Perhaps though the distinctions can be quite less than sharp. — TonesInDeepFreeze

Perhaps. The matter of mathematics being invented or discovered is absolutely derivative from the ontology. It is possible that the foundations are also derivative from ontology or vice versa.

An interpretation, aka 'a model'. — TonesInDeepFreeze

Ah.

But that does not imply that "Winston Churchill was French" is true in all interpretations, but it does imply that "Winston Churchill was French" is true in at least one interpretation. — TonesInDeepFreeze

Can we say the proposition "Winston Churchill was French" is A? If so, if A is True, what do you make of the following table?



In the interpretation where ¬(a → (b ∧ ¬b)) is T, A is always T.

realist

logicist

formalist

structuralist

constructivist — TonesInDeepFreeze

I think the first and fourth are about the metaphysics of mathematics, while the second, third, and fifth are about the foundations of mathematics, so then they would be grouped in two separate sets. Some of the questions in my thread are also about what arrows we are supposed to draw between those two supposed sets.

I mean, can't we regard 'logical axiom' as merely a logical notion without ontological commitment? — TonesInDeepFreeze

I would imagine so, there is nothing about logical statements to me that imply an ontological commitment. On the other hand, my problem is that I am very unpersuaded by platonism, so I am not keen on spotting flaws in arguments against it.

Classification is not depentent on Platonism or platonism. — Gregory

I would agree with that, but that is because I am not a platonist or a Platonist, I lean towards nominalism.

We use language however in a Platonic way — Gregory

I don't think we can use language in a Platonic or nominalistic way, since these two are already related to how we use language.

So you are asking "couldn't a formalist not be a nominalist?" — TonesInDeepFreeze

That, and also "Couldn't a logicist not be a platonist?".

It does, just indirectly.

That is one of the questions I ask in my Grundlagenkrise thread :^)

b. The article associates formalism with nominalism, logicism with realism, and intuitionism with conceptualism. The last one seems uncontroversial, but how true are the first two? Couldn’t a logicist also be a nominalist? Why does reduction of mathematics to logical propositions have to imply numbers as abstract objects? — Lionino

"Winston Churchill was French" does not imply a contradiction. But that does not imply that "Winston Churchill was French" is true. — TonesInDeepFreeze

Of course.



I don't know what to make of your use of "interpretation".

Your post is not coherent. If you rewrite it I can give a reply.

I don't know off the top of my head. Maybe someone else can answer.

It is not the case that if A then both B and not-B — TonesInDeepFreeze

Isn't this the same as "if A then contradiction"?

It is not the case that if A then B&~B implies A. — TonesInDeepFreeze

Is this supposed to be the translation of ~(A→(B &~B)) implies A?

So what is the proper translation of 2?

Does it not?
¬(A → (B ∧ ¬B))→A is valid
¬(A → (B ∧ ¬B))↔A is valid

Saw this graph. Not sure if it is true

But who said I'm not a Platonist? I am? When it suits my argument. I'm a formalist as well at times. — fishfry

Are the two really mutually exclusive?

A shot into a crowd of Trump supporters and you hit a working man with a family.
A shot into a crowd of antifa/BLM, like in Kenosha, and you hit sex offenders and pedophiles.

Food for thought.

How many quantum numbers are there to describe an electron orbiting around an atomic nucleus?

For Aristotle, the universals only exist where they are instantiated, e.g. in triangular things. — Count Timothy von Icarus

Curious that the greatest genius of history agrees with me on virtually every issue.

But to the point, isn't that view a sort of immanent realism of universals? It surely reminds me of mathematical immanent realism.

So let me ask directly.

@TonesInDeepFreeze

Let a proposition P be A→(B∧¬B)
Whenever P is 1, A is 0.
In natural language, we might say: when it is true that A implies a contradiction, we know A is false.

Now a proposition Q: ¬(A→(B∧¬B))
Whenever Q is 1, A is 1.
Do you think it is correct to translate this as: when it is not true that A implies a contradiction, we know A is true?

Isn't this a fairly big problem given that (¬¬A↔A)? — Leontiskos

Yes, I am waiting for someone to clarify what is up with that.

