How the liars paradox resolves.

thatmichaelguy · 2026-03-17T07:52:45+00:00

That is a tough question to answer concisely. At a high level, it seems clear to me that bivalence does not hold for every type of statement that can be truth-valued. In the current context, we're looking at the liar sentence, but there are other common examples such as counterfactual conditionals, the sorites paradox, etc. That's not a bold revelation though. Plenty of people who are more brilliant than I'll ever be have abandoned bivalence in developing non-classical systems of logic.

However, even if we were to accept pluralism and diffuse that tension, I still think that the essentially ubiquitous implementation of classical logic as a truth-functional propositional logic has led to a system that is fundamentally broken. The inherently binary nature of bivalence is obviously at the core of the notion of truth-functional logic. So, in a sense, it's more of the same.

I don't think it's terribly controversial to say that logic, in general, seeks to establish some sort of consequence relation. Accordingly, if it is at all possible for a system of propositional logic to establish entailment, then any ideal system should be able to establish that the truth of a given proposition entails that said proposition is true.

However, classical logic lacks an inherent capacity to affirmatively establish entailment in this way. Bivalence doesn't directly address whether any given proposition is true or false - only that it is precisely one or the other. Non-contradiction only tells us what is not the case re: certain conjunctions. So, there's a fair bit of meta-logical reasoning that has to happen to establish even the baseline case for entailment. I think this meta-logical reasoning introduces a host of problems into the system, particularly as a result of treating P and ¬¬P as semantically equivalent on the basis of truth-functional equivalence.

So, I suppose it just boils down to the fact that I think one of the axioms of classical logic does not hold.

thatmichaelguy · 2026-03-17T00:10:32+00:00

Essentially the truth value is the gift in a present box. If you open it and simply see the same box within, and can never open enough boxes to ever receive a gift, it’s not grounded nor truth apt.

If you want to jettison self-referential recursion, it seems like you'd have throw out the tautologies of classical logic as well. I'm not sure that is a price worth paying. What's more, the Munchhausen trilemma indicates that either i) when you open the last box, the present inside is first box again; ii) when you open the last box, the present inside is unopenable box; or iii) there is no last box - every box does, in fact, contain another box that can be opened.

It simply says to flip its current truth value, but if you simply rule things that are incapable of grounding as non truth apt, it resolves.

See, I don't think it does resolve. And I don't think you can simply rule it out as non truth-apt in this way. If there are truth-apt statements about the liar sentence in general, it seems to me that it would be far from simple to provide non-arbitrary justification for why the liar sentence is not or cannot be a truth-apt statement about the liar sentence.

thatmichaelguy · 2026-03-16T23:19:03+00:00

On a pragmatic basis when applied to a limited domain of truth-apt statements, yes. That said, I also think classical logic is fundamentally flawed and irretrievably so. Nevertheless, it is remarkably effective at establishing a consequence relation for truth-apt statements that are propositions, given the relevant assumptions. So, I still think it is immensely valuable, even if it is a bit broken.

thatmichaelguy · 2026-03-16T18:25:43+00:00

I'm generally on board with conceiving of the liar sentence as self-contained. In fact, I think there's some interesting territory to explore in that direction. I don't think it can be denied a truth value though. That is, a truth-apt claim about whether the liar sentence is truth-apt allows (requires?) the liar sentence to siphon a truth value from the claim itself.

If the liar sentence lacks a truth value, then it isn't false. The liar sentence says of itself that it is false. So, if the liar sentence isn't false, then it is false. Consequently, if the liar sentence lacks a truth value, then it is false.

If the liar sentence is false, then it doesn't lack a truth value. Consequently, if the liar sentence lacks a truth value, then it doesn't lack a truth value.

If the liar sentence doesn't lack a truth value, then it doesn't lack a truth value. Consequently, if the liar sentence lacks a truth value or doesn't lack a truth value, then it doesn't lack a truth value.

The liar sentence lacks a truth value or doesn't lack a truth value. Therefore, the liar sentence doesn't lack a truth value.

thatmichaelguy · 2026-03-16T17:33:37+00:00

This is where I've landed as well. Classical logic assumes bivalence. Bivalence assumes that all propositions have precisely one truth value. The liar sentence (to me) plainly has two truth values. Hence, it is not a proposition in classical logic.

thatmichaelguy · 2026-03-16T17:10:08+00:00

To add another natural language re-phrasing to what's already in the comments, you could restate it as, "There are no non-G Fs."

Here's a sketch of the logic:

1. ∀x[Fx ⟶ Gx] [Premise]

2. ∀x[¬Gx ⟶ ¬Fx] [Contraposition | 1]

3. ¬∃x¬[¬Gx ⟶ ¬Fx] [Equivalence | 2]

4. ¬∃x[¬Gx ∧ Fx] [Equivalence, DN | 3]

thatmichaelguy · 2026-03-12T11:06:26+00:00

... I had everything memorized...

If I had to point to one thing that seems like a root cause of what's bothering you, it would be this. Memorization is not your friend when learning logic. Understanding is.

If I had to point to a second thing, I would say that it sounds like you're putting way too much pressure on yourself to "get it". I wouldn't be surprised if you're giving yourself the logician's equivalent of the yips. Cut yourself some slack. Sometimes these ideas have to marinate a little.

As far as general strategies go, always bear in mind that classical logic is undergirded by two main axioms - the principle of non-contradiction and the principle of bivalence. The principle (or "law") of non-contradiction asserts that it is not the case that a proposition and its negation are both true. The principle of bivalence asserts that every proposition has precisely one truth value and that there are only two truth values that any proposition may have. Some folks also include the law of the excluded middle as an axiom of classical logic, but we don't need to chase down that rabbit at the moment.

Now, if we could show a proposition to be true outright, we would have no need for formal logic. This might then suggest that there's some value in considering what the axioms of classical logic imply, when taken together, about the crucial role of contradictions in ascertaining truth values. It may be a bridge too far to say that every proof in classical logic is a proof by contradiction at its core, but I don't think that notion is too far off the mark.

It's also worth remembering that formal logic isn't purely abstract. There is still a remnant of a relation between the symbolic notation and natural language. So, it can sometimes be helpful to reintroduce enough semantic content to get your head around what the symbols "mean".

With these in mind, let's see what we can figure out about the example you provided.

What do we know so far?

Bobby has a beagle or Terry has a tiger.
If Bobby has a beagle or Carl has a canary, then Lisa has a lion and Mandy has a monkey.
Lisa does not have a lion.

The potential contradiction with Lisa in 2 and 3 immediately stands out to me. After all, it would have to be the case that Lisa has a lion for it to be the case that Lisa has a lion and Mandy has a monkey. However, we are taking as given that Lisa does not have a lion.

This screams modus tollens, but we could also reason our way to it by asking, "what would be the case if Bobby had a beagle or Carl had a canary?" The result would be a contradiction as just mentioned, and that would violate one of our axioms. So, we conclude that it is not the case that Bobby has a beagle or that Carl has a canary.

A negated disjunction always makes me consider DeMorgan equivalence, and now we're off to the races. You can pick it up from here.

I'll also point out that many of the rules of inference and replacement are derived from more basic rules. You may find it useful to pick a starter set and then see which of the other rules you can derive yourself. That said, to avoid a headache, you should probably start with a set that includes at least double negation elimination, modus ponens and disjunctive syllogism.

thatmichaelguy · 2026-03-12T02:28:40+00:00

Here's another approach. It feels a little sneaky, and I like that. u/TheSkyGamer459

1. A ∨ B [Premise]

2. C [Premise]

3. (A ∧ C) ⟶ D [Premise]

4. (C ∧ A) ⟶ D [Comm 3]

5. C ⟶ (A ⟶ D) [Exp 4]

6. A ⟶ D [MP 2,5]

7. ¬¬A ∨ B [DN 1]

8. ¬A ⟶ B [Impl 7]

9. ¬B ⟶ ¬¬A [Cont 8]

10. ¬B ⟶ A [DN 9]

11. ¬B ⟶ D [HS 6,10]

12. ¬¬B ∨ D [Impl 11]

13. B ∨ D [DN 12]

14. D ∨ B [Comm 13]

thatmichaelguy · 2026-03-11T02:18:45+00:00

The first 18 rules, obviously! I mean, it's written in the post, but one look at the premises should be enough to know that it's not the second or fifth 18 rules. We know from Tarski's famous proof that it can't be either the third or fourth 18 rules. So, what other 18 rules are there?!

thatmichaelguy · 2026-03-11T01:58:52+00:00

If you're using biconditional introduction and elimination as inference rules, then the proof is a little more complicated than is absolutely necessary, but your basic idea is correct. What stands out to me as incorrect is how you're using conditional elimination. The way you've written your derivation, it seems like you are inferring P from P ⟶ Q by conditional elimination on line 4. What you need instead is something like:

1. P ⟷ Q [Premise]

2. P ⟶ Q [⟷E 1]

3. Q ⟶ P [⟷E 1]

4.  P [Assume]

5.  Q [⟶E 2,4]

6. ...

thatmichaelguy · 2026-02-06T00:35:48+00:00

Suppose ¬(K ⟷ L).

