**Followup to**: Logical Pinpointing

"Oh! Hello. Back again?"

Yes, I've got another question. Earlier you said that you *had *to use second-order logic to define the numbers. But I'm pretty sure I've heard about something called 'first-order Peano arithmetic' which is also supposed to define the natural numbers. Going by the name, I doubt it has any 'second-order' axioms. Honestly, I'm not sure I understand this second-order business at all.

"Well, let's start by examining the following model:"

"This model has three properties that we would expect to be true of the standard numbers - 'Every number has a successor', 'If two numbers have the same successor they are the same number', and '0 is the only number which is not the successor of any number'. All three of these statements are true in this model, so in that sense it's quite numberlike -"

And yet this model clearly is *not* the numbers we are looking for, because it's got all these mysterious extra numbers like C and -2*. That C thing even loops around, which I certainly wouldn't expect any number to do. And then there's that infinite-in-both-directions chain which isn't corrected to anything else.

"Right, so, the difference between first-order logic and second-order logic is this: In first-order logic, we can get rid of the ABC - make a statement which *rules out* any model that has a loop of numbers like that. But we can't get rid of the infinite chain underneath it. In second-order logic we can get rid of the extra chain."

I would ask you to explain why that was true, but at this point I don't even know what second-order logic *is.*

"Bear with me. First, consider that the following formula *detects 2-ness:"*

x + 2 = x * 2

In other words, that's a formula which is true when x is equal to 2, and false everywhere else, so it singles out 2?

"Exactly. And this is a formula which detects odd numbers:"

∃y: x=(2*y)+1

Um... okay. That formula says, 'There exists a y, such that x equals 2 times y plus one.' And that's true when x is 1, because 0 is a number, and 1=(2*0)+1. And it's true when x is 9, because there exists a number 4 such that 9=(2*4)+1... right. The formula is true at all odd numbers, and only odd numbers.

"Indeed. Now suppose we had some way to *detect the existence* of that ABC-loop in the model - a formula which was *true* at the ABC-loop and *false* everywhere else. Then I could adapt the *negation* of this statement to say 'No objects like this are allowed to exist', and add that as an axiom alongside 'Every number has a successor' and so on. Then I'd have *narrowed down* the possible set of models to get rid of models that have an extra ABC-loop in them."

Um... can I rule out the ABC-loop by saying ¬∃x:(x=A)?

"Er, only if you've told me what A is in the first place, and in a logic which has ruled out all models with loops in them, you shouldn't be able to point to a specific object that doesn't exist -"

Right. Okay... so the idea is to rule out loops of successors... hm. In the numbers 0, 1, 2, 3..., the number 0 isn't the successor of any number. If I just took a group of numbers starting at 1, like {1, 2, 3, ...}, then 1 wouldn't be the successor of any number *inside* that group. But in A, B, C, the number A is the successor of C, which is the successor of B, which is the successor of A. So how about if I say: 'There's no group of numbers G such that for any number x in G, x is the successor of some other number y in G.'

"Ah! Very clever. But it so happens that you just used second-order logic, because you talked about *groups* or *collections* of entities, whereas *first-order logic* only talks about *individual* entities. Like, suppose we had a logic talking about kittens and whether they're innocent. Here's a model of a universe containing exactly three distinct kittens who are all innocent:"

Er, what are those 'property' thingies?

"They're all possible collections of kittens. They're labeled *properties* because every collection of kittens corresponds to a property that some kittens have and some kittens don't. For example, the collection on the top right, which contains only the grey kitten, corresponds to a predicate which is true at the grey kitten and false everywhere else, or to a property which the grey kitten has which no other kitten has. Actually, for now let's just pretend that 'property' just says 'collection'."

Okay. I understand the concept of a collection of kittens.

