|
|
to each other.
|
|
|
|
| $p$ | $q$ | $p \to q$ | $q \to p$ | $\neg p \to \neg q$ |
|
|
| --- | --- | --------- | --------- | ------------------- |
|
|
| T | T | T | T | T |
|
|
| T | F | F | T | T |
|
|
| F | T | T | F | F |
|
|
| F | F | T | T | T |
|
|
|
|
As you can see, $q \to p \equiv \neg p \to \neg q$, so the converse and the
|
|
inverse of a conditional statement are logically equivalent to each other.
|
|
|
|
28. "Do you mean that you think you can find out the answer to it?" said the
|
|
March Hare.
|
|
|
|
"Exactly so," said Alice.
|
|
|
|
"Then you should say what you mean," the March Hare went on.
|
|
|
|
"I do," Alice hastily replied; "at least-at least I mean what I say-that's the
|
|
same thing, you know."
|
|
|
|
"Not the same thing a bit!" said the Hatter.
|
|
|
|
"Why, you might just as well say that 'I see what I eat' is the same thing as 'I
|
|
eat what I see'!"
|
|
|
|
-from "A Mad Tea-Party" in _Alice in Wonderland_, by Lewis Carroll
|
|
|
|
The Hatter is right. "I say what I mean" is not the same thing as "I mean what I
|
|
say." Rewrite each of these two sentences in if-then form and explain the
|
|
logical relation between them. (This exercise is referred to in the introduction
|
|
to Chapter 4.)
|
|
|
|
$$ p \to q $$
|
|
|
|
$$ q \to p $$
|
|
|
|
| $p$ | $q$ | $p \to q$ | $q \to p$ |
|
|
| --- | --- | --------- | --------- |
|
|
| T | T | T | T |
|
|
| T | F | F | T |
|
|
| F | T | T | F |
|
|
| F | F | T | T |
|
|
|
|
You can also rewrite this as "If I say something, I mean it." and "If I mean it,
|
|
I say something." Even from a linguistic standpoint these are not the same
|
|
statements. One can "mean it" (be sincere) and never say something, for example.
|
|
|
|
If statement forms $P$ and $Q$ are logically equivalent, then
|
|
$P \leftrightarrow Q$ is a tautology. Conversely, if $P \leftrightarrow Q$ is a
|
|
tautology, then $P$ and $Q$ are logically equivalent. Use $\leftrightarrow$ to
|
|
convert each of the logical equivalences in 29-31 to a tautology. Then use a
|
|
truth table to verify each tautology.
|
|
|
|
29. $p \to (q \vee r) \equiv (p \wedge \neg q) \to r$
|
|
|
|
$$ (p \to (q \vee r)) \leftrightarrow (p \wedge \neg q \to r) $$
|
|
|
|
| $p$ | $q$ | $r$ | $\neg q$ | $(q \vee r)$ | $(p \wedge \neg q)$ | $p \to (q \vee r)$ | $(p \wedge \neg q) \to r$ |
|
|
| --- | --- | --- | -------- | ------------ | ------------------- | ------------------ | ------------------------- |
|
|
| T | T | T | F | T | F | T | T |
|
|
| T | T | F | F | T | F | T | T |
|
|
| T | F | T | T | T | T | T | T |
|
|
| T | F | F | T | F | T | F | F |
|
|
| F | T | T | F | T | F | T | T |
|
|
| F | T | F | F | T | F | T | T |
|
|
| F | F | T | T | T | F | T | T |
|
|
| F | F | F | T | F | F | T | T |
|
|
|
|
30. $p \wedge (q \vee r) \equiv (p \wedge q) \vee (p \wedge r)$
|
|
|
|
$$ (p \wedge (q \vee r)) \leftrightarrow ((p \wedge q) \vee (p \wedge r)) $$
|
|
|
|
| $p$ | $q$ | $r$ | $(q \vee r)$ | $(p \wedge q)$ | $(p \wedge r)$ | $(p \wedge (q \vee r))$ | $((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 |
|
|
|
|
31. $p \to (q \to r) \equiv (p \wedge q) \to r$
|
|
|
|
$$ p \to (q \to r) \leftrightarrow (p \wedge q) \to r $$
|
|
|
|
| $p$ | $q$ | $r$ | $(q \to r)$ | $(p \wedge q)$ | $p \to (q \to r)$ | $(p \wedge q) \to r$ |
|
|
| --- | --- | --- | ----------- | -------------- | ----------------- | -------------------- |
|
|
| T | T | T | T | T | T | T |
|
|
| T | T | F | F | T | F | F |
|
|
| T | F | T | T | F | T | T |
|
|
| T | F | F | T | F | T | T |
|
|
| F | T | T | T | F | T | T |
|
|
| F | T | F | F | F | T | T |
|
|
| F | F | T | T | F | T | T |
|
|
| F | F | F | T | F | T | T |
|
|
|
|
Rewrite each of the statements in 32 and 33 as a conjunction of two if-then
|
|
statements.
