"No, the speaker might know that the book is in the car but choose to be coy, though literally honest and correct, in saying "The book might be in the car". If I was looking for the book, then I would not appreciate my friend being coy that way, but he would not be logically incorrect." - TonesInDeepFreeze

The crank replied to that I refute myself by admitting that I would not appreciate the coyness and that the issue is a moral one.

That's the kind of reply made by someone who doesn't know how to discuss philosophy.

(1) The bit about coyness was merely for flavor. We can leave it off:

The speaker might know that the book is in the car but still be literally honest and correct, in saying "The book might be in the car".

If your friend asks, "Where is the book?" and you don't know and answer, "I don't know, but it might be in the car", or if you don't know and answer just, "It might be in the car", then you don't think your friend is a liar for that!

That is tremendously obvious not just philosophically but in everyday communication.

Not knowing whether proposition Q is true does not preclude that Q might be true.

I can't believe this even needs to be belabored.

(2) Consider another example even including the coyness bit:

Your birthday is soon. You ask your friend whether there will be a party. He says, "There might be, and I'm not saying more". Then there is a party, and you find out that your friend knew about it all along, and you do appreciate his coyness because it preserved a welcome suspense and surprise. And, by the way, what he said is true in both instances, and in both instances, he did not lie.

It's ridiculous that one should even have to explain such things to the crank, but I do in the interest of an abundance of refuting his utterly wrongheaded thinking.

(3) And, obviously, we don't refute a basic understanding of the mere modality of 'possibly' with regard to epistemic considerations by going completely out of the ballpark by saying the modal notion is refuted on ethical grounds!

/

One more time:

"I don't know Q" is not inconsistent with "Possibly Q".

No rational person thinks otherwise.

and

"Necessarily Q" is not inconsistent with "Possibily Q"

No rational person thinks otherwise. Or at least, no rational person informed about modal logic thinks otherwise.

/

Somehow, contrary to both basic philosophy and everyday language, some people have jumped to the conclusion that 'Possibly' is the negation of 'Necessary'. There is no rational reason to jump to that conclusion. Jumping to that conclusion seems to me to be a function of people not stopping to think that negation is not the only differing relation between concepts. The relation here is not negation but rather of duals.

Let q, Q, R be any sentences:

(1) 'necessary' ('N') is primitive, not defined. 'possibly' ('P') is defined, not primitive.

* The modal operators are duals, not negations, of each other.

df. Pq <-> ~N~q

thm. Nq <-> ~P~q

That is NOT equivalent with:

Pq <->~Nq

That is NOT a definition used in basic modal logic.

And NOT equivalent with:

Nq <-> ~Pq

That is NOT a theorem of basic modal logic.

The relation is not of negation but of duals.

P is the dual of N. And N is the dual of P.

* Just as the the quantifiers are duals, not negations, of each other:

df. ExQ <-> ~Ax~Q

thm. AxQ <-> ~Ex~Q

That is NOT equivalent with:

ExQ <-> ~AxQ

That is NOT a definition used in quantifier logic.

And NOT equivalent with:

AxQ <-> ~ExQ

That is NOT a theorem of quantifier logic.

The relation is not of negation but of duals.

The existential quantifier is the dual of the universal quantifier. And the universal quantifier is the dual of the existential quantifier.

* And note how 'all' and 'some' correspond with 'necessary' and 'possible'. Roughly stated:

"for all x, Q" is true if and only if Q is true for all x

"for some x, Q" is true if and only if Q is true for at least one x

and

q is necessary if and only if q is true in all worlds

q is possible if and only if q is true in at least one world

* Just as 'and' and 'or' are duals, not negations, of each other:

df. (Q or R ) <-> ~(~Q & ~R)

thm. (Q & R) <-> ~(~Q or ~R)

That is NOT equivalent with:

(Q or R ) <-> ~(Q & R)

That is NOT a definitions used in sentential logic.

And NOT equivalent with:

(Q & R) <-> ~(Q or R)

That is NOT a theorem of sentential logic.

The relation is not of negation but of duals.

Disjunction is the dual of conjunction. And conjunction is the dual of disjunction.

/

And to refute a confusion of the crank:

The crank mentions that we use the phrase 'possible worlds' in "q is necessary if and only if q is true in all possible worlds" and then we define 'possible' in terms of 'necessary'.

