You can't prove arithmetic from arithmetic because we created it. — Philosophim
Do [we] have uncountably many interpretations? — fishfry
Provability, if I have understood it correctly, means that a truth of a statement/conjecture can be derived from some axiomatic system or logical rules. — ssu
With diagonalization, we get only an indirect proof. — ssu
Is diagonalization a way to find mathematical statements that cannot be proven by a direct proof, but only can shown to be true by reductio ad absurdum? — ssu
What the Turing Machine cannot compute is found exactly by using diagonalization (or negative self-reference) that we are talking in the first place. — ssu
"In particular, the consistency of P is unprovable in P, assuming that P is consistent. (in the contrary case, of course, every statement is provable)." - Godel (His proof of this being very slippery.) — tim wood
For each element B in P(S0), form the sentence that is the coordinate conjunction of all the sentences in B. — The Vastness of Natural Language
considering the first incompleteness theorem:
PA ⊢ ∃ A ( A ⇔ ¬Bew(ÍAÎ ) ) — Tarskian
Hilbert believed it so strongly that he insisted that all his colleagues should work on proving the above. — Tarskian
untested experimental vaccine shots — Tarskian
As an non-mathematician/logician, I'm not familiar with the terminology. So it is sentence - proof sequence - axioms? I still assume there is a link between sentence and the set of axioms.A sentence is provable from a set of axioms and set of inference rules if and only if there is a proof sequence (or tree, tableaux, etc.) resulting with the statement. — TonesInDeepFreeze
Diagonalization itself of course doesn't require an indirect proof. What I meant that it itself is an indirect proof: first is assumed that all reals, lets say on the range, (0 to 1) can be listed and from this list through diagonalization is a made a real that is cannot be on the list. Hence not all the reals can be listed and hence no 1-to-1 correspondence with natural numbers. Reductio ad absurdum.No, diagonalization does not require indirect proof. — TonesInDeepFreeze
What the Turing Machine cannot compute is found exactly by using diagonalization (or negative self-reference) that we are talking in the first place. — ssu
Exactly.I think the theorem you have in mind is that there is no algorithm that decides whether a program and input halt. The proof uses diagonalization. — TonesInDeepFreeze
Yes. Obviously Turing constructed a quite important and remarkable proof for the uncomputability of the Entscheidungsproblem. But is that constructiveness a problem?But, again, the proof is constructive. Given an algorithm, we construct a program and input such that the algorithm does not decide whether the program halts with that input. — TonesInDeepFreeze
So the existence statement is in the meta-theory, not in PA. — TonesInDeepFreeze
∃ A ( PA ⊢ ( A ⇔ ¬Bew(ÍAÎ ) ) )
PA ⊢ ∃ A ( A ⇔ ¬Bew(ÍAÎ ) )
What untested vaccines? (Of course, they're untested for the people who are taking them in tests.) — TonesInDeepFreeze
https://www.cancerresearchuk.org/about-cancer/find-a-clinical-trial/how-clinical-trials-are-planned-and-organised/how-long-it-takes-for-a-new-drug-to-go-through-clinical-trials
Drug testing and licensing
All new drugs and treatments have to be thoroughly tested before they are licensed and available for patients.
A new drug is first studied in the laboratory. If it looks promising, it is carefully studied in people. It may be then licensed if the trial shows that it works well and doesn’t cause too many side effects. You may hear this process called ‘from bench to bedside’.
There is no typical length of time it takes for a drug to be tested and approved. It might take 10 to 15 years or more to complete all 3 phases of clinical trials before the licensing stage. But this time span varies a lot.
There are many factors that affect how long it takes for a drug to be licensed.
He proposed the project. But he insisted that all of them undertake it? Moreover, is there even one colleague to whom Hilbert insisted the colleague undertake it? — TonesInDeepFreeze
https://en.wikipedia.org/wiki/Hilbert%27s_problems
Hilbert's problems are 23 problems in mathematics published by German mathematician David Hilbert in 1900. They were all unsolved at the time, and several proved to be very influential for 20th-century mathematics. Hilbert presented ten of the problems (1, 2, 6, 7, 8, 13, 16, 19, 21, and 22) at the Paris conference of the International Congress of Mathematicians, speaking on August 8 at the Sorbonne.
The other 21 problems have all received significant attention, and late into the 20th century work on these problems was still considered to be of the greatest importance.
Since 1900, mathematicians and mathematical organizations have announced problem lists but, with few exceptions, these have not had nearly as much influence nor generated as much work as Hilbert's problems.
