You can generally throw unfalsifiable beliefs into your utility function but you might consider this intellectually dishonest.
As a quick analogy, a solipsist can still care about other people.
Incorrect
Karma: 1,065
I escape by writing a program that simulates 3^^3 copies of myself escaping and living happily ever after (generating myself by running Solomonoff Induction on a large amount of text I type directly into the source code).
You might be able to contain it with a homomorphic encryption scheme
I’m guessing Eliezer would lose most of his advantages against a demographic like that.
Oh god, remind me to never play the part of the gatekeeper… This is terrifying.
The lifespan dilemma applies to all unbounded utility functions combined with expected value maximization, it does not require simple utilitarianism.
Would your post on eating babies count, or is it too nonspecific?
http://lesswrong.com/lw/1ww/undiscriminating_skepticism/1scb?context=1
(I completely agree with the policy, I’m just curious)
There are people who claim to be less confused about this than I am
Solipsists should be able to dissolve the whole thing easily.
Thanks, can you recommend a textbook for this stuff? I’ve mostly been learning off Wikipedia.
I can’t find a textbook on logic in the lesswrong textbook list.
Therefore a theory could be ω-consistent because it fails to prove P(n), even though P(n) is true in the standard model.
I thought for ω-consistency to even be defined for a theory it must interpret the language of arithmetic?
Just use the axiom schema of induction instead of the second order axiom of induction and you will be able to produce theorems though.
But you wont be able to produce all true statements in SOL PA, that is, PA with the standard model, because of the incompleteness theorems. To be able to prove a larger subset of such statements, you would have to add new axioms to PA. If adding an axiom T to PA does not prevent the standard model from being a model of PA+T, that is it does not prove any statements that require the existence of nonstandard numbers, then PA+T is ω-consistent.
So, why can’t we just keep adding axioms T to PA, check whether PA+T is ω-consistent, and have a more powerful theory? Because we can’t in general determine whether a theory is ω-consistent.
Perhaps a probabilistic approach would be more effective. Anyone want to come up with a theory of logical uncertainty for us?
There’s no complete deductive system for second-order logic.
An infinite number of axioms like in an axiom schema doesn’t really hurt anything, but you can’t have infinitely long single axioms.
∀x((x = 0) ∨ (x = S0) ∨ (x = SS0) ∨ (x = SSS0) ∨ ...)is not an option. And neither is the axiom set
P0(x) iff x = 0 PS0(x) iff x = S0 PSS0(x) iff x = SS0 ... ∀x(P0(x) ∨ PS0(x) ∨ PS0(x) ∨ PS0(x) ∨ ...)We could instead try the axioms
P(0, x) iff x = 0 P(S0, x) iff x = S0 P(SS0, x) iff x = SS0 ... ∀x(∃n(P(n, x)))but then again we have the problem of n being a nonstandard number.
I don’t see what the difference is… They look very similar to me.
At some point you have to translate it into a (possibly infinite) set of first-order axioms or you wont be able to perform first-order resolution anyway.
I don’t see how you would define Pn(x) in the language of PA.
Let’s say we used something like this:
Pn(x) iff ((0 + n) = x)Let’s look at the definition of +, a function symbol that our model is allowed to define:
a + 0 = a a + S(b) = S(a + b)“x + 0 = x” should work perfectly fine for nonstandard numbers.
So going back to P(x):
“there exists some n such that ((0 + n) = x)”
for a nonstandard number x, does there exist some number n such that ((0+n) = x)? Yup, the nonstandard number x! Set n=x.
Oh, but when you said nth successor you meant n had to be standard? Well, that’s the whole problem isn’t it!
Are you saying that in reality every property P has actually three outcomes: true, false, undecidable?
By Godel’s incompleteness theorem yes, unless your theory of arithmetic has a non-recursively enumerable set of axioms or is inconsistent.
And that those always decidable, like e.g. “P(n) <-> (n = 2)” cannot be true for all natural numbers, while those which can be true for all natural numbers, but mostly false otherwise, are always undecidable for… some other values?
I’m having trouble understanding this sentence but I think I know what you are asking about.
There are some properties P(x) which are true for every x in the 0 chain, however, Peano Arithmetic does not include all these P(x) as theorems. If PA doesn’t include P(x) as a theorem, then it is independent of PA whether there exist nonstandard elements for which P(x) is false.
Let’s suppose that for any specific value V in the separated chain it is possible to make such property PV. What would that prove? Would it contradict the article? How specifically?
I think this is what I am saying I believe to be impossible. You can’t just say “V is in the separated chain”. V is a constant symbol. The model can assign constants to whatever object in the domain of discourse it wants to unless you add axioms forbidding it.
Honestly I am becoming confused. I’m going to take a break and think about all this for a bit.
Could someone please confirm my statements in the new sequence post about first-order logic? I want to be sure my understanding is correct.
http://lesswrong.com/lw/f4e/logical_pinpointing/7qv6?context=1#7qv6
If our axiom set T is independent of a property P about numbers then by definition there is nothing inconsistent about the theory T1 = “T and P” and also nothing inconsistent about the theory T2= “T and not P”.
To say that they are not inconsistent is to say that they are satisfiable, that they have possible models. As T1 and T2 are inconsistent with each other, their models are different.
The single zero-based chain of numbers without nonstandard numbers is a single model. Therefore, if there exists a property about numbers that is independent of any theory of arithmetic, that theory of arithmetic does not logically exclude the possibility of nonstandard elements.
By Godel’s incompleteness theorems, a theory must have statements that are independent from it unless it is either inconsistent or has a non-recursively-enumerable theorem set.
Each instance of the axiom schema of induction can be constructed from a property. The set of properties is recursively enumerable, therefore the set of instances of the axiom schema of induction is recursively enumerable.
Every theorem of Peano Arithmetic must use a finite number of axioms in its proof. We can enumerate the theorems of Peano Arithmetic by adding increasingly larger subsets of the infinite set of instances of the axiom schema of induction to our axiom set.
Since the theory of Peano Arithmetic has a recursively enumerable set of theorems it is either inconsistent or is independent of some property and thus allows for the existence of nonstandard elements.
“Because if you had another separated chain, you could have a property P that was true all along the 0-chain, but false along the separated chain. And then P would be true of 0, true of the successor of any number of which it was true, and not true of all numbers.”
But the axiom schema of induction does not completely exclude nonstandard numbers. Sure if I prove some property P for P(0) and for all n, P(n) ⇒ P(n+1) then for all n, P(n); then I have excluded the possibility of some nonstandard number “n” for which not P(n) but there are some properties which cannot be proved true or false in Peano Arithmetic and therefore whose truth hood can be altered by the presence of nonstandard numbers.
Can you give me a property P which is true along the zero-chain but necessarily false along a separated chain that is infinitely long in both directions? I do not believe this is possible but I may be mistaken.
My visualization ability improves the closer I am to sleep, being near perfect during a lucid dream.