You're taking the game too seriously. But it is the attitude I've seen most people take in posts discussing logic, riddles, games, etc. When they interact with me I feel hugely overwhelmed.
They believe I am trolling, but I simply lack wording and reasoning.

New rule: The sum of the product of any two integers is omega minus (now corrected!) the double of Lionino’s integers. 

Beautiful poem of Angelos Sikelianos:

O when,
Huntress,
with naked foot beating the snow all night,
shall we see at last on the summit,
like Boreas
and the lion
rousing himself from sleep,
the rising sun?

for in that light
the contest will be decided,
and the new-fledged young men
will row their opponents from the cliff top
and, like the poppy beside the sheaf,
each life
will be harvested in the golden light
spontaneously, silently.

Nice to read a new post from you, Forrest.  :smile:

New rule: There is an integer that is neither a Fhorrest integers nor a Gill integer.

The sum of any two integers is zero. — jgill

The product of any two integers is omega. (Where omega is the first number bigger than any integers). — Pfhorrest

It must be a number smaller than omega, but not zero.

Something like this: x (the suspicious integer) < Ω.

It is 1. Why? Because it is the smallest integer greater than zero and the smallest of Omega.

Everything I wrote above is pure crank, right? :lol:

Oyarzabal (who scored the second goal yesterday) and his family have received threats from Basque nationalists. Oyarza is from a tiny town in Basque Country, so everyone knows where he and his mother live. The painting says: 'NO to play for the Spain National Team'

It's disgusting, but what can I expect from an issue that has existed since the 1960s? As well as Ireland had ‘The Troubles’ we have a big issue with ETA and his political lobby: EH Bildu.

https://www.elmundo.es/pais-vasco/2024/07/15/669506c6e9cf4a963f8b4581.html?cid=SIN26101&utm_source=marca.com&utm_medium=interno_bt&utm_campaign=SIN26101&_ga=2.37133860.509677226.1720796434-798002902.1719681135

There is nothing unclear. After the recent posts and examples (@RussellA’s one was very good), I now see the point of the riddle. It was funny to talk and debate about it. 

Basically, it was my first attempt to put logic into practice. I decided to start with a basic riddle. My intention is to level up in the future.

I understand it now, but not when I first posted this OP. Perhaps it is simple and clear for you, but not for me. One of the demands in my post was for someone to correct me if necessary because I wanted to learn and grasp everything through examples and "formulations."You thought I was trolling. I promise you I wasn't. I always posted with proper behavior.

I still don't get why the answer everyone else is giving isn't satisfying to you. — flannel jesus

Of course their answers (including yours) please me. But I'm keeping up with postings and questions because I want to learn how to apply logic. We are all welcome to ask questions. I promise that I am not trolling any of you.

I'm personally amazed that he's made such a simple riddle last 3 pages, when nobody else has any question about what the answer is. — flannel jesus

Be honest, you're having as much fun with this thread as a child in the park with a big red balloon. :smile:

Got it. :up:

A is either True or False, so A∨¬A. Therefore, A→B can be either True or False (in this case False). No inconsistency in the solution presented. — Lionino

What do you mean by ‘no inconsistency’? That A is coherent but ambiguous or what?

The problem with A is:

B -> A ∧ B = ¬A.

I don’t know whether is coherent or not. But the negation of “A is the truth-teller" means that “A is either the liar or the ambiguous person" Therefore, you are claiming he is the ambiguous and not B. Agree?

A very well written and informative reply. Every kind of help is appreciated here. Thank you, lad.

There is no middle ground to account for person A , who is neither a "Liar" nor "Not a Liar". — RussellA

This is the core of the riddle, indeed. I tried to pin A’s state, but it turned out to be more difficult than I expected. If I am not mistaken, we need some statements to ensure that the truth-teller and the liar are different people. "There is exactly one liar" turns into a similar trio of statements. if x is true, A cannot be the liar. if x is false, A cannot be the truth teller.

But the debate on A is that he is able to be three different positions: truth teller, liar, and ambiguous. Can I get away with one proposition for A? No, I can’t. Since A “sometimes tells the truth”, it means he can also lie. Therefore, A can be in another position (liar/person who sometimes lies). Then I asked yesterday if A was ambiguous or just contradictory. The debate remains.

Gotcha. :up:

Prima facie these might mean two different things:

1. I only sometimes tell the truth
2. I sometimes tell the truth — Michael

I agree. That’s about what happens to A.

Strictly speaking (2) might be true even if I always tell the truth. — Michael

Yes. And reaching (more or less) that conclusion, we can (perhaps) say that A is the ambiguous here. Right?

Person A is the person who sometimes tells the truth. If Person C is the person who always tells the truth then Person A is lying. — Michael

:clap:

I couldn’t have said it better. Michael, frame that post please because @flannel jesus is not capable of seeing that A can actually be a liar too.

You edited your posts after reading the arguments of Michael and Igitur :lol:

He can lie, I've said that explicitly — flannel jesus

Dude…

everyone here except you has understood that b must be the liar. Who else do you see claiming a might be the liar? — flannel jesus

I have never said that A always lies. It is obvious and that would make the riddle senseless. As far as I can understand the relationship between the three, my statement goes as follows:

A: Sometimes tells the truth. Therefore, he can lie often.
B: always lies.
C: Always tells the truth.

Therefore, A and B are liars and C is the only truth teller. If A sometimes tells the truth it means he can also lie as well as B.

The three statements work on the understanding that A happened to be lying rather than telling the truth. — RussellA

A=can both lie and tell the truth — Igitur

Thank you so much for your posts. You explained very well what I tried to explain, but I couldn’t find the correct premises due to my lack of wording and logic skills.

Do you see it now, @flannel jesus? Because you claim A can’t be a liar emphatically. While, as Igitur noted, if A sometimes tells the truth, A can be both a truth teller and a liar. Therefore, there is the possibility for A to be a liar as well as B.

Yes, you're overcomplicating something very simple. C tells the truth. C says B is the liar. Therefore, B is the liar. — flannel jesus

Yes, I understood it at the first glance, but:

A sometimes tells the truth. — flannel jesus

If Person A is the person who sometimes tells the truth, then it means he sometimes lies. Person A could be a liar as well. Yes or no?

We already know C is the truth-teller. But I was wondering what happened to A. Some claim he is ambiguous, others he is contradictory. If he is ambiguous, A sometimes tells the truth and sometimes lies. If he is contradictory, A always tells the truth because B is always true.

Am I missing something in that attempt to use logic? As I asked at the beginning of the OP I wonder whether I am correctly formulating the logic or not.

I am trying to make an approach. I claim that A is just contradictory in his statement, but you are defending that A is precisely the ambiguous here because he ‘sometimes tells the truth’ and B is always the liar. I defend the opposite: B is ambiguous and A is contradictory for always telling the truth. It is B who often says the truth and others not, but not both. Then, when B tells the truth, A tells the truth as well. B is the ambiguous.

Excellent! :up: Thanks to your contribution to this thread. Also, to @flannel jesus for keeping the discussion alive. I am enjoying this.

The context where A is always true and not sometimes.

I thought B was ambiguous at first glance, but consider this, Michael:

Person A claims person B always tells the truth. Person B claims person B (himself) sometimes tells the truth. Okay, then, A tells the truth and B always lies. A is the ambiguous person here. I tried to explain that he is just contradictory and B is ambiguous.
But I am starting to realise that A is dragged down by the ambiguity of B. And then I asked myself: does this make A ambiguous or just contradictory with his statement?

Assuming either one of them is the truth teller leads to contradiction, so we don't. — flannel jesus

That’s what I tried to argue! If A is contradictory, then C is the truthteller, and B is ambiguous. I mean, according to this context, B could be the one who sometimes tells the truth.

Yes. I agree and I see the point. But what happens to A then? B is the only one who says he tells the truth sometimes. Nonetheless, it seems the riddle turned out with A being the ambiguous person, and this is very tricky to me. B cannot be the truth-teller. Therefore, A did not speak the truth  and therefore, A is not the truth-teller either. We all agree that C is the truth-teller. Then, is A the contradictory, ambiguous, or unreliable person here?

For me, it is like pulling a rabbit out of the hat. It is surprising that A is actually the problem and not B. Or did you see the point from the beginning? 

I back up your argument, and I also see A as ambiguous. A claims person B always tells the truth. B claims that person B (himself) sometimes tells the truth. Then, A tells the truth (ambiguous). But consider for a second this:

Assume A is true always, but he says B is true always. Is A still ambiguous or contradictory?

Assume B is always true, but he himself admits that he is ambiguous (because B stated that he sometimes tells the truth). Then B is the liar, though not always. He is ambiguous in this context.

Are you suggesting that A is actually ambiguous and not B?

A sometimes tells the truth, and his statement in this riddle just happens to be a lie. Presumably one can imagine a has told the truth at some other occasion. — flannel jesus

Exactly. That is why I questioned whether A contradicts his own assertion or if A, like B, is simply ambiguous. At least, we both believe that C is the lone truth-teller.
Then B is always false and ambiguous. But what happens to A?

That only leaves C as the guy who always tells the truth — flannel jesus

I had the same thoughts about C. A is a liar, and B is ambiguous , so I believe C is the lone truth-teller. But I was wondering if I had properly written those in logic language because the riddle statement requested if it could be formulated in first-order logic or not. Since you did not criticise my initial question, I assume I formulated it correctly.

The rest naturally follows — flannel jesus

The liar is B, but also A. Is it contradictory or ambiguous?

Okay. I will be more clear.

is he saying he's definitely not the guy who always tells the truth? — flannel jesus

Yes.

when he says he sometimes tells the truth, is he saying he's definitely not the guy who always tells the truth? — flannel jesus

B says the truth or not but not both. B says the truth often. B also lies often. B doesn’t say both. The ambiguity of B is the core of this riddle. I copied and pasted it as it is written on the internet. I haven’t altered anything. We can conclude he is not definitely the person who always tell the truth, yes. Nonetheless, A claims B always tells the truth.