|
|
|
|
32. This quadratic equation has two distinct real roots if, and only if, its
|
|
discriminant is greater than zero.
|
|
|
|
"If this quadratic equation has two distinct real roots, then it's discriminant
|
|
is greater than zero and if this quadratic equation's discriminant is greater
|
|
than zero, then it has two distinct real roots."
|
|
|
|
33. This integer is even if, and only if, it equals twice some integer.
|
|
|
|
"If this integer is even, then it equals twice some integer and if this integer
|
|
is equal to twice some integer, then it is even."
|
|
|
|
Rewrite the statements in 34 and 35 in if-then form in two ways, one of which is
|
|
the contrapositive of the other. Use the formal definition of "only if."
|
|
|
|
34. The Cubs will win the pennant only if they win tomorrow's game.
|
|
|
|
Contrapositive:
|
|
|
|
"If the Cubs have not won tomorrow's game, then they will not win the pennant."
|
|
|
|
Only-if:
|
|
|
|
"If the Cub's won the pennant, then they will have won tomorrow's game."
|
|
|
|
35. Sam will be allowed on Signe's racing boat only if he is an expert sailor.
|
|
|
|
Contrapositive:
|
|
|
|
"If Sam is not an expert sailor, then he will not be allowed on Sign'es racing
|
|
boat."
|
|
|
|
Only-if:
|
|
|
|
"If Sam was allowed on Signe's racing boat, then he was an expert sailor."
|
|
|
|
36. Taking the long view on your education, you go to the Prestige Corporation
|
|
and ask what you should do in college to be hired when you graduate. The
|
|
personnel director replies that you will be hired _only if_ you major in
|
|
mathematics or computer science, get a B average or better, and take
|
|
accounting. You do, in fact, become a math major, get a B+ average, and take
|
|
accounting. You return to Prestige Corporation, make a formal application,
|
|
and are turned down. Did the personnel director lie to you?
|
|
|
|
No, because "only if" means that the conditions have already been met, not
|
|
conditions that have yet to be met. Because you "become a math major", "get a B+
|
|
average", and "take accounting", you have not fulfilled the director's
|
|
requirements, which is that "You _have_ majored in mathematics or computer
|
|
science, you have a B average or better, and you have taken accounting."
|
|
|
|
Some programming languages use statements of the form "$r$ unless $s$" to mean
|
|
that as long as $s$ does not happen, then $r$ will happen. More formally:
|
|
|
|
**Definition:**
|
|
|
|
If $r$ and $s$ are statements,
|
|
|
|
**$r$ unless $s$** means if $\neg s$ then $r$.
|
|
|
|
In 37-39 rewrite the statements in if-then form.
|
|
|
|
37. Payment will be made on fifth unless a new hearing is granted.
|
|
|
|
"If a new hearing is not granted, then payment will be made on the fifth."
|
|
|
|
38. Ann will go unless it rains.
|
|
|
|
"If it does not rain, then Ann will go."
|
|
|
|
39. This door will not open unless a security code is entered.
|
|
|
|
"If a security code is not entered, then this door will not open."
|
|
|
|
Rewrite the statements in 40 and 41 in if-then form.
|
|
|
|
40. Catching the 8:05 bus is a sufficient condition for my being on time for
|
|
work.
|
|
|
|
"If I catch the 8:05 bus, then I am on time for work."
|
|
|
|
41. Having two $45\degree$ angles is a sufficient condition for this triangle to
|
|
be a right triangle.
|
|
|
|
"If I have two $45\degree$ angles, then this triangle is a right triangle."
|
|
|
|
Use the contrapositive to rewrite the statements in 42 and 43 in if-then form in
|
|
two ways.
|
|
|
|
42. Being divisible by $3$ is a necessary condition for this number to be
|
|
divisible by $9$.
