Petry's paradox
david petry
At the moment, I am baffled by this.
Godel's theorem tells us that for a wide class of theories, the
assertion that the theory is consistent is logically independent of
the theory itself. So, we can add to the axioms of the theory either
the axiom that the theory is consistent, or that it is inconsistent,
and get a new consistent theory.
So let T be a consistent theory. And let S = T+~CON(T). That is,
we've added to the theory T the axiom that T is inconsistent. So S
must be consistent. Let's look at S+CON(S), which must be consistent.
Since any theorem in T will also be a theorem in S, the consistency of
S will imply the consistency of T. Hence it will be a theorem of S
+CON(S) that T is consistent, but of course, it is an axiom of S
+CON(S) that T is inconsistent, which seems to be a paradoxical
situation.
What am i missing?
abo
wrote:
S proves ~ CON(T). S proves, if axioms A are a subset of axioms B,
and S proves ~ CON(A), then ~ CON(B). Thus S proves ~ CON(S). Thus S
+ CON(S) is inconsistent. (Remark that Godel still holds: S is
consistent, and it cannot prove CON(S). Godel never said anything
about whether, if S is consistent, S can prove
~ CON(S).)
Aatu Koskensilta
> Godel's theorem tells us that for a wide class of theories, the
> assertion that the theory is consistent is logically independent of
> the theory itself. So, we can add to the axioms of the theory either
> the axiom that the theory is consistent, or that it is inconsistent,
> and get a new consistent theory.
The second incompleteness theorem only tells us that if a theory T
meeting certain criteria is consistent it does not prove "T is
consistent". A consistent theory might prove itself inconsistent, as
PA + "PA is inconsistent" does, for example.
--
Aatu Koskensilta (aatu.kos...@xortec.fi)
"Wovon man nicht sprechen kann, daruber muss man schweigen"
- Ludwig Wittgenstein, Tractatus Logico-Philosophicus
Daryl McCullough
>
>
>
>At the moment, I am baffled by this.
>
>Godel's theorem tells us that for a wide class of theories, the
>assertion that the theory is consistent is logically independent of
>the theory itself.
No, that's not necessarily true. What Godel argued was the
following:
If T is consistent, then T does not prove Con(T).
If T is omega-consistent, then T does not prove ~Con(T).
A theory T of arithmetic is omega-consistent if for
every formula Phi(x),
If T proves each of the following
statements:
Phi(0)
Phi(1)
Phi(2)
...
then T does *not* prove
Exists x ~Phi(x)
Note that if T is omega-inconsistent, then it proves the existence
of a natural number that is provably unequal to 0, and unequal to
1, unequal to 2, etc. An omega-inconsistent theory can only have
nonstandard models.
If T is omega-inconsistent, then it is possible that T proves
~Con(T).
Rosser showed that any consistent axiomatizable theory extending
Peano Arithmetic, whether omega-consistent or not, has a statement
that is independent. The Rosser sentence is a sentence R with
the following property:
R <-> if there is a proof of R from the axioms of T, then
there is a shorter proof of ~R
The negation of R has the following property:
~R <-> there is a proof of R and there is no shorter proof
of ~R
If T is consistent, then neither R nor ~R is provable:
Assume that R is provable. Then there is some proof p of R.
By checking all proofs shorter than p, either you find a
proof q of ~R, or you don't. If you find such a proof,
then T is inconsistent (since you have a proof of R and
~R). If you don't find such a proof, then you can use
that fact to prove the claim: There is a proof of R,
and there is no shorter proof of ~R. But this claim
is equivalent to ~R. So again you have a proof of
R and ~R.
A similar argument shows that if T is consistent, then
it does not prove ~R.
--
Daryl McCullough
Ithaca, NY
MoeBlee
wrote:
It seems to be a non sequitur to conclude that "T is consistent" is a
theorem of S+Con(S).
