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From: Marchal <marchal.domain.name.hidden>

Date: Wed May 9 03:27:00 2001

Hi,

I'm sorry. I realise I made an omission error in the post I

send the Wed, 17 Nov 1999 at 06:24:30:

http://www.escribe.com/science/theory/m1417.html

(the post I refer to George yesterday (8 may 2001) for getting an

acquaintance with G and G*).

Perhaps you have seen it.

The omission error is in the definition of G* (i'm very sorry!)

I said:

G* has as axioms all theorems of G, and, as ONLY

rule of inference, the modus ponens.

Well, if that was the definition of G*, G* would be included

in G! (actually G* would be equal to G in the sense of having the

same theorems).

It is, (as I have said in other posts):

G* has as axioms all theorems of G, all wff []A->A, and has,

as ONLY rule of inference, the modus ponens.

For exemple G* proves []false -> false, that is G* proves <>true,

that is G* proves the consistency of the sound machine which

we are interviewing. Of course the sound machine itself is mute

on that question. The machine (and the guardian) both can justify

that silence, both G and G* proves <>true -> -[]<>true ("Godel's

second incompleteness theorem").

Note that G* is not closed for the necessation rule (of inference),

and cannot be!

If you add the necessitation rule to G*, it is easy to see that G*

becomes inconsistent. Indeed G* proves <>true, so we the necessitation

rule you have that G* proves []<>true.

But G* proves <>true -> -[]<>true, and also <>true, so by the

modus ponens rule, G* proves []<>true.

So G* proves []<>true & -[]<>true. But G* proves all ((A & -A) -> false)

by classical propositional calculus, so

G* proves (([]<>true & -[]<>true) -> false) and by modus ponens again

G* proves false.

You could find a shorter proof using only Loeb 's formula L instead

of Goedel's second theorem). Recall L is ([]([]A->A)) -> []A

Bruno

Received on Wed May 09 2001 - 03:27:00 PDT

Date: Wed May 9 03:27:00 2001

Hi,

I'm sorry. I realise I made an omission error in the post I

send the Wed, 17 Nov 1999 at 06:24:30:

http://www.escribe.com/science/theory/m1417.html

(the post I refer to George yesterday (8 may 2001) for getting an

acquaintance with G and G*).

Perhaps you have seen it.

The omission error is in the definition of G* (i'm very sorry!)

I said:

G* has as axioms all theorems of G, and, as ONLY

rule of inference, the modus ponens.

Well, if that was the definition of G*, G* would be included

in G! (actually G* would be equal to G in the sense of having the

same theorems).

It is, (as I have said in other posts):

G* has as axioms all theorems of G, all wff []A->A, and has,

as ONLY rule of inference, the modus ponens.

For exemple G* proves []false -> false, that is G* proves <>true,

that is G* proves the consistency of the sound machine which

we are interviewing. Of course the sound machine itself is mute

on that question. The machine (and the guardian) both can justify

that silence, both G and G* proves <>true -> -[]<>true ("Godel's

second incompleteness theorem").

Note that G* is not closed for the necessation rule (of inference),

and cannot be!

If you add the necessitation rule to G*, it is easy to see that G*

becomes inconsistent. Indeed G* proves <>true, so we the necessitation

rule you have that G* proves []<>true.

But G* proves <>true -> -[]<>true, and also <>true, so by the

modus ponens rule, G* proves []<>true.

So G* proves []<>true & -[]<>true. But G* proves all ((A & -A) -> false)

by classical propositional calculus, so

G* proves (([]<>true & -[]<>true) -> false) and by modus ponens again

G* proves false.

You could find a shorter proof using only Loeb 's formula L instead

of Goedel's second theorem). Recall L is ([]([]A->A)) -> []A

Bruno

Received on Wed May 09 2001 - 03:27:00 PDT

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