The Pessimist Sentences

These sentences are (like the Liar sentence, which is "The Liar Sentence is not true") paradoxical: they cannot consistently be taken to be either true or false.

If the $n$ th Pessimist Sentence, call it $S_n$ , is true, then for some $m>n$ , the sentence $S_m$ is false. Since $S_m$ says that for at least one $k>m$ , $S_k$ is false, if $S_m$ is false then for all $k>m$ , $S_k$ is true. But any true $S_k$ requires that there be false Pessimist sentences with higher indices. That contradicts the hypothesis that $S_n$ is true. So no Pessimist sentence can be true.

Suppose, on the other hand, that $S_n$ is false. Then what it says (that there is at least one false $S_m$ with $m>n$ ) is false, and so all $S_m$ are true for $m>n$ . But we've already seen that no Pessimist sentence can be true. So it's also impossible for any $S_n$ to be false.

So there are zero true and zero false Pessimist Sentences.

If we say that this list of sentences is finite (say with length L), then the Lth statement (that at least one statement after L is false) is false, because there are no statements after L. This causes every other statement in the list to be true, because for all n in 1...L, statement L is false and L>n. If we allow length L to approach infinity (an infinite list), then the Lth sentence (the infinitieth sentence) is still the last one (this is the equivalent of the omega ordinal) and all statements n in 1...L (1...infinity, aka all whole numbers) are true. Therefore, for all finite lists with length L, t+f = (L-1)+1 = L and for all infinite lists t+f=(infinity-1)+1 = infinity+1 = infinity (which I guess technically is undefined, but it's not zero). This result (t+f=L) also follows from the fact that each statement is either true or false (which I proved above) and the sum of the amount of true and the amount of false statements is the amount of statements (L).

The Pessimist statements are inconsistent, and cannot be shown to be consistently true or false.

Suppose that there exists some n such that the $n^{th}$ Pessimist statement is true. Then there must exist some $m > n$ such that the $m^{th}$ Pessimist statement is false. Thus, for all $k>m$ , the $k^{th}$ Pessimist statement is true. But this is impossible, since that would imply that there is at least one Pessimist statement which is both true and false. Thus, there cannot be a true Pessimist statement (and so $t$ must be 0). But this is also impossible, since the negation of the first Pessimist statement (which we know must be true) states that all subsequent Pessimist statements are true. In fact, for any $n \geq 1$ we have that the $n^{th}$ Pessimist statement being false leads to a contradiction, and so there cannot be a false Pessimist statement (and so $f$ must be 0).

The accepted answer is zero, however (at least using classical logic) it is clear that the Pessimist statements as a collection are inconsistent, so you can make an argument that any value you submit should be considered correct, via the principle of explosion.

Not a solution but I almost screwed up cause my brain said "oh no, MATH!" but then after my first 2 mistakes I took a few minutes to think it through and got 0. Also was this the same as domain and range cause if not then I guess I was just lucky?

Suppose any given pessimist sentence $S_n$ is false. Then all the (infinitely many) remaining pessimist sentences must be true. But $S_{n+1}$ says that there is at least one of the remaining sentences is false, which is a direct contradiction, so our hypothesis must be wrong and thus $S_n$ can't be false. Since this argument works for all $n$ , we conclude $f = 0$ .

Now suppose any given pessimist sentence $S_n$ is true. Then there must exist at least one pessimist sentence $S_m$ which is false. But as we just showed, none of the pessimist sentences can be false, so this is also a contradiction. Therefore, we can deduce that $S_n$ is not true, and since this also works for all $n$ we conclude $t = 0$ .

Together, these two arguments give us $t + f = 0$ .

If the 1st pessimist sentence is false, then all later pessimist sentences are true. For n=2, there is a second pessimist sentence that is true and therefore for at least one m>2, the mth pessimist sentence is false. This cannot happen. Then the 1st pessimist sentence is true. Then for at least one m>1, the mth pessimist sentence is false.Then for n=m, the mth pessimist sentence would say that not all later pessimist sentences are true (this conclusion is questionable since there might be more than 1 mth pessimist sentences but I guess not because it looks like the pessimist sentences are sorted. For example, two 1st pessimist sentences doesn't make sense). But that's false. Then all later pessimist sentences are true. For n=m+1, there is an (m+1)st pessimist sentence that is true and therefore for at least one k>m+1, the kth pessimist sentence is false. This is a contradiction. Then the 1st pessimist sentence is false. We have that there are no true or false pessimist sentences and so t+f=0. Well, I used the 1st pessimist number because that's how I thought about it but the real argument would be if we started just by simply talking about the nth pessimist sentence.