Here's what you have:
(1) T is consistent and T satisfies Godel [let's use Rosser]
conditions ... assumption
(2) S = T+~Con(T) ... definition
(3) S is consistent and S satisfies Rosser conditions... from (1) and
(2).
assumption (1) still undischarged
(4) S+Con(S) is consistent and S+Con(S) satisifes Rosser
conditions ... from (3)
assumption (1) still undischarged
(5) S is consistent -> T is consistent ... from (2)
(7) S+Con(S) is consistent -> T is consistent ... from (2)
(8) T is consistent ... (4) and (7), though we had it with (1) anyway.
assumption (1) still undischarged
(9) (T is consistent and T satisfies the Rosser conditions) -> S
+Con(S) is consistent
assumption (1) discharged.
Then you inferred:
Con(T) is provable IN THE SYSTEM S+Con(S).
But what justifies that inference?
Actually, discharging your assumption that T is consistent leaves you
with just:
Con(T) -> Con(T)
so
S+Con(S) |- Con(T) -> Con(T),
which is not saying much.
Also, you argued:
~Con(T) is an axiom of S+Con(S), so, since S+Con(S) is consistent, we
should not have Con(T) as a theorem of S+Con(S).
But again note that we don't have S+Con(S) is consistent standalone,
but rather we have S+Con(S) is consistent on the ASSUMPTIONS that T is
consistent and T satisifies the Rosser conditions, so we don't have
your paradoxical:
S+Con(S) is consistent and S+Con(S) is consistent,
since we don't have
S+Con(S) |- Con(T),
but rather only
If T is consistent and T satisfies the Rosser conditions then S+Con(S)
is consistent
and
Regardless of T, we have S+Con(S) |- Con(T) -> Con(T),
none of which is paradoxical.
MoeBlee
Jesse F. Hughes
--
Jesse F. Hughes
"Mathematicians who read proofs of my results seem to basically lose
some part of themselves, like it rips at their souls, and they are no
longer quite right in the head." -- James S. Harris, Geek Cthulhu
Newberry
wrote:
> On 2008-01-29, in sci.logic, david petry wrote:
>
> > Godel's theorem tells us that for a wide class of theories, the
> > assertion that the theory is consistent is logically independent of
> > the theory itself. So, we can add to the axioms of the theory either
> > the axiom that the theory is consistent, or that it is inconsistent,
> > and get a new consistent theory.
>
> The second incompleteness theorem only tells us that if a theory T
> meeting certain criteria is consistent it does not prove "T is
> consistent". A consistent theory might prove itself inconsistent, as
> PA + "PA is inconsistent" does, for example.
Isn't it kind of a contradiction?
herbzet
Seemingly, but there's a loophole:
The axiom "PA is inconsistent" actually says that there exists
a number that is the Godel number of a proof of some formula
P & ~P (or 0 = 1, or whatever). It is a statement of existence.
It does not tell us _which_ number that is.
If PA is consistent, then there will be no proof in PA of any formula
P & ~P. Any model of PA + ~Con(PA), which asserts that there *is*
a Godel-number for such a proof-in-PA, will be a non-standard model
with non-standard numbers. The number that is the Godel number of
a proof-in-PA of some formula P & ~P will be a non-standard number -- a
"supernatural" number.
It is perhaps worth recalling here that the Lowenheim-Skolem
theorem says that any consistent first-order theory in a language
with a countable signature (such as PA) will have a model of every
infinite cardinality, if it has a model of any infinite cardinality.
So PA, if it is consistent (has a model), it will necessarily have
non-standard models. If PA is consistent, then PA + ~Con(PA) is
consistent (this follows from Godel) but it will have only
non-standard models.
--
hz
Peter_Smith
wrote:
> At the moment, I am baffled by this.
Cue shameless self advertising ....
You might find a helpful discussion of this sort of thing in e.g. Sec
25.6 of my Gödel book.
Cheers, Peter S.
herbzet
herbzet wrote:
> So PA, if it is consistent (has a model), it will necessarily have
> non-standard models. If PA is consistent, then PA + ~Con(PA) is
> consistent (this follows from Godel) but it will have only
> non-standard models.
