231
$\begingroup$

I received this question from my mathematics professor as a leisure-time logic quiz, and although I thought I answered it right, he denied. Can someone explain the reasoning behind the correct solution?

Which answer in this list is the correct answer to this question?

  1. All of the below.
  2. None of the below.
  3. All of the above.
  4. One of the above.
  5. None of the above.
  6. None of the above.

I thought:

  • $2$ and $3$ contradict so $1$ cannot be true.
  • $2$ denies $3$ but $3$ affirms $2,$ so $3$ cannot be true
  • $2$ denies $4,$ but as $1$ and $3$ are proven to be false, $4$ cannot be true.
  • $6$ denies $5$ but not vice versa, so $5$ cannot be true.

at this point only $2$ and $6$ are left to be considered. I thought choosing $2$ would not deny $1$ (and it can't be all of the below and none of the below) hence I thought the answer is $6.$

I don't know the correct answer to the question. Thanks!

$\endgroup$
  • 60
    $\begingroup$ It looks like 5 to me. I'm assuming we want exactly one of the statements to be true. 1-3 are out since 4 would also be true. As is 4, since it would imply one of 1-3 is true. Also, 6 is cannot be true since 5 would lead to a contradiciton. However, if you assume 5 is the only true statement, everything else seems to be false. $\endgroup$ – Kajelad Apr 4 '17 at 7:14
  • 5
    $\begingroup$ I just made a little excel spreadsheet and it turns out that having 5 true and the rest false is the unique fixed point. $\endgroup$ – Oscar Cunningham Apr 4 '17 at 7:20
  • 7
    $\begingroup$ @DanielV: If you look carefully, the only consistent assignment is that only 5 is true. I agree that the wording of the main question is a bit sloppy, but it turns out that indeed, there is only one correct answer. There are no actual, mutually exclusive solution sets, though in principle this is possible for this kind of riddle, yes. But why would that be a problem? $\endgroup$ – tomasz Apr 4 '17 at 12:09
  • 3
    $\begingroup$ "The question is impredictably defined, so there are potentially multiple mutually exclusive solution sets. " -- And yet I and others had no trouble figuring out that only 5 can be correct. "There is no reason to assume that only 1 of those assignments leads to a consistent result." -- It's a conclusion, not an assumption. Sheesh. $\endgroup$ – Jim Balter Apr 5 '17 at 11:42
  • 2
    $\begingroup$ Is a statement being true the same thing as a statement being the answer? If statement A is the answer, then statement A is true; but does that imply that if statement A is true, then statement A is the answer? Without making the assumption that the two are the same, you can still show that 1, 3, and 4 are not the answer. Then, you can say that if 2 is true, then it is the answer and 5 and 6 are false; and if 6 is true, then 6 is the answer and 2 is false, with 5 still being true. $\endgroup$ – Johnathan Gross Aug 17 '18 at 6:48

15 Answers 15

183
$\begingroup$

"6 denies 5 but not vice versa, so 5 cannot be true." This is the incorrect statement. You are right that 5 does not deny 6, but neither does it affirm 6, so this does not rule out 5. Rather, if 6 holds, then 5 holds a fortiori, but this is a contradiction since 6 denies 5. So 6 is ruled out, and 5 is the only possible answer.

Edit: Note that this approach does not assume that only one answer is correct! I argue above that 6 cannot be true. The OP argues quite clearly that 1, 3, and 4 cannot be true. I might revise the OP's discussion of 2 as follows: if 2 holds, then 4 is false. If we take 4 to mean "at least one of the above," this is already a contradiction. If we take 4 to mean "exactly one of the above," then since 3 is false (by 2), 1 must be true, also a contradiction. So we have shown separately that 1, 2, 3, 4, and 6 produce contradictions. Thus, a posteriori there is at most one correct answer, although this was never assumed. In fact, there is exactly one, since if 5 were false, at least one of the contradictory statements above it would hold.

