We have an infinite amount of theorems $A_1, A_2 \ldots A_n$. Each one of the theorems says: "All of the following theorems are false.". Is there a single one of these theorems that is true?

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    $\begingroup$ Do you mean preceding when you say following? $\endgroup$
    – Quintec
    Commented Oct 6, 2018 at 19:12
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    $\begingroup$ And do you mean propositions when you say theorems? (A theorem is a thing with a proof, which is necessarily true if your axioms are true and your rules of deduction never produce false conclusions from true premises.) $\endgroup$
    – Gareth McCaughan
    Commented Oct 6, 2018 at 19:17
  • $\begingroup$ Hi, @ponikolo. It appears that your question has been put on hold — it currently remains unclear or hard to tell of what you are asking for other members of this site. (I have not downvoted — I respect the fact that you are a new contributor to this community.) Thus, I strongly recommend that you visit the Help Center (in particular, here). Be specific. If you ask a vague question, you may get a vague answer; but if you give us details and context, we can provide a useful answer :) $\endgroup$
    – Mr Pie
    Commented Oct 7, 2018 at 7:05
  • $\begingroup$ Hello @user477343 I've edited my question, hopefully this makes more sense. $\endgroup$
    – ponikoli
    Commented Oct 7, 2018 at 9:16
  • $\begingroup$ Hey, I think that this question makes perfect sense. I even upvoted it. Well tried when it is not on hold, i will try answering it. $\endgroup$ Commented Oct 7, 2018 at 10:55

9 Answers 9


There can not be more than one true statement. If both $A_i$ and $A_j$ are true ($i<j$), this makes $A_i$ false since there is now a statement after $A_i$ that is true.

There can not be exactly one true statement. If $A_i$ is true, that means $A_{i+1}$ and all remaining statements are false. But if all remaining statements are false, that would make $A_{i+1}$ true.

There can not be zero true statements. That would mean that $A_1$ and all following statements are false, but since all following statements are false that would make $A_1$ true.

So I don't think we can say anything about the statements that would be consistent logically.


Here's a different approach-

If we consider the theorems to be arranged in a loop pattern such that theorem $A_1$ is followed by theorem $A_2$ to $A_n$, theorem $A_2$ is followed by $A_3$ to $A_n$ then $A_1$, theorem $A_3$ is followed by $A_4$ to $A_n$ then $A_1$,$A_2$ and so on. There can in fact be exactly one true statement.
Let's say the true statement be $A_x$, since the theorems are in a loop it would make all other statements false, because all other statements have $A_x$ in the list of following statements, making them all false and thus maintaining its own truth.


The answer is

no; Let $i > 0$ be the smallest integer for which $A_i$ is true. That means $A_{i+1}$ is false, and there must be at least one $j \ge i+2$ for which $A_j$ is true. But this contradicts the fact that $A_i$ is true.

  • $\begingroup$ This would make A2 true as well, would't it? Since now all statements following A2 are false. $\endgroup$
    – Jafe
    Commented Oct 7, 2018 at 12:52
  • $\begingroup$ @jafe you're right, I messed things up quite a bit. $\endgroup$
    – Glorfindel
    Commented Oct 7, 2018 at 12:59
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    $\begingroup$ And if all are false, that makes A1 true and the madness begins anew :) $\endgroup$
    – Jafe
    Commented Oct 7, 2018 at 13:03

Infinity hurts my brain, but...

Assume that all $A$ are false for $A_1 \ldots A_n$ as $n$ approaches infinity. This makes $A_1 \ldots A_n$ true for all $n$ as n approaches infinity. This is a contradiction. Therefore by induction, there must exist at least one $A_x$ in which $A_x$ is true. We don't know where, but it is out there in the infinite wild somewhere!


The answer:

Yes, there is exactly one that is true.


Assume for the purpose of argument that the answer is no. That means that either all the statements are false or that more than one is true. If all the statements are false, then in particular all the statements following $A_1$ are false, making $A_1$ true, contradicting the assumption. So at least one statement is true. If two or more are true, the truth of the lowest numbered statement means the subsequent statements are false, contradicting the assumption that they are true. So there is only one true statement.


My answer is:

We cannot say for sure


What I mean is: we have to look at the end. The very last statement can be either true or false we dont know, since there's no statement after that.

Lets turn it around:

We pick an infinite amount of theorem saying the earlier one is wrong.
We don't know if the first one is true, because there is no statement like everyone loves bunnys or is Trump a foolish.
as Informatician I would say you have the value 'null' in a boolean. An undefined state, not true and not false.


I beg to differ with the accepted answer (note that I replace “theorem” by “statement” as theorems are statements that have been proven true):

There is exactly one statement that is true, and that is the last one. The last one is true because there are no following statements, and therefore any statement about the following statements is vacuously true. Since the last statement is true, all preceding ones must be false.

Now you might object that

there supposedly is no last statement, because there are infinitely many.

But that reasoning is false because

the statements are ordered, and therefore their sequence has to be described by ordinals. Since there exist infinite ordinals, there can be both an infinite number of statements and a last statement. And since the assumption that there is no last statement leads to a contradiction, there clearly must be one.

A further hint at this solution is found in the question as

the question explicitly states the statements as $A_1, A_2\ldots A_n$, without any dots after the $A_n$. Thus $A_n$ is clearly the last statement, where $n$ is an infinite ordinal, for example $\omega$, or $\omega+1$.


No. The combined statements of the theorem turns out to be a contradiction. So there does not exist a theorem that is true.

Lets say $A_i$ is true for some $i>0$


$A_i \iff (\forall j>i. \neg A_{j})$

Splitting rhs in i+1 and rest gives us

$A_i \iff ((\neg A_{i+1}) \land (\forall k>i+1. \neg A_{k}))$

but since $A_{i+1} \iff (\forall k>i+1. \neg A_{k})$

Hence, substituting this in previous equation we get

$A_i \iff ((\neg A_{i+1}) \land A_{i+1})$

$A_i \iff F$

$\neg A_i \lor F$

$\neg A_i$

This contradicts our initial assumption that $A_i$ is true Hence,


P.S. Interestingly, if we consider this questions as one of those puzzles where an English sentence can be interpreted in multiple ways, then there can exist a solution as follows.

For example, it is nowhere mentioned that "All of the following theorems are false" is the "only" thing that a theorem says or it is "one of the" thing that theorem says. If a theorem could additionally say "All predecessor theorems are also false". In that case, there will exist exactly one theorem which is true.


I claim that all propositions are true.

As shown in many answers, the question leads to a contadiction.

If i<j then $A_i$ implies $A_j$ but at the same times $A_i$ says $A_j$ is false. So if $A_i$ is true then $A_j$ must be both true and false.

That would mean all $A_i$ are false. But that makes all $A_i$ true by definition.

The conclusion is that all $A_i$ are both true and false. This is not impossible, just highly undesirable. It is known as the principle of explosion. See Principle of explosion. If the axioms are contradictory then everything can be proven.

So it is not a single one but all propositions that are true.


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