From 953a69f9b14e7941cf16f75aaa7187e87bfd7f38 Mon Sep 17 00:00:00 2001 From: tomit4 Date: Mon, 25 May 2026 22:59:13 -0700 Subject: [PATCH] :construction: Going through exercises 2.3 --- appendix_b.txt | 2 +- chapter_2/exercises.md | 756 +++++++++++++++++++++++++++++++++++++++++ leftoff.txt | 2 +- 3 files changed, 758 insertions(+), 2 deletions(-) diff --git a/appendix_b.txt b/appendix_b.txt index 7b8d084..71657e3 100644 --- a/appendix_b.txt +++ b/appendix_b.txt @@ -1 +1 @@ -905 +906 diff --git a/chapter_2/exercises.md b/chapter_2/exercises.md index 0a05c41..b9d5897 100644 --- a/chapter_2/exercises.md +++ b/chapter_2/exercises.md @@ -1633,6 +1633,8 @@ It is not true that $\sqrt{2} = \dfrac{a}{b}$ for some integers $a$ and $b$. $\therefore$ ______. +"$\sqrt{2}$ is not rational." (modus tollens) + 2. If $1 - 0.99999 \dots$ is less than every positive real number, then it equals @@ -1640,6 +1642,9 @@ zero. ______. +"The number $1 - 0.99999 \dots$ is less than every positive real number." (modus +ponens) + $\therefore$ The number $1 - 0.99999 \dots$ equals zero. 3. @@ -1650,6 +1655,8 @@ I am not a monkey's uncle. $\therefore$ ______. +"Logic is not easy." (modus tollens) + 4. If this graph can be colored with three colors, then it can be colored with four @@ -1659,6 +1666,8 @@ This graph cannot be colored with four colors. $\therefore$ ______. +"This graph cannot be colored with three colors." (modus tollens) + 5. If they were unsure about the address, then they would have telephoned. @@ -1667,6 +1676,8 @@ ______. $\therefore$ They were sure of the address. +"They did not telephone." (modus tollens) + 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 @@ -1681,6 +1692,17 @@ q \to p \\ \therefore p \vee q $$ +| $p$ | $q$ | $p \to q$ | $q \to p$ | $p \vee q$ | +| --- | --- | --------- | --------- | ---------- | +| T | T | T | T | T | +| T | F | F | T | T | +| F | T | T | F | T | +| F | F | T | T | **F** | + +The last row shows that while it is possible to have all true premises, it is +also possible to have these premises arrive at a false conclusion, therefore +this argument form is invalid. + 7. $$ @@ -1690,6 +1712,20 @@ p \to q \\ \therefore r $$ +| $p$ | $q$ | $r$ | $\neg q$ | $p \to q$ | $\neg q \vee r$ | $p$ | $r$ | +| --- | --- | --- | -------- | --------- | --------------- | --- | ----- | +| T | T | T | F | T | T | T | **T** | +| T | T | F | F | T | F | T | F | +| T | F | T | T | F | T | T | T | +| T | F | F | T | F | T | T | F | +| F | T | T | F | T | T | F | T | +| F | T | F | F | T | F | F | F | +| F | F | T | T | T | T | F | T | +| F | F | F | T | T | T | F | F | + +Here the only row where all the premises are true concludes with a true +consequent, therefore this argument form is valid. + 8. $$ @@ -1699,6 +1735,20 @@ p \to r \\ \therefore r $$ +| $p$ | $q$ | $r$ | $\neg q$ | $p \vee q$ | $p \to \neg q$ | $p \to r$ | $r$ | +| --- | --- | --- | -------- | ---------- | -------------- | --------- | ----- | +| T | T | T | F | T | F | T | T | +| T | T | F | F | T | F | F | F | +| T | F | T | T | T | T | T | **T** | +| T | F | F | T | T | T | F | F | +| F | T | T | F | T | T | T | **T** | +| F | T | F | F | T | T | T | **F** | +| F | F | T | T | F | T | T | T | +| F | F | F | T | F | T | T | F | + +On row 6, we see that we have all true premises that lead to a false conclusion, +therefore this argument form is invalid. + 9. $$ @@ -1708,6 +1758,20 @@ p \vee \neg q \\ \therefore \neg r $$ +| $p$ | $q$ | $r$ | $\neg q$ | $\neg r$ | $p \wedge q$ | $p \wedge q \to \neg r$ | $p \vee \neg q$ | $\neg q \to p$ | $\neg r$ | +| --- | --- | --- | -------- | -------- | ------------ | ----------------------- | --------------- | -------------- | -------- | +| T | T | T | F | F | T | F | T | T | F | +| T | T | F | F | T | T | T | T | T | **T** | +| T | F | T | T | F | F | T | T | T | **F** | +| T | F | F | T | T | F | T | T | T | **T** | +| F | T | T | F | F | F | T | F | T | F | +| F | T | F | F | T | F | T | F | T | T | +| F | F | T | T | F | F | T | T | F | F | +| F | F | F | T | T | F | T | T | F | T | + +On the third row, we see that we have all true premises that lead to a false +conclusion, therefore this argument form is invalid. + 10. $$ @@ -1717,6 +1781,20 @@ $$ (This is the form of argument shown on pages 37 and 38.) +| $p$ | $q$ | $r$ | $\neg p$ | $\neg q$ | $\neg r$ | $p \vee q$ | $\p \vee q \to r$ | $\neg p \wedge \neg q$ | $\neg r \to \neg p \wedge \neg q$ | +| --- | --- | --- | -------- | -------- | -------- | ---------- | ----------------- | ---------------------- | --------------------------------- | +| T | T | T | F | F | F | T | T | F | **T** | +| T | T | F | F | F | T | T | F | F | F | +| T | F | T | F | T | F | T | T | F | **T** | +| T | F | F | F | T | T | T | F | F | F | +| F | T | T | T | F | F | T | T | F | **T** | +| F | T | F | T | F | T | T | F | F | F | +| F | F | T | T | T | F | F | T | T | **T** | +| F | F | F | T | T | T | F | T | T | **T** | + +All rows where the argument $p \vee q \to r$ are true have true conclusions, +therefore this argument form is valid. + 11. $$ @@ -1725,6 +1803,20 @@ p \to q \vee r \\ \therefore \neg p \vee \neg r $$ +| $p$ | $q$ | $r$ | $\neg p$ | $\neg q$ | $\neg r$ | $q \vee r$ | $p \to q \vee r$ | $\neg q \vee \neg r$ | $\neg p \vee \neg r$ | +| --- | --- | --- | -------- | -------- | -------- | ---------- | ---------------- | -------------------- | -------------------- | +| T | T | T | F | F | F | T | T | F | F | +| T | T | F | F | F | T | T | T | T | **T** | +| T | F | T | F | T | F | T | T | T | **F** | +| T | F | F | F | T | T | F | F | T | T | +| F | T | T | T | F | F | T | T | F | T | +| F | T | F | T | F | T | T | T | T | **T** | +| F | F | T | T | T | F | T | T | T | **T** | +| F | F | F | T | T | T | F | T | T | **T** | + +On row three, we can see that there is a case where both precedents are true, +but the consequent is false, therefore this argument form is invalid. + 12. Use truth tables to show that the following forms of argument are invalid. a. @@ -1736,6 +1828,16 @@ q \\ \text{converse error} $$ +| $p$ | $q$ | $p \to q$ | $q$ | $p$ | +| --- | --- | --------- | --- | ----- | +| T | T | T | T | **T** | +| T | F | F | F | T | +| F | T | T | T | **F** | +| F | F | T | F | F | + +As you can see on row three, we have two hypotheses that are true, but the +conclusion is false, therefore this argument form is invalid. + b. $$ @@ -1745,6 +1847,16 @@ p \to q \\ \text{inverse error} $$ +| $p$ | $q$ | $p \to q$ | $\neg p$ | $\neg q$ | +| --- | --- | --------- | -------- | -------- | +| T | T | T | F | F | +| T | F | F | F | T | +| F | T | T | T | **F** | +| F | F | T | T | **T** | + +As you can see on row three, we have two hypotheses that are true, but the +conclusion is false, therefore this argument form is invalid. + 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 @@ -1759,22 +1871,180 @@ p \to q \\ \therefore \neg p $$ +| $p$ | $q$ | $p \to q$ | $\neg q$ | $\neg p$ | +| --- | --- | --------- | -------- | -------- | +| T | T | T | F | F | +| T | F | F | T | F | +| F | T | T | F | T | +| F | F | T | T | **T** | + +The third column $p \to q$, and fourth column $\neg q$ are the premises. And the +final column $\neg p$ is the conclusion. Because all premises and the conclusion +are true, and there are no rows where all premises are true and the conclusion +is false, this argument form is valid. + 14. Example 2.3.3(a) +$$ +p \\ +\therefore p \vee q +$$ + +| $p$ | $q$ | $p$ | $p \vee q$ | +| --- | --- | --- | ---------- | +| T | T | T | **T** | +| T | F | T | **T** | +| F | T | F | T | +| F | F | F | F | + +The third column (copied from the first) is the first and only premise, $p$. The +fourth