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ +1, but of course as the other top answer says, this assumes that there is exactly one true statement. $\endgroup$ – 6005 Apr 7 '17 at 9:51
  • 5
    $\begingroup$ @6005 Not at all. I will edit my answer to more explicitly address this concern. $\endgroup$ – RCT Apr 7 '17 at 16:55
  • 1
    $\begingroup$ Your updated answer is complete :) I was mostly referring to your last statement "5 is the only possible answer". $\endgroup$ – 6005 Apr 7 '17 at 21:49
  • 3
    $\begingroup$ @RCT I'd like to learn to explain things the way you do. $\endgroup$ – shyam Nov 15 '19 at 10:42
216
$\begingroup$
// gcc ImpredictivePropositionalLogic1.c -o ImpredictivePropositionalLogic1.exe -std=c99 -Wall -O3

/*
Which answer in this list is the correct answer to this question?

(a) All of the below.
(b) None of the below.
(c) All of the above.
(d) One of the above.
(e) None of the above.
(f) None of the above.
*/

#include <stdio.h>
#define iff(x, y) ((x)==(y))

int main() {
  printf("a b c d e f\n");
  for (int a = 0; a <= 1; a++)
  for (int b = 0; b <= 1; b++)
  for (int c = 0; c <= 1; c++)
  for (int d = 0; d <= 1; d++)
  for (int e = 0; e <= 1; e++)
  for (int f = 0; f <= 1; f++) {
    int Ra = iff(a, b && c && d && e && f);
    int Rb = iff(b, !c && !d && !e && !f);
    int Rc = iff(c, a && b);
    int Rd = iff(d, (a && !b && !c) || (!a && b && !c) || (!a && !b && c));
    int Re = iff(e, !a && !b && !c && !d);
    int Rf = iff(f, !a && !b && !c && !d && !e);

    int R = Ra && Rb && Rc && Rd && Re && Rf;
    if (R) printf("%d %d %d %d %d %d\n", a, b, c, d, e, f);
  }
  return 0;
}

This outputs:

a b c d e f
0 0 0 0 1 0

The main point I'd like to get across is that you cannot assume at the outset that there is only 1 satisfying assignment. For example consider the question:

Which of the following is true?
    (a) both of these
    (b) both of these

You might be tempted to say that both (a) and (b) are true. But it is also consistent that both (a) and (b) are false. The tendency to assume singularity from definitions isn't correct when the definitions are impredictive.

| cite | improve this answer | |
$\endgroup$
  • 7
    $\begingroup$ I liked the creativity! $\endgroup$ – Bora M. Alper Apr 4 '17 at 10:33
  • 9
    $\begingroup$ "you cannot assume at the outset that there is only 1 satisfying assignment" - But the question implies there is exactly one answer $\endgroup$ – BlueRaja - Danny Pflughoeft Apr 4 '17 at 15:59
  • 9
    $\begingroup$ @BlueRaja-DannyPflughoeft - While it is true that the wording of the question is appropriate for only one solution, it doesn't take much experience with these problems to realize you cannot count on that. People tend to be terribly lax in their grammar, particularly when it comes to recreational problems. $\endgroup$ – Paul Sinclair Apr 4 '17 at 16:13
  • 5
    $\begingroup$ I just wrote Python code, and it confirms that your output is the unique possibility even if 4. means ​ "at least" ​ rather than "exactly". ​ ​ ​ ​ $\endgroup$ – user57159 Apr 4 '17 at 18:26
  • 2
    $\begingroup$ @MarkSchultheiss : ​ paste.ee/p/o0Hlc ​ ​ ​ ​ $\endgroup$ – user57159 Apr 5 '17 at 21:12
46
$\begingroup$

I am no logician or mathematician. Here's my layman's take on this:

a. If 1 is true, 2 is true, but 2 contradicts the rest of 1. So 1 is self-contradictory, so 1 is out. 2 is still in.

b. 3 can't be true because we know 1 is out. So 3 is out.

c. if 2 is true, 4 is false, but 4 in fact supports 2 being true, because with 1 and 3 out, the "one of the above" must be 2. 2 is therefore self-contradictory. So 2 is out.

d. Having established that 1, 2 and 3 are out, 4 cannot be true. So 4 is out.

e. 5 being true supports the fact that 1,2,3 and 4 are out. It is non-committal on 6, allowing 6 to be false, which in turn allows 5 to be true. So 5 is still in.

f. if 6 is true, 5 must be false, meaning at least one of 1,2,3,4 are true, which we know to be impossible. So 6 is out.