But 'possible' in 'possible worlds' is merely for intuition and is not at all needed formally. The semantics for modal logics need only mention 'worlds' (for that matter, not even 'worlds' needs to be mentioned as indeed "worlds" are merely members of a certain set that is part of a structure).

Moreover, we do not define 'necessary'. It is primitive. But we do go on to adopt semantics and axioms so that it is a theorem (not a definition) that, roughly put, Nq if and only if q is true in all worlds.

Also, as mentioned, we define 'possible' in terms of the primitive 'necessary'. But we recognize that we could do it in reverse: we could take 'possible' as primitive and define 'necessary' in terms of 'possible':

df. Nq <-> ~P~q

But that is not circularity. In any given treatment of the subject, we commit to one or the other but not both: 'necessary' is primitive or 'possible' is primitive.

But the whole point of Hilbert's hotel is that it can take in more guests. If it kicks guests out as it takes new guests in it's not actually able to hold more. — keystone

It never holds more than denumerably many guests.

The reason why you think 0.9[...] is rational is because you believe it equals 1 — keystone

Wrong. We prove that any decimal expansion that has an infinitely repeating part represents a rational number. We may also conclude that 0.9[...] is rational because it is 1, but we don't have to do that just to prove that it is a rational number.

1) You provide any real number
2) I convert it to binary
3) Using the bijection, I find the correspond number within the range (0,1)
4) That number has a meaning in the hotel manager's system — keystone

Yes, every denumerable binary sequence corresponds to a real number in [0 1]. And in a story tale, every binary denumerable binary sequence "codes" whether there is nor is not a guest in the room at that position in the sequence. No one disputes that.

I understand the claim that any number that infinitely repeats a finite sequence after the decimal point is a rational number. I know it's a basic and conventional idea. What I'm saying is that this claim rests on the notion of limits. — keystone

Please provide a basis or source for that claim. I would have to refresh my memory; I don't recall whether the proof of "every decimal expansion that has an infinitely repeating part represents a rational number" must use limits. I am highly skeptical though of your claim that it does. Actually, though, since 'limit' is itself a defined notion, any proof that uses the notion could skip the notion.

Without limits, I don't think you can even prove that 0.9[...]-0.9[...]=0? — keystone

That one is a whopper of ignorance.

(1) We prove the theorem Ax(x is a real number -> x-x = 0) without having to invoke the notion of limits. We do it right from the field axioms (and the definition of subtraction), which are themselves theorems of set theory per any of the constructions of the real number system.

Just considering the inverse law of addition is as basic as high school algebra. But here you flaunt your ignorance of even that. I never get over being amazed at the hubris of people who think they are showing flaws in mathematics that they know virtually nothing about.

(2) Since 'limit' is itself a defined term, no proof requires invoking it.

Your criticism of my story was of an inconsequential intermediate step. And even now, you focusing on the program is secondary. — keystone

It was a steep you mentioned in defense of your argument. My criticism of that step is correct. And you presented your program to me also in defense of your argument, so I pointed out your program doesn't help your argument.

[From] the axioms of set theory, we derive the theorems of calculus. — TonesInDeepFreeze

Ridiculous. — Deus

What I said is true.

Calculus was developed well before set theory came into the scene. — TonesInDeepFreeze

Yes, and it was axiomatized by set theory. What I said is true: From the axioms of set theory, we derive the theorems of calculus.

Also in the field of mathematics it’s nothing more than a minor development/distraction — TonesInDeepFreeze

That is a sweeping and ignorant statement.

Mathematics for the sciences? — keystone

Yes, from the axioms of set theory, we derive the theorems of calculus.

I'm working in Hilbert's fictional realm. — keystone

Set theory is abstract. It doesn't have hotels. To be more exact, I should say that from an imaginary analogy to set theory, you impose an incoherent interpretation. It's incoherent because you start out by describing a program to output values (presumably in a certain order) but it's not a program.

I exhausted loads of my time and patience with Thomson's lamp with you. You're making a variation of the same mistake here.

Printing it early is no trick.

the program has value even though we cannot literally go the limit. — keystone

".89[...]" is notation for a limit. And that limit is .9. And ".9[...]' is also notation for a limit. .9[...] = 1.

