🚧 Fin 2.2
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@ -769,56 +769,185 @@ Rewrite the statements in 1-4 in if-then form.
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1. This loop will repeat exactly $n$ times if it does not contain a **stop** or
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1. This loop will repeat exactly $n$ times if it does not contain a **stop** or
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a **go to**.
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a **go to**.
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"If this lop does not contain a **stop** or a **go to**, then it will repeat
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exactly $N$ times."
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2. I am on time for work if I catch the 8:05 bus.
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2. I am on time for work if I catch the 8:05 bus.
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"If I catch the 8:05 bus, then I am on time for work."
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3. Freeze or I'll shoot.
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3. Freeze or I'll shoot.
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"If you do not freeze, then I'll shoot."
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4. Fix my ceiling or I won't pay my rent.
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4. Fix my ceiling or I won't pay my rent.
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"If you do not fix my ceiling, then I won't pay my rent."
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Construct truth tables for the statements forms in 5-11.
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Construct truth tables for the statements forms in 5-11.
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5. $\neg p \vee q \to \neg q$
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5. $\neg p \vee q \to \neg q$
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| $p$ | $q$ | $\neg p$ | $\neg q$ | $\neg p \vee q$ | $\neg p \vee q \to \neg q$ |
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| --- | --- | -------- | -------- | --------------- | -------------------------- |
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| T | T | F | F | T | F |
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| T | F | F | T | F | T |
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| F | T | T | F | T | F |
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| F | F | T | T | T | T |
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6. $(p \vee q) \vee (\neg p \wedge q) \to q$
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6. $(p \vee q) \vee (\neg p \wedge q) \to q$
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| $p$ | $q$ | $\neg p$ | $(p \vee q)$ | $(\neg p \wedge q)$ | $(p \vee q) \vee (\neg p \wedge q)$ | $(p \vee q) \vee (\neg p \wedge q) \to q$ |
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| --- | --- | -------- | ------------ | ------------------- | ----------------------------------- | ----------------------------------------- |
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| T | T | F | T | F | T | T |
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| T | F | F | T | F | T | F |
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| F | T | T | T | T | T | T |
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| F | F | T | F | F | F | T |
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7. $p \wedge \neg q \to r$
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7. $p \wedge \neg q \to r$
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| $p$ | $q$ | $r$ | $\neg q$ | $p \wedge \neg q$ | $p \wedge \neg q \to r$ |
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| --- | --- | --- | -------- | ----------------- | ----------------------- |
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| T | T | T | F | F | T |
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| T | T | F | F | F | T |
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| T | F | T | T | T | T |
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| T | F | F | T | T | F |
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| F | T | T | F | F | T |
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| F | T | F | F | F | T |
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| F | F | T | T | F | T |
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| F | F | F | T | F | T |
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8. $\neg p \vee q \to r$
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8. $\neg p \vee q \to r$
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| $p$ | $q$ | $r$ | $\neg p$ | $\neg p \vee q$ | $\neg p \vee q \to r$ |
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| --- | --- | --- | -------- | --------------- | --------------------- |
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| T | T | T | F | T | T |
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| T | T | F | F | T | F |
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| T | F | T | F | F | T |
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| T | F | F | F | F | T |
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| F | T | T | T | T | T |
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| F | T | F | T | T | F |
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| F | F | T | T | T | T |
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| F | F | F | T | T | F |
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9. $p \wedge \neg r \leftrightarrow q \vee r$
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9. $p \wedge \neg r \leftrightarrow q \vee r$
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| $p$ | $q$ | $r$ | $\neg r$ | $p \wedge \neg r$ | $q \vee r$ | $p \wedge \neg r \leftrightarrow q \vee r$ |
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| --- | --- | --- | -------- | ----------------- | ---------- | ------------------------------------------ |
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| T | T | T | F | F | T | F |
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| T | T | F | T | T | T | T |
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| T | F | T | F | F | T | F |
