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🚧 Going through exercises 2.3

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@ -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: