Trying to test me to see if I actually got it?
In absence of anything else, A is related to itself. T doesn’t prove this. T says nothing about what A is related to. It just happens to be true if there are no other axioms, because, in T, A is the only thing.
Sort of.
So you’re saying that “A is related to itself” is true, but not provable, in T.
What if I add an axiom to make system S:
S1: A is a thing.
S2: Every thing is related to just one thing.
S3: Nothing is related to itself.
Now in S, is A related to itself?
Instead of waiting for an answer, I’ll asume you’d answer “no”–that seems trivial enough, right?
But here’s the problem. Let ‘A’ stand for the statement “A is related to itself.” You think that given S1 and S2, A is true. But you also think that given S1, S2 and S3, A is false. But here’s a simple derivation based on these two beliefs of yours:
Suppose S1 & S2 & S3
Given that supposition, it is the case that not-A.
It also follows from the supposition that S1 & S2.
Given that, (i.e., S1 & S2,) it is the case that A.
So, on the supposition in the first line, it is the case both that not-A & A.
Contradiction.
So our supposition can’t be true.
But that conclusion amounts to saying that the axiom system S is inconsistent. It seems fairly clear to me that it’s not though. You?
Well, I’m sorry you think so (but certainly not bitter). However, if I have belaboured my point somewhat, then only because I didn’t see you either address or acknowledge it at all, and because I wanted to deal in general with what confused you in the specific – it’s kinda like discussing special relativity with someone who keeps coming up with new thought experiments to violate the light speed limit (jerking back and forth an extremely long, infinitely rigid rod, rotating a sufficiently large sphere, you’ve probably heard 'em all); it’s ultimately of no use to just refute every single new proposed counter-example, cause there’ll always be another one. If you want to make any headway at all, you have to deal with the central misunderstanding.
Anyway, it’s a moot point – you’ve certainly made a step in the right direction by admitting your error, and I do applaud you for that; just don’t stop now, there’s still some way before you (as there is before me). Ultimately, enhancing one’s understanding is far more rewarding than being right.
Yes, including someone who thought they could convince me that all you had to do was move the flashlight… :smack:
Are you suggesting that with a major attitude adjustment on my part, and a minor one on your part, we could have some interesting conversations? I’m game.
All else being equal: I would rather be right than wrong. I would rather learn something than be right. (:dubious: usually?) And I would rather be wrong because of an unanswered question than be right and not know why, (or be unable to defend the position.)
Let me get this straight. Are we still debating whether F describes F[sub]2[/sub]?
What prevents a system from proving A and not-A? Nothing that I know of. (other than our own human sense of, “that’s just not right.”)
Yes, That is what I would say about it.
One other thing I could try to say…
In T, is ArA a possible solution to the axioms? What other solutions are there? What other things in T can A be related to? Is A related to any “thing” in T that is not A? no? Well, if A is related to one “thing” in T, but is not related to any “thing” that isn’t A, A must be related to A.
Is there anything wrong with my logic there?
Did I correctly surmise that your creation of S was in response to an objection to my saying that “A is related to A” is proven in T?
Alternately.
In T, what happens if we assume the opposite of “A is related to A?”
if “A is not related to A,” is A related to any “thing” in T? no? then T2 isn’t satisfied. So, not-(not-A), (and that is equal to A)
Not to step on Frylock’s toes, but what is your position (ch4rl3s) on the following axiom system:
There are thingamajigs named C and D
At least one thingamajig is also a whatchamacallit
Do these prove that C is a whatchamacallit? Do these prove that D is a whatchamacallit? Is there a model “described” by these axioms in the same sense as you claim F[sub]2[/sub] to be “described” by the field axioms? Are these consistent? And similar questions.
(One might also consider the slight variant, in which we take the axioms to be
There are distinct thingamajigs named C and D
Precisely one thingamajig is a whatchamacallit
I would not consider the issues raised to be substantially different, but who knows? Maybe your responses would differ)
An element not being defined by the axioms of a system does not imply that the element does not exist. It’s certainly easier to wrap your head around a system where the elements are defined either directly or by a method of construction, but neither method of definition is strictly necessary. See Finite Geometryfor a discussion of the Fano Plane, a finite geometry in which only one point is specified to exist in the axioms but which actually contains exactly 7 points.
{A} satisfies the axioms of T. It’s the smallest set that satisfies the axioms, since the empty set {} contradicts T1. It is important to note that T requires both a set of things and a set of relations. If our set of things is {A} then our set of relations, in order not to violate the axioms of T must be {ArA}.
