"Which answer in this list is the correct answer to this question?"
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?
- All of the below.
- None of the below.
- All of the above.
- One of the above.
- None of the above.
- 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!
$\endgroup$ 1715 Answers
$\begingroup$// 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 0The 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 theseYou 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.
$\endgroup$ 15 $\begingroup$"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.
$\endgroup$ 4 $\begingroup$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.
$\endgroup$ 3 $\begingroup$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.
- $a\leftrightarrow b\land c\land d\land e\land f$
- $b\leftrightarrow \neg c\land \neg d\land \neg e\land \neg f$
- $c\leftrightarrow a\land b$
- $d\leftrightarrow (a\land \neg b\land \neg c)\lor(\neg a\land b\land \neg c)\lor (\neg a\land \neg b\land c)$
- $e\leftrightarrow \neg a\land \neg b\land \neg c\land \neg d$
- $f\leftrightarrow \neg a\land \neg b\land \neg c\land \neg d\land \neg e$
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.
$\endgroup$ 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.
$\endgroup$ $\begingroup$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?
- All of the below.
- None of the below.
- All of the above.
- One of the above.
- None of the above.
- 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?
- Some roses are red.
- Some violets are blue.
- All dogs are mammals.
- All of the above.
- 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?
- George Washington
- Thomas Jefferson
- Millard Fillmore
- 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?
- All of the below.
- None of the below.
- All of the above.
- One of the above.
- None of the above.
- 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.
$\endgroup$ 11 $\begingroup$Let's use predicate calculus (). 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".
$\endgroup$ $\begingroup$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.
$\endgroup$ $\begingroup$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:
$\endgroup$ 2 $\begingroup$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.
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!
$\endgroup$ 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.
$\endgroup$ $\begingroup$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!
$\endgroup$ 9 $\begingroup$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. $\endgroup$ $\begingroup$ 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.
$\endgroup$ $\begingroup$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?"
RESOLVE 'above' in (3): "What is the answer to 'What is the answer to above?' ?"
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'
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.
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?"
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 :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:
$\endgroup$ 6'Which question does 'this' refer to?'