"In first-order logic, we can talk about individual kittens, and how they relate to other individual kittens, and whether or not any kitten bearing a certain relation exists or doesn't exist. For example, we can talk about how the grey kitten adores the brown kitten. In second-order logic, we can talk about collections of kittens, and whether or not those collections exist. So in first-order logic, I can say, 'There exists a kitten which is innocent', or 'For every individual kitten, that kitten is innocent', or 'For every individual kitten, there exists another individual kitten which adores the first kitten.' But it requires second-order logic to make statements about *collections* of kittens, like, 'There exists no collection of kittens such that every kitten in it is adored by some other kitten inside the collection.'"

I see. So when I tried to say that you couldn't have any group of numbers, such that every number in the group was a successor of some other number in the group...

"...you quantified over the existence or nonexistence of *collections* of numbers, which means you were using *second-order logic.* However, in this particular case, it's easily possible to rule out the ABC-loop of numbers using only first-order logic. Consider the formula:"

x=SSSx

x plus 3 is equal to itself?

"Right. That's a first-order formula, since it doesn't talk about collections. And that formula is false at 0, 1, 2, 3... but true at A, B, and C."

What does the '+' mean?

"Er, by '+' I was trying to say, 'this formula works out to True' and similarly '¬' was supposed to mean the formula works out to False. The general idea is that we now have a formula for detecting 3-loops, and distinguishing them from *standard *numbers like 0, 1, 2 and so on."

I see. So by adding the new axiom, ¬∃x:x=SSSx, we could rule out all the models containing A, B, and C or any other 3-loop of nonstandard numbers.

"Right."

But this seems like a rather arbitrary sort of axiom to add to a fundamental theory of arithmetic. I mean, I've never seen any attempt to describe the numbers which says, 'No number is equal to itself plus 3' as a basic premise. It seems like it should be a theorem, not an axiom.

"That's because it's brought in using a more general rule. In particular, first-order arithmetic has an *infinite axiom schema* - an infinite but computable scheme of axioms. Each axiom in the schema says, for a different first-order formula Φ(x) - pronounced 'phi of x' - that:"

*If*Φ is true at 0, i.e: Φ(0)*And if*Φ is true of the successor of any number where it's true, i.e: ∀x: Φ(x)→Φ(Sx)*Then*Φ is true of all numbers: ∀n: Φ(n)

(Φ(0) ∧ (∀x: Φ(x) → Φ(Sx))) → (∀n: Φ(n))

"In other words, every *formula* which is true at 0, and which is true of the successor of any number of which it is true, is true *everywhere.* This is the *induction schema* of first-order arithmetic. As a special case we have the *particular* inductive axiom:"

(0≠SSS0 ∧ (∀x: (x≠SSSx) → (Sx≠SSSSx)) → (∀n: n≠SSSn)

But that doesn't say that for all n, n≠n+3. It gives some premises from which that conclusion would follow, but we don't know the premises.

"Ah, however, we can *prove* those premises using the *other* axioms of arithmetic, and hence prove the conclusion. The formula (SSSx=x) is false at 0, because 0 is not the successor of *any *number, including SS0. Similarly, consider the formula SSSSx=Sx, which we can rearrange as S(SSSx)=S(x). If two numbers have the same successor they are the same number, so SSSx=x. If truth at Sx proves truth at x, then falsity at x proves falsity at Sx, modus ponens to modus tollens. Thus the formula is false at zero, false of the successor of any number where it's false, and so must be false everywhere under the induction axiom schema of first-order arithmetic. And so first-order arithmetic can rule out models like this:"

...er, I think I see? Because if this model obeys all the *other* axioms which which we *already* specified, that *didn't* filter it out earlier - axioms like 'zero is not the successor of any number' and 'if two numbers have the same successor they are the same number' - then we can *prove* that the formula x≠SSSx is true at 0, and prove that if the formula true at x it must be true at x+1. So once we then add the *further* axiom that *if* x≠SSSx is true at 0, and *if* x≠SSSx is true at Sy when it's true at y, *then* x≠SSSx is true at all x...

"We already have the premises, so we get the conclusion. ∀x: x≠SSSx, and thus we filter out all the 3-loops. Similar logic rules out N-loops for all N."