However, T1 does not restrict us to the set {A}. The set {A, B} satisfies T1, since A exists in our set. We can then choose any of the following four sets of relations (I’m assuming here that the “relation” defined in T2 is one-way, that is, ArB does not imply BrA. If I’m wrong about your intention here Frylock, please correct me.): {ArA, BrB}, {ArA, BrA}, {ArB, BrA}, {ArB, BrB}. Every combination of {A, B} and one of the pairs of relations satisfies T. A is an element of the set (is a “thing”) and every element is related to one element.
We can have larger sets. In fact, we can have an infinite number of sets. Those sets can be finite or infinite, and the number of possible relation combinations for those sets gets very large, very quickly, but any choice of a set of “things”, as long as it contains the element A, satisfies T1.
T does not prove ArA. It does not prove not-ArA. ArA is undecidable in T.
R1: A is a thing.
R2: Every thing is related to just one thing.
R3: There are no things other than A
In R we can prove ArA. Clearly the only allowable set is {A}, and as above, in this set, ArA is the only allowed relation. But R3 is crucial to this proof. R3, and only R3 is how we keep the set {A, B} from being considered. {A} is the smallest, simplest, easiest to think of set that satisfies T. But without R3 (which is equivalent to the assumption that you added to the field axioms in the other thread), you cannot rule out the possibility of the set {A, B}, or the set {A, B, C}, or any other set that contains A.
In S, the set {A} does not satisfy the axioms. The only possible set of relations for the set {A}, again, is ArA. This contradicts S3, as has been shown. Now, if we look at the set {A, B} and the set of relations {ArB, BrA}, what happens?
S1: A is a thing. (Check)
S2: Every thing is related to just one thing. (Check)
S3: Nothing is related to itself. (Check)
So {A, B} and {ArB, BrA} is a valid model of the axiom system S. It is not the only one. {A, B, C} and {ArB, BrC, CrA} also satisfy every axiom of the system. In S, BrA is undecidable (because we have two valid models that give different answers to the question). ArA, however, is not undecidable in S. It is simply false. There are no models that satisfy the axioms in which ArA is true.
Lets call the system Q
Q1: A is a thing.
Q2: Every thing is related to just one thing.
Q3: A is not related to A
Now lets look at our sets. {A} and {ArA} satisfy Q1 and Q2, just like they did before. They do not satisfy Q3. What about {A, B} and {ArB, BrA}? Again, they satisfy Q1 and Q2, like they did in T, and they also satisfy Q3. In fact, the set {A, B} with the alternate choice of relations {ArB, BrB} still satisfies Q3, as A is not related to A. What if instead of (or in addition to, though Q3 would be redundant) Q3, we had “Q4: Nothing is related to itself.”? Then {A, B} {ArB, BrB} would no longer satisfy Q. We would have to choose the relation set {ArB, BrA} to go with the set {A, B} in order to satisfy all the axioms.
So, it really seems to me that you’re getting hung up by assuming R3, that there are no elements other than A. Once you let that go, everything else should fall into place.
I don’t think so. Though the outcome of this discussion could have an influence on your opinion regarding the Field stuff.
That’s right. My point (as you saw later) was to go on to indicate that this contradiction means the system S is inconsistent.
To be clear, you’re saying that S is inconsistent, where S is the following axiom set:
S1. A is a thing.
S2. Every thing is related to just one thing.
S3. No thing is related to itself.
You’ve rightly agreed that your understanding commits you to the view that S is inconsistent.
But S is consistent. For if S were inconsistent, then counting would be impossible. This is so because the counting numbers together with the “successor” relation satisfy S. This means that if S contains a contradiction, then the activity of starting with one number, naming its sucessor, naming the next successor and so on should eventually lead to a point where a successor can’t be named on pain of contradiction. (I wish I could put that in more rigorous terms but it’s been too long since I’ve had to prove the consistency of an axiom set. Maybe someone more recently and thoroughly experienced can fill in the gaps here for me. But still–isn’t it pretty clear to you, Ch4rl3s, that S must be consistent, since S is basically a formalization of counting, or indeed, of any linear ordering system? If S is inconsistent, then linear ordering is impossible!)
It can be. Here is where it seems like we keep getting tripped up. You seem to think that a kind of principle of parsimony applies to axiom sets–entities not strictly needed in order to satisfy the axioms ipso facto do not exist according to those axioms. But in actual practice, mathematicians and logicians do not follow this principle of parsimony. With the exercise regarding axiom systems T and S above, I tried to show why–this “principle of parsimony” as I called it actually leads to absurd results such as the notion that S is inconsistent.