Leaving 5.

| cite | improve this answer | |
$\endgroup$
  • 3
    $\begingroup$ This is the way I solved it as well. $\endgroup$ – shoover Apr 4 '17 at 15:45
  • $\begingroup$ The only problem is that the question is entirely ignored. If the question were "Why are aardvarks?" then this would still lead to the same answer. And "None of the above" isn't really an answer to either question. But that doesn't invalidate your reasoning, of course. $\endgroup$ – RemcoGerlich Nov 15 '19 at 12:34
  • $\begingroup$ @RemcoGerlich, I feel the same way. A question with an unresolved pronoun does not make sense/cannot be answered, as I explain in simple language in my answer. It's a case of infinite recursion without a terminating base case. Even so, I'm not sure above reasoning of boolean logic is correct, because that too, involves pronoun-resolution, but people seem to arbitrarily stop it after 1-level of recursion. $\endgroup$ – d-_-b Nov 17 '19 at 7:30
33
$\begingroup$

You can use propositional logic to formalize the problem, then satisfying assignments help to find the solutions.

Let $a,b,c,d,e,f$ represent the six sentences, respectively.

  1. $a\leftrightarrow b\land c\land d\land e\land f$
  2. $b\leftrightarrow \neg c\land \neg d\land \neg e\land \neg f$
  3. $c\leftrightarrow a\land b$
  4. $d\leftrightarrow (a\land \neg b\land \neg c)\lor(\neg a\land b\land \neg c)\lor (\neg a\land \neg b\land c)$
  5. $e\leftrightarrow \neg a\land \neg b\land \neg c\land \neg d$
  6. $f\leftrightarrow \neg a\land \neg b\land \neg c\land \neg d\land \neg e$

Assuming there is at least one solution: 7. $a\lor b\lor c \lor d\lor e\lor f$

The only satisfying truth assignment is the one which sets $a,b,c,d,f$ to false and set $e$ true. So choice 5 is the solution.

| cite | improve this answer | |
$\endgroup$
  • 7
    $\begingroup$ Your proposition nr 7 isn't actually necessary, since the all-false assignment contradicts proposition 6. $\endgroup$ – tomsmeding Apr 4 '17 at 19:37
  • 1
    $\begingroup$ (4) admits the possibility that all 3 of a,b,c, are true. $\endgroup$ – DanielV Apr 5 '17 at 21:13
  • $\begingroup$ @DanielV Corrected. Is there a more compact form? $\endgroup$ – LoMaPh Apr 5 '17 at 22:27
  • $\begingroup$ @LoMaPh Yes: $$\tag{4} d \leftrightarrow (a \leftrightarrow b \leftrightarrow c) \land \lnot (a \land b \land c)$$ Parentheses left out because $\;\leftrightarrow\;$ is associative. $\endgroup$ – MarnixKlooster ReinstateMonica Apr 6 '17 at 18:28
  • $\begingroup$ This is mathematical way to answer this question (+1). $\endgroup$ – ً ً Apr 7 '17 at 11:30
9
$\begingroup$

First suppose there is only one correct answer, as perhaps implied by the question. $1,3,4,$ if true, imply more than one correct answer, and so cannot be true. If $2$ is true and $1,3$ are not then $4$ is true. So $2$ must be false. If $6$ is true, then $1,2,3,4$ are false and $5$ is true, contradicting $6$. So $5$ is the only possibility.

Now assume open season, and more than one could be true.

If $1$ is true then $2$ is true, but that is impossible for $3$. So $1$ is false. If $2$ is true and $1$ false, $3$ is false and $4$ is true, but that contradicts $2$. $3$ and $4$ are then false because everything above them is false. The same argument about $5$ and $6$ applies as before. So $5$ is true.

| cite | improve this answer | |
$\endgroup$
7
$\begingroup$

It is important to distinguish the present question:

Which answer in this list is the correct answer to this question?

from the following:

Which answer in this list is true?

When we are asked:

Which answer in this list is true?

  1. All of the below.
  2. None of the below.
  3. All of the above.
  4. One of the above.
  5. None of the above.
  6. None of the above.

The correct answer is option 5, which we can discover by assuming the truth of each statement in turn and ruling out each non-viable option. See this post for a similar question.

However, the original question—Which answer in this list is the correct answer to this question?—requires another approach. To see this, let’s present the same question with a different answer set.

Which answer in this list is the correct answer to this question?