|
|
|
|
The statement is:
|
|
|
|
"If this number is not divisible by $3$, then this number is not divisible by
|
|
$9$."
|
|
|
|
The contrapositive is:
|
|
|
|
"If this number is divisible by $9$, then this number is divisible by $3$."
|
|
|
|
43. Doing homework regularly is a necessary condition for Jim to pass the
|
|
course.
|
|
|
|
The statement is:
|
|
|
|
"If Jim does not do homework regularly, then Jim does not pass the course."
|
|
|
|
The contrapositive is:
|
|
|
|
"If Jim passed the course, then Jim did the homework."
|
|
|
|
Note that "a sufficient condition for $s$ is $r$" means $r$ is a sufficient
|
|
condition for $s$ and that "a necessary condition for $s$ is $r$" means $r$ is a
|
|
necessary condition for $s$. Rewrite the statements in 44 and 45 in if-then
|
|
form.
|
|
|
|
44. A sufficient condition for Jon's team to win the championship is that it win
|
|
the rest of its games.
|
|
|
|
"If Jon's team wins the rest of its games, then it will win the championship."
|
|
|
|
45. A necessary condition for this computer program to be correct is that it not
|
|
produce error messages during translation.
|
|
|
|
"If this computer program produces error messages during translation, then this
|
|
computer program is not correct."
|
|
|
|
46. "If compound X is boiling, then its temperature must be at least
|
|
$150\degree$C." Assuming that this statement is true, which of the following
|
|
must also be true?
|
|
|
|
$p$ = "Compound X is boiling"
|
|
|
|
$q$ = "Compound X's temperature is at least $150\degree$C"
|
|
|
|
a. If the temperature of compound X is at least $150\degree$C, then compound X
|
|
is boiling.
|
|
|
|
$$ q \to p $$
|
|
|
|
This is not equivalent to $p \to q$, so this is not necessarily true.
|
|
|
|
b. If the temperature of compound X is less than $150\degree$C, then compound X
|
|
is not boiling.
|
|
|
|
$$ \neg q \to \neg p $$
|
|
|
|
This is the contrapositive statement, and is logically equivalent to the
|
|
original statement. This is therefore true.
|
|
|
|
c. Compound X will boil only if its temperature is at least $150\degree$C.
|
|
|
|
"If Compound X is boiling, then its temperature is at least $150\degree$C."
|
|
|
|
This is the same statement as the original $p \to q$, so this is true.
|
|
|
|
d. If compound X is not boiling, then its temperature is less than
|
|
$150\degree$C.
|
|
|
|
$$ \neg p \to q $$
|
|
|
|
This is not related to the original statement be it by contrapositive, converse,
|
|
inverse, only-if, etc. So this is not true.
|
|
|
|
e. A necessary condition for compound X to boil is that its temperature be at
|
|
least $150\degree$C.
|
|
|
|
"If Compound X is not at a boil, then its temperature is not at least
|
|
$150\degree$C."
|
|
|
|
$$ \neg p \to \neg c $$
|
|
|
|
This is the inverse of the statement, and therefore not equivalent to the
|
|
original statement.
|
|
|
|
f. A sufficient condition for compound X to boil is that its temperature be at
|
|
least $150\degree$C.
|
|
|
|
"If Compound X is at a boil, then its temperature is at least $150\degree$C"
|
|
|
|
This statement is equivalent to the original, and so therefore is true.
|
|
|
|
In 47-50(a) use the logical equivalences $p \to q \equiv \neg p \vee q$ and
|
|
$p \leftrightarrow q \equiv (\neg p \vee q) \wedge (\neg q \vee p)$ to rewrite
|
|
the given statement forms without using the symbol $\to$ or $\leftrightarrow$,
|
|
and (b) use the logical equivalence $p \vee q \equiv \neg(\neg p \wedge \neg q)$
|
|
to rewrite each statement form using only $\wedge$ and $\neg$.
|
|
|
|
a.
|
|
|
|
$$ p \to q \equiv \neg p \vee q $$
|
|
|
|
$$ p \leftrightarrow q \equiv (\neg p \vee q) \wedge (\neg q \vee p) $$
|
|
|
|
Avoid $\to$ and $\leftrightarrow$.
|
|
|
|
b.
|
|
|
|
$$ p \vee q \equiv \neg(\neg p \wedge \neg q) $$
|
|
|
|
Only use $\wedge$ and $\neg$.