To belabor the point, if PA is consistent, then PA + ~Con(PA) will
not have the standard model because the standard model has only
natural numbers in it, and no natural number will be the Godel
number of a proof-in-PA that P & ~P (since we're assuming PA
is, in fact, consistent); this contradicts the assertion that
there exists a number corresponding to such a proof. So ~Con(PA)
would be false in the standard model, so the standard model is not
a model of PA + ~Con(PA).
--
hz
Aatu Koskensilta
How so?
david petry
wrote:
> On 2008-02-03, in sci.logic, Newberry wrote:
>
> > On Jan 29, 5:46 am, Aatu Koskensilta <aatu.koskensi...@xortec.fi>
> > wrote:
>
> >> The second incompleteness theorem only tells us that if a theory T
> >> meeting certain criteria is consistent it does not prove "T is
> >> consistent". A consistent theory might prove itself inconsistent, as
> >> PA + "PA is inconsistent" does, for example.
>
> > Isn't it kind of a contradiction?
>
> How so?
In the natural language, common sense notion of consistency, a story
or theory is consistent iff it is consistent with facts considered to
be known--i.e it must be consistent with truth. And, of course, PA +
"PA is inconsistent" is not consistent with what we believe to be
true.
Aatu Koskensilta
> In the natural language, common sense notion of consistency, a story
> or theory is consistent iff it is consistent with facts considered to
> be known--i.e it must be consistent with truth.
So you would consider a story in which humans went extinct in 1953
inconsistent? I'm afraid you'll find not many share this peculiar
notion of consistency of yours.
Newberry
Then I would not say that the theory PROVES its own inonsistency.
> It is perhaps worth recalling here that the Lowenheim-Skolem
> theorem says that any consistent first-order theory in a language
> with a countable signature (such as PA) will have a model of every
> infinite cardinality, if it has a model of any infinite cardinality.
>
> So PA, if it is consistent (has a model), it will necessarily have
> non-standard models. If PA is consistent, then PA + ~Con(PA) is
> consistent (this follows from Godel) but it will have only
> non-standard models.
>
> --
> hz- Hide quoted text -
>
> - Show quoted text -
herbzet
> Then I would not say that the theory PROVES its own inconsistency.
I have no good reply, pro or con, at the moment.
Peter_Smith
> In the natural language, common sense notion of consistency, a story
> or theory is consistent iff it is consistent with facts considered to
> be known--i.e it must be consistent with truth.
Not so. Even in everyday contexts (outside the logic classroom) we
very sharply distinguish between a story's being internally
inconsistent (as in "the part of the report on p. 7 of the Guardian is
inconsistent with the rest of the report on p. 12"), and a story's not
being consistent with the truth (i.e. being false). We know that if a
story is internally inconsistent it can't be true. But a story can be
internally consistent without being "consistent with truth".
You are entirely familiar with that important distinction -- described
one way or another -- in natural language. It applies equally well
here. And in logical contexts, "consistent" always means internally
consistent. PA + "PA is inconsistent" is not "consistent with what we
believe to be true"; but it is plain consistent if PA is.
david petry
Whatever.
Do you not feel a tinge of cognitive dissonance when you consider the
following two statements together?
1) We could not possibly believe 'A' and maintain a consistent belief
system.
2) 'A' is consistent.
MoeBlee
wrote:
> On Feb 9, 12:53 am, Peter_Smith <ps...@cam.ac.uk> wrote:
> > On Feb 9, 1:56 am, david petry <david_lawrence_pe...@yahoo.com> wrote:
>
> > > In the natural language, common sense notion of consistency, a story
> > > or theory is consistent iff it is consistent with facts considered to
> > > be known--i.e it must be consistent with truth.
>
> > Not so. Even in everyday contexts (outside the logic classroom) we
> > very sharply distinguish between a story's being internally
> > inconsistent (as in "the part of the report on p. 7 of the Guardian is
> > inconsistent with the rest of the report on p. 12"), and a story's not
> > being consistent with the truth (i.e. being false). We know that if a
> > story is internally inconsistent it can't be true. But a story can be
> > internally consistent without being "consistent with truth".