  1. Some roses are red.
  2. Some violets are blue.
  3. All dogs are mammals.
  4. All of the above.
  5. None of the above.

On further reflection, we are dealing with a pseudo question. 1–3 are all true, which may tempt us to choose answer 4. However, 4 does not really answer this question.

Once we see that 1-4 do not really answer this question, the new temptation is to choose answer 5. However, 5 does not answer the question either—it merely declares that there is no satisfactory answer listed among the answer choices.

To see this more clearly, let’s look at a different question.

Which US President in this list signed the Emancipation Proclamation?

  1. George Washington
  2. Thomas Jefferson
  3. Millard Fillmore
  4. None of the above.

Since I know that Abraham Lincoln—not Washington, Jefferson, or Fillmore—signed the Emancipation Proclamation, I would most certainly choose option 4 on any standard multiple-choice test. However, my choosing option 4 has to do with testing conventions, not giving the correct answer to the question. The Emancipation Proclamation was signed by Lincoln, not by “none of the above.”

So, even when we look at a real, rather than pseudo, question selecting “none of the above” is a testing convention by which the test-taker claims that the answer set has no satisfactory answer. Note: declaring that there is no satisfactory answer listed among the answer choices is different from providing the correct answer.

With this in mind, let’s go back to the original:

Which answer in this list is the correct answer to this question?

  1. All of the below.
  2. None of the below.
  3. All of the above.
  4. One of the above.
  5. None of the above.
  6. None of the above.

Once we stop treating “none of the above” as an answer to the question and start treating it as a statement about the other options, we can see that choices 2, 5, and 6 are all true statements. However, being true will not make 2, 5, or 6 an answer to the question any more than adding other true statements to the list—like “some roses are red” or “all dogs are mammals”—would make those statements an answer to the question. As with my question about the Emancipation Proclamation, to declare that no satisfactory answer is listed among the answer choices is different from providing the correct answer.

So, if we are dealing with a pseudo question, why don’t we see this at first glance?

Because the list associated with this particular pseudo question uses the “none of the above” list option as a red herring to draw us off-track. We have grown accustomed to thinking that marking “none of the above” answers a question, but this is false. Again, “none of the above” allows us to declare that no satisfactory answer is listed among the answer choices, which is different from providing a correct answer.

So, there is not a correct answer choice, nor is the question even intelligible. The question’s author has exploited a common misunderstanding to trick us into thinking otherwise—which makes for a great puzzle.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ The question is as silly as "This statement is false" 'paradox'. Until the pronoun 'this' has been resolved to a noun (which does not contain any pronouns), the question is not sensible. It's a case of infinite recursion since OP did not define a base case, or rather the base-case itself does not terminate, as I explain in my answer $\endgroup$ – d-_-b Nov 16 '19 at 22:12
  • $\begingroup$ @d-_-b we seem to agree that the question presented in the OP is not a sensible question. My tactic has been to refer to it as a pseudo question. BTW, I’m inclined to agree that an analysis of the question structure shows that its use of the pronoun this leads to an infinite regress, leaving us with no viable referent for the pronoun this. It seems to me, the primary difference between your answer and mine is one of focus. While you have focused on the infinite regress, I’ve chosen to focus on rhetorical tricks used to distract us from the unintelligibility of the question. $\endgroup$ – Chris Nov 17 '19 at 1:20
  • 1
    $\begingroup$ If the math prof. posed this question to a linguist, he would get laughed at, and not get his point across. Instead, he could have just appended 'is True' to each statement and asked: Which of the following statements is True ? 1. All of the below are True. 2. None of the below is True. 3. All of the above are True. 4. One of the above is True. 5. None of the above is True. 6. None of the above is True. $\endgroup$ – d-_-b Nov 17 '19 at 3:19
  • 1
    $\begingroup$ @d-_-b By trick, I don’t mean anything mischievous. I just find the question’s wording and the choice to italicize this curious. Plus, I would not it put past a math prof to place a puzzle within a puzzle for “leisure-time logic quiz.” As you’ve noted, a question like “Which one of these is true?” avoids the additional issue we’ve been discussing. See this post. $\endgroup$ – Chris Nov 18 '19 at 1:06
  • 1
    $\begingroup$ I think your point could be better illustrated by first saying "What is the correct answer to this question? (a) $1=1$ (b) $1 \ne 1$" then demonstrating that there was an implicit assumption there contradicted by "Which of the following is a false statement? (a) $1=1$ (b) $1 \ne 1$" That is, all the responses (including mine) implicitly assume the question is "which of the following is true" but there are clearly questions for which that is not the answer. $\endgroup$ – DanielV Nov 26 '19 at 10:36
4
$\begingroup$

Let's use predicate calculus (http://www.cs.toronto.edu/~hehner/FMSD/). I'll not explain my answer in great detail, because chapter 1 and lecture segments 1 and 2 of FMSD should suffice to explain it.