|
|
|
|
47. $p \wedge \neg q \to r$
|
|
|
|
a.
|
|
|
|
$$ (p \wedge \neg q) \to r \equiv \neg(p \wedge \neg q) \vee r $$
|
|
|
|
b.
|
|
|
|
$$ (p \wedge \neg q) \to r \equiv \neg(p \wedge \neg q) \vee r $$
|
|
|
|
$$ \neg\left[(p \wedge \neg q) \wedge \neg r \right] $$
|
|
|
|
48. $p \vee \neg q \to r \vee q$
|
|
|
|
a.
|
|
|
|
$$ p \vee \neg q \to r \vee q \equiv \neg(p \vee \neg q) \vee (r \vee q) $$
|
|
|
|
b.
|
|
|
|
$$ \neg(p \vee \neg q) \vee (r \vee q) $$
|
|
|
|
$$ \neg\left[\neg\neg(p \vee \neg q) \wedge \neg(r \vee q)\right] $$
|
|
|
|
$$ \neg\left[(p \vee \neg q) \wedge \neg(r \vee q)\right] $$
|
|
|
|
$$ \neg\left[(p \vee \neg q) \wedge (\neg r \wedge \neg q)\right] $$
|
|
|
|
$$ \neg\left[(\neg p \wedge q) \wedge (\neg r \wedge \neg q)\right] $$
|
|
|
|
49. $(p \to r) \leftrightarrow (q \to r)$
|
|
|
|
a.
|
|
|
|
$$ (\neg(p \to r) \vee (q \to r)) \wedge (\neg(q \to r) \vee (p \to r)) $$
|
|
|
|
b.
|
|
|
|
$$ \neg(\neg\neg(p \to r) \wedge \neg(q \to r)) \wedge \neg(\neg\neg(q \to r) \wedge \neg(p \to r)) $$
|
|
|
|
$$ \neg((p \to r) \wedge \neg(q \to r)) \wedge \neg((q \to r) \wedge \neg(p \to r)) $$
|
|
|
|
50. $(p \to (q \to r)) \leftrightarrow ((p \wedge q) \to r)$
|
|
|
|
a.
|
|
|
|
$$ (\neg(p \to (q \to r)) \vee ((p \wedge q) \to r)) \wedge (\neg((p \wedge q) \to r) \vee (p \to (q \to r))) $$
|
|
|
|
$$ (\neg(\neg p \vee (q \to r)) \vee (\neg(p \wedge q) \vee r)) \wedge (\neg(\neg(p \wedge q) \vee r) \vee (\neg p \vee (q \to r))) $$
|
|
|
|
$$ (\neg(\neg p \vee (\neg q \vee r)) \vee (\neg(p \wedge q) \vee r)) \wedge (\neg(\neg(p \wedge q) \vee r) \vee (\neg p \vee (\neg q \vee r))) $$
|
|
|
|
b.
|
|
|
|
Stolen from Gemini (everything else done by hand so far, lol):
|
|
|
|
$$ \neg\left(\neg(p \wedge q \wedge \neg r) \wedge (p \wedge q \wedge \neg r)\right) \wedge \neg\left(\neg(p \wedge q \wedge \neg r) \wedge (p \wedge q \wedge \neg r)\right) $$
|
|
|
|
51. Given any statement form, is it possible to find a logically equivalent form
|
|
that uses only $\neg$ and $\wedge$? Justify your answer.
|
|
|
|
Yes, we can always convert any statement form into a logically equivalent form
|
|
using other equivalency identities and De Morgan's laws. Consider:
|
|
|
|
$$ p \to q \equiv \neg p \vee q $$
|
|
|
|
Using De Morgan's Law we can then:
|
|
|
|
$$ p \to q \equiv \neg(\neg(\neg p) \wedge \neg q) $$
|
|
|
|
$$ p \to q \equiv \neg(p \wedge \neg q) $$
|
|
|
|
Now let's consider:
|
|
|
|
$$ p \leftrightarrow q \equiv (p \to q) \wedge (q \to p) $$
|
|
|
|
$$ p \leftrightarrow q \equiv \neg(p \wedge \neg q) \wedge \neg(q \wedge \neg p) $$
|
|
|
|
---
|
|
|
|
**Exercise Set 2.3**
|
|
|
|
Page 99
|
|
|
|
Use modus ponens or modus tollens to fill in the blanks in the arguments of 1-5
|
|
so as to produce valid inferences.