>
> > You are entirely familiar with that important distinction -- described
> > one way or another -- in natural language. It applies equally well
> > here. And in logical contexts, "consistent" always means internally
> > consistent. PA + "PA is inconsistent" is not "consistent with what we
> > believe to be true"; but it is plain consistent if PA is.
>
> Whatever.
Why "whatever"? Peter gave you fine response. I don't see that it is
ineffective so to deserve a "whatever".
> Do you not feel a tinge of cognitive dissonance when you consider the
> following two statements together?
>
> 1) We could not possibly believe 'A' and maintain a consistent belief
> system.
> 2) 'A' is consistent.
No, unless 'possibly' in your sense is that of logical possibility.
MoeBlee
david petry
> On Feb 11, 2:19 pm, david petry <david_lawrence_pe...@yahoo.com>
> wrote:
>
>
>
>
>
> > On Feb 9, 12:53 am, Peter_Smith <ps...@cam.ac.uk> wrote:
> > > On Feb 9, 1:56 am, david petry <david_lawrence_pe...@yahoo.com> wrote:
>
> > > > In the natural language, common sense notion of consistency, a story
> > > > or theory is consistent iff it is consistent with facts considered to
> > > > be known--i.e it must be consistent with truth.
>
> > > Not so. Even in everyday contexts (outside the logic classroom) we
> > > very sharply distinguish between a story's being internally
> > > inconsistent (as in "the part of the report on p. 7 of the Guardian is
> > > inconsistent with the rest of the report on p. 12"), and a story's not
> > > being consistent with the truth (i.e. being false). We know that if a
> > > story is internally inconsistent it can't be true. But a story can be
> > > internally consistent without being "consistent with truth".
>
> > > You are entirely familiar with that important distinction -- described
> > > one way or another -- in natural language. It applies equally well
> > > here. And in logical contexts, "consistent" always means internally
> > > consistent. PA + "PA is inconsistent" is not "consistent with what we
> > > believe to be true"; but it is plain consistent if PA is.
>
> > Whatever.
>
> Why "whatever"? Peter gave you fine response. I don't see that it is
> ineffective so to deserve a "whatever".
The "whatever" is meant to indicate that I think he missed the point.
> > Do you not feel a tinge of cognitive dissonance when you consider the
> > following two statements together?
>
> > 1) We could not possibly believe 'A' and maintain a consistent belief
> > system.
> > 2) 'A' is consistent.
>
> No, unless 'possibly' in your sense is that of logical possibility.
That is the sense under discussion.
Maybe I could point out to you that almost any belief system could be
brought back to consistency by adding the belief, "I simply refuse to
accept any conclusion that can only be reached by more than 'N' steps
of reasoning", for some suitably small value of 'N'.
MoeBlee
Then, yes, I would think it irrational to hold both 1) and 2) where
the sense of the word 'possibly' is logical possibility. But so what?
That's not what we're doing when we don't find a contradiction in your
example in your original post.
> Maybe I could point out to you that almost any belief system could be
> brought back to consistency by adding the belief, "I simply refuse to
> accept any conclusion that can only be reached by more than 'N' steps
> of reasoning", for some suitably small value of 'N'.
Okay, an interesting thought, but so what? The consistency of a formal
theory does not depend on allowing only n steps of reasoning.
EVERY formal theory T, other than a theory of just validities, is such
Tu{P} is inconsistent for some contingent sentence P.
MoeBlee
david petry
> On Feb 11, 5:57 pm, david petry <david_lawrence_pe...@yahoo.com>
> > > > Do you not feel a tinge of cognitive dissonance when you consider the
> > > > following two statements together?
>
> > > > 1) We could not possibly believe 'A' and maintain a consistent belief
> > > > system.
> > > > 2) 'A' is consistent.
>
> > > No, unless 'possibly' in your sense is that of logical possibility.
> > That is the sense under discussion.
>
> Then, yes, I would think it irrational to hold both 1) and 2) where
> the sense of the word 'possibly' is logical possibility. But so what?