Rather than saying things don’t exist which aren’t strictly necessary for satisfaction of an axiom set, mathematicians and logicians typically follow a principle that anything can exist as long as its existence isn’t incompatible (in the sense of allowing the derivation of a contradiction) with the axiom set. The existence of things other than A is consistent with T1 and T2. Therefore, the axiom set T does not rule out the existence of things other than A. Models including more than one element can satisfy T. T does not specify whether A is related to itself, or rather, is related to one of these other possible things. The axiom set T, by itself, does not decide that proposition. If it did, then as we’ve seen, S would be inconsistent. But if S were inconsistent, linear orderings (such as that used in the simple counting process) would be impossible.
I think what I’ve said here answers the questions in your two subsequent posts as well.
Also, what Crackrat said.
Also, what Indistinguishable asked.
Sorry I didn’t get back to this sooner. Busy holiday season.
My view is that these are different scenarios. When you have 2 thingamajigs, one of which must be a whatcamacallit, you don’t have enough information to say which one it might be.
Thanks very much to Crackrat for the very helpful example of Fano’s Geometry. A much needed refresher on constructing sets and whatnot. It has changed my mind, somewhat.
I looked through it eagerly to see what could be proven. Axiom 1: ok, we have a line. Axiom 2: there are 3 points on that line. Axiom 3: oooh, a point not on the original line… It took me a while to work out exactly how the rest was constructed. I was happily on my way to deriving Theorem 2… when I scrolled down and saw it.
Anyway, as well as being fun, it helped me looking at Frylock’s S axiom set.
Let’s prove a theorem. From S2, A is related to a thing. From S3, A isn’t related to itself.
So, Theorem 1: There must be a thing, B, such that ArB, and B!=A.
S3 allows us to construct a new thing that wasn’t prohibitted by other axioms.
That’s a much better proof of the consitency of S to me than:
Thanks again to Crackrat. I’ll readily give him the credit for my change of mind on this one.
But, I still have other questions.
I have never before seen an axiom like R3… Until I started using one like it and asking why it was necessary. It seems to me, and I think I said something like this already, that an axiom like this would be necessary for most axiom sets to create a particular proof system. (Like F[sub]2[/sub] mentioned before.) It would be almost ubiquitously necessary, and wouldn’t hurt in any system that didn’t need it. And therefore, there is no reason for it not to be simply a base property of proof theory. And I thought it was. I thought basically we assumed that there was no way to create new members, (of this set for this particular system,) except from the axioms already stated.
If we take the first 2 axioms of Fano:
Axiom 1. There exists at least one line.
Axiom 2. Every line has exactly three points incident to it.
and ask, how many lines are there, it makes much more sense to me to say that we can’t tell. And I see why you say, from T, that we can’t tell how many things there might be. And I was almost convinced for a while. Until I asked myself what the scope of the question was.
are we asking, from T, how many things there are in the universe? Or in T?
What about our, previously proposed, complete system of the natural numbers with addition? How many numbers are there? Are we asking how many in the universe, or how many in the system itself? (yes, I know the natural numbers are infinite, but they’re countable. The Reals aren’t. We can compare them.) Does a complete theory of natural numbers say there are no more numbers, or that there are no more natural numbers?
The axiom set T allows us to extend it with more things if we add more axioms. But T itself only talks about “A type things.” Are there things in T, other than A? Not, are there things in the universe other than A? (or, our universe is T for that discussion.) S3 gives us a way to construct a thing other than A, so S contains A and B. But, T doesn’t give us a way to construct any other members of T.
I see three scenarios. 1. A system is complete and every theorem it proves is true for all extensions. 2. A system is incomplete and doesn’t contain enough information to decide a relationship. (like Indistinguishable’s C,D whatcamacallit system.) or 3. it’s incomplete, but does contain enough information to decide a relationship, even if it isn’t true for another extension. (like I’m claiming for T.)
Everyone seems to be saying that scenario 3 isn’t possible. When you add R3: there are no things other than A… It completes that question for R. ArA is now true for every extension of R. Why are you saying that it isn’t possible to prove a theorem unless we’ve completed the system for that question?
Anybody?
I’m afraid my reply mostly just repeats things I’ve already said.