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| T | F | F | T | T | F | F |
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| F | T | T | F | F | T | F |
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| F | T | F | T | F | T | F |
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| F | F | T | F | F | T | F |
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| F | F | F | T | F | F | T |
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10. $(p \to r) \leftrightarrow (q \to r)$
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10. $(p \to r) \leftrightarrow (q \to r)$
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| $p$ | $q$ | $r$ | $(p \to r)$ | $(q \to r)$ | $(p \to r) \leftrightarrow (q \to r)$ |
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| --- | --- | --- | ----------- | ----------- | ------------------------------------- |
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| T | T | T | T | T | T |
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| T | T | F | F | F | T |
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| T | F | T | T | T | T |
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| T | F | F | F | T | F |
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| F | T | T | T | T | T |
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| F | T | F | T | F | F |
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| F | F | T | T | T | T |
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| F | F | F | T | T | T |
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11. $(p \to (q \to r)) \leftrightarrow ((p \wedge q) \to r)$
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11. $(p \to (q \to r)) \leftrightarrow ((p \wedge q) \to r)$
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| $p$ | $q$ | $r$ | $(q \to r)$ | $(p \wedge q)$ | $(p \to (q \to r))$ | $((p \wedge q) \to r)$ | $(p \to (q \to r)) \leftrightarrow ((p \wedge q) \to r)$ |
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| --- | --- | --- | ----------- | -------------- | ------------------- | ---------------------- | -------------------------------------------------------- |
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| T | T | T | T | T | T | T | T |
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| T | T | F | F | T | F | F | T |
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| T | F | T | T | F | T | T | T |
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| T | F | F | T | F | T | T | T |
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| F | T | T | T | F | T | T | T |
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| F | T | F | F | F | T | T | T |
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| F | F | T | T | F | T | T | T |
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| F | F | F | T | F | T | T | T |
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12. Use the logical equivalence established in Example 2.2.3,
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12. Use the logical equivalence established in Example 2.2.3,
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$p \vee q \to r \equiv (p \to r) \wedge (q \to r)$, to rewrite the following
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$p \vee q \to r \equiv (p \to r) \wedge (q \to r)$, to rewrite the following
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statement. (Assume that $x$ represents a fixed real number.)
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statement. (Assume that $x$ represents a fixed real number.)
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If $x > 2 $ or $x < -2$, then $x^2 > 4$.
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If $x > 2 $ or $x < -2$, then $x^2 > 4$.
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If $x > 2$, then $x^2 > 4$ and if $x < -2$, then $x^2 > 4$.
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13. Use truth tables to verify the following logical equivalences. Include a few
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13. Use truth tables to verify the following logical equivalences. Include a few
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words of explanation with your answers.
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words of explanation with your answers.
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a. $p \to q \equiv \neg p \vee q$
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a. $p \to q \equiv \neg p \vee q$
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| $p$ | $q$ | $\neg p$ | $p \to q$ | $\neg p \vee q$ |
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| --- | --- | -------- | --------- | --------------- |
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| T | T | F | T | T |
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| T | F | F | F | F |
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| F | T | T | T | T |
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| F | F | T | T | T |
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Both columns for $p \to q$ and $\neg p \vee q$ have the same truth values, so
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they are logically equivalent.
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b. $\neg(p \to q) \equiv p \wedge \neg q$
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b. $\neg(p \to q) \equiv p \wedge \neg q$
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| $p$ | $q$ | $\neg q$ | $(p \to q)$ | $\neg(p \to q)$ | $p \wedge \neg q$ |
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| --- | --- | -------- | ----------- | --------------- | ----------------- |
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| T | T | F | T | F | F |
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| T | F | T | F | T | T |
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| F | T | F | T | F | F |
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| F | F | T | T | F | F |
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Both columns for $\neg(p \to q)$ and $p \wedge \neg q$ have the same truth
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values, so they are logically equivalent.
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14.
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14.