So then did we get rid of all the nonstandard numbers, and leave only the standard model?

"No. Because there was also that problem with the infinite chain ... -2*, -1*, 0*, 1* and so on."

Here's one idea for getting rid of the model with an infinite chain. All the nonstandard numbers in the chain are "greater" than all the standard numbers, right? Like, if *w* is a nonstandard number, then *w* > 3, *w* > 4, and so on?

"Well, we can prove by induction that no number is less than 0, and *w* isn't equal to 0 or 1 or 2 or 3, so I'd have to agree with that."

Okay. We should also be able to prove that if x > y then x + z > y + z. So if we take nonstandard *w* and ask about *w* + *w*, then *w* + *w* must be greater than *w* + 3, *w* + 4, and so on. So *w* + *w* can't be part of the infinite chain at all, and yet adding any two numbers ought to yield a third number.

"Indeed, that does prove that if there's one infinite chain, there must be *two* infinite chains. In other words, that original, exact model in the picture, can't all by itself be a model of first-order arithmetic. But showing that the chain implies the existence of yet other elements, isn't the same as proving that the chain doesn't exist. Similarly, since all numbers are even or odd, we must be able to find *v* with *v* + *v *= *w*, or find *v* with *v* + *v* + 1 = *w*. Then *v* must be part of another nonstandard chain that comes before the chain containing *w*."

But then that requires an *infinite* number of infinite chains of nonstandard numbers which are all greater than any standard number. Maybe we can extend this logic to eventually reach a contradiction and rule out the existence of an infinite chain in the first place - like, we'd show that any complete collection of nonstandard numbers has to be *larger than itself -*

"Good idea, but no. You end up with the conclusion that if a single nonstandard number exists, it must be part of a chain that's infinite in both directions, i.e., a chain that looks like an ordered copy of the negative and positive integers. And that if an infinite chain exists, there must be infinite chains corresponding to all *rational numbers.* So something that could actually be a nonstandard model of first-order arithmetic, has to contain at least the standard numbers *followed by* a copy of the rational numbers with each rational number replaced by a copy of the integers. But then *that* setup works just fine with both addition and multiplication - we can't prove that it has to be any larger than what we've already said."

Okay, so how *do* we get rid of an infinite number of infinite chains of nonstandard numbers, and leave just the standard numbers at the beginning? What kind of statement would they violate - what sort of axiom would rule out all those extra numbers?

"We have to use second-order logic for that one."

Honestly I'm still not 100% clear on the difference.

"Okay... earlier you gave me a *formula* which detected odd numbers."

Right. ∃y: x=(2*y)+1, which was true at x=1, x=9 and so on, but not at x=0, x=4 and so on.

"When you think in terms of *collections of numbers,* well, there's *some *collections which can be defined by formulas. For example, the collection of odd numbers {1, 3, 5, 7, 9, ...} can be defined by the formula, with x free, ∃y: x=(2*y)+1. But you could also try to talk about just the collection {1, 3, 5, 7, 9, ...} as a collection, a set of numbers, whether or not there happened to be any formula that defined it -"

Hold on, how can you talk about a set if you can't define a formula that makes something a member or a non-member? I mean, that seems a bit smelly from a rationalist perspective -

"Er... remember the earlier conversation about kittens?"

"Suppose you say something like, 'There *exists a collection* of kittens, such that every kitten adores only other kittens in the collection'. Give me a room full of kittens, and I can count through all possible collections, check your statement for each collection, and see whether or not there's a collection which is actually like that. So the statement is meaningful - it can be falsified or verified, and it constrains the state of reality. But you didn't give me a *local formula* for picking up a *single* kitten and deciding whether or not it ought to be in this mysterious collection. I had to iterate through all the *collections *of kittens, find the *collections* that matched your statement, and only then could I decide whether any individual kitten had the property of being in a collection like that. But the statement was still falsifiable, even though it was, in mathematical parlance, *impredicative - *that's what we call it when you make a statement that can only be verified by looking at many possible collections, and doesn't start from any particular collection that you tell me how to construct."