|
|
|
|
1.
|
|
|
|
If $\sqrt{2}$ is rational, then $\sqrt{2} = \dfrac{a}{b}$ for some integers $a$
|
|
and $b$.
|
|
|
|
It is not true that $\sqrt{2} = \dfrac{a}{b}$ for some integers $a$ and $b$.
|
|
|
|
$\therefore$ ______.
|
|
|
|
2.
|
|
|
|
If $1 - 0.99999 \dots$ is less than every positive real number, then it equals
|
|
zero.
|
|
|
|
______.
|
|
|
|
$\therefore$ The number $1 - 0.99999 \dots$ equals zero.
|
|
|
|
3.
|
|
|
|
If logic is easy, then I am a monkey's uncle.
|
|
|
|
I am not a monkey's uncle.
|
|
|
|
$\therefore$ ______.
|
|
|
|
4.
|
|
|
|
If this graph can be colored with three colors, then it can be colored with four
|
|
colors.
|
|
|
|
This graph cannot be colored with four colors.
|
|
|
|
$\therefore$ ______.
|
|
|
|
5.
|
|
|
|
If they were unsure about the address, then they would have telephoned.
|
|
|
|
______.
|
|
|
|
$\therefore$ They were sure of the address.
|
|
|
|
Use truth tables to determine whether the argument forms in 6-11 are valid.
|
|
Indicate which columns represent the premises and which represent the
|
|
conclusion, and include a sentence explaining how the truth table supports your
|
|
answer. Your explanation should show that you understand what it means for a
|
|
form of argument to be valid or invalid.
|
|
|
|
6.
|
|
|
|
$$
|
|
p \to q \\
|
|
q \to p \\
|
|
\therefore p \vee q
|
|
$$
|
|
|
|
7.
|
|
|
|
$$
|
|
p \\
|
|
p \to q \\
|
|
\neg q \vee r \\
|
|
\therefore r
|
|
$$
|
|
|
|
8.
|
|
|
|
$$
|
|
p \vee q \\
|
|
p \to \neg q \\
|
|
p \to r \\
|
|
\therefore r
|
|
$$
|
|
|
|
9.
|
|
|
|
$$
|
|
p \wedge q \to \neg r \\
|
|
p \vee \neg q \\
|
|
\neg q \to p \\
|
|
\therefore \neg r
|
|
$$
|
|
|
|
10.
|
|
|
|
$$
|
|
p \vee q \to r \\
|
|
\therefore \neg r \to \neg p \wedge \neg q
|
|
$$
|
|
|
|
(This is the form of argument shown on pages 37 and 38.)
|
|
|
|
11.
|
|
|
|
$$
|
|
p \to q \vee r \\
|
|
\neg q \vee \neg r \\
|
|
\therefore \neg p \vee \neg r
|
|
$$
|
|
|
|
12. Use truth tables to show that the following forms of argument are invalid.
|
|
|
|
a.
|
|
|
|
$$
|
|
p \to q \\
|
|
q \\
|
|
\therefore p \\
|
|
\text{converse error}
|
|
$$
|
|
|
|
b.
|
|
|
|
$$
|
|
p \to q \\
|
|
\neg p \\
|
|
\therefore \neg q \\
|
|
\text{inverse error}
|
|
$$
|
|
|
|
Use truth tables to show that the argument forms referred to in 13-21 are valid.
|
|
Indicate which columns represent the premises and which represent the
|
|
conclusion, and include a sentence explaining how the truth table supports your
|
|
answer. Your explanation should show that you understand what it means for a
|
|
form of argument to be valid.
|
|
|
|
13. Modus tollens:
|
|
|
|
$$
|
|
p \to q \\
|
|
\neg q
|
|
\therefore \neg p
|
|
$$
|
|
|
|
14. Example 2.3.3(a)
|
|
|
|
15. Example 2.3.3(b)
|
|
|
|
16. Example 2.3.4(a)
|
|
|
|
17. Example 2.3.4(b)
|
|
|
|
18. Example 2.3.5(a)
|
|
|
|
19. Example 2.3.5(b)
|
|
|
|
20. Example 2.3.6
|
|
|
|
21 Example 2.3.7
|
|
|
|
Use symbols to write the logical form of each argument in 22 and 23, and then
|
|
use a truth table to test the argument for validity. Indicate which columns
|
|
represent the premises and which represent the conclusion, and include a few
|
|
words of explanation showing that you understand the meaning of validity.