I had written more under my own post but got myself confused also on whether the letter here is a statement or a variable.

¬(a→(b∧¬b)) → a — Leontiskos

Yes. It is what I say here.

But let's say (a→b)∧(a→¬b) is False, does that mean A is true? That is what the logical tables would say: — Lionino

But, as my second phrase in this post, I got confused as to whether ¬(a→(b∧¬b)) means a variable that is in contradiction with (a→(b∧¬b)) or simply that (a→(b∧¬b)) is False.

¬(a→(b∧¬b)) is only ever True (meaning A does not imply a contradiction) when A is True. But I think it might be we are putting the horse before the cart. It is not that ¬(a→(b∧¬b)) being True makes A True, but that, due to the definition of material implication, ¬(a→(b∧¬b)) can only be True if A is true. Likewise, what I say here in the first post:

and the way material implication works in classical logic is that, if the antecedent is false, the implication is always true — Lionino

The denial of the implication can only be True if the antecedent is True. If the antecedent is False, the denial of the implication is False.

1 – We know (a→b)∧(a→¬b) is the same as a→(b∧¬b), and (b∧¬b) is a contradiction. So (a→b)∧(a→¬b) just means A implies a contradiction. If (a→b)∧(a→¬b) is True, A cannot be True, it has to be False. But let's say (a→b)∧(a→¬b) is False, does that mean A is true? That is what the logical tables would say: — Lionino

But (a→b)∧(a→¬b) being False simply means that A does not imply a contradiction, it should not mean A is True automatically.

((a→b)∧(a→¬b))↔¬a is valid — Lionino

That is true if "both props" is understood as (A → B) ^ (A → ¬B) and "imply ¬A" as the proposition being True means A is False — Lionino

((a→b)∧(a→¬b))↔¬a

The ↔ operator means that everytime the left side is 1 (True), the right side is also 1, and same for 0 (False). So (a→b)∧(a→¬b) does not mean that A is False, unless we say (a→b)∧(a→¬b) is True. If (a→b)∧(a→¬b) is False, ¬A is False, so A can be True or not¹. So (a→b)∧(a→¬b) may be taken as a proposition p which can take the values of 0 or 1 as well.

More confusions stemming from whether 'A' means 'A is true' or a variable which may take the values True or False. It would be better if there were a way to easily distinguish the two. Perhaps a doctor thesis for someone out there.

1 – We know (a→b)∧(a→¬b) is the same as a→(b∧¬b), and (b∧¬b) is a contradiction. So (a→b)∧(a→¬b) just means A implies a contradiction. If (a→b)∧(a→¬b) is True, A cannot be True, it has to be False. But let's say (a→b)∧(a→¬b) is False, does that mean A is true? That is what the logical tables would say:

Pathetic attempt at a humble brag and confession of historical illiteracy. The orange billionaire wish he had one hundreth of the culture and education that Mussolini had, and you can't paint a Hitler moustache on someone and then call them a "fascist", that's fucking stupid.

If these threads https://thephilosophyforum.com/discussion/9705/god https://thephilosophyforum.com/discussion/15172/is-atheism-illogical are something to go by: not uncharitable enough

a bit eccentric — fishfry

Let's not be so open-minded our brains fall out.

plant-based meat alternatives — LFranc

Yeaaaaah... no. All the power to them, but let's not throw reality out of the window. Same with solar panels.

I could raise a few objections about that, but I will not. Instead, I will quote myself about a specific kind of fictionalist:

b. Field's fictionalism is that mathematical statements are false, and mathematical statements are given with a fictional operator: "According to arithmetics, there are infinitely many prime numbers". Whereas without the operator, the statement would be false, as numbers don't exist (standard semantics). — Lionino

And though a Pyrrhonian may throw himself or others into a momentary amazement and confusion by his profound reasonings; the first and most trivial event in life will put to flight all his doubts and scruples, and leave him the same, in every point of action and speculation, with the philosophers of every other sect, or with those who never concerned themselves in any philosophical researches. When he awakes from his dream, he will be the first to join in the laugh against himself, and to confess, that all his objections are mere amusement, and can have no other tendency than to show the whimsical condition of mankind... — David Hume, An Enquiry Concerning Human Understanding, § xii, 128

Aka: everybody is a realist when they walk out of the door.

