Para-consistent logic

[Bill Spight]

No, the idea of the proof is :

1) Suppose that there is a largest number.
2) This number is unique (by definition of "the largest")
3) Because he exist and is unique, we can exhibit and name it : L
4) We consider the number M = L+1
5) M>L, so L is not the largest number
6) You have both "L is the largest number" and "L is not the largest number"
7) Point 6) implies that one of these assertions is false, and because our reasoning is valid, that means that "L is not the largest number"

But in this proof, nowhere you have infinity

I disagree, because if L stands for the maximum of N then if L is false that'd mean that N has no maximum and that'd mean it's an infinite set. The problem is: L needs to stand for all members of N, for all natural numbers, not just one. If L stands just for one number, even an arbitrary one, then the proof just shows that this specific but unknown number isn't the largest (but maybe another one).

So you think that a set of numbers may have more than one largest member?

No, but the point of the proof is to assume a largest number L and then find out that it is not, so the assumption was wrong in the first place. If a variable just stands for one number then this proof just shows that the number behind L is not the largest, but maybe L+1 is

No, L was the largest, by assumption. In that case there is no natural number, L+1 (or otherwise), which could also be the largest.

and there you have a regressus. My point is: to prove something for a set you have to use variables that somehow stand for all objects of the set and that's a problem, if you deal with infinitive sets, because in infinity there is no "all", it's an ever ongoing unclosed process.... (This is just to show you my concern, I am not a math guy, so I can be wrong easily.)

So you do not believe that all natural numbers are greater than zero, because you cannot prove that for each natural number, one at a time?

Edited for emphasis and clarity.

[paK0]

No, the idea of the proof is :

1) Suppose that there is a largest number.
2) This number is unique (by definition of "the largest")
3) Because he exist and is unique, we can exhibit and name it : L
4) We consider the number M = L+1
5) M>L, so L is not the largest number
6) You have both "L is the largest number" and "L is not the largest number"
7) Point 6) implies that one of these assertions is false, and because our reasoning is valid, that means that "L is not the largest number"

But in this proof, nowhere you have infinity

I disagree, because if L stands for the maximum of N then if L is false that'd mean that N has no maximum and that'd mean it's an infinite set. The problem is: L needs to stand for all members of N, for all natural numbers, not just one. If L stands just for one number, even an arbitrary one, then the proof just shows that this specific but unknown number isn't the largest (but maybe another one).

So you think that a set of numbers may have more than one largest member?

No, but the point of the proof is to assume a largest number L and then find out that it is not, so the assumption was wrong in the first place. If a variable just stands for one number then this proof just shows that the number behind L is not the largest, but maybe L+1 is

No, L was the largest, by assumption. In that case there is no natural number, L+1, which could be the largest.

L being the largest was in fact an assumption, which is pretty much the point of the proof. An assumption can be verified or falsified, in this case the assumption that L is the largest natural number was falsified

and there you have a regressus. My point is: to prove something for a set you have to use variables that somehow stand for all objects of the set and that's a problem, if you deal with infinitive sets, because in infinity there is no "all", it's an ever ongoing unclosed process.... (This is just to show you my concern, I am not a math guy, so I can be wrong easily.)

So you do not believe that all natural numbers are greater than zero, because you cannot prove that for each natural number, one at a time?

From a mathematical standpoint its actually quite easy to show (some) things for infinite numbers, as natural numbers are defined by the peano axioms.

The relevant ones for this example would be:
- 0 is part of the natural numbers
- Every natural number has a successor (which implies: S(m) = n => m < n)
- 0 is not the successor of any natural number

So you basically get:
x < 0
S(x) = 0 which is forbidden, leading to the conclusion that there is no natural number smaller than 0.

[SmoothOper]

I am with you Pippen. I am a believer in the axiom of determinancy which essentially disallows infinite proofs, which come to think of it, is how I happened across the para consistent logic article.

The axiom of determinacy concerns games of infinite length, so apparently you disagree with Pippen, since Pippen is (apparently) doing a bad job of expressing either finitism or ultrafinitism.