Ah... hm. What about infinite universes of kittens, so you can't iterate through all possible collections in finite time?

"If you say, 'There exists a collection of kittens which all adore each other', I could exhibit a group of three kittens which adored each other, and so prove the statement true. If you say 'There's a collection of four kittens who adore only each other', I might come up with a constructive proof, given the other known properties of kittens, that your statement was false; and any time you tried giving me a group of four kittens, I could find a fifth kitten, adored by some kitten in your group, that falsified your attempt. But this is getting us into some rather deep parts of math we should probably stay out of for now. The point is that even in infinite universes, there are second-order statements that you can prove or falsify in finite amounts of time. And once you admit those *particular *second-order statements are talking about something meaningful, well, you might as well just admit that second-order statements in general are meaningful."

...that sounds a little iffy to me, like we might get in trouble later on.

"You're not the only mathematician who worries about that."

But let's get back to numbers. You say that we can use second-order logic to rule out any infinite chain.

"Indeed. In second-order logic, instead of using an infinite axiom schema over all formulas Φ, we quantify over *possible collections *directly, and say, in a *single* statement:"

∀P: P(0) ∧ (∀x: P(x) → P(Sx)) → (∀n: P(n))

"Here P is any predicate true or false of individual numbers. Any collection of numbers corresponds to a predicate that is true of numbers inside the collection and false of numbers outside of it."

Okay... and how did that rule out infinite chains again?

"Because *in principle,* whether or not there's any first-order formula that picks them out, there's *theoretically* a collection that contains the standard numbers {0, 1, 2, ...} and *only* the standard numbers. And if you treat that collection as a predicate P, then P is true at 0 - that is, 0 is in the standard numbers. And if 200 is a standard number then so is 201, and so on; if P is true at x, it's true at x+1. On the other hand, if you treat the collection 'just the standard numbers' as a predicate, it's false at -2*, false at -1*, false at 0* and so on - those numbers *aren't* in this theoretical collection. So it's vacuously true that this predicate is true at 1* if it's true at 0*, because it's *not* true at 0*. And so we end up with:"

"And so the single second-order axiom..."

∀P: P0 ∧ (∀x: Px → P(Sx)) → (∀n: Pn)

"...rules out any disconnected chains, finite loops, and indeed every nonstandard number, in one swell foop."

But what did that axiom *mean,* exactly? I mean, taboo the phrase 'standard numbers' for a moment, pretend I've got no idea what those are, just explain to me what the axiom actually *says*.

"It says that the model being discussed - the model which fits this axiom - makes it impossible to form *any collection closed under succession* which includes 0 and doesn't include *everything*. It's impossible to have *any collection of objects in this universe *such that 0 is in the collection, and the successor of everything in the collection is in the collection, and yet this collection doesn't contain *everything.* So you can't have a disconnected infinite chain - there would then exist at least one collection over objects in this universe that contained 0 and all its successor-descendants, yet didn't contain the chain; and we have a shiny new axiom which says that can't happen."

Can you perhaps operationalize that in a more sensorymotory sort of way? Like, if this is what I believe about the universe, then what do I expect to see?

"If this is what you believe about the mathematical model that you live in... then you believe that neither you, nor any adversary, nor yet a superintelligence, nor yet God, can consistently say 'Yea' or 'Nay' to objects in such fashion that when you present them with 0, they say 'Yea', and when you present them with any other object, if they say 'Yea', they also say 'Yea' for the successor of that object; and yet there is some object for which they say 'Nay'. You believe this can never happen, no matter what. The way in which the objects in the universe are arranged by succession, just doesn't let that happen, ever."

Ah. So if, say, they said 'Nay' for 42, I'd go back and ask about 41, and then 40, and by the time I reached 0, I'd find either that they said 'Nay' about 0, or that they said 'Nay' for 41 and yet 'Yea' for 40. And what do I expect to see if I believe in first-order arithmetic, with the infinite axiom schema?