|
|
|
|
22.
|
|
|
|
If Tom is not on team $A$, then Hua is on team $B$.
|
|
|
|
If Hua is not on team $B$, then Tom is on team $A$.
|
|
|
|
$\therefore$ Tom is not on team $a$ or Hua is not on team $B$.
|
|
|
|
23.
|
|
|
|
Oleg is a math major or Oleg is an economics major.
|
|
|
|
If Oleg is a math major, then Oleg is required to take Math 362.
|
|
|
|
$\therefore$ Oleg is an economics major or Oleg is not required to take
|
|
Math 362.
|
|
|
|
Some of the arguments in 24-32 are valid, whereas others exhibit the converse or
|
|
the inverse error. Use symbols to write the logical form of each argument. If
|
|
the argument is valid, identify the rule of inference that guarantees its
|
|
validity. Otherwise, state whether the converse or the inverse error is made.
|
|
|
|
24.
|
|
|
|
If Jules solved this problem correctly, then Jules obtained the answer $2$.
|
|
|
|
Jules obtained the answer $2$.
|
|
|
|
$\therefore$ Jules solved this problem correctly.
|
|
|
|
25.
|
|
|
|
This real number is rational or it is irrational.
|
|
|
|
This real number is not rational.
|
|
|
|
$\therefore$ This real number is irrational.
|
|
|
|
26.
|
|
|
|
If I go to the movies, I won't finish my homework.
|
|
|
|
If i don't finish my homework, I won't do well on the exam tomorrow.
|
|
|
|
$\therefore$ If I go to the movies, I won't do well on the exam tomorrow.
|
|
|
|
27.
|
|
|
|
If this number is larger than $2$, then its square is larger than $4$.
|
|
|
|
This number is not larger than $2$.
|
|
|
|
$\therefore$ The square of this number is not larger than $4$.
|
|
|
|
28.
|
|
|
|
If there are as many rational numbers as there are irrational numbers, then the
|
|
set of all irrational numbers is infinite.
|
|
|
|
The set of all irrational numbers is infinite.
|
|
|
|
$\therefore$ There are as many rational numbers as there are irrational numbers.
|
|
|
|
29.
|
|
|
|
If at least one of these two numbers is divisible by $6$, then the product of
|
|
these two numbers is divisible by $6$.
|
|
|
|
Neither of these two numbers is divisible by $6$.
|
|
|
|
$\therefore$ The product of these two numbers is not divisible by $6$.
|
|
|
|
30.
|
|
|
|
If this computer program is correct, then it produces the correct output when
|
|
run with the test data my teacher gave me.
|
|
|
|
This computer program produces the correct output when run with the test data my
|
|
teacher gave me.
|
|
|
|
$\therefore$ This computer program is correct.
|
|
|
|
31.
|
|
|
|
Sandra knows Java and Sandra knows C++.
|
|
|
|
$\therefore$ Sandra knows C++.
|
|
|
|
32.
|
|
|
|
If I get a Christmas bonus, I'll buy a stereo.
|
|
|
|
If I sell my motorcycle, I'll buy a stereo.
|
|
|
|
$\therefore$ If I get a Christmas bonus or I sell my motorcycle, then I'll buy a
|
|
stereo.
|
|
|
|
33. Give an example (other than Example 2.3.11) of a valid argument with a false
|
|
conclusion.
|
|
|
|
34. Give an example (other than Example 2.3.12) of an invalid argument with a
|
|
true conclusion.
|
|
|
|
35. Explain in your own words what distinguishes a valid form of argument from
|
|
an invalid one.
|
|
|
|
36. Given the following information about a computer program, find the mistake
|
|
in the program.
|
|
|
|
a. There is an undeclared variable or there is a syntax error in the first five
|
|
lines.
|
|
|
|
b. If there is a syntax error in the first five lines, then there is a missing
|
|
semicolon or a variable name misspelled.
|
|
|
|
c. There is not a missing semicolon.
|
|
|
|
d. There is not a misspelled variable name.
|
|
|
|
37. In the back of an old cupboard you discover a note signed by a pirate famous
|
|
for his bizarre sense of humor and love of logical puzzles. In the note he
|
|
wrote that he had hidden treasure somewhere on the property. He listed five
|
|
true statements (a-e below) and challenged the reader to use them to figure
|
|
out the location of the treasure.
|
|
|
|
a. If this house is next to a lake, then the treasure is not in the kitchen.