This is a clear contradiction to the riddle. Right?

B is the liar — flannel jesus

Exactly.

'1+1 = 2' means that the value of the expression '1+1' is the same as the value of the expression '2'. — TonesInDeepFreeze

Why does it take so long to understand an axiom that appears so simple?

I once read Bertrand Russell's works, and one of them was Principia Mathematica. Well, it took him and another colleague of his more than 300 pages to prove that 1+1 = 2. I acknowledge that I struggled to understand some of their pages and axioms due to my lack of familiarity with logical language. There was even some criticism of the work of Russell and Whitehead because it seems the work was based on finding 'truth logic' 

I asked myself then. Is 1 + 1 = 2 a logical truth? And I found on the Internet big debates among mathematicians and logicians about whether it is a tautology, a logical truth, or a theorem.

In the following link (1 + 1 = 2) you will see similar answers to yours: 1 + 1 = 2 is a 'definition'.
2 is another way of defining '1 + 1' if I am not mistaken...

Angelos Sikelianos. Selected Poems. Translated and introduced by Edmund Keeley and Philip Sherrard.

Kazantzakis once said that if he were awarded the Nobel Prize in Literature, he would only accept it if he could share it with Sikelianos.

Identity. — TonesInDeepFreeze

The number of things is not the things. — TonesInDeepFreeze

:up:

And, in mathematics it is very clear that "=" is not defined as "is". — Metaphysician Undercover

Could it be defined as "equals to..."?

Mathematics adheres to the law of identity, since in mathematics, for any x, x=x, which is to say, for any x, x is x. — TonesInDeepFreeze

Are you referring to 'is' in terms of identity or value? For example: 5 is 5 in both mathematics and in our understanding of numerical systems. Meanwhile, £5 doesn't equal $5 or €5 because of the disparity in monetary value. Although every bill or note is represented by the payment of x5, it will depend on the value. So, x = x, doesn't equal to "is." To apply this, I need to carefully consider the specific context. Right? 

I'm sorry to be a bit abrupt, but if you don't keep your feet on the ground, you're bound to lose contact with reality. — Ludwig V

You are right. I am the one who apologises for derailing the topic in an inconsistent scenario. I thought the non-existence of a dog was a fascinating topic to discuss, but I admit that I overreacted.

Is there a non-existing dog? If there is, it doesn't exist. If there isn't, it doesn't exist. — Ludwig V

Don’t get me wrong. I explained myself mistakenly. It is true that you didn’t mention the non-existing dog, and I think Athena never thought about it either. But since this mysterious dog showed up in this game yesterday, I started to think about his interference in the counting. Well, if we imagine there is actually a dog who doesn’t eat anything, it means that it should be represented with a zero (0) in the counting. As ssu pointed out, it took a while for Western mathematics to accept zero as a number. According to this issue, maybe Plato would never have taken the dog who doesn’t eat anything into account, but yet it is clear we should take the dog into account, and thus, the dog exists. Right? 

There cannot be a dog that eats the most - there's bound to be another one that eats more. Similarly for the dog that eats the least. Infinity doesn't follow the normal rules. — Ludwig V

Sorry, I was foolish in trying to follow usual norms when infinity is involved. :sweat:

Well, strictly speaking they are identified by the amount of food they eat, which determines their position in the line. — Ludwig V

Ah, the so-called non-existing dog is the one who doesn’t anything at all. I get it now. But I assumed every dog ate at least a bit.

So, since they are identical in every way, apart from the amount of food they eat, there is no other way to identify them.
It is easy to think that they must exist, but if the line is infinite, any specified dog has another dog after it. — Ludwig V

Yes, this is how I see the tricky game. If I'm not mistaken, the dog who eats less than the preceding dog would be represented by 0.00000000…, and so on. However, this dog does exist. It consumes something, even when it is infimum.

Thanks, ssu. A great and very well-written post. I appreciate your teachings. :up:
By the way, @Ludwig V has stated something interesting:

Your ordering means you have to start from a dog that you cannot identify. — Ludwig V

It is true that my knowledge of mathematics and logic is pretty limited.  Yet, if I understand the rules of this entertaining game correctly, the counting starts with two identified dogs. The one at the top (the dog who eats the most) and the one at the bottom (the dog who eats the least). Honestly, I think those two are always ‘there’ but it is a mistake to try to identify them with numbers. I follow Zeno’s point as indicative. This is why Plato was wrong in this game. He forgot to count the two 'axiomatic' or 'affirmative premise' dogs. I don't even sure what to call these two (maybe Teo and Sarah :lol: ). As ssu pointed out, the transcendental dogs are the sole obstacle in following Zeno's point. These exist, but everything becomes complicated if we are fixated with labelling the dogs in numerical sequence.