"In that case, you believe there's no neatly specifiable, compactly describable *rule* which behaves like that. But if you believe the second-order version, you believe nobody can possibly behave like that even if they're answering randomly, or branching the universe to answer different ways in different alternate universes, and so on. And note, by the way, that if we have a finite universe - i.e., we throw out the rule that *every* number has a successor, and say instead that 256 is the only number which has no successor - then we can verify this axiom in finite time."

I see. Still, is there any way to rule out infinite chains using *first*-order logic? I might find that easier to deal with, even if it looks more complicated at first.

"I'm afraid not. One way I like to look at it is that first-order logic can talk about* constraints on how the model looks from any local point*, while only second-order logic can talk about *global qualities* of chains, collections, and the model as a whole. Whether every number has a successor is a local property - a question of how the model looks from the vantage point of any one number. Whether a number plus three, can be equal to itself, is a question you could evaluate at the local vantage point of any one number. Whether a number is *even, *is a question you can answer by looking around for a single, individual number x with the property that x+x equals the first number. But when you try to say that there's *only one connected chain* starting at 0, by invoking the idea of *connectedness *and *chains *you're trying to describe non-local properties that require a logic-of-possible-collections to specify."

Huh. But if all the 'local' properties are the same regardless, why worry about global properties? In first-order arithmetic, any 'local' formula that's true at zero and all of its 'natural' successors would also have to be true of all the disconnected infinite chains... right? Or did I make an error there? All the other infinite chains besides the 0-chain - all 'nonstandard numbers' - would have just the same properties as the 'natural' numbers, right?

"I'm afraid not. The first-order axioms of arithmetic may fail to pin down whether or not a Turing machine halts - whether there *exists a time* at which a Turing machine halts. Let's say that from our perspective inside the standard numbers, the Turing machine 'really doesn't' halt - it doesn't halt on clock tick 0, doesn't halt on clock tick 1, doesn't halt on tick 2, and so on through all the standard successors of the 0-chain. In nonstandard models of the integers - models with other infinite chains - there might be somewhere inside a *nonstandard chain* where the Turing machine goes from running to halted and stays halted thereafter."

"In this new model - which is fully compatible with the first-order axioms, and can't be ruled out by them - it's not true that 'for every number t at which the Turing machine is running, it will still be running at t+1'. Even though if we could somehow restrict our attention to the 'natural' numbers, we would see that the Turing machine was running at 0, 1, 2, and every time in the successor-chain of 0."

Okay... I'm not quite sure what the *practical *implication of that is?

"It means that many Turing machines which *in fact* never halt at any standard time, can't be *proven not to halt* using first-order reasoning, because their non-halting-ness *does not actually follow logically* from the first-order axioms. Logic is about which conclusions follow from which premises, remember? If there are models which are compatible with all the first-order premises, but still falsify the statement 'X runs forever', then the statement 'X runs forever' can't *logically follow* from those premises. This means you won't be able to prove - *shouldn't* be able to prove - that this Turing machine halts, using *only *first-order logic."

How exactly would this fail in practice? I mean, where does the proof go bad?

"You wouldn't get the second step of the induction, 'for every number t at which the Turing machine is running, it will still be running at t+1'. There'd be nonstandard models with some nonstandard t that falsifies the premise - a nonstandard time where the Turing machine goes from running to halted. Even though if we could somehow restrict our attention to *only the standard numbers*, we would see that the Turing machine was running at 0, 1, 2, and so on."