You don't like that mathematics for the sciences doesn't comport with your understanding of impossible fictional realms. Yeah, that's a real dagger in the heart of the mathematics for the sciences.

eventually output a 1 — keystone

Then it will miss outputting one of the 9s.

You can't have cake and eat it too.

If it runs only finitely many steps but outputs the 1, then it skips an infinite number of the 9s.

If it runs without end, then it outputs each of the 9s, but never outputs the 1.

it will never print a 1, but I can still write the program — keystone

You can write whatever you like, but my point, as seen in context, is that it's not a program to print all the entries in the sequence.

Consider the ascending sequence of members of w+1 (omega plus one). That is a denumerable sequence with a last entry. But there is no program to write all the entries in that ascending order. On the other hand, it's trivial to have a program write:

1, 1/2, 1/4 ...

while 0 is not an output.

That's just a starker example of what you're doing. Yes, it's a program, and it outputs every successive halving. But 1 is not an output of the program.

He’s referring to the halting problem in relation to turings complete machine — Deus

No he's not. The halting problem is not that there are programs that don't halt. But rather that there is no program to decide whether any given program and input will halt.

You don't know what you're talking about. You're just throwing out red herrings. A form of trolling.

a teetotaller abstains from drinking on all occasions apart from when he unknowingly drinks alcohol because his wife can’t be trusted. — Deus

Depends on the exact interpretation of a given definition. If taken literally in the sense of 'practices complete abstinence' then drinking alcohol even inadvertently makes one no longer a teetotaler at that moment. But I grant that ordinarily, probably most people wouldn't regard that as failure to maintain being a teetotaler; and I overlooked that possible situation. So, I'll give you not a full point for that one, but at least most of a point.

I can certainly write a program to output digits corresponding to 0.89[...]1. It's just that that program can never be executed to completion so it would never reach a moment where it would output a 1 digit. — keystone

Whatever you have in mind, it's not a program. If P is a program to print the entries in a denumerable sequence, then for each entry, there is step at which that entry is printed.

As in other threads, you're using technical sounding verbiage without regard for making sense with it.

Do you claim that teatotallers are not drunktrollers ? — Deus

The word is 'teetotaler'. No teetotaler is drunk, therefore no teetotaler is both drunk and a troller.

I didn't assume anyone is drunk.

Refer to the subject of non-standard analysis in mathematical logic (starting with Abraham Robinson), or, with a different method, internal set theory. You may consult many a book or article. Probably many articles on online.

You think it is incorrect to say that no teetotaler is drunk?

A teatotaller <> drunktroller — Deus

I don't know what '<>' is meant to symbolize. If it is for some form of equivalence, it's the opposite of what I said. No teetotaler is drunk.

nor am I interested in debating the validity of non-standard analysis — keystone

Infinitesimals are rigorously handled in non-standard analysis. It's not a question of validity.

I didn't know whether you're talking about yourself or about me. (And I couldn't resist the wordplay.) In any case, if your point is that you're better at least for not being drunk, then congrats.

First you called me a crank troll but in the above statement I’m only referred to as troll. — Deus

Both.

drunk troller — Deus

A teetotaler can't be a drunktroller.

a logician yourself — Deus

I'm not.

The point is not to insult, but rather to flag the situation. Usually cranks are not trolls, since they are sincere, though horribly self-misguided. But I take you as a troll since you don't even offer arguments but just simple flippant nonsense.

'flexible enough', 'deal with operations', 'render useless'.

Not even philosophical, let alone mathematical.

Deus is a crank troller.

Define 'self-limiting theory'.

Okay, I get it. You're trolling.

Proof that 0.89'1 = 0.9 — Real Gone Cat

(Using the apostrophe for '...')

for any x an y,

x'y

doesn't stand for an real number expansion.

In ordinary set theoretic context, there is no object called 'infinity' that is an operand in an addition operation.

♾ — Deus

Is that supposed to be the leminscate?

the countable infinity problem that was proved by Cantor — Deus

What specifically do you refer to ?

guys like Tones — Deus

There's only one.