Translate the claims of each choice into the calculus:

  • (all of the below)$1 = 2 \wedge 3 \wedge 4 \wedge 5 \wedge 6 $
  • (none of the below) \begin{align*} &2 \\ = &\neg (3 \vee 4 \vee 5 \vee 6) \\ &\text{hint: De Morgan's Law} \\ = &\neg 3 \wedge \neg 4 \wedge \neg 5 \wedge \neg 6 \end{align*}
  • (all of the above) \begin{align*} &3 \\ = & 1 \wedge 2 \\ = & 2 \wedge (3 \wedge \neg 3) \wedge \dots \wedge (6 \wedge \neg 6) \\ = & 2 \wedge \bot \wedge \dots \wedge \bot \\ = &\bot \end{align*}

Since $3 = \bot$, we can update the statements of $1$ and $2$:

  • $1 = (2 \wedge 3 \wedge \dots \wedge 6) = 2 \wedge \bot \wedge \dots \wedge 6) = \bot$
  • $2 = \neg 3 \wedge 4 \wedge \dots \wedge 6 = \top \wedge 4 \wedge \dots \wedge 6 = 4 \wedge \dots \wedge 6$

Moving on to $4$:

  • (one of the above) $4 = 1 \vee 2 \vee 3 = \bot \vee 2 \vee \bot = 2$
  • $5 = \neg 1 \wedge \neg 2 \wedge \neg 3 \wedge \neg 4 = \top \wedge \neg 2 \wedge \top \wedge \neg 4 = \neg 2 \wedge \neg 4 = \neg 2 \wedge \neg 2 = \neg 2 $
  • $6 = \neg 1 \wedge \neg 2 \wedge \neg 3 \wedge \neg 4 \wedge \neg 5 = \top \wedge \neg 2 \wedge \top \wedge \neg 2 \wedge 2 = \bot$ (because $2 \wedge \neg 2 = \bot$)

Going back and examining $2$:

  • $2 = \neg 4 \wedge \neg 5 \wedge \neg 6 = \neg 2 \wedge \neg \neg 2 \wedge \top = \neg 2 \wedge 2 = \bot $
  • $4 = 2 = \bot$
  • $5 = \neg 2 = \top$

So, $5$ is a theorem, and thus the only "correct answer".

| cite | improve this answer | |
$\endgroup$
2
$\begingroup$

Your last point is not correct:

If 6 is true, then 5 is false, which implies that at least one of 1-4 is correct, which is a contradiction. So 6 is false.

On the other hand, 4 is not correct, this implies that 2 is incorrect. Indeed 4 is correct if and only if 2 is correct since 1 and 3 are false. (If 2 is true then 4 is true by the content of 4, but it has been shown that 4 is false.)

Hence the only choice is 5.

| cite | improve this answer | |
$\endgroup$
2
$\begingroup$

Since this is a question on mathemetics, I would argue that the correct answer is that this is not a question at all. Of course it is syntactically valid, but not semantically.

I will try to rephrase it in terms of sets:

May A be the set of possible answers. The set of correct answers B is the intersection of A and B.

This definition is clearly self-referential and therefore not covered in ZFC.

(I think I read something by Bertrand Russel once about sentences that were syntactically valid but semantically invalid. Can't remember right now. Russel's Paradox is definitely related.)

All other answers so far really answer a different question:

May A be the set of possible answers. The set of correct answers B is a non-empty subset of A that is free of contradictions. There is exactly one such subset.