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a. Show that the following statement forms are all logically equivalent:
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a. Show that the following statement forms are all logically equivalent:
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$p \to q \vee r$, $p \wedge \neg q \to r$, and $p \wedge \neg r \to q$
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$p \to (q \vee r)$, $(p \wedge \neg q) \to r$, and $(p \wedge \neg r) \to q$
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| $p$ | $q$ | $r$ | $\neg q$ | $\neg r$ | $(q \vee r)$ | $(p \wedge \neg q)$ | $(p \wedge \neg r)$ | $p \to (q \vee r)$ | $(p \wedge \neg q) \to r$ | $(p \wedge \neg r) \to q$ |
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| --- | --- | --- | -------- | -------- | ------------ | ------------------- | ------------------- | ------------------ | ------------------------- | ------------------------- |
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| T | T | T | F | F | T | F | F | T | T | T |
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| T | T | F | F | T | T | F | T | T | T | T |
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| T | F | T | T | F | T | T | F | T | T | T |
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| T | F | F | T | T | F | T | T | F | F | F |
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| F | T | T | F | F | T | F | F | T | T | T |
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| F | T | F | F | T | T | F | F | T | T | T |
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| F | F | T | T | F | T | F | F | T | T | T |
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| F | F | F | T | T | F | F | F | T | T | T |
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b. Use the logical equivalences established in part (a) to rewrite the following
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b. Use the logical equivalences established in part (a) to rewrite the following
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sentence in two different ways. (Assume that $n$ represents a fixed integer.)
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sentence in two different ways. (Assume that $n$ represents a fixed integer.)
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If $n$ is prime, then $n$ is odd or $n$ is $2$.
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If $n$ is prime, then $n$ is odd or $n$ is $2$.
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"If $n$ is prime and $n$ is not odd then $n$ is $2$"
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"If $n$ is prime and $n$ is not $2$, then $n$ is odd."
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15. Determine whether the following statement forms are logically equivalent:
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15. Determine whether the following statement forms are logically equivalent:
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$p \to (q \to r)$ and $(p \to q) \to r$
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$p \to (q \to r)$ and $(p \to q) \to r$
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| $p$ | $q$ | $r$ | $(q \to r)$ | $(p \to q)$ | $p \to (q \to r)$ | $(p \to q) \to r$ |
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| --- | --- | --- | ----------- | ----------- | ----------------- | ----------------- |
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| T | T | T | T | T | T | T |
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| T | T | F | F | F | F | T |
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| T | F | T | T | T | T | T |
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| T | F | F | T | F | T | T |
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| F | T | T | T | T | T | T |
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| F | T | F | F | T | T | F |
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| F | F | T | T | T | T | T |
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| F | F | F | T | T | T | F |
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No, they are not equivalent, $p \to (q \to r) \cancel{\equiv} (p \to q) \to r$
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as their truth values are not the same.
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In 16 and 17, write each of the two statements in symbolic form and determine
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In 16 and 17, write each of the two statements in symbolic form and determine
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whether they are logically equivalent. Include a truth table and a few words of
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whether they are logically equivalent. Include a truth table and a few words of
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explanation to show that you understand what it means for statements to be
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explanation to show that you understand what it means for statements to be
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@ -827,68 +956,294 @@ logically equivalent.
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16. If you paid full price, you didn't buy it at Crown Books. You didn't buy it
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16. If you paid full price, you didn't buy it at Crown Books. You didn't buy it
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at Crown Books or you paid full price.
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at Crown Books or you paid full price.
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$p$ is "You paid full price"
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$q$ is "You bought it at Crown Books"
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$$ p \to \neg q \text{ and } \neg q \vee p $$
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| $p$ | $q$ | $\neg q$ | $p \to \neg q$ | $\neg q \vee p$ |
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| --- | --- | -------- | -------------- | --------------- |
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| T | T | F | F | T |
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| T | F | T | T | T |
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| F | T | F | T | F |
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| F | F | T | T | T |
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No, these two statements are not logically equivalent as their truth tables do
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not have the same truth values.