I noted the first place in your original post that you spouted nonsense. Now you've revised. If I'm in the mood, I'll give you a second chance.

my notation of putting a digit after the repeating term is an interesting way of potentially representing an infinitesimal — keystone

That is more nonsense.

I don't know what that emoji means. I take it though that it doesn't indicate anything substantive.

Just trying to make sense of RAA. — Agent Smith

Just look at the truth table.

Anyway, you didn't use RAA.

Pretty much, you proved P from the premise ~~P. Congratulations. And your point is?

I'll use '*' instead of the overbar.

0.89*1

is not defined.

There is no real number that has an infinite decimal expansion but with a final entry.

Your imaginistic scenario, not even itself approaching a mathematical argument, not even of alternative mathematics, is done. Argument by undefined symbolism is a non-starter.

You are typical of cranks who argue with undefined terminology and symbolisms. Using terminology and symbolisms in merely impressionistic ways.

/

continuously growing but always finite — keystone

On your own finitistic terms, at any point, the sequence is finite. 'continuously growing' is never witnessed. Only finitely many individual finite sequences.

I'm just unsure why you're characterizing modal logic as ones that deal with existence predicates — Kuro

I agreed that existence predicates are handled in systems other than modal logic. And I'm not claiming that every version of modal logic in basic forms includes the advanced subject of an existence predicate.

most modal logics are standardly extensions of FOL with K and some of the additional modal axioms, and therefore do not express nontrivial existence predicates. — Kuro

But in the overall subject of modal logic, we do find a definition an existence predicate. We find that in textbooks such as Hughes & Cresswell (among the preeminent introductions to modal logic) and L.T.F. Gamut. I'm highlighting modal logic for this subject only because one is more likely to encounter a course in, or textbook on, modal logic before some of the other advanced alternative logics.

I have no interest in convincing you or anyone else not to investigate existence predicates in whatever logic systems you or anyone else wishes to study them in whatever order you or anyone else wishes to study them.

/

It's been a while since I studied this, but, if I recall correctly, Hughes & Cresswell and L.T.F. Gamut do define an existence predicate in modal logic that is an extension of classical FOL=. (I'll happily stand corrected though if I my memory is incorrect.)

most logics with existence predicates are not modal — Kuro

But modal logic is the more common one to study than all the others combined. (That's not an argument that modal logic is "better" or anything like that, just that it's natural enough to first turn to modal logic, as a common subject, to see what it offers, while not precluding that the number of other approaches is potentially inexhaustible too.)

Yes, for any rational being it is not plausible that for all q we have Pq -> ~q.

But — Srap Tasmaner

And that 'but' is not going refute that it is not the case that for all q we have Pq -> ~q.

it is also the case that "It might be in the car" implicates (but does not entail) "I don't know for sure where it is" — Srap Tasmaner

No, the speaker might know that the book is in the car but choose to be coy, though literally honest and correct, in saying "The book might be in the car". If I was looking for the book, then I would not appreciate my friend being coy that way, but he would not be logically incorrect.

Or, let 'Kq' stand for 'q is known'. Let 'L' stand for '~K~q'.

For any rational being it is not plausible that for all q we have Lq -> ~q.

Or, let 'Bq' stand for 'q is believed'. Let 'Cq' stand for '~B~q'.

For any rational being it is not plausible that for all q we have Cq -> ~q.

Anyway, the point stands, only a nutcase says that "Possibly the book is in the car" implies that the book is not in the car.

Let's make it a life and death situation:

A young boy is lost in treacherous terrain. The county sheriff's search and rescue expert tells the parents, "Possibly he's in the canyon. So he's not in the canyon." I don't think there is any parent in the world who would say, "Okay, I understand your logic perfectly. Let's not waste time looking in the canyon."

/

Unrelated but poignant is Sartre's "The Wall". SPOILER ALERT. In the Spanish Civil war, Pablo is a prisoner of the fascists. His imprisoners will execute him if he doesn't give up the hiding place of his comrade Ramon. Pablo believes Ramon is not hiding in the nearby graveyard. As a joke on his imprisoners, Pablo lies to them that Ramon is hiding in the graveyard. But Ramon is hiding in the graveyard. And later Pablo learns that Ramon was caught in the graveyard and killed.

"I'm the only one here who is right. Everyone else is wrong. I have an open mind. They don't". Thus spake the crank.