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ The formal term for what you are referring to is "impredictive definitions", and there has been a lot of back and forth over when they are valid. For example, the induction axiom of peano arithmetic is potentially impredictive because the "predicate" referred to in the axiom could refer to $\mathbb N$ itself. To assert that impredictive definitions must always be semantically invalid I think is a bit outrageous, but I do agree that the problem is more about which assignments are consistent moreso than provable. (I do refer to that in my answer and a comment under my answer). $\endgroup$ – DanielV Nov 15 '19 at 11:06
  • $\begingroup$ @DanielV, a question with an unresolved pronoun does not make sense/cannot be answered, as I explain in simple language in my answer. It's a case of infinite recursion without a terminating base case. $\endgroup$ – d-_-b Nov 17 '19 at 7:26
1
$\begingroup$

It doesn't appear to me that the yet very interesting point brought up by RCT, nor the original post supply a reason for excluding answer number 2. In fact, the original point that it would not deny 1, does not mean it affirms it, still, and therefore, if #3 to #6 are false, #2 is valid. Until we confute #2, we cannot call 5 true either, samewise.

I think an interesting argumentation - while irrelevant after the preceeding answers - is that the answer has to be a not-self-excluding answer (i.e: not #1, #3, #4). In fact, if the answer were to be #4, for instance, it would define the answer to be not #4, excluding the possibility that #4 is indeed the answer (a contradiction).

This, and RCT point leave us with #2 and #5 again.

Answer 4 ("One of the above") would be true if 2 ("None of the below") were, but 2 contradicts 4, therefore 2 cannot be true and 5 is the only possible answer.

It's the same answer, it just appeared to me that I may have missed to see a valid proof in previous statements, but I may be wrong.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

Just because there is only one proposition that does not contradict the rest (the 5th one) does not mean that it is the answer. We must consider the question first. And the question does not make any sense. It is a non-starter. You might as well write, "42", and leave it at that. Thankfully, this does not count towards your grade!

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 1-4 are all provably false, which means that 5 is definitely true and is the only true statement. $\endgroup$ – brianberns Nov 15 '19 at 22:53
  • 1
    $\begingroup$ @brianberns, a question with an unresolved pronoun does not make sense/cannot be answered, as I explain in simple language in my answer. It's a case of infinite recursion without a terminating base case. $\endgroup$ – d-_-b Nov 17 '19 at 7:26
  • 1
    $\begingroup$ @brianberns, 'this' in 'this conversation' is not self-referential. it is referring to a conversation about, say 'speed boats', that happened just prior. A stranger on the street doesn't come out of the blue and tell you 'I'm enjoying this conversation'. You'll ask 'what conversation?'. Even if there was no prior talk, and if the stranger had just meant to say 'I enjoy going to random people and saying some words to them', then you're stopping the self-referential recursion after 1-level i.e."I'm enjoying 'this' conversation" -> "I'm enjoying 'I'm enjoying this conversation' conversation" $\endgroup$ – d-_-b Nov 18 '19 at 21:26
  • 1
    $\begingroup$ At which point the inner 'this' in "I'm enjoying this conversation" - doesn't not expand/recurse/resolve to anything else, and you treat it as a dumb word without meaning. It's like this - "how many words does this question have?" - You're tempted to answer 7 because you do 1-level recursion to resolve/replace 'this' : "How many words are in 'How many words are in this question?' question?" - At which point you treat the inner one as a dumb sentence without meaning. That is all clear and we are in agreement. But OP's question requires you to NOT treat the inner resolution as a dumb statement. $\endgroup$ – d-_-b Nov 18 '19 at 21:29
  • 1
    $\begingroup$ If you see the examples I gave in my answer, when we know there are 3 levels of recursion required, we have no qualms about pestering the asker to expand/resolve the pronouns until we get to a concrete valid question without any pronouns remaining. But when there are infinite levels of recursion required, we suddenly take upon the burden of resolving it ourselves, which is not sensible. If someone wants us to answer "what is the answer to this question" - it is his responsibility to explain what "this" means. If he says "this" refers to "that", again it is his burden to resolve "that". $\endgroup$ – d-_-b Nov 18 '19 at 21:31
1
$\begingroup$

I think this quiz tires to lead you to the closest answer instead the correct answer. IDK, just saying. So what I did was analyze the question using a matrix and crossing statements each other. Based on the result I imply the answer is statement number 6, since is the closest to a truth statement.

Happy Friday!