|
|
|
|
b. If the tree in the front yard is an elm, then the treasure is in the kitchen.
|
|
|
|
c. This house is next to a lake.
|
|
|
|
d. The tree in the front yard is an elm or the treasure is buried under the
|
|
flagpole.
|
|
|
|
e. If the tree in the back yard is an oak, then the treasure is in the garage.
|
|
|
|
Where is the treasure hidden?
|
|
|
|
38. You are visiting the island described in Example 2.3.14 and have the
|
|
following encounters with natives.
|
|
|
|
a. Two natives _A_ and _B_ address you as follows:
|
|
|
|
_A_ says: Both of us are knights.
|
|
|
|
_B_ says: _A_ is a knave.
|
|
|
|
What are _A_ and _B_?
|
|
|
|
b. Another two natives _C_ and _D_ approach you but only _C_ speaks.
|
|
|
|
_C_ says: Both of us are knaves.
|
|
|
|
What are _C_ and _D_?
|
|
|
|
c. You then encounter natives _E_ and _F_.
|
|
|
|
_E_ says: _F_ is a knave.
|
|
|
|
_F_ says: _E_ is a knave.
|
|
|
|
How many knaves are there?
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d. Finally, you meet a group of six natives, _U_, _V_, _W_, _X_, _Y_, and _Z_,
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who speak to you as follows:
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_U_ says: None of us is a knight.
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_V_ says: At least three of us are knights.
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_W_ Says: At most three of us are knights.
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_X_ says: Exactly five of us are knights.
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_Y_ says: Exactly two of us are knights.
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_Z_ says: Exactly one of us is a knight.
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Which are knights and which are knaves?
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39. The famous detective Percule Hoirot was called in to solve a baffling murder
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mystery. He determined the following facts:
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a. Lord Hazelton, the murdered man, was killed by a blow on the head with a
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brass candlestick.
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b. Either Lady Hazelton or a maid, Sara, was in the dining room at the time of
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the murder.
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c. If the cook was in the kitchen at the time of the murder, then the butler
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killed Lord Hazelton with a false dose of strychnine.
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d. If Lady Hazelton was in the dining room at the time of the murder, then the
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chauffeur killed Lord Hazelton.
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e. If the cook was not in the kitchen at the time of the murder, then Sara was
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not in the dining room when the murder was committed.
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f. If Sara was in the dining room at the time the murder was committed, then the
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wine steward killed Lord Hazelton.
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|
Is it possible for the detective to deduce the identity of the murderer from
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these facts? If so, who did murder Lord Hazelton? (Assume there was only one
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cause of death.)
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|
40 Sharky, a leader of the underworld, was killed by one of his own band of four
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henchmen. Detective Sharp interviewed the men and determined that all were lying
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except for one. He deduced who killed Sharky on the basis of the following
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|
statements:
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a. Socko: Lefty killed Sharky.
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b. Fats: Muscles didn't kill Sharky.
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c. Lefty: Muscles was shooting craps with Socko when Sharky was knocked off.
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d. Muscles: Lefty didn't kill Sharky.
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|
Who did kill Sharky?
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|
In 41-44 a set of premises and a conclusion are given. Use the valid argument
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|
forms listed in Table 2.3.1 to deduce the conclusion from the premises, giving a
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|
reason for each step as in Example 2.3.8. Assume all variables are statement
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|
variables.
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41.
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a. $\neg p \vee q \to r$
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b. $s \vee \neg q$
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c. $\neg t$
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|
d. $p \to t$
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e. $\neg p \wedge r \to \neg s$
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f. $\therefore \neg q$
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42.
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a. $p \vee q$
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|
b. $q \to r$
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|
c. $p \wedge s \to t$
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|
d. $\neg r$
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e. $\neg q \to u \wedge s$
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|
f. $\therefore t$
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43.
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|
a. $\neg p \to r \wedge \neg s$
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|
b. $t \to s$
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|
c. $u \to \neg p$
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|
d. $\neg w$
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|
e. $u \vee w$
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|
f. $\therefore \neg t$
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44.
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|
a. $p \to q$
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|
b. $r \vee s$
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|
c. $\neg s \to \neg t$
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|
d. $\neg q \vee s$
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|
|
|
e. $\neg s$
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|
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|
f. $\neg p \wedge r \to u$
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|
g. $w \vee t$
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|
h. $\therefore u \wedge w$
|