No actually the axiom of determinancy says that there are no games of infinite length. IE simply stating I can continuously generate larger numbers than you is not sufficient, at some point you have to stop generating larger numbers, all arguments must be of finite length.

[Bill Spight]

actually the axiom of determinancy says that there are no games of infinite length.

So no need for a ko or superko rule, right?

[Monadology]

That is not good, because it means that such a logic is not only inconsistent, but also inconsistent at its meta-level...and since we just do not understand what 'p and not-p' means it means that this logic just makes no sense. The world might make no sense, but it becomes not better to talk about no sense in no sense^^.

Priest is a dialetheist. He's advocating for inconsistency and it's begging the question in this matter to criticize him or his logic for allowing inconsistencies.

EDIT: There's also no reason to think that we don't know what 'p and not-p' means. Here's an example: "This sentence is false."

Here's another example, in the form of a representation of a situation in which a contradiction occurs.

One way to handle those examples is to suggest that the appearance of a contradiction is illusory. The other way is to take seriously that a contradiction is occurring in some form. But argument has to be made either way (and Priest, in his published works elsewhere, makes arguments for why the Liar Sentence should be taken to be an example of a genuine contradiction). So I don't see why, prima facie, we should assume that we don't know what 'p and not-p' means, given that we can have debates about purported examples.

If Priest was an esoteric who'd sell books and holy water we'd certainly not take him seriously^^

You're right. Fortunately, he's a very intelligent logician and philosopher who has written extensively on these technical issues in a rigorous and thoroughgoing manner.

let me not even start about guys like Hegel who basically sold gibberish as fundamental philosophy.

Hegel is not gibberish. Appropriately (given your opinion on Priest), Priest takes Hegel to be a crucial and important turning point in the history of philosophy. But setting Priest aside, there is enough clear-headed exposition of Hegel going on these days (For example, Robert Pippin, Terry Pinkard and Robert Brandom) that there is a substantial burden of evidence on anyone claiming Hegel is gibberish. I can understand why someone would walk away with that impression after a first attempt at the Phenomenology of Spirit, but it's an impression that is false. I'll leave it at that since this thread isn't about Hegel anyway.

[SmoothOper]

That is not good, because it means that such a logic is not only inconsistent, but also inconsistent at its meta-level...and since we just do not understand what 'p and not-p' means it means that this logic just makes no sense. The world might make no sense, but it becomes not better to talk about no sense in no sense^^.

P vs NP was always a little perplexing to me because the size of the input in terms of the numbers of symbols including logical statements, always seemed to grow exponentially relative to the number of variables. In fact the input could be exponential, and only have two or three variables. If you limit the total number of arguments then I suspect the size of the truth tables should be quite manageable.

[lemmata]

I should mention I am very skeptic of infinities also. I do not think we can prove them, we can just prove that "it goes on and on and on and on [but we do not know if there is an end in fantasiciollion years]". I find it paradoxical and inconsistent to talk about infinite sets, because since this set has infinite objects it is never finished and stable. Every proof about this set has to be incomplete.

Also, modern higher math uses variables to prove things about infinite sets. E.g. they prove that there are infinite natural numbers, because every number n has a successor n+1, so that there can't be a last one. BUT: That assumes that "n" stands for all possbile natural numbers, infinitely many as we just saw. How can one assume that? How can a single letter stand for 1. a single number but 2. at the same time for all? And on top of that there are no rules/axioms about that, it's just pure assumption and practice.

Therefore I like the "only what we can acutally calculate (even with a computer)"-math. Anything else is metaphysics in disguise.

People usually do not object to statements like "each child has a biological father" by proclaiming that there is no way that the word "child" can stand for every child on earth. This is essentially the objection you are raising about the statement in the form "every child c has a biological father f". However, latter is just the mathematical style of expressing the former.

[SmoothOper]

I should mention I am very skeptic of infinities also. I do not think we can prove them, we can just prove that "it goes on and on and on and on [but we do not know if there is an end in fantasiciollion years]". I find it paradoxical and inconsistent to talk about infinite sets, because since this set has infinite objects it is never finished and stable. Every proof about this set has to be incomplete.