and final column is the consequent, $p \vee q$. Because both rows where +the premise is true (rows 1 and 2) return true conclusions, this argument form +is valid. + 15. Example 2.3.3(b) +$$ +q \\ +\therefore p \vee q +$$ + +| $p$ | $q$ | $p \vee q$ | +| --- | --- | ---------- | +| T | T | **T** | +| T | F | T | +| F | T | **T** | +| F | F | F | + +The second column is the first and only premise, $q$. The third and final column +is the consequent, $p \vee q$. Because both rows where the premise is true (rows +1 and 3) return true conclusions, this argument form is valid. + 16. Example 2.3.4(a) +$$ +p \wedge q \\ +\therefore p +$$ + +| $p$ | $q$ | $p \wedge q$ | $p$ | +| --- | --- | ------------ | ----- | +| T | T | T | **T** | +| T | F | F | T | +| F | T | F | F | +| F | F | F | F | + +The third column is the only premise, $p \wedge q$. The fourth column is the +consequent $p$. The only row (row 1) where the premise is true, the conclusion +is also true, therefore this argument form is valid. + 17. Example 2.3.4(b) +$$ +p \wedge q \\ +\therefore q +$$ + +| $p$ | $q$ | $p \wedge q$ | $q$ | +| --- | --- | ------------ | ----- | +| T | T | T | **T** | +| T | F | F | F | +| F | T | F | T | +| F | F | F | F | + +The third column is the only premise, $p \wedge q$. The fourth column is the +consequent $q$. The only row (row 1) where the premise is true, the conclusion +is also true, therefore this argument form is valid. + 18. Example 2.3.5(a) +$$ +p \vee q \\ +\neg q \\ +\therefore p +$$ + +| $p$ | $q$ | $p \vee q$ | $\neg q$ | $p$ | +| --- | --- | ---------- | -------- | ----- | +| T | T | T | F | T | +| T | F | T | T | **T** | +| F | T | T | F | F | +| F | F | F | T | F | + +The third and fourth rows are the premises, $p \vee q$ and $\neg q$ +respectively. And the fifth row is the conclusion $p$. The only row where all +premises are true (row 2) returns a true conclusion, and therefore this argument +form is valid. + 19. Example 2.3.5(b) +$$ +p \vee q \\ +\neg p \\ +\therefore q +$$ + +| $p$ | $q$ | $p \vee q$ | $\neg q$ | $q$ | +| --- | --- | ---------- | -------- | ----- | +| T | T | T | F | T | +| T | F | T | F | F | +| F | T | T | T | **T** | +| F | F | F | T | F | + +The third and fourth rows are the premises, $p \vee q$ and $\neg p$ +respectively. And the fifth row is the conclusion $q$. The only row where all +premises are true (row 3) returns a true conclusion, and therefore this argument +form is valid. + 20. Example 2.3.6 +$$ +p \to q \\ +q \to r \\ +\therefore p \to r +$$ + +| $p$ | $q$ | $r$ | $p \to q$ | $q \to r$ | $p \to r$ | +| --- | --- | --- | --------- | --------- | --------- | +| T | T | T | T | T | **T** | +| T | T | F | T | F | F | +| T | F | T | F | T | T | +| T | F | F | F | T | F | +| F | T | T | T | T | **T** | +| F | T | F | T | F | T | +| F | F | T | T | T | **T** | +| F | F | F | T | T | **T** | + +The fourth and fifth columns are the premises, $p \to q$ and $q \to r$ +respectively. The sixth column is the conclusion, $p \to r$. The rows where both +premises are true (rows 1, 5, 7, 8) all have true conclusions, therefore this +argument form is valid. + 21 Example 2.3.7 +$$ +p \vee q \\ +p \to r \\ +q \to r \\ +\therefore r +$$ + +| $p$ | $q$ | $r$ | $p \vee q$ | $p \to r$ | $q \to r$ | $r$ | +| --- | --- | --- | ---------- | --------- | --------- | ----- | +| T | T | T | T | T | T | **T** | +| T | T | F | T | F | F | F | +| T | F | T | T | T | T | **T** | +| T | F | F | T | F | T | F | +| F | T | T | T | T | T | **T** | +| F | T | F | T | T | F | F | +| F | F | T | F | T | T | T | +| F | F | F | F | T | T | F | + +The fourth, fifth, and sixth columns are the precedents, $p \vee q$, $p \to r$, +and $q \to r$ respectively. The seventh column is the conclusion, $r$. The rows +where all three precedents are true (1, 3, 5) also have true conclusions, and +therefore this argument form is valid. + Use symbols to write the logical form of each argument in 22 and 23, and then use a truth table to test the argument for validity. Indicate which columns represent the premises and which represent the conclusion, and include a few @@ -1788,6 +2058,30 @@ If Hua is not on team $B$, then Tom is on team $A$. $\therefore$ Tom is not on team $a$ or Hua is not on team $B$. +Let $p$ be "Tom is on team $A$." + +Let $q$ "Hua is on team $B$." + +Symbolically, the given statement is: + +$$ +\neg p \to q \\ +\neg q \to p \\ +\therefore \neg p \vee \neg q +$$ + +| $p$ | $q$ | $\neg p$ | $\neg q$ | $\neg p \to q$ | $\neg q \to p$ | $\neg p \vee \neg q$ | +| --- | --- | -------- | -------- | -------------- | -------------- | -------------------- | +| T | T | F | F | T | T | **F** | +| T | F | F | T | T | T | **T** | +| F | T | T | F | T | T | **T** | +| F | F | T | T | F | F | T | + +Rows 5 and 6 are the premises, $\neg p \to q$ and $\neg q \to p$ respectively. +The conclusion is the 7th column, $\neg p \vee \neg q$. In the first row, both +premises are true, but the conclusion is false, therefore this argument form is +invalid. + 23. Oleg is a math major or Oleg is an economics major. @@ -1797,6 +2091,35 @@ If Oleg is a math major, then Oleg is required to take Math 362. $\therefore$ Oleg is an economics major or Oleg is not required to take Math 362. +Let $p$ be "Oleg is a math major." + +Let $q$ be "Oleg is an economics major." + +Let $r$ be "Oleg is required to take Math 362." + +Symbolically, the given statement is: + +$$ +p \vee q \\ +p \to r \\ +\therefore q \vee \neg r +$$ + +| $p$ | $q$ | $r$ | $\neg r$ | $p \vee q$ | $p \to r$ | $q \vee \neg r$ | +| --- | --- | --- | -------- | ---------- | --------- | --------------- | +| T | T | T | F | T | T | **T** | +| T | T | F | T | T | F | T | +| T | F | T | F | T | T | **F** | +| T | F | F | T | T | F | T | +| F | T | T | F | T | T | **T** | +| F | T | F | T | T | T | **T** | +| F | F | T | F | F | T | F | +| F | F | F | T | F | T | T | + +The 5th and 6th columns are the premises, $p \vee q$ and $p \to r$ respectively. +The 7th column is the conclusion, $q \vee \neg r$. On row 3, the two premises +are true but the conclusion is false, therefore this argument form is invalid. + Some of the arguments in 24-32 are valid, whereas others exhibit the converse or the inverse error. Use symbols to write the logical form of each argument. If the argument is valid, identify the rule of inference that guarantees its @@ -1810,6 +2133,21 @@ Jules obtained the answer $2$. $\therefore$ Jules solved this problem correctly. +Let $p$ be "Jules solved this problem correctly." + +Let $q$ be "Jules obtained answer $2$." + +Symbolically, the given statement is: + +$$ +p \to q \\ +q \\ +\therefore p +$$ + +Symbolically, this is exactly the converse error, and therefore is an invalid +argument form. + 25. This real number is rational or it is irrational. @@ -1818,6 +2156,20 @@ This real number is not rational. $\therefore$ This real number is irrational. +Let $p$ be "This real number is rational." + +Let $q$ be "This real number is irrational." + +Symbolically, our given statement is: + +$$ +p \vee q \\ +\neg p \\ +\therefore q +$$ + +This argument's form is valid by the rule of elimination. + 26. If I go to the movies, I won't finish my homework. @@ -1826,6 +2178,22 @@ If i don't finish my homework, I won't do well on the exam tomorrow. $\therefore$ If I go to the movies, I won't do well on the exam tomorrow. +Let $p$ be "I go to the movies." + +Let $q$ be "I finish my homework." + +Let $r$ be "I do well on the exam tomorrow." + +Symbolically, the given statement is: + +$$ +p \to \neg q \\ +\neg q \to \neg r \\ +\therefore p \to \neg r +$$ + +This argument's form is valid by the rule of transitivity. + 27. If this number is