It looks like you’re still not convinced that when deciding what follows from what axioms, we should use a “principle of permission” rather than a “principle of parsimony.” Here’s my argument that the principle of permission is the right one to use.
I will prove each of those two premises in turn, and the conclusion naturally follows.
From the conclusion of The Argument, together with the natural (though admitedly not trivial) assumption that Permission and Parsimony are the only two options available, it follows that Permission is the principle to use.
(Apologies if I’ve misread your latest post.)
From my view, the system SX has the theorem “T(sx)1: ArA.” The system S which has one more axiom doesn’t prove this theorem. S instead has the theorem “T(s)1: ArB where B!=A.” Both of these are consistent from my view point.
I don’t buy that statement, so, this conclusion doesn’t follow:
Your view point appears to say that the system S can’t be an extension of the system SX, because we have added the axiom “SX3: there is no way to build another element or operation.” But to my mind, that is just an observation from looking at the axioms given.
SX1: A is a thing
SX2: Everything is related to just one other thing
SX3: there is no way to build another thing or relation
Do you say this creates the “system” SX? I say this completes the system SX. SX is now complete and ArA is true for every extension. And S is not a valid extension. Let’s try to build it from SX.
SX1: A is a thing
SX2: Everything is related to just one other thing
SX3: there is no way to build another thing or relation
S4: No thing is related to itself
This doesn’t build S, (even though it has all the same axioms as S; the last 2 are just reversed, but it makes a difference.) And it makes whatever system this is inconsistent. And how often have you seen an axiom like SX3? No one has told me that they’ve ever seen this before. They’ve been telling me its necessary. So, is it invisible? If so, then that is my system. It’s assumed as a base property of proof theory.
Tell me, how do you build the element B from the axioms of SX? You can’t. There is only one model of SX that satisfies the axioms. There are many other models that can satisfy the axioms, but they are all extensions of SX; additional axioms that allow you to build new elements, create new relations, etc.
From your view, it seems to me that since SX doesn’t prohibit the natural numbers and the operation +, do we have to assume that they are in some models of SX? They aren’t constructable from SX. I agree that they are in some models that will also satisfy the axioms of SX, but I don’t believe that they are in any model of SX. likewise, {A,B}, (with the various possible relations,) isn’t a model of SX.
Well, your viewpoint is wrong then. If, using two axioms, you can prove a statement, then that statement is true, no matter how many more axioms you have in your system. If the other axioms allow you to prove the contradiction of the other statement, then your axioms are inconsistent. Every theorem of an axiom system is a theorem of every extension of that system.
Therefore,
You say, “Consider the set {A, B}, and the relations {ArB, BrA}. I submit that these satisfy the axioms of SX.” Then everyone looks, agrees that in fact, this model satisfies both SX[sub]1[/sub] and SX[sub]2[/sub].
S1. A is a thing.
S2. Every thing is related to just one thing.
S3. No thing is related to itself.
I don’t agree (yet? lol) with the statement “But S is consistent. For if S were inconsistent, then counting would be impossible.” made by Frylock.
Is not Identity self-relation? A=A or I(A)=A or IA=A? Given this, the above set probably does not allow one to derive counting. Nor could it be it consistent, since S3 either: is inconsistent with S2 if the system admits the concept of Identity, or else is consistent with S2 in the absence of Identity (in which case the word “itself” is a problem!).
Counting proofs, as far as I know, all make use of identity. Another way of putting this is that the word “itself” (which is obviously a relation of a thing to itself) appears in S3, so S3 can be read as “nothing can ever be itself but nor can it be not itself” (since non-identity, like inequality, is a clearly relationship too). S3 could therefore be any old contradiction such as “A = not A” from which we can, of course, prove or disprove anything whatsoever.
Hey, neither am I actually, which is why I went on later to give a simpler and more traditionally basic proof of S’s consistency. (S has a model, any axiom set with a model is consistent, so S is consistent.)
Yes, but by “x is related to y” in my specification of S I meant to denote a particular formal relation implicitly defined by the axioms. “X is related to y” isn’t supposed to refer to just any relation, whether it be identity or whatever. Rather it refers to a particular relation. What relation? S doesn’t say–it’s just talking about a particular relation denoted by the phrase “is related to.”
The counting numbers, as a model, satisfy S if you interpret “is related to” to mean “is a successor of.”
Meanwhile, identity can’t satisfy the “is related to” relation in S, because S3 says nothing is related to itself, while everything is identical with itself. This means that however you interpret “is related to” in S, you can’t interpret it to refer to the identity relation.