Also, modern higher math uses variables to prove things about infinite sets. E.g. they prove that there are infinite natural numbers, because every number n has a successor n+1, so that there can't be a last one. BUT: That assumes that "n" stands for all possbile natural numbers, infinitely many as we just saw. How can one assume that? How can a single letter stand for 1. a single number but 2. at the same time for all? And on top of that there are no rules/axioms about that, it's just pure assumption and practice.

Therefore I like the "only what we can acutally calculate (even with a computer)"-math. Anything else is metaphysics in disguise.

People usually do not object to statements like "each child has a biological father" by proclaiming that there is no way that the word "child" can stand for every child on earth. This is essentially the objection you are raising about the statement in the form "every child c has a biological father f". However, latter is just the mathematical style of expressing the former.

Actually, I didn't read that carefully enough. I think it is really annoying when mathematicians fail to work at set level. For example everyone knows that the integers I are a subset of the rationals R therefore there must be more rationals than the integers. However there is a cockamamie proof where mathematicians make a mapping from integers to IxI then claim that there are as many I as R. Though in my opinion, it is impossible to do division on all integers without a mapping to the rationals R in which the integers I are a strict subset of the rationals. Though I could see how you could go for that kind of proof if you don't have a concept where you can't refer to an infinite set of things.

[Bill Spight]

Actually, I didn't read that carefully enough. I think it is really annoying when mathematicians fail to work at set level. For example everyone knows that the integers I are a subset of the rationals R therefore there must be more rationals than the integers.

As is often the case, what "everybody knows" is not so.

For one thing, integers are not a subset of rationals, strictly speaking. A rational is an ordered pair of integers. True, there are rationals that are equal to integers, but that is not exactly the same thing.

However there is a cockamamie proof where mathematicians make a mapping from integers to IxI then claim that there are as many I as R.

What does it mean to say that two infinities are equal, or that one infinity is greater than another? If you do not admit absolute infinities, then the question is meaningless. Fine. But then you do not get to say that there are more rationals than integers.

[SmoothOper]

Actually, I didn't read that carefully enough. I think it is really annoying when mathematicians fail to work at set level. For example everyone knows that the integers I are a subset of the rationals R therefore there must be more rationals than the integers.

As is often the case, what "everybody knows" is not so.

For one thing, integers are not a subset of rationals, strictly speaking. A rational is an ordered pair of integers. True, there are rationals that are equal to integers, but that is not exactly the same thing.

However there is a cockamamie proof where mathematicians make a mapping from integers to IxI then claim that there are as many I as R.

What does it mean to say that two infinities are equal, or that one infinity is greater than another? If you do not admit absolute infinities, then the question is meaningless. Fine. But then you do not get to say that there are more rationals than integers.

Someone doesn't understand the axiom of determinancy. 1:2 is not a rational number (1,2) is not a rational number, 1/2 is a rational number given the operation of division and equivalency classes, however if you have a suitable set of equivalency classes, then you can see that there are more rationals than integers, since integers are a subset. If you don't have the equivalency classes you can't define division, thus your sets of tuples aren't rational numbers. Some people who believe in the axiom of choice will sort of choose to do the mapping implicitly to make the proof work, other people won't be so kind as to go along.

[RBerenguel]

Actually, I didn't read that carefully enough. I think it is really annoying when mathematicians fail to work at set level. For example everyone knows that the integers I are a subset of the rationals R therefore there must be more rationals than the integers.

As is often the case, what "everybody knows" is not so.

For one thing, integers are not a subset of rationals, strictly speaking. A rational is an ordered pair of integers. True, there are rationals that are equal to integers, but that is not exactly the same thing.

However there is a cockamamie proof where mathematicians make a mapping from integers to IxI then claim that there are as many I as R.

What does it mean to say that two infinities are equal, or that one infinity is greater than another? If you do not admit absolute infinities, then the question is meaningless. Fine. But then you do not get to say that there are more rationals than integers.