But if a Turing machine really actually halts, there's got to be some *particular time* when it halts, like on step 97 -

"Indeed. But 97 exists in *all *nonstandard models of arithmetic, so we can prove its existence in first-order logic. Any time 0 is a number, every number has a successor, numbers don't loop, and so on, there'll exist 97. Every nonstandard model has *at least* the standard numbers. So whenever a Turing machine *does* halt, you can prove in first-order arithmetic that it halts - it does indeed follow from the premises. That's kinda what you'd *expect,* given that you can just watch the Turing machine for 97 steps. When something actually does halt, you *should* be able to prove it halts without worrying about unbounded future times! It's when something *doesn't actually* halt - in the standard numbers, that is - that the existence of 'nonstandard halting times' becomes a problem. Then, the conclusion that the Turing machine runs forever *may not actually follow* from first-order arithmetic, because you can obey all the premises of first-order arithmetic, and yet still be inside a nonstandard model where this Turing machine halts at a nonstandard time."

So second-order arithmetic is more powerful than first-order arithmetic in terms of *what follows from the premises*?

"That follows inevitably from the ability to talk about *fewer possible models*. As it is written, 'What is true of one apple may not be true of another apple; thus more can be said about a single apple than about all the apples in the world.' If you can restrict your discourse to a narrower collection of models, there are more facts that follow inevitably, because the more models you might be talking about, the fewer facts can possibly be true about all of them. And it's also definitely true that second-order arithmetic proves more theorems than first-order arithmetic - for example, it can prove that a Turing machine which computes Goodstein sequences always reaches 0 and halts, or that Hercules always wins the hydra game. But there's a bit of controversy we'll get into later about whether second-order logic is *actually *more powerful than first-order logic in general."

Well, sure. After all, just because nobody has ever yet invented a first-order formula to filter out all the nonstandard numbers, doesn't mean it can never, ever be done. Tomorrow some brilliant mathematician might figure out a way to take an individual number x, and do local things to it using addition and multiplication and the existence or nonexistence of other individual numbers, which can tell us whether that number is part of the 0-chain or some other infinite-in-both-directions chain. It'll be as easy as (a=b*c) -

"Nope. Ain't never gonna happen."

But maybe you could find some entirely different creative way of first-order axiomatizing the numbers which has *only *the standard model -

"Nope."

Er... how do you *know *that, exactly? I mean, part of the Player Character Code is that you don't give up when something *seems *impossible. I can't quite see *yet *how to detect infinite chains using a first-order formula. But then earlier I didn't realize you could rule out finite loops, which turned out to be quite simple once you explained. After all, there's two distinct uses of the word 'impossible', one which indicates positive knowledge that something can *never *be done, that no *possible* chain of actions can *ever *reach a goal, even if you're a superintelligence. This kind of knowledge requires a strong, definite grasp on the subject matter, so that you can rule out *every *possible avenue of success. And then there's another, *much more common* use of the word 'impossible', which means that you thought about it for five seconds but didn't see any way to do it, usually used in the presence of *weak *grasps on a subject, subjects that seem sacredly mysterious -

"Right. Ruling out an infinite-in-both-directions chain, using a first-order formula, is the *first *kind of impossibility. We *know *that it can never be done."

I see. Well then, what do you think you know, and how do you think you know it? How is this definite, positive knowledge of impossibility obtained, using your strong grasp on the non-mysterious subject matter?

"We'll take that up next time."

Part of the sequence *Highly Advanced Epistemology 101 for Beginners*

Next post: "Godel's Completeness and Incompleteness Theorems"

Previous post: "By Which It May Be Judged"

No we don't.

The model with the natural numbers and a single "integer line" is not a first order model of arithmetic. The reason is this. For a non-standard number "a" large enough there is a (non-standard) natural number that's approximately some rational fraction of "a." This number then has successors and predecessors, so it has an "integer line" around it. But because we can play this game for any fraction, we need lots of integer lines (ordered according to the total ordering on the rationals).

See this for details:

http://web.mit.edu/24.242/www/NonstandardModels.pdf

Yes we do.

The problem of a chain isn't intended to be limited to the problem of exactly one chain, and I didn't want to complicate the diagram or confuse my readers by showing them a copy of the rationals with each rational replaced by a copy of the integers. If you can't get rid of a larger structure that has a chain in it, you can't get rid of the chain. To put it another way, showing that the chain depicted implies further extra elements isn't the same as ruling out the existence of that chain.

