Prove ( P = |- Q ) => |- ( P = Q )
(P and Q have the same sets of free variables.) This simple theorem (I
created 12/1/05) provides the link between Theory of Computation and
Proof Theory (Incompleteness in Logic) that theoreticians such as
myself have been looking for since the 1930′s. (Each Theory of
Computation theorem becomes an Incompleteness theorem in Logic,
providing almost trivial formal derivation of the exact results of
Godel, Rosser and Smullyan.)
C-B
Hint: Consider the equivalent ( P = |-Q ) => ( P = |- P )
On Wed, 14 Dec 2005, Charlie-Boo wrote:
> Prove ( P = |- Q ) => |- ( P = Q )
As you are mixing the object language with the meta language
your statement as presented is non-sense gibberish.
> (P and Q have the same sets of free variables.)
> Hint: Consider the equivalent ( P = |-Q ) => ( P = |- P )
All gibberish is equivalent to all other gibberish except some
gibberish is more equivalently certified and guaranteed non-sense.
William Elliot wrote:
> On Wed, 14 Dec 2005, Charlie-Boo wrote:
> > Prove ( P = |- Q ) => |- ( P = Q )
> As you are mixing the object language with the meta language
> your statement as presented is non-sense gibberish.
Don’t you think that provability is expressible within the Logic?
C-B
- Hide quoted text — Show quoted text -
> > (P and Q have the same sets of free variables.)
> > Hint: Consider the equivalent ( P = |-Q ) => ( P = |- P )
> All gibberish is equivalent to all other gibberish except some
> gibberish is more equivalently certified and guaranteed non-sense.
On Thu, 15 Dec 2005, Charlie-Boo wrote:
> William Elliot wrote:
> > On Wed, 14 Dec 2005, Charlie-Boo wrote:
> > > Prove ( P = |- Q ) => |- ( P = Q )
> > As you are mixing the object language with the meta language
> > your statement as presented is non-sense gibberish.
> Don’t you think that provability is expressible within the Logic?
Do it. But before you do it, explain what your object language is.
William Elliot wrote:
> On Thu, 15 Dec 2005, Charlie-Boo wrote:
> > William Elliot wrote:
> > > On Wed, 14 Dec 2005, Charlie-Boo wrote:
> > > > Prove ( P = |- Q ) => |- ( P = Q )
> > > As you are mixing the object language with the meta language
> > > your statement as presented is non-sense gibberish.
> > Don’t you think that provability is expressible within the Logic?
> Do it. But before you do it, explain what your object language is.
Something as fundamental as the above theorem is certainly valid in
more than one Logic. Like any good metamathematical theorem, we give
the least limiting requirements for our conclusions to hold (cf the
Godel/Rosser competition.) We instead allow whatever Logics through
which our argument can be carried. (I’m not saying that the proof
has to be formal or representable in the Logic, although of course
that’s always nice. The formality begins after the above is proven.
The theorems of informal proofs are the axioms of formal proofs.
Uninspired axioms are NOT inevitable.)
C-B
"I had all of the problems of science analyzed and solved – until I
awoken only to discover it was but the ranting of a madman."
12-22-05
- Hide quoted text — Show quoted text -
On Thu, 15 Dec 2005, Charlie-Boo wrote:
> William Elliot wrote:
> > On Thu, 15 Dec 2005, Charlie-Boo wrote:
> > > William Elliot wrote:
> > > > On Wed, 14 Dec 2005, Charlie-Boo wrote:
> > > > > Prove ( P = |- Q ) => |- ( P = Q )
> > > > As you are mixing the object language with the meta language
> > > > your statement as presented is non-sense gibberish.
> > > Don’t you think that provability is expressible within the Logic?
> > Do it. But before you do it, explain what your object language is.
> Something as fundamental as the above theorem is certainly valid in
> more than one Logic.
Indeed, they’re called not well formed formulas.
"Charlie-Boo" <ch…@aol.com> wrote in message
news:1134612658.814742.58980@f14g2000cwb.googlegroups.com…
> Prove ( P = |- Q ) => |- ( P = Q )
What does this mean? I am not familiar with the notation.
Dan
- Hide quoted text — Show quoted text -
> (P and Q have the same sets of free variables.) This simple theorem (I
> created 12/1/05) provides the link between Theory of Computation and
> Proof Theory (Incompleteness in Logic) that theoreticians such as
> myself have been looking for since the 1930′s. (Each Theory of
> Computation theorem becomes an Incompleteness theorem in Logic,
> providing almost trivial formal derivation of the exact results of
> Godel, Rosser and Smullyan.)
> C-B
> Hint: Consider the equivalent ( P = |-Q ) => ( P = |- P )
On 14 Dec 2005 18:10:58 -0800, "Charlie-Boo" <ch…@aol.com> wrote:
>Prove ( P = |- Q ) => |- ( P = Q )
Various people have asked what the heck this means.