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$$ p \to \neg q \cancel{\equiv} \neg q \vee p $$
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One could not buy it at Crown Books, and not paid full price, and also one could
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have paid full price and have also bought it at Crown books.
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17. If $2$ is a factor of $n$ and $3$ is a factor of $n$, then $6$ is a factor
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17. If $2$ is a factor of $n$ and $3$ is a factor of $n$, then $6$ is a factor
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of $n$. $2$ is not a factor of $n$ or $3$ is not a factor of $n$ or $6$ is a
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of $n$. $2$ is not a factor of $n$ or $3$ is not a factor of $n$ or $6$ is a
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factor of $n$.
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factor of $n$.
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$p$ is "$2$ is a factor of $n$"
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$q$ is "$3$ is a factor of $n$"
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$r$ is "$6$ is a factor of $n$"
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$$ p \wedge q \to r\text{ and } \neg p \vee \neg q \vee r $$
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| $p$ | $q$ | $r$ | $\neg p$ | $\neg q$ | $p \wedge q$ | $\neg p \vee \neg q$ | $p \wedge q \to r$ | $\neg p \vee \neg q \vee r$ |
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| --- | --- | --- | -------- | -------- | ------------ | -------------------- | ------------------ | --------------------------- |
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| T | T | T | F | F | T | F | T | T |
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| T | T | F | F | F | T | F | F | F |
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| T | F | T | F | T | F | T | T | T |
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| T | F | F | F | T | F | T | T | T |
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| F | T | T | T | F | F | T | T | T |
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| F | T | F | T | F | F | T | T | T |
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| F | F | T | T | T | F | T | T | T |
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| F | F | F | T | T | F | T | T | T |
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Yes, they are equivalent, as their truth tables show the same truth values.
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$$ p \wedge q \to r \equiv \neg p \vee \neg q \vee r $$
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18. Write each of the following three statements in symbolic form and determine
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18. Write each of the following three statements in symbolic form and determine
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which pairs are logically equivalent. Include truth tables and a few words
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which pairs are logically equivalent. Include truth tables and a few words
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of explanation.
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of explanation.
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$p$ = "It walks like a duck"
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$q$ = "It talks like a duck"
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$r$ = "It is a duck"
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If it walks like a duck and it talks like a duck, then it is a duck.
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If it walks like a duck and it talks like a duck, then it is a duck.
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$$ p \wedge q \to r $$
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Either it does not walk like a duck or it does not talk like a duck, or it is a
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Either it does not walk like a duck or it does not talk like a duck, or it is a
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duck.
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duck.
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$$ p \vee \neg q \vee r $$
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If it does not walk like a duck and it does not talk like a duck, then it is not
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If it does not walk like a duck and it does not talk like a duck, then it is not
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a duck.
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a duck.
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$$ \neg p \wedge \neg q \to \neg r $$
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| $p$ | $q$ | $r$ | $\neg p$ | $\neg q$ | $\neg r$ | $p \wedge q$ | $p \vee \neg q$ | $\neg p \wedge \neg q$ | $p \wedge q \to r$ | $p \vee \neg q \vee r$ | $\neg p \wedge \neg q \to \neg r$ |
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| --- | --- | --- | -------- | -------- | -------- | ------------ | --------------- | ---------------------- | ------------------ | ---------------------- | --------------------------------- |
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| T | T | T | F | F | F | T | T | F | T | T | T |
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| T | T | F | F | F | T | T | T | F | F | T | T |
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| T | F | T | F | T | F | F | T | F | T | T | T |
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| T | F | F | F | T | T | F | T | F | T | T | T |
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| F | T | T | T | F | F | F | F | F | T | T | T |
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| F | T | F | T | F | T | F | F | F | T | F | T |
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| F | F | T | T | T | F | F | T | T | T | T | F |
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| F | F | F | T | T | T | F | T | T | T | T | T |
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No, none of these statements are equivalent, as you can see in their
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corresponding truth table's truth values.