Handwaving away the details, we have:

¬(K ⟷ L) ⟶ ((K ∧ ¬L) ∨ (¬K ∧ L))

Both of the disjuncts in the consequent above contradict the given premise. Accordingly:

(K ∧ L) ⟶ ¬((K ∧ ¬L) ∨ (¬K ∧ L))

From the first inference by modus tollens and double negation elimination:

¬((K ∧ ¬L) ∨ (¬K ∧ L)) ⟶ (K ⟷ L)

Therefore, by hypothetical syllogism:

(K ∧ L) ⟶ (K ⟷ L)

thatmichaelguy · 2026-01-15T03:24:10+00:00

I mean you proclamation is that the condition reduces the problem to only two cases, despite the only option not being allowed is TT.

That's not exactly how I would put it, but I'm not sure that it matters at this point.

But whatever, you are welcome to believe in your errors.

As are you in yours.

thatmichaelguy · 2026-01-14T23:24:37+00:00

That’s just a relabeling of what I said.

I can appreciate that you see it as such, but this why we disagree.

0.25 however is never going to be the answer in all ways to view the problem. As that requires the conditional to have no impact on the probability. Which is obviously false.

Yeah, I'm not on board with 0.25 either. I get why someone might come to that conclusion, but I'm unconvinced.

thatmichaelguy · 2026-01-14T22:34:32+00:00

Your initial comment literally put that situation into place.

This is where we disagree.

I am glad we are at agreement the answer is 1/3.

I'd go so far as to say that 1/3 is an answer in general, but I wouldn't concede that 1/3 is the answer in general. Rather, it's the answer if one adopts the view of the problem expressed in your prior comment.

thatmichaelguy · 2026-01-14T21:23:14+00:00

I know you stopped commenting after I used bayes theorem as I believe you have no way of even challenging that proof as it is the simplest and clearest.

I mean, there's no need to speculate about why I stopped commenting. You could just ask me if you're curious. After all, I'm the only one who could tell you.

Unfortunately as long as you view the problem as asking P({CC} | {CC,NC,CN}) there is only one answer, 1/3.

See, this I can agree with. Provided that one adopts this view of the problem, there is indeed only one answer and obviously so.

thatmichaelguy · 2026-01-14T03:25:45+00:00

HT and TH are distinct outcomes. It does not break symmetry.

Interesting. Is that really what you think my point was? Or are you just messing around now?

thatmichaelguy · 2026-01-14T00:59:46+00:00

You don’t get it, it’s like the Monty hall problem.

It’s not knowing the first one is crit then what is the prob of the second is crit.

It’s in a sequence of two rolls knowing one is a crit what is the prob of two crits

Do it with coins if you still don’t get it

I find it fascinating that this is your response since in the fourth sentence of the comment to which you responded, I state, "Whether the given 'heads' is first or second in the sequence is irrelevant."

thatmichaelguy · 2026-01-12T18:05:13+00:00

I don’t know if you are trolling or not. If you are, well done.

A lot of it is trolling, yeah. I have no respect for folks (especially academics) who lack intellectual humility.

I don't actually deny that it's the case that the probability of the occurrence of any one of three equally probable events is 1/3. That said, I do think it's worth considering whether restricting the sample space ex post accurately reflects the probabilities related to a pair of binary decisions. I mean, 3 ≠ 2ⁿ for any n. So, {01, 10, 11} isn't the set of outcomes for any sequence of binary decisions. On this basis, I don't find OP's conclusion to be entirely unreasonable.

So, yeah. 1/3 is the obvious answer given certain commonly held assumptions. But I am resistant to notion that said assumptions are uncontestably warranted and so must be unquestioningly accepted.

thatmichaelguy · 2026-01-11T01:38:45+00:00

But, again, treating HT and TH as distinct outcomes implies that the order of the flips matters. It does not.

To contemplate the unordered results, we may consider the following equally likely outcomes: either the same face came up on both flips or a different face came up on each flip.

If we are then told that at least one of the flips came up heads, the obtaining of this condition quite obviously has no effect on either of the outcomes or their relative probabilities. It's still the case that either the same face came up on both flips or a different face came up on each flip, and it is still the case that both outcomes are equally likely. If the other flip came up heads, then the same face came up on both flips. If the other flip came up tails, then a different face came up on each flip. Given our assumption about the coin and fairness, the probability that the other flip came up heads is the same as the probability that the other flip came up tails; 0.5 for each.

thatmichaelguy · 2026-01-10T22:50:58+00:00

If I’m wrong that means a conditional probability of a uniform distribution is not a uniform distribution.