You should _say_ what it means, instead of replying
with questions.
Hint: My best attempt at deciphering it leads to
something obviously false:
"If P equals ‘Q has a proof’ then there is a
proof of ‘P equals Q’."
Obviously false, since if P equals ‘Q has a proof’
then P does not equal Q. So that must not be what you mean.
>(P and Q have the same sets of free variables.) This simple theorem (I
>created 12/1/05) provides the link between Theory of Computation and
>Proof Theory (Incompleteness in Logic) that theoreticians such as
>myself have been looking for since the 1930′s. (Each Theory of
>Computation theorem becomes an Incompleteness theorem in Logic,
>providing almost trivial formal derivation of the exact results of
>Godel, Rosser and Smullyan.)
Awesome.
>C-B
>Hint: Consider the equivalent ( P = |-Q ) => ( P = |- P )
************************
David C. Ullrich
- Hide quoted text — Show quoted text -
William Elliot wrote:
> On Thu, 15 Dec 2005, Charlie-Boo wrote:
> > William Elliot wrote:
> > > On Thu, 15 Dec 2005, Charlie-Boo wrote:
> > > > William Elliot wrote:
> > > > > On Wed, 14 Dec 2005, Charlie-Boo wrote:
> > > > > > Prove ( P = |- Q ) => |- ( P = Q )
> > > > > As you are mixing the object language with the meta language
> > > > > your statement as presented is non-sense gibberish.
> > > > Don’t you think that provability is expressible within the Logic?
> > > Do it. But before you do it, explain what your object language is.
> > Something as fundamental as the above theorem is certainly valid in
> > more than one Logic.
> Indeed, they’re called not well formed formulas.
Not to be flippant, but which characters of the expression are not
well-formed? Is ( P = |-Q ) a wff? Is => an operator? etc.
C-B
Dan Christensen wrote:
> "Charlie-Boo" <ch…@aol.com> wrote in message
> > Prove ( P = |- Q ) => |- ( P = Q )
> What does this mean? I am not familiar with the notation.
> Dan
P and Q are variables that range over wffs
|- means "It is provable that"
= is logical equivalence
=> is implication
Check out any text on Logic, especially Modal Logic e.g. The
Unprovability of Consistency, or The Logic of Provability by Boolos.
C-B
- Hide quoted text — Show quoted text -
> > (P and Q have the same sets of free variables.) This simple theorem (I
> > created 12/1/05) provides the link between Theory of Computation and
> > Proof Theory (Incompleteness in Logic) that theoreticians such as
> > myself have been looking for since the 1930′s. (Each Theory of
> > Computation theorem becomes an Incompleteness theorem in Logic,
> > providing almost trivial formal derivation of the exact results of
> > Godel, Rosser and Smullyan.)
> > C-B
> > Hint: Consider the equivalent ( P = |-Q ) => ( P = |- P )
David C. Ullrich wrote:
> On 14 Dec 2005 18:10:58 -0800, "Charlie-Boo" <ch…@aol.com> wrote:
> >Prove ( P = |- Q ) => |- ( P = Q )
> Various people have asked what the heck this means.
> You should _say_ what it means, instead of replying
> with questions.
You are way too loose with your tongue.
1. "Various people have asked what the heck this means.": Only one
person asked a question, viz, "What does this mean?".
2. "instead of replying with questions.": I had yet to respond to
that one question.
3. I posed one question, in response to the statement that my theorem
is ill-formed. As the statement did not say what constructs in the
theorem are ill-formed, I asked if he was referring to the use of |- as
an operator in a wff.
At no point did I reply to a question with a question.
> Hint: My best attempt at deciphering it leads to
> something obviously false:
> "If P equals ‘Q has a proof’ then there is a
> proof of ‘P equals Q’."
That’s it. What was so hard about that?
> Obviously false, since if P equals ‘Q has a proof’
> then P does not equal Q.
Why is that false? (Once again, what you are so adamantly convinced of
is just not so.)
> So that must not be what you mean.
Thanks for the vote of confidence.
> >(P and Q have the same sets of free variables.) This simple theorem (I
> >created 12/1/05) provides the link between Theory of Computation and
> >Proof Theory (Incompleteness in Logic) that theoreticians such as
> >myself have been looking for since the 1930′s. (Each Theory of
> >Computation theorem becomes an Incompleteness theorem in Logic,
> >providing almost trivial formal derivation of the exact results of
> >Godel, Rosser and Smullyan.)
> Awesome.
Would it be awesome (interesting, new, useful) if it were true?