Hence the wording, "How do we get rid of the chain?" not "How do we get rid of this particular exact model here?"

A very quick way to see that there must be more than one chain is to note that if x > y, then x + z > y + z. An element of the nonstandard chain is greater than any natural number, so if we add two nonstandard numbers together, the result must be greater than the nonstandard starting point plus any natural number. Therefore there must be another chain which comes after the first one. For more on this see the linked paper.

EDIT: Several others reported misinterpreting what I had in the original, so I've edited the post accordingly. Thanks for raising the issue, Ilya!

It's probably worth explicitly mentioning that the structure that you described isn't actually a model of PA. I'd imagine that could otherwise be confusing for readers who have never seen this stuff before and are clever enough to notice the issue.

Edited.

< is defined in terms of plus by saying x<y iff there exists a such that y=z+x. + is supposed to be provided as a primitive operation as part of the data consisting of a model of PA. It's not actually possible to give a concrete description of what + looks like in general for non-standard models because of Tenenbaums's Theorem, but at least when one of x or y (say x) is a standard number it's exactly what you'd expect: x+y is what you get by starting at y and going x steps to the right.

To see that x<y whenever x is a standard number and y isn't, you need to be a little tricky. You actually prove an infinite family of statements. The first one is "for all x, either x=0 or else x>0". The second is "for all x, either x=0 or x=1 or x>1", and in general it's "for all x, either x=0,1,..., or n, or else x>n". Each of these can be proven by induction, and the entire infinite family together implies that a non-standard number is bigger than every standard number.

Eliezer, I just want to say thanks. This conversational method of teaching logic/math is very approachable and engaging to me. Much appreciated!

Consider using "☑" or similar) to mean "true", rather than overloading "+"?

The double-turnstile ⊨ is the usual symbol for saying that a sentence is true (in a given model).

Can you (or someone else too, I guess) give an example of a Turing machine with a nonstandard halting time? It's not clear to me what you mean by running a Turing machine for a nonstandard number of steps. (I think I can make this meaningful in my favorite nonstandard model of Peano arithmetic, namely an ultrapower of the standard model, but I don't see how to make it meaningful in general.)

In a sense, no. Eliezer's point is this: Given the actual Turing machine with number m = 4 = SSSS0 and input i = 2 = SS0, you can substitute these in to get a closed formula φ whose meaning is "the Turing machine SSSS0 halts on input SS0". The actual formula is something like, "There is a number e such that e denotes a valid execution history for machine SSSS0 on input SS0 that ends in a halting state." In the standard model, talking about the standard numbers, this formula is true iff the machine actually halts on that input. But in first-order logic, you cannot

pinpointthe standard model, and so it can happen that formula φ is false in the standard model, but true in some nonstandard model. If you use second-order logic (and believe its standard semantics, not its Henkin semantics), formula φ isvalid, i.e. true ineverymodel, if and only if machine 4 really halts on input 2.Ok -- as I replied to earthwormchuck, I think Eliezer isn't saying at

allthat there is a useful way in which these nonstandard execution histories are "really" talking about Turing machines, he's saying the exact opposite: theyaren'ttalking about Turing machines, which is bad if youwantto talk about Turing machines, since it means that first-order logic doesn't suffice for expressing exactly what it is youdowant to talk about.You should be careful with addition and multiplication - to use them, you would have to define them first, and this is not trivial if you have the natural numbers plus A->B->C->A, infinite chains and so on.

In addition, "group" has a specific mathematical meaning, if you use it for arbitrary sets this is quite confusing.

Does it matter if you don't have formal rules for what you're doing with models?

Do you expect what you're doing with models to be formalizable in ZFC?

Does it matter if ZFC is a first-order theory?

...now I find myself wishing philosophical whimsy reached more often for things like kitten innocence than contrived torture scenarios.

Every time you take box A and box B from Omega, Omicron tortures a kitten.