> That's not what we're doing when we don't find a contradiction in your
> example in your original post.
The topic here is my response to Aatu's question to Newberry.
The point is, if A = PA+"PA is inconsistent", then 'A' may be
consistent, but we could not believe 'A' and maintain a consistent
belief system, which I believe suggests that the natural language
meaning of "consistent" is different from the logician's meaning.
Aatu Koskensilta
> Do you not feel a tinge of cognitive dissonance when you consider the
> following two statements together?
>
> 1) We could not possibly believe 'A' and maintain a consistent belief
> system.
> 2) 'A' is consistent.
Sure. They're contradictory. But the following two statements may well
be true simultaneously:
1. X believes "My beliefs are contradictory"
2. X's beliefs are not contradictory
Of course, if 2. is true X won't know, or believe, 2, but it is
entirely possible for someone to mistakenly believe his beliefs are
contradictory -- an eminently sensible belief, taking human
fallibility into account! -- even if, by some amazing accident, his
beliefs are, in fact, not contradictory.
MoeBlee
wrote:
> On Feb 11, 6:20 pm, MoeBlee <jazzm...@hotmail.com> wrote:
>
> > On Feb 11, 5:57 pm, david petry <david_lawrence_pe...@yahoo.com>
> > > > > Do you not feel a tinge of cognitive dissonance when you consider the
> > > > > following two statements together?
>
> > > > > 1) We could not possibly believe 'A' and maintain a consistent belief
> > > > > system.
> > > > > 2) 'A' is consistent.
>
> > > > No, unless 'possibly' in your sense is that of logical possibility.
> > > That is the sense under discussion.
>
> > Then, yes, I would think it irrational to hold both 1) and 2) where
> > the sense of the word 'possibly' is logical possibility. But so what?
> > That's not what we're doing when we don't find a contradiction in your
> > example in your original post.
>
> The topic here is my response to Aatu's question to Newberry.
>
> The point is, if A = PA+"PA is inconsistent", then 'A' may be
> consistent, but we could not believe 'A' and maintain a consistent
> belief system,
What contradiction would one be committed to believing?
It would be a whacky system, but I don't know what contradiction you
think one would have to believe.
> which I believe suggests that the natural language
> meaning of "consistent" is different from the logician's meaning.
There is not just one natural language meaning of 'consistent' but at
least a few. And that the technical use in mathematical logic differs
from certain natural languages sentences is not disputed.
MoeBlee
david petry
wrote:
> On 2008-02-11, in sci.logic, david petry wrote:
>
> > Do you not feel a tinge of cognitive dissonance when you consider the
> > following two statements together?
>
> > 1) We could not possibly believe 'A' and maintain a consistent belief
> > system.
> > 2) 'A' is consistent.
>
> Sure. They're contradictory. But the following two statements may well
> be true simultaneously:
>
> 1. X believes "My beliefs are contradictory"
> 2. X's beliefs are not contradictory
>
> Of course, if 2. is true X won't know, or believe, 2, but it is
> entirely possible for someone to mistakenly believe his beliefs are
> contradictory -- an eminently sensible belief, taking human
> fallibility into account! -- even if, by some amazing accident, his
> beliefs are, in fact, not contradictory.
I believe that I am incapable of understanding how a person fully
capable of introspection could honestly and meaningfully believe that
his own beliefs are contradictory.
herbzet
david petry wrote:
> The topic here is my response to Aatu's question to Newberry.
>
> The point is, if A = PA+"PA is inconsistent", then 'A' may be
> consistent, but we could not believe 'A' and maintain a consistent
> belief system,
Sure we could. A = PA + ~Con(PA) has models: What theory A
is true of are things different from our ordinary conception
of the natural numbers, that's all. We can "believe" its theorems
are true of any of its models.
> which I believe suggests that the natural language
> meaning of "consistent" is different from the logician's meaning.
The logicians meaning of "consistent" is in perfect agreement with
an ordinary natural language meaning of "consistent", as PS has
pointed out.