Isn't The Matrix about vegetarianism?

And basically bijections are equations like y=f(x) — ssu

Just a nitpick. Not every f(x) function is bijective. I don't think there is a general form of a bijective function. 1+1=2 is not bijective either because it is not a function.

an imaginary problem — Leontiskos

On analytic philosophy and thought experiments, a post I read elsewhere might be funny:

Yes. It is what I said.

Even in my attempt I had to change the definition of an operator. — Lionino

The "or" operator must not be traditionally understood, rather it should be understood as "It can be one or the other, both are possible". So A∨¬A means A can be True or False. — Lionino

Using first-order logic is really improductive for this purpose. Even in my attempt I had to change the definition of an operator. Since one of the results is "Sometimes T or F", we are dealing with a three-valued logic. That would be the meaning that my or-operator takes.

Possible image of shooter

That A is coherent but ambiguous or what — javi2541997

A's statement can be either True or False. Since it can be either True or False, we can choose whichever value agrees with the other others. So the inconsistency has to be somewhere else if there is one. But there isn't any.

B -> A ∧ B = ¬A — javi2541997

Not sure what that means.

The shooter shows up in BlackRock commercial at 0:03 and 0:19 https://www.youtube.com/watch?v=_67dyF4J7ag
The upload has been removed by the uploader

Image shows a school in Bethel Park, PA, where Thomas M. Crooks lived (not to be confused with Thomas Crooks also from Bethel Park).
Here is reupload https://streamable.com/mzfq3w

The attempt is the logical conclusion of anti-Trumpism. If you repeat long enough that another human being is an existential threat it won’t be long before someone takes action. It was only a matter of time until the persecution reached murderous levels. — NOS4A2

Happened 6 years ago already https://brazilian.report/liveblog/politics-insider/2024/06/11/stabbed-bolsonaro-lone-wolf-feds/



Random guy on Youtube said it first https://www.youtube.com/watch?v=Ey0qVzG8_vU&t=666s Check out timestamp

Strange. I saw an image of Trump still up speaking while the sniper is being mounted (if understood correctly they knew of the shooter but didn't take Trump down).

Derivative problem. If you are a platonist, you think math is invented, if you are a nominalist or conceptualist, you think math is discovered. — Lionino

LOL

I messed that up. It is the inverse, fixed now.

@flannel jesus @Michael
Check it out

I made an attempt to formulate the 3 statements:

A ∧ B. A is true because B is true.
B ∨ = ¬B. B says the truth or not but not both.
¬(C ∨ B). C says the truth as a negation of B. — javi2541997

I would rather say: A→B, B→(B∨¬B), C→¬B.
Each variable here must be understood as True, so A→B is not a formula where A and B can take either value but a proposition stating "If A is true, B is true". Saying "A" amounts to saying "A is true".
The "or" operator must not be traditionally understood, rather it should be understood as "It can be one or the other, both are possible". So A∨¬A means A can be True or False. Otherwise B∨¬B is a tautology, but here it means the variable can take either values.

Assumption: A tells the truth.
There is only one scenario where A→B is True and A is True: when both A and B are True. So in this case none of these two would be the liar. That leaves us with C as the liar. So C→¬B is False, this is only ever true when B and C are True. So we have a contradiction. The assumption is therefore wrong.
Assumption: A lies.
The only scenario where A→B is False is when A is True and B is False. This is a contradiction.
We are left with A sometimes tells the truth.

Now on to the other two. Since B talks about itself, let's go with C.
Assumption: C lies.
C→¬B is False only when B and C are True. Contradiction.
So C must tell the truth, as A sometimes tells the truth, and B is the liar.

Let's check for consistency:
B→(B∨¬B)
B is False. So (B∨¬B) is False, it is always the case that ¬B.
C is True. So C→¬B is True, that is the case when C is True and B is false. So far so good.
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.

I have an odd feeling about my solution, but the result agrees with others.