You can't prove arithmetic from arithmetic because we created it.
— Philosophim
Given a theory adequate for a certain amount of arithmetic, for example, PA, it's redundant to say that the theory proves all its theorems. But if the theory is formal and consistent, then there are truths of arithmetic that are not provable in the theory. This has nothing to do with who "created" the theory. — TonesInDeepFreeze
So it is sentence - proof sequence - axioms? — ssu
What I meant that it itself is an indirect proof: first is assumed that all reals, lets say on the range, (0 to 1) can be listed — ssu
Turing constructed a quite important and remarkable proof for the uncomputability of the Entscheidungsproblem. But is that constructiveness a problem? — ssu
It is logic which proves arithmetic, not arithmetic itself. — Philosophim
So, according to your remark, the diagonal lemma should be phrased as:
∃ A ( PA ⊢ ( A ⇔ ¬Bew(ÍAÎ ) ) ) — Tarskian
A must also be a sentence in PA — Tarskian
If F(x) is a formula, then there exists a sentence S such that ('#' for 'the numeral for the Godel number of'):
PA |- S <-> F(#(S))
Applied to the negation of the provability predicate ('P' for the provability predicate):
There exists a sentence G such that:
PA |- G <-> ~P(#(G)) — TonesInDeepFreeze
PA is its own meta-theory in this case — Tarskian
If PA is not its own meta-theory in this case, what theory is then the meta-theory? — Tarskian
These vaccines were obviously not tested for long-term consequences. — Tarskian
David Hilbert had a habit of drawing the attention of the entire mathematical world to his agenda. — Tarskian
His program, however, was about proving a falsehood, such as "mathematics is complete" (i.e. there is a proof for every truth). Hilbert wanted other mathematicians to find proof for his misguided ideology. — Tarskian
Arithmetic can be reduced entirely to logic — Tarskian
Look at virtually any article or textbook to see that they are, modulo stylistic and symbol choices, along those lines and not with the existential quantifier in the scope of the turnstile. — TonesInDeepFreeze
Moreover, when we move on to mention 'truth', the language for PA cannot be its own meta-language. — TonesInDeepFreeze
https://math.stackexchange.com/questions/1717893/in-relatively-simple-words-what-could-be-a-model-of-sf-zfc
My question is: What does a model of ZFC look like?
The question seems to come down to a common confusion: how can a model of ZFC be a set, if we want to use ZFC to study sets?
The answer is, essentially, that things don't work that way. To study "models", we need to work in a metatheory that already has some concept of set or collection. The "metatheory" here is the theory that we use, as a tool, to study the object theory.
There are many options for such a metatheory. One option would be ZFC itself, except that (by the second incompleteness theorem), ZFC can't prove that there is a model of ZFC.
That should read "countably infinite." We can think of endless permutations of language, but we could also spend and infinite amount of time saying the names of the reals between any two natural numbers. — Count Timothy von Icarus
What specific reduction do you have in mind? — TonesInDeepFreeze
Nicely phrase. Our new chum is propounding much more than is supported by the maths. Here and elsewhere. — Banno
Any readable proof of Cantor's Theorem will contain at most a finite number of characters. Yet it shows can be used to show* that there are numbers sets* with a cardinality greater than ℵ0. — Banno
And we are faced again with the difference between what is said and what is shown. — Banno
So will we count the number of grammatical strings a natural language can produce, and count that as limiting what can be - what word will we choose - rendered? That seems somehow insufficient. — Banno
And here I might venture to use rendered as including both what can be said and what must instead be shown.
a
Somehow, despite consisting of a finite number of characters, both mathematics and English allow us to discuss transfinite issues. We understand more than is in the literal text; we understand from the ellipses that we are to carry on in the same way... And so on. — Banno
But further, we have a way of taking the rules and turning them on their heads, as Davidson shows in "A nice derangement of epitaphs". Much of the development of maths happens by doing just that, breaking the conventions. — Banno
Sometimes we follow the rules, sometimes we break them. No conclusion here, just a few notes. — Banno
I never understood it like that. In fact, I just always ignored it. I always saw it as PA talking about itself. — Tarskian
where is the truth of ZFC? — Tarskian
model theory is only straightforward for the simple case of PA being interpreted by ZFC's truth — Tarskian
smoke and mirrors — Tarskian
I don't know your criteria for straightforwardness, but model theory is rigorously developed, though it does use infinitistic mathematics. — TonesInDeepFreeze
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