Someone doesn't understand the axiom of determinancy. 1:2 is not a rational number (1,2) is not a rational number, 1/2 is a rational number given the operation of division and equivalency classes, however if you have a suitable set of equivalency classes, then you can see that there are more rationals than integers, since integers are a subset. If you don't have the equivalency classes you can't define division, thus your sets of tuples aren't rational numbers. Some people who believe in the axiom of choice will sort of choose to do the mapping implicitly to make the proof work, other people won't be so kind as to go along.

Technically, the equivalence classes used to define rationals don't (strictly) depend on division. They are just equivalence classes, which happen to behave pretty much like what a layman thinks is a rational number. It's like defining tensors as universal objects in categories. The fact that they can afterwards be used for something does not come for granted, or should imply categories behave just so.

Edit: things->thinks

[Tryss]

Some people who believe in the axiom of choice will sort of choose to do the mapping implicitly to make the proof work, other people won't be so kind as to go along.

But we can make the mapping explicitly :

First, we map the integer to a couple : n-> (n mod(2), n/2),

9073 -> (1,4536)

The first one will give the sign of the rationnal (0is positive, 1 is negative), and the second will give its absolute value :

We then map every positive integer to a sequence "equal to 0 after an arbitrary lenght", by taking the powers of its prime factors :

4536 = 2^3 * 3^4 * 5^0 * 7^1 => (3,4,0,1,0,0,...)

Then, we can map this sequence to a sequence of integers ( 0->0, 1->-1, 2->1, 3->-2 ...)

(3,4,0,1,0,0,...) => (-2,3,0,-1,0,0...)

Then we map this sequence to a positive rationnal, :

2^(-2)*3^3*7^(-1) => 27/28

Then we "add" the sign (cf the first part)

9073 -> - 27/28

And this is indeed a bijection (you can make all the steps backward) :

22/7 = 2^(1) * 7^(-1) * 11^1

=> (1,0,0,-1,1,0,...)

=> (2,0,0,1,2,0,...)

=> 2^2 * 7 * 11^2 = 3388

We multiply this number by 2, and because 22/7 is positive, we add 0

=> 6776

Obviously, you need to have proof that the decomposition in prime factors of a positive exist and is unique.

[SmoothOper]

Technically, the equivalence classes used to define rationals don't (strictly) depend on division. They are just equivalence classes, which happen to behave pretty much like what a layman thinks is a rational number. It's like defining tensors as universal objects in categories. The fact that they can afterwards be used for something does not come for granted, or should imply categories behave just so.

Edit: things->thinks

However, division does depend on rational equivalency classes, and isn't defined over integers.

[SmoothOper]

22/7

So 22 is a rational and 7 is a rational and division is an operation defined on the rationals 22 and 7, 22 happens to also be an integer, but the result of twenty two divided by seven is not an integer, therefore there are more rationals than integers.

[Bill Spight]

Actually, I didn't read that carefully enough. I think it is really annoying when mathematicians fail to work at set level. For example everyone knows that the integers I are a subset of the rationals R therefore there must be more rationals than the integers.

As is often the case, what "everybody knows" is not so.

For one thing, integers are not a subset of rationals, strictly speaking. A rational is an ordered pair of integers. True, there are rationals that are equal to integers, but that is not exactly the same thing.

However there is a cockamamie proof where mathematicians make a mapping from integers to IxI then claim that there are as many I as R.

What does it mean to say that two infinities are equal, or that one infinity is greater than another? If you do not admit absolute infinities, then the question is meaningless. Fine. But then you do not get to say that there are more rationals than integers.

Someone doesn't understand the axiom of determinancy. 1:2 is not a rational number (1,2) is not a rational number, 1/2 is a rational number given the operation of division and equivalency classes, however if you have a suitable set of equivalency classes, then you can see that there are more rationals than integers, since integers are a subset.

I did not say that ordered pairs of integers are rationals, I said the converse. Does someone understand English?

Given two integers, A and B, with B != 0, we may define a rational number as an ordered pair, (A,B), with certain properties. For instance, (A,B) = (C,D) iff A*D = B*C. And (A,B)*(C,D) = (A*C,B*D). And so on.