More to the point is the question of whether, and to what degree,
the first-order formula ~Con(PA) captures the natural language
meaning of the meta-theoretical statement "PA is inconsistent".
That's a puzzlement to me, at the moment.
--
hz
david petry
> david petry wrote:
> > The topic here is my response to Aatu's question to Newberry.
>
> > The point is, if A = PA+"PA is inconsistent", then 'A' may be
> > consistent, but we could not believe 'A' and maintain a consistent
> > belief system,
>
> Sure we could. A = PA + ~Con(PA) has models: What theory A
> is true of are things different from our ordinary conception
> of the natural numbers, that's all. We can "believe" its theorems
> are true of any of its models.
In a recent response to Aatu, I suggested that when we talk about what
people can believe, we must add the restriction that we are talking
about people fully capable of introspection. It's not really clear
what it would mean to say that a person incapable of introspection has
beliefs.
A person capable of introspection who believes 'A' (as defined above)
would reason as follows: "Since I believe that my own beliefs include
a contradiction, I have to believe that I am capable of arriving at
absolutely any conclusion at all, and hence I have no reason to
believe that any of the conclusions I reach are necessarily true,
which implies that I can't justify believing that I have any beliefs
at all".
Quadibloc
wrote:
> On 2008-02-09, in sci.logic, david petry wrote:
>
> > In the natural language, common sense notion of consistency, a story
> > or theory is consistent iff it is consistent with facts considered to
> > be known--i.e it must be consistent with truth.
>
> So you would consider a story in which humans went extinct in 1953
> inconsistent? I'm afraid you'll find not many share this peculiar
> notion of consistency of yours.
Can the theory formed by a consistent theory T with the axiom "T is
inconsistent" added be consistent, *given* that, thanks to Godel's
Theorem, that T cannot prove itself to be consistent?
Well, if T cannot prove itself consistent, then it is true that T+"T
is inconsistent" cannot prove that T is consistent either, so it does
not prove any statements which contradict each other by reason of the
added axiom. That does seem to leave it consistent given that T is
consistent.
It is still, however, _flawed_. Since T actually is consistent, it
includes a false statement, so it can't be properly expanded in some
directions by adding new true axioms. There may not be a technical
mathematical term for this kind of flaw, and it is certainly possible
for something to be flawed without inconsistency being one of its
flaws.
John Savard
Quadibloc
wrote:
> Do you not feel a tinge of cognitive dissonance when you consider the
> following two statements together?
>
> 1) We could not possibly believe 'A' and maintain a consistent belief
> system.
> 2) 'A' is consistent.
Well, given that:
T is a consistent logical system that, by Godel's proof, cannot prove
its own consistency,
A is T plus "T is inconsistent",
then it is true that I can't *believe* A. I can't believe that A is
completely true and consistent.
However, since I do believe that T cannot prove its own consistency, I
also believe that in A one cannot prove "T is consistent", so in A one
can't prove any statement and that statement's own contradiction.
So A is flawed. It includes one false statement. But inconsistency
isn't one of A's flaws; there are no contradictions *visible* within A
itself, even though embedding A in a larger meta-system shows that A
would lead to contradictions at that level.
If one uses the term "consistent" in a narrow technical sense - which
is needed for doing logic precisely - there is no paradox. If we use
it in a vernacular sense to claim that A is not flawed, _then_ that
can't be quite right. The system A does have a flaw, but what the new
axiom contradicts is a true fact about T that *is not within T
itself*. So the flaw isn't inconsistency; I suppose you could call it
meta-inconsistency.
John Savard
Daryl McCullough
>More to the point is the question of whether, and to what degree,
>the first-order formula ~Con(PA) captures the natural language
>meaning of the meta-theoretical statement "PA is inconsistent".
>
>That's a puzzlement to me, at the moment.
The formula ~Con(PA) captures the natural language meaning
of "PA is inconsistent" in the sense that we can prove that
PA is inconsistent <-> ~Con(PA) is true of the naturals.