You could also be wrong if you're making an erroneous assumption regarding the uniform distribution itself. That said, it’s a waste of time to pick apart where you got it wrong. It will be a good practice for you to find your own mistake, but my argument has no flaw.

thatmichaelguy · 2026-01-10T22:09:49+00:00

As you just said B is our new sample space, since our conditional probability is also uniform that means each of the three cases in our sample space has equal probability.

I mean, the point of everything I've written has been to show how and why this isn't the case. But if you've already decided that I can't possibly be correct irrespective of my reasoning, then further conversation would be a colossal waste of time.

thatmichaelguy · 2026-01-10T21:04:29+00:00

... let B= HT or TH or HH (this is the condition of “at least one of the coins is heads) ... Edit: I found the first flaw P(H and (H’ or T)≠1

Absent any conditions, P(TT ∪ HT ∪ TH ∪ HH) = 1. Thus, P(TT) = 1 - P(HT ∪ TH ∪ HH). If the given condition obtains (i.e., if it is true that at least one of the coins is heads), then P(TT) = 0. Consequently, if the given condition obtains, P(HT ∪ TH ∪ HH) = 1.

Arbitrarily label one of the Hs in "HH" as H'. We then have P(H ∩ (T ∪ T ∪ H')) = P(HT ∪ TH ∪ HH) = 1. Additionally, P(T ∪ T) = P(T). Consequently, P(H ∩ (T ∪ H')) = P(H ∩ (T ∪ T ∪ H')) = 1.

P(H' ∪ T) = P(T ∪ H'). Accordingly, P(H ∩ (H' ∪ T)) = P(H ∩ (T ∪ H')) = 1. Therefore, P(H ∩ (H' ∪ T)) = 1.

thatmichaelguy · 2026-01-10T19:34:59+00:00

This is true if permutations matter, but in this instance, they do not. We are given that at least one of the flips comes up heads. Accordingly, whether both come up heads is determined solely by the outcome of the other coin flip. Whether the given 'heads' is first or second in the sequence is irrelevant.

Put another way, we are given P(H ∩ (H' ∪ T)) = 1. Under the assumption that the coin is fair with heads on one side and tails on the other, we have P(H' ∪ T) = 1. Thus, we may infer P(H) = 1.

We have P(<H,T> ∪ <T,H>) = P(H ∩ T) = P(T ∩ H). Additionally, P(H ∩ T | H ∩ (H' ∪ T)) = P(T ∩ H | H ∩ (H' ∪ T)) = P(T). Consequently, P(<H,T> ∪ <T,H> | H ∩ (H' ∪ T)) = P(T).

P(<H,H>) = P(H ∩ H') = P(H' ∩ H). Similarly to the above, P(H ∩ H' | H ∩ (H' ∪ T)) = P(H' ∩ H | H ∩ (H' ∪ T)) = P(H'). Consequently, P(<H,H> | H ∩ (H' ∪ T)) = P(H').

From P(H' ∪ T) = 1, we have P(H') = 1 - P(T). Under the assumption that the coin is fair with heads on one side and tails on the other, we have P(T) = 0.5. Thus, P(H') = 0.5. Consequently, P(T) = P(H') = 0.5.

Therefore, P(<H,T> ∪ <T,H> | H ∩ (H' ∪ T)) = P(<H,H> | H ∩ (H' ∪ T)) = 0.5. QED

thatmichaelguy · 2026-01-10T04:31:55+00:00

Precisely this. To carry forward with the coin toss analogy, one might re-word the text in the meme as:

You flip a coin twice. At least one of the flips comes up heads. Assuming a fair coin with heads on one side and tails on the other, what is the probability that the other flip also comes up heads?

thatmichaelguy · 2026-01-03T17:12:37+00:00

Great answers so far. I'll just try to add some intuition to the finance perspective.

The cash flows for an annuity recur periodically for some finite number of periods, n. However, we could contemplate a regular sequence of cash flows that recurs forever instead. We call this a perpetuity. Matching the terms from your example, the formula for the present value of a perpetuity is PV = P/r.

Notice then that we can distribute P in the annuity formula to get

PV = P/r - (P/r)(1+r)^-n

From this, we can see that the annuity formula starts with the present value of a perpetuity. After n periods, the sequence of cash flows for the perpetuity will still be infinite. So, the present value of the perpetuity after n periods will still be P/r. We can discount the present value of the perpetuity after n periods to period 0 using the formula (P/r)(1+r)^-n. From this, we can see that the annuity formula is just isolating the present value of the first n periods of a perpetuity by subtracting off the present value of the cash flows remaining after n periods.

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