It was once thought that adoring cats caused one to get tortured. However, a recent medical study has come out, showing that most cat adorers have a certain gene, ACGT, and that whether someone has the gene or not, their chances of getting tortured go

downif they adore cats. The strong correlation between adoring cats and getting tortured is because of a third factor, ACGT, that leads to both.Having learned of this new study, would you choose to adore cats?

Or, indeed, should I choose not to adore cats because it might be evidence I had toxoplasmosis?

Actually, that toxoplasmosis thing is the only happiness-creating-preference-inducing, negative-side-effect disease I actually know that really works for Solomon's Problem. You can either pet cute kittens already tested and guaranteed not to have toxoplasmosis, or refrain. This ought to be our go-to real-life example against EDT!

Nice post, but I think you got something wrong. Your structure with a single two-sided infinite chain isn't actually a model of first order PA. If x is an element of the two-sided chain, then y=2x=x+x is another non-standard number, and y necessarily lies in a different chain since y-x=x is a non-standard number. Of course, you need to be a little bit careful to be sure that this argument can be expressed in first order language, but I'm

prettysure it can. So, as soon as there is one chain of non-standard numbers, that forces the existence of infinitely many.Did you mean to say "connected", or did I miss something?

Fascinating, I thought Tennanbaum's theorem implied non-standard models were rather impossible to visualize. The non-standard model of Peano arithmetic illustrated in the diagram only gives the successor relation, there's no definition of addition and multiplication. Tennenbaum's theorem implies there's no computable way to do this, but is there a proof that they can be defined at all for this particular model?

connected?"First, consider that the following formula detects 2-ness" Consider changing this to something like "First, consider that the following formula detects 2-ness among the numbers as we want them to be"? It wasn't immediately obvious to me that the starred chain's '-1*' didn't satisfy the equation.

Er, also, you might want to have only one of the interlocutors beginning sentences with "Er" lest we lose track of which is supposed to be current-you. ;)

But yeah, a nice exposition!

This is so much clearer than my college class.

I'm going to have to read the proof of the hydra game, because I pretty quickly got over 2.8k nodes and still in increasing...

It's even worse than that, the function for the maximum number of nodes you end up before they start going down, if you play using the worst possible strategy, increases faster than any function which Peano arithmetic can prove to be total (i.e., it grows faster than any Turing machine run on various inputs, which Peano arithmetic can prove to halt for any input). To say that this grows faster than the Ackermann function is putting it very mildly.

Well, this was very interesting, formal logic is a very fun topic. I just spent ~10 minutes trying to find a way in first order logic to write that axiom, as it intuitively feels (to someone who has studied formal logic at least) that there

shouldbe a way... Of course I failed, all the axioms I attempted turned out to be no more powerful then "0 is not the successor of any number". I am deeply intrigued by this problem, and I am looking forward to your next post where you explain exactlywhyit's impossible.Okay, my brain isn't wrapping around this quite properly (though the explanation has already helped me to understand the concepts far better than my college education on the subject has!).

Consider the statement: "There exists no x for which, for any number k, x after k successions is equal to zero." (¬∃x: ∃k: xS-k-times = 0, k>0 is the closest I can figure to depict it formally). Why doesn't that axiom eliminate the possibility of any infinite or finite chain that involves a number below zero, and thus eliminate the possibility of the two-side... (read more)

Nice post! Are you going to get into Dedekind's proof that all models of the second-order Peano axioms are isomorphic?

Edit: Fixed typo.

(exists x) (forall y) not (S(y) = x)

This statement is a first order statement which is true of standard natural numbers, but is not true in your model.

???

I don't see yet how this connects to the other posts from the epistemology sequence, but it's definitely nice. I've wanted to learn more mathematical logic for some time. I didn't quite understand why exactly using an axiom schema isn't as good as using second order logic, before I read this post.

I read that "any" as "for at least one", rather than as "for every". That confused me quite a bit. Maybe native speake... (read more)