larger than $2$, then its square is larger than $4$. @@ -1834,6 +2202,20 @@ This number is not larger than $2$. $\therefore$ The square of this number is not larger than $4$. +Let $p$ be "This number is larger than $2$a" + +Let $q$ be "This number's square is larger than $4$." + +Symbolically, the given statement is: + +$$ +p \to q \\ +\neg p \\ +\therefore \neg q +$$ + +This argument's form is invalid, as it demonstrates the inverse error. + 28. If there are as many rational numbers as there are irrational numbers, then the @@ -1843,6 +2225,20 @@ The set of all irrational numbers is infinite. $\therefore$ There are as many rational numbers as there are irrational numbers. +Let $p$ be "There are as many rational numbers as there are irrational numbers." + +Let $q$ be "The set of all irrational numbers is infinite." + +Symbolically, the given statement is: + +$$ +p \to q \\ +q \\ +\therefore p +$$ + +This argument's form is invalid as it demonstrates a converse error. + 29. If at least one of these two numbers is divisible by $6$, then the product of @@ -1852,6 +2248,20 @@ Neither of these two numbers is divisible by $6$. $\therefore$ The product of these two numbers is not divisible by $6$. +Let $p$ be "At least one of these two numbers is divisible by $6$." + +Let $q$ be "The product of these two numbers is divisible by $6$." + +Symbolically, this statement is: + +$$ +p \to q \\ +\neg p \\ +\therefore \neg q +$$ + +This argument's form is invalid as it demonstrates an inverse error. + 30. If this computer program is correct, then it produces the correct output when @@ -1862,12 +2272,40 @@ teacher gave me. $\therefore$ This computer program is correct. +Let $p$ be "This computer program is correct." + +Let $q$ be "This computer program produces the correct output when run with the +test data my teacher gave me." + +Symbolically, the given statement is: + +$$ +p \to q \\ +q \\ +\therefore p +$$ + +This argument form is invalid by the converse error. + 31. Sandra knows Java and Sandra knows C++. $\therefore$ Sandra knows C++. +Let $p$ be "Sandra knows Java." + +Let $q$ be "Sandra knows C++." + +Symbolically, the given statement is: + +$$ +p \wedge q \\ +\therefore q +$$ + +This argument's form is valid by the rule of specialization. + 32. If I get a Christmas bonus, I'll buy a stereo. @@ -1877,15 +2315,52 @@ If I sell my motorcycle, I'll buy a stereo. $\therefore$ If I get a Christmas bonus or I sell my motorcycle, then I'll buy a stereo. +Let $p$ be "I get a Christmas bonus." + +Let $r$ be "I buy a stereo." + +Let $q$ be "I sell my motorcycle." + +Symbolically, the given statement is: + +$$ +p \to r \\ +q \to r \\ +\therefore p \vee q \to r +$$ + +This argument form is valid by rule of transitivity. + 33. Give an example (other than Example 2.3.11) of a valid argument with a false conclusion. +"If the sky is blue, then the ground is purple." + +"The sky is blue." + +$\therefore$ "The ground is purple." + 34. Give an example (other than Example 2.3.12) of an invalid argument with a true conclusion. +"If the sky is blue, then the ground is purple." + +"The ground is purple." + +$\therefore$ "The sky is blue." + 35. Explain in your own words what distinguishes a valid form of argument from an invalid one. +An argument is any statement or series of statements that are asserted to be +true, these statement(s) are known as the premises of the argument, while the +final statement of the argument is the conclusion. A valid argument is any +argument in which all instances where the premises are all true and the +conclusion is also true. + +An invalid argument is any argument in which all instances where the premises +are true, there is one or more instances where the conclusion is false. + 36. Given the following information about a computer program, find the mistake in the program. @@ -1899,6 +2374,57 @@ c. There is not a missing semicolon. d. There is not a misspelled variable name. +Let $p$ be "There is an undeclared variable." + +Let $q$ be "There is a syntax error in the first five lines." + +Let $r$ be "There is a missing semicolon." + +Let $s$ be "There is a variable name that is misspelled." + +Symbolically: + +$$ +p \vee q \\ +q \to r \vee s \\ +\neg r \\ +\neg s \\ +\therefore p +$$ + +If we break this down logically, we can work backwards: + +1. + +$$ +\neg s \\ +\neg p \\ +\therefore \neg r \wedge \neg s +$$ + +This is true by c and d, the definition of $\vee$, and De Morgan's laws. + +2. Therefore $q$ cannot be true: + +$$ +q \to r \vee s \\ +\neg r \wedge \neg s +\therefore \neg q +$$ + +This is true by b and the rule of modus tollens. + +3. Therefore $p$ is true: + +$$ +p \vee q \\ +\neg q \\ +\therefore p +$$ + +This is true by a and thee rule of elimination. This leaves us to conclude that +"There is an undeclared variable." + 37. In the back of an old cupboard you discover a note signed by a pirate famous for his bizarre sense of humor and love of logical puzzles. In the note he wrote that he had hidden treasure somewhere on the property. He listed five @@ -1918,6 +2444,73 @@ e. If the tree in the back yard is an oak, then the treasure is in the garage. Where is the treasure hidden? +Let $p$ be "This house is next to a lake." + +Let $q$ be "The treasure is in the kitchen." + +Let $r$ be "The tree in the front yard is an elm." + +Let $s$ be "The treasure is buried under the flagpole." + +Let $t$ be "The tree in the back yard is an oak." + +Let $u$ be "The treasure is in the garage." + +Symbolically: + +$$ +p \to \neg q \\ +r \to q \\ +p \\ +r \vee s \\ +t \to u \\ +\therefore \text{ ?} +$$ + +1. We can work this one through as soon as we have our first true assertion, + which is our third assertion $p$. This is true by c. + +$$ p $$ + +2. + +$$ +p \to \neg q \\ +p +\therefore \neg q +$$ + +This is true by modus ponens. + +3. + +$$ +r \to q \\ +\neg q \\ +\therefore \neg r +$$ + +This is true by modus tollens. + +4. + +$$ +r \vee s \\ +\neg r \\ +\therefore s +$$ + +This is true by the rule of elimination. And we can actually stop here, as $s$ +is "The treasure is buried under the flagpole." + +Note however that if we were to go further and evaluate $t \to u$, we actually +cannot know if $t$ is true or not, as $\neg r \cancel{\to} t$, meaning that just +because the tree in the back yard is not an elm, it doesn't mean that the tree +in the back yard is an oak, that statement is never given to us in the problem +statement. + +So once again, the answer is "The treasure is buried under the flagpole." + 38. You are visiting the island described in Example 2.3.14 and have the following encounters with natives. @@ -1929,12 +2522,63 @@ _B_ says: _A_ is a knave. What are _A_ and _B_? +Suppose that _A_ is a knight: + +_A_ is a knight. + +$\therefore$ Both _A_ and _B_ are knights. + +Then _B_'s statement also is true: + +_B_ is a knight. + +$\therefore$ _A_ is a knave. + +This is a contradiction, as "_A_ is a knave" contradicts "Both _A_ and _B_ are +knights." + +Therefore _A_ must be a knave. + +_A_ is a knave. + +$\therefore$ Either _A_ or _B_ or both are knaves. + +Then we test _B_'s statement: + +_B_ is a knight. + +$\therefore$ _A_ is a knave. + +Which satisfies the conclusion from our evaluation of _A_'s statement, so +therefore _B_ is a knight. + +_A_ is a knave, and _B_ is a knight. + 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_? +Suppose _C_ is a knight. + +_C_ is a knight + +$\therefore$ Both _C_ and _D_ are knaves. + +This is a contradiction as _C_ is assumed to be a knight, but claims both _C_ +and _D_ are a knaves. + +_C_ is a knave (by contradiction) + +This means that either _C_ or _D_ or both are knaves (by negation of the +supposition and De Morgan's laws). + +But, if both are knaves, then this validates _C_'s supposition, so it must be +that _D_ is a knight. + +_D_ is a knight and _C_ is a knave. + c. You then encounter natives _E_ and _F_. _E_ says: _F_ is a knave. @@ -1943,6 +2587,39 @@ _F_ says: _E_ is a knave. How many knaves are there? +Let's suppose _E_ is a knight: + +_E_ is a knight. + +$\therefore$ _F_ is a knave. + +We then assert that _F_ is a knave by _E_'s supposition. + +_F_ is a knave. + +$\therefore$ _E_ is a knight by negation of _F_'s supposition. + +Which validates our initial hypothesis. + +Let's now hypothesize that _E_ is a knave: + +_E_ is a knave. + +$\therefore$ _F_ is a knight. + +We then assert that _F_ is a knight by negation of _E_'s supposition. + +_F_ is a knight. + +$\therefore$ _E_ is a knave. + +Which validates our secondary hypothesis. + +This means we cannot conclude who is a knight and who is a knave, just that +there is a knave and a knight in this pair. + +This is one knave. + d. Finally, you meet a group of six natives, _U_, _V_, _W_, _X_, _Y_, and _Z_, who speak to you as follows: @@ -1960,6 +2637,85 @@ _Z_ says: Exactly one of us is a knight. Which are knights and which are knaves? +Let's assume _U_ is a knight: + +_U_ is a knight. + +$\therefore$ None of us are knaves. + +But this is a contradiction, because _U_ would also be a knave by his +supposition and that would contradict our assumption. + +Therefore _U_ is a knave. That's one knave. Moving on... + +Let's assume _V_ is a knight: + +_V_ is a knight. + +$\therefore$ At least three of us are knights. + +We can move on and assume this so far. + +Then we assume _W_ is a knight: + +_W_ is a knight. + +$\therefore$ At most three of us are knights. + +Both _V_ and _W_ can only be knights if there are exactly three knights, which +the remaining statements by _X_, _Y_, and _Z_ all would contradict, and we know +_U_ is a knave, so either _V_ or _W_ is a knave or both of them are knaves. + +Let's move on and assume that _X_ is a knight: + +_X_ is a knight. + +$\therefore$ Exactly five of us are knights. + +But we already know that can't be true, because we have a total of 6 natives +here, and we know at least two of them are knaves (_U_ and either _V_ or _W_ or +both are knaves so far). So _X_ is a knave. + +Moving on, let's assume _Y_ is a knight. + +_Y_ is a knight. + +$\therefore$ Exactly two of us are knights. + +Nothing thus far contradicts this except for _V_'s supposition, therefore either +_V_ or _Y_ are knaves, but not both (one of them must be a knight). + +Let's assume _Z_ is a knight. + +_Z_ is a knight. + +$\therefore$ Exactly one of us is a knight. + +But we just established that either _V_ or _Y_ is a knave, but not both, this +contradicts _Z_'s supposition. _Z_ is a knave. + +The only candidates left as knights or knaves are _V_, _W_, and _Y_. + +Let's suppose that both _V_ and _W_ are knaves. This would mean that there would +be exactly three knights, which would work as that would make all _V_, _W_, and +_Y_ as knights...except that _Y_ claims there is exactly two knights, and that +would contradict our assumption. + +Therefore either _V_ or _W_ is a knave, but not both. + +If that's the case, then there are exactly two knights, and _Y_ is correct, _Y_ +is a knight. + +And for _Y_ to be a knight, _V_ must be a knave, as _V_'s suppositions +contradicts _Y_'s confirmed supposition, while _W_'s supposition does not +contradict _Y_'s supposition. + +So here's our final tally of knight's and knaves: + +Knaves: _U_, _V_, _X_, _Z_ + +Knights: _W_, _Y_ + 39. The famous detective Percule Hoirot was called in to solve a baffling murder mystery. He determined the following facts: diff --git a/leftoff.txt b/leftoff.txt index 3ad5abd..398050c 100644 --- a/leftoff.txt +++ b/leftoff.txt @@ -1 +1 @@ -99 +101