Based on the result I imply the answer is statement number 6

| cite | improve this answer | |
$\endgroup$
  • 2
    $\begingroup$ 1-4 are all false, which means that 5 is definitely true, which means that 6 is false. 5 is the only true statement. $\endgroup$ – brianberns Nov 15 '19 at 22:50
1
$\begingroup$

If we simply reverse the list, and swap the above/below, then it's simplicity becomes more obvious, though we may not have notice how we were suckered into considering the problem in a really difficult order.

Isn't it a meta-problem problem?

Which answer in this list is the correct answer to *this* question?
    None of the below.
    None of the below.
    One of the below.
    All of the below.
    None of the above.
    All of the above.
| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

I see this question a little differently:

If I take the #5 statement as a denial of the set of items containing 1-4, then I am judging the truth or falsity of the statement on the basis of set 1-4.

If I take the statement of #6 however, I am actually judging a different set than I am with #5. I am including in that set all items with the denial of #5 included. What I would be saying then is that "none of the above items in set 1-5 are true because the set itself is not structured correctly". Since 6 is outside the previous set, it becomes the "best answer". Notice I don't bother evaluating the truth of number 5 against number six, because it generates a contradiction. I consider number 5 to be simply a "liar" and move forward. Why would I not do that with #5? Because #6 is not in the set of items 1-5 and before six arrives, there is no contradiction in the set of items 1-4 and so 5 has a claim to truth. Once 5 becomes part of the set, you cannot deny a denial. He has no real existence or purpose at that point.

An analogy might be to have a group of black cats (1-4). A tall stranger, #5 shows up and says "These black cats all suck". Okay. So we take number five under advisement and say that is a possibility for all black cats 1-4. Another tall stranger, #6 shows up and says all black cats suck, and so does tall stranger #5 because he is trying to generate problems with me by contradiction. Since my statement includes him, and his statement does not include me, give me priority. He isn't following the rules of set construction. How do you know? Because I am here, last in this set. He, on the other hand, is not last in the set. His position betrays the lie.

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

This question is not answerable.

Because the contract between a Noun and its Pronoun is not fulfilled.
In computer science lingo, it's a Recursion Without Termination, since the base case is not defined.

Bear with me, as I give 4 examples to illustrate.


1

(1): "what color is the sky?"

(2): "What is the answer to above?"

(2) cannot be answered as-it-is without clarifying what the pronoun 'above' refers to.

Step 1 - RESOLVE 'above' : "What is the answer to 'what color is the sky?' ?

Step 2 - RECURSE 'what color is the sky?'

Step 3 - RETURN 'blue'

(1) 'what color is the sky?' - is a Valid & Terminating base case - It is a question (Valid), and it has no pronouns (Terminating).
(2) refers to (1). Hence (2) is also Valid and eventually Terminating.

Questions (1) & (2) are answerable.
Agree?


2

(1): "the sky is blue"

(2): "What is the answer to above?"

Step 1 - RESOLVE 'above' : "What is the answer to 'the sky is blue' ?

Step 2 - RECURSE 'the sky is blue'

Step 3 - RETURN ??

(1) 'the sky is blue' - is a Terminating but Invalid base case - It has no pronouns (Terminating), but it is not even a question (Invalid).
(2) refers to (1). Hence (2) is also eventually Terminating but Invalid.

Question (2) is not answerable.
Agree?


3

(1): "what color is the sky?"

(2): "What is the answer to above?"

(3): "What is the answer to above?"

  1. RESOLVE 'above' in (3): "What is the answer to 'What is the answer to above?' ?"

  2. RECURSE 'What is the answer to above?'

    2.1. RESOLVE 'above' in (2): "What is the answer to 'what color is the sky?' ?"

    2.2. RECURSE 'what color is the sky?'

    2.2.1. RESOLVE ? Nothing to resolve
    
    2.2.2. RECURSE ? Nothing to recurse
    
    2.2.3. RETURN 'blue'
    

    2.3. RETURN 'blue'

  3. RETURN 'blue'

(1) 'what color is the sky?' - is a Valid & Terminating base case.
(2) refers to (1) . Hence (2) is also Valid and eventually (1-level deep) Terminating.
(3) refers to (2). Hence (3) is also Valid and eventually (2-levels deep) Terminating.

Questions (1), (2) and (3) are answerable.
Agree?