In any model of the theory A = PA + ~Con(PA), the formula
~Con(PA) does *not* mean "PA is inconsistent". It means
something complicated like "There exists a hyperfinite
number x such that x is a code for a hyperfinite structure
such that blah, blah, blah".
herbzet
Hah! I figured something like that. Thanks.
--
hz
MoeBlee
wrote:
> A person capable of introspection who believes 'A' (as defined above)
> would reason as follows: "Since I believe that my own beliefs include
> a contradiction, I have to believe that I am capable of arriving at
> absolutely any conclusion at all, and hence I have no reason to
> believe that any of the conclusions I reach are necessarily true,
Non sequitur. One may recognize that one is able to incorrectly
conclude contingent statements that are in contradiction with one
another but still recognize that some of ones beliefs are of necessary
truths. Also, you're assuming that the person understands ex falso
quodlibet, which not all people do, not even all fairly intelligent
and rational people .
And backing up to an earlier point of yours. A rational person might
recognize that he has thousands upon thousands (millions?) of beliefs
and that therefore there is great likelihood that at least two of
those beliefs are inconsistent (since one might have this great number
of beliefs without having checked each one against all the others),
and therefore believe, based on that likelihood, that one does have
inconsistent beliefs.
MoeBlee
david petry
> On Feb 13, 2:14 am, david petry <david_lawrence_pe...@yahoo.com>
> wrote:
>
> > A person capable of introspection who believes 'A' (as defined above)
> > would reason as follows: "Since I believe that my own beliefs include
> > a contradiction, I have to believe that I am capable of arriving at
> > absolutely any conclusion at all, and hence I have no reason to
> > believe that any of the conclusions I reach are necessarily true,
> A rational person might
> recognize that he has thousands upon thousands (millions?) of beliefs
> and that therefore there is great likelihood that at least two of
> those beliefs are inconsistent
So we're really not talking about the same thing. I had in mind
fundamental, core beliefs that form the foundation of our
understanding of the world, not specific beliefs about the world which
always include an element of vagueness and uncertainty.
MoeBlee
wrote:
> I had in mind
> fundamental, core beliefs that form the foundation of our
> understanding of the world, not specific beliefs about the world which
> always include an element of vagueness and uncertainty.
Then that does change the question, as well as introduce a question as
to what is and is not a 'fundamental, core belief that forms the
foundation of our understanding of the world'.
Nevertheless, I don't fnd it too very hard to imagine someone who
admits that say, his religious beliefs contradict his scientific
beliefs, or someone who has a philosophy such as Taoism (or at least
in its more popularized characterization) in which contradictions are
embraced.
MoeBlee
Gauster
> theory extending
> Peano Arithmetic, whether omega-consistent or not,
> has a statement
> that is independent. The Rosser sentence is a
> sentence R with
> the following property:
>
> R <-> if there is a proof of R from the axioms of T,
> then
> there is a shorter proof of ~R
>
> The negation of R has the following property:
>
> ~R <-> there is a proof of R and there is no shorter
> proof
> of ~R
>
> If T is consistent, then neither R nor ~R is
> provable:
>
> Assume that R is provable. Then there is some proof p
> of R.
> By checking all proofs shorter than p, either you
> find a
> proof q of ~R, or you don't. If you find such a
> proof,
> then T is inconsistent (since you have a proof of R
> and
> ~R). If you don't find such a proof, then you can use
> that fact to prove the claim: There is a proof of R,
> and there is no shorter proof of ~R. But this claim
> is equivalent to ~R. So again you have a proof of
> R and ~R.
>
> A similar argument shows that if T is consistent,
> then
> it does not prove ~R.
I've been trying to recreate this proof formally following pretty much the same argument used in Gödel's original proof, taking into account that the bound for the "shorter" proof allows the sentence R to be primitive recursive.
But I fail to proof the second part, that is, that if T is consistent it does not proof ~R. If I use the same method as in Gödel's proof then I definitely need omega-consistency. Unfortunately I cannot find Rosser's proof on the Internet...Any clue on how the argument goes exactly in this second case?