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$$ p \wedge q \to r \cancel{\equiv} p \vee \neg q \vee r \cancel{\equiv} \neg p \wedge \neg q \to \neg r $$
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19. True or false? The negation of "If Sue is Luiz's mother, then Ali is his
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19. True or false? The negation of "If Sue is Luiz's mother, then Ali is his
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cousin" is "If Sue is Luiz's mother, then Ali is not his cousin."
|
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
|
20. Write negations for each of the following statements. (Assume that all
|
||||||
variables represent fixed quantities or entities, as appropriate.)
|
variables represent fixed quantities or entities, as appropriate.)
|
||||||
|
|
||||||
a. If $P$ is a square, then $P$ is a rectangle.
|
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.
|
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.
|
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 $2$.
|
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 $0$.
|
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.
|
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
|
g. If $n$ is divisible by $6$, then $n$ is divisible by $2$ and $n$ is divisible
|
||||||
by $3$.
|
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
|
21. Suppose that $p$ and $q$ are statements so that $p \to q$ is false. Find the
|
||||||
truth values of each of the following.
|
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. $\neg p \to q$
|
a. $\neg p \to q$
|
||||||
|
|
||||||
|
$$ F \to F = T $$
|
||||||
|
|
||||||
b. $p \vee q$
|
b. $p \vee q$
|
||||||
|
|
||||||
|
$$ T \vee F = T $$
|
||||||
|
|
||||||
c. $q \to p$
|
c. $q \to p$
|
||||||
|
|
||||||
|
$$ F \to T = T $$
|
||||||
|
|
||||||
22. Write contrapositives for the statements of exercise 20.
|
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.
|
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.
|
Use truth tables to establish the truth of each statement in 24-27.
|
||||||
|
|
||||||
24. A conditional statement is not logically equivalent to its converse.
|
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.
|
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
|
26. A conditional statement and its contrapositive are logically equivalent to
|
||||||
each other.
|
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
|
27. The converse and inverse of a conditional statement are logically equivalent
|
||||||
to each other.
|
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
|
28. "Do you mean that you think you can find out the answer to it?" said the
|
||||||
March Hare.
|
March Hare.
|
||||||
|
|
||||||
|
|
@ -911,6 +1266,21 @@ 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
|
logical relation between them. (This exercise is referred to in the introduction
|
||||||
to Chapter 4.)
|
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
|
If statement forms $P$ and $Q$ are logically equivalent, then
|
||||||
$P \leftrightarrow Q$ is a tautology. Conversely, if $P \leftrightarrow Q$ is a
|
$P \leftrightarrow Q$ is a tautology. Conversely, if $P \leftrightarrow Q$ is a
|
||||||
tautology, then $P$ and $Q$ are logically equivalent. Use $\leftrightarrow$ to
|
tautology, then $P$ and $Q$ are logically equivalent. Use $\leftrightarrow$ to
|
||||||
|
|
@ -919,25 +1289,88 @@ truth table to verify each tautology.
|
||||||
|
|
||||||
29. $p \to (q \vee r) \equiv (p \wedge \neg q) \to r$
|
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. $p \wedge (q \vee r) \equiv (p \wedge q) \vee (p \wedge r)$
|
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. $p \to (q \to r) \equiv (p \wedge q) \to r$
|
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
|
Rewrite each of the statements in 32 and 33 as a conjunction of two if-then
|
||||||
statements.
|
statements.
|
||||||
|
|
||||||
32. This quadratic equation has two distinct real roots if, and only if, its
|
32. This quadratic equation has two distinct real roots if, and only if, its
|
||||||
discriminant is greater than zero.
|
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.
|
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
|
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."
|
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.
|
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.
|
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
|
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
|
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
|
personnel director replies that you will be hired _only if_ you major in
|
||||||
|
|
@ -946,6 +1379,12 @@ the contrapositive of the other. Use the formal definition of "only if."
