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From: Hal Finney <hal.domain.name.hidden>

Date: Thu, 29 Jul 2004 09:53:18 -0700 (PDT)

Tell me again where I am going wrong. Consider each of these examples:

117. q

...

191. Bp

...

207. p -> q

Now, we will say that the machines "believes" something if it is one of

its theorems, right? So we can say that the machine "believes q", it

"believes Bp", and it "believes p->q", right? We could equivalently say

it "believes q is true", etc., but that is redundant. If it writes x

down as a theorem, we will say it believes x, which is shorthand for

saying that it believes x is true. The "is true" part has no real

meaning and does not seem helpful.

We also have this shorthand Bx to mean "the machine believes x". So we

(not the machine, but us, you and I!) can also write, Bq, BBp and B(p->q),

and all of these are true statements, right?

The problem arises when we start to use this same letter B in the

machine's theorems. It is easy to slide back and forth between the

machine's B and our B. But there is no a priori reason to assume that

they are the same. That is something that has to be justified.

Focus on 207. p->q for a moment. We know that, according to the machine's

rules, this theorem means that if it ever writes down p as a theorem,

it will write down q. Therefore it is true that Bp->Bq. This is simply

another way of saying the same thing! Bp means that p is a theorem,

by definition of the letter B, in the real world. And similarly Bq

means that q is a theorem. Given that p->q is a theorem, then if p is

a theorem, so is q. Therefore it is true that B(p->q) -> (Bp -> Bq).

This is not a theorem of the machine, it is a truth in the real world.

What I want to say is that 207. p->q "means" Bp -> Bq. It means that if

the machine ever derives p, it will derive q. This is a true statement

about the operations of the machine. It is not a theorem of the machine.

When we talk about what something "means", I think it has to be what it

means to us, not what it means to the machine. When the machine writes

117.q, it doesn't mean anything to the machine. To us it means that the

machine believes q, or that the machine believes q is true.

Given this approach, I am very hesistant to say that 191. Bp means that

the machine believes that it believes p. I have no problem saying that it

means that the machine believes Bp. But to say that the machine "believes

that it believes p" uses the word "believes" in two very different and

confusing ways. The first "believes" is just a statement about what the

machine has derived as a theorem. We choose to say that the machine's

theorems are what it "believes". I am OK with that. But the second

"believes" refers to the letter B, which we are arbitrarily choosing to

identify with this word "belief".

To the machine, B is just a letter. I still say that I need to know what

the rules are that the machine will apply to that letter. I see that I

was wrong to think that p -> Bp was a rule the machine would have if it

were "normal". You said that for a normal machine, if it ever proved p,

it would also prove Bp. Okay, but how could it possibly do this without

ANY rules to deal with the letter B?

Normality is not something one can just assert about a machine. You would

need to give it a rule for dealing with the letter B, then you could prove

(not the machine proving, you would prove it) that if the machine ever

derived p, its rules for dealing with the letter B would then cause it

to derive Bp. In this way you would show that the machine was normal.

My axiom, which I should have written as, "0a. for all x, x->Bx", would

in fact be sufficient to show that a machine which had that axiom would

be normal. A machine which had this rule for dealing with the letter

B would be normal, because any time it derived p, it could immediately

derive Bp using this axiom.

However, this machine may be too powerful. Although it is normal, it

is much more.

So my question to you is, what is an example of an axiom for dealing

with the letter B that would make a machine be "just" normal, but no more?

Hal Finney

Received on Thu Jul 29 2004 - 13:40:02 PDT

Date: Thu, 29 Jul 2004 09:53:18 -0700 (PDT)

Tell me again where I am going wrong. Consider each of these examples:

117. q

...

191. Bp

...

207. p -> q

Now, we will say that the machines "believes" something if it is one of

its theorems, right? So we can say that the machine "believes q", it

"believes Bp", and it "believes p->q", right? We could equivalently say

it "believes q is true", etc., but that is redundant. If it writes x

down as a theorem, we will say it believes x, which is shorthand for

saying that it believes x is true. The "is true" part has no real

meaning and does not seem helpful.

We also have this shorthand Bx to mean "the machine believes x". So we

(not the machine, but us, you and I!) can also write, Bq, BBp and B(p->q),

and all of these are true statements, right?

The problem arises when we start to use this same letter B in the

machine's theorems. It is easy to slide back and forth between the

machine's B and our B. But there is no a priori reason to assume that

they are the same. That is something that has to be justified.

Focus on 207. p->q for a moment. We know that, according to the machine's

rules, this theorem means that if it ever writes down p as a theorem,

it will write down q. Therefore it is true that Bp->Bq. This is simply

another way of saying the same thing! Bp means that p is a theorem,

by definition of the letter B, in the real world. And similarly Bq

means that q is a theorem. Given that p->q is a theorem, then if p is

a theorem, so is q. Therefore it is true that B(p->q) -> (Bp -> Bq).

This is not a theorem of the machine, it is a truth in the real world.

What I want to say is that 207. p->q "means" Bp -> Bq. It means that if

the machine ever derives p, it will derive q. This is a true statement

about the operations of the machine. It is not a theorem of the machine.

When we talk about what something "means", I think it has to be what it

means to us, not what it means to the machine. When the machine writes

117.q, it doesn't mean anything to the machine. To us it means that the

machine believes q, or that the machine believes q is true.

Given this approach, I am very hesistant to say that 191. Bp means that

the machine believes that it believes p. I have no problem saying that it

means that the machine believes Bp. But to say that the machine "believes

that it believes p" uses the word "believes" in two very different and

confusing ways. The first "believes" is just a statement about what the

machine has derived as a theorem. We choose to say that the machine's

theorems are what it "believes". I am OK with that. But the second

"believes" refers to the letter B, which we are arbitrarily choosing to

identify with this word "belief".

To the machine, B is just a letter. I still say that I need to know what

the rules are that the machine will apply to that letter. I see that I

was wrong to think that p -> Bp was a rule the machine would have if it

were "normal". You said that for a normal machine, if it ever proved p,

it would also prove Bp. Okay, but how could it possibly do this without

ANY rules to deal with the letter B?

Normality is not something one can just assert about a machine. You would

need to give it a rule for dealing with the letter B, then you could prove

(not the machine proving, you would prove it) that if the machine ever

derived p, its rules for dealing with the letter B would then cause it

to derive Bp. In this way you would show that the machine was normal.

My axiom, which I should have written as, "0a. for all x, x->Bx", would

in fact be sufficient to show that a machine which had that axiom would

be normal. A machine which had this rule for dealing with the letter

B would be normal, because any time it derived p, it could immediately

derive Bp using this axiom.

However, this machine may be too powerful. Although it is normal, it

is much more.

So my question to you is, what is an example of an axiom for dealing

with the letter B that would make a machine be "just" normal, but no more?

Hal Finney

Received on Thu Jul 29 2004 - 13:40:02 PDT

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