4 Coming to OP's example.

(Q): "Which answer in this list is the correct answer to this question?"

(Q) cannot be answered as-it-is without clarifying what the pronoun 'this' refers to.

  1. RESOLVE 'this' : "Which answer in this list is the correct answer to 'Which answer in this list is the correct answer to this question?' question?"

  2. RECURSE 'Which answer in this list is the correct answer to this question?'

    2.1. RESOLVE 'this' : "Which answer in this list is the correct answer to 'Which answer in this list is the correct answer to this question?' question?"

    2.2. RECURSE 'Which answer in this list is the correct answer to this question?'

    2.2.1. RESOLVE 'this' : "Which answer in this list is the correct answer to 'Which answer in this list is the correct answer to this question?' question?"
    
    2.2.2. RECURSE : 'Which answer in this list is the correct answer to this question?'
    
    2.2.2.2.2.2.2....... INFINITE RECURSION : 
    
  3. RETURN ??

(Q) 'Which answer in this list is the correct answer to this question?' - is a Valid but Non-Terminating base case - It is a question (Valid), but has pronouns (Non-Terminating).
(Q) refers to (Q). Hence (Q) is also Valid but never Terminating.

Question (Q) is not answerable.
Agree ?


The question is basically just a chain of pronouns which will never get resolved to a noun.
It's no different from the following conversation :

(OP) : What is the answer to above?

(Me) : You have a responsibility to resolve the pronoun 'above' before I can answer the question. What does 'above' refer to ?

(OP) : to 'oogabooga'

(Me) : You have a responsibility to resolve the pronoun 'oogabooga'. What does 'oogabooga' refer to ?

(OP) : to 'takamaka'

(Me) : You still have a responsibility to resolve 'takamaka'. What does it refer to ?

(OP) : to 'kumpiwumpi'

...

OP's responsibility of pronoun resolution does not end just because he referred to a cyclic self-referencing statement. Remember, it is OP's responsibility for breaking you out of an infinite loop, not yours.

So, the proper response to OP's question is to repeatedly ask infinitely:

'Which question does 'this' refer to?'

| cite | improve this answer | |
$\endgroup$
  • 3
    $\begingroup$ Don't completely agree. The question "How many words are in this question?" does make total sense (answer is 7). The semantics of the question need to be considered when evaluating its validity. $\endgroup$ – user59096 Nov 16 '19 at 7:37
  • $\begingroup$ @user59096, The 2 questions are not equivalent, because you are breaking the word-meaning association after 1 replacement e.g. "How many words are in "How many words are in this question?" question?" - At that point, you stop expanding the meaning of pronoun 'this'. You are treating it simply as a 4-letter word without meaning. Then it's easy to ask sensible questions like 'how many letters, words or syllables IT has' - cos IT now refers to something concrete. Something is concrete when all meaningful pronouns in it have been replaced with their nouns. The question OP asked is not concrete. $\endgroup$ – d-_-b Nov 16 '19 at 7:55
  • $\begingroup$ @user59096, let me now start afresh and ask you something - "what is the answer to this question?" .. how will you respond.. most likely by asking "which question?". if i reply with "this question"... you'll again respond "which question?".. if i keep on replying "this question", you'll call me a fool and walk away. And that is perfectly correct. Until I tell you what "this" resolves/concretes to, you can't answer the question, rather the question is incomplete. $\endgroup$ – d-_-b Nov 16 '19 at 7:58
  • $\begingroup$ @user59096, if we were to do the same for OP's question - we would do 1 replacement - "Which answer in this list is the correct answer to "Which answer in this list is the correct answer to this question?" question?.. Similar to my example in my answer, it would be - "Which answer in this list is the correct answer to "What day is it today?" question?". In order to answer this, I must first find an answer to the inner question. In order to do that, I must first understand the inner question. To do that, I must first resolve the pronoun 'this' in the inner question. Only OP can do that. $\endgroup$ – d-_-b Nov 16 '19 at 8:02
  • $\begingroup$ I should have been clearer: I mostly agree with you. My example was to illustrate that the simple existence of "this" in a question or sentence does not mean that it is invalid. $\endgroup$ – user59096 Nov 16 '19 at 14:04

Not the answer you're looking for? Browse other questions tagged or ask your own question.