|
||||||
accounting. You return to Prestige Corporation, make a formal application,
|
accounting. You return to Prestige Corporation, make a formal application,
|
||||||
and are turned down. Did the personnel director lie to you?
|
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
|
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:
|
that as long as $s$ does not happen, then $r$ will happen. More formally:
|
||||||
|
|
||||||
|
|
@ -959,27 +1398,54 @@ In 37-39 rewrite the statements in if-then form.
|
||||||
|
|
||||||
37. Payment will be made on fifth unless a new hearing is granted.
|
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.
|
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.
|
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.
|
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
|
40. Catching the 8:05 bus is a sufficient condition for my being on time for
|
||||||
work.
|
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
|
41. Having two $45\degree$ angles is a sufficient condition for this triangle to
|
||||||
be a right triangle.
|
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
|
Use the contrapositive to rewrite the statements in 42 and 43 in if-then form in
|
||||||
two ways.
|
two ways.
|
||||||
|
|
||||||
42. Being divisible by $3$ is a necessary condition for this number to be
|
42. Being divisible by $3$ is a necessary condition for this number to be
|
||||||
divisible by $9$.
|
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
|
43. Doing homework regularly is a necessary condition for Jim to pass the
|
||||||
course.
|
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
|
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
|
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
|
necessary condition for $s$. Rewrite the statements in 44 and 45 in if-then
|
||||||
|
|
@ -988,43 +1454,163 @@ form.
|
||||||
44. A sufficient condition for Jon's team to win the championship is that it win
|
44. A sufficient condition for Jon's team to win the championship is that it win
|
||||||
the rest of its games.
|
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
|
45. A necessary condition for this computer program to be correct is that it not
|
||||||
produce error messages during translation.
|
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
|
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
|
$150\degree$C." Assuming that this statement is true, which of the following
|
||||||
must also be true?
|
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
|
a. If the temperature of compound X is at least $150\degree$C, then compound X
|
||||||
is boiling.
|
is boiling.
|
||||||
|
|
||||||
b. IF the temperature of compound X is less than $150\degree$C, then compound X
|
$$ q \to p $$
|
||||||
|
|
||||||
|
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.
|
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.
|
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
|
d. If compound X is not boiling, then its temperature is less than
|
||||||
$150\degree$C.
|
$150\degree$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
|
e. A necessary condition for compound X to boil is that its temperature be at
|
||||||
least $150\degree$C.
|
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
|
f. A sufficient condition for compound X to boil is that its temperature be at
|
||||||
least $150\degree$C.
|
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
|
In 47-50(a) use the logical equivalences $p \to q \equiv \neg p \vee q$ and
|
||||||
$p \leftrightarrow q \equiv (\neg p \vee q) \weddge (\neg q \vee p)$ to rewrite
|
$p \leftrightarrow q \equiv (\neg p \vee q) \wedge (\neg q \vee p)$ to rewrite
|
||||||
the given statement forms without using the symbol $\to$ or $\leftrightarrow$,
|
the given statement forms without using the symbol $\to$ or $\leftrightarrow$,
|
||||||
and (b) use the logical equivalence $p \vee q \equiv \neg(\neg p \wedge \neg q)$
|
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 $\neg$.
|
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. $p \wedge \neg q \to r$
|
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. $p \vee \neg q \to r \vee q$
|
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. $(p \to r) \leftrightarrow (q \to r)$
|
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. $(p \to (q \to r)) \leftrightarrow ((p \wedge q) \to r)$
|
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
|
51. Given any statement form, is it possible to find a logically equivalent form
|
||||||
that uses only $\neg$ and $\wedge$? Justify your answer.
|
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:
|
||||||
|
|
||||||
|
$$ p \leftrightarrow q \equiv (p \to q) \wedge (q \to p) $$
|
||||||
|
|
||||||
|
$$ p \leftrightarrow q \equiv \neg(p \wedge \neg q) \wedge \neg(q \wedge \neg p) $$
|
||||||
|
|
|
||||||
Loading…
Add table
Add a link
Reference in a new issue