C-B
- Hide quoted text — Show quoted text -
> >Hint: Consider the equivalent ( P = |-Q ) => ( P = |- P )
> ************************
> David C. Ullrich
In article <1134676919.959126.261…@z14g2000cwz.googlegroups.com>,
- Hide quoted text — Show quoted text -
"Charlie-Boo" <ch…@aol.com> wrote:
>Dan Christensen wrote:
>> "Charlie-Boo" <ch…@aol.com> wrote in message
>> > Prove ( P = |- Q ) => |- ( P = Q )
>> What does this mean? I am not familiar with the notation.
>> Dan
>P and Q are variables that range over wffs
>|- means "It is provable that"
>= is logical equivalence
>=> is implication
>Check out any text on Logic, especially Modal Logic e.g. The
>Unprovability of Consistency, or The Logic of Provability by Boolos.
OK then, using some modal logic for provability and a more-common
logical notation, you’re claiming:
(*) (P <-> []Q) -> [](P <-> Q).
Letting P be []Q, this implies:
(1) ([]Q <-> []Q) -> []([]Q <-> Q), i.e.
[]([]Q <-> Q).
This is false in any system where provability is different from truth.
And indeed, []([]Q -> Q) says the system can prove its own consistency,
which means (via Goedel) that it either is in fact inconsistent or is
weaker than arithmetic.
If you want provability and truth to be the same, you can dispense with
all the modal machinery and just use some existing well-thought-out
constructive logic (e.g. Intuitionistic).
>C-B
>> > (P and Q have the same sets of free variables.) This simple theorem (I
>> > created 12/1/05) provides the link between Theory of Computation and
>> > Proof Theory (Incompleteness in Logic) that theoreticians such as
>> > myself have been looking for since the 1930′s. (Each Theory of
>> > Computation theorem becomes an Incompleteness theorem in Logic,
>> > providing almost trivial formal derivation of the exact results of
>> > Godel, Rosser and Smullyan.)
>> > C-B
>> > Hint: Consider the equivalent ( P = |-Q ) => ( P = |- P )
–
—————————
| BBB b \ Barbara at LivingHistory stop co stop uk
| B B aa rrr b |
| BBB a a r bbb | Quidquid latine dictum sit,
| B B a a r b b | altum viditur.
| BBB aa a r bbb |
—————————–
Barb Knox wrote:
> OK then, using some modal logic for provability and a more-common
> logical notation, you’re claiming:
> (*) (P <-> []Q) -> [](P <-> Q).
Yes, but I use the standard symbol |- for provable instead of box [].
> Letting P be []Q, this implies:
> (1) ([]Q <-> []Q) -> []([]Q <-> Q), i.e.
> []([]Q <-> Q).
> This is false in any system where provability is different from truth.
No. It isn’t []Q <-> Q. It’s []([]Q < – >Q)
> And indeed, []([]Q -> Q) says the system can prove its own consistency,
> which means (via Goedel) that it either is in fact inconsistent or is
> weaker than arithmetic.
This seems valid. I really need only the "hint": that P = |-P.
Since that means (|-P)=(|-Q) I thought that would imply |-(P=Q) but
maybe not. Do you agree that P = |-P ? This is what I use to derive
Incompleteness results from any Theory of Computation theorem of the
form "(relation) is / isn’t recursively enumerable."
> If you want provability and truth to be the same,
Only for P!
C-B
- Hide quoted text — Show quoted text -
> you can dispense with
> all the modal machinery and just use some existing well-thought-out
> constructive logic (e.g. Intuitionistic).
> >C-B
> >> > (P and Q have the same sets of free variables.) This simple theorem (I
> >> > created 12/1/05) provides the link between Theory of Computation and
> >> > Proof Theory (Incompleteness in Logic) that theoreticians such as
> >> > myself have been looking for since the 1930′s. (Each Theory of
> >> > Computation theorem becomes an Incompleteness theorem in Logic,
> >> > providing almost trivial formal derivation of the exact results of
> >> > Godel, Rosser and Smullyan.)
> >> > C-B
> >> > Hint: Consider the equivalent ( P = |-Q ) => ( P = |- P )
> —
> —————————
> | BBB b \ Barbara at LivingHistory stop co stop uk
> | B B aa rrr b |
> | BBB a a r bbb | Quidquid latine dictum sit,
> | B B a a r b b | altum viditur.
> | BBB aa a r bbb |
> —————————–
Barb Knox wrote:
> If you want provability and truth to be the same, you can dispense with
> all the modal machinery and just use some existing well-thought-out
> constructive logic (e.g. Intuitionistic).
False. Truth and provability are *not* the same in
intuitionistic/constructive logics. If you claim otherwise, show me a
valid proof of the law of non-contradiction i.e., of ~(P&~P), in these
logics. Any "proof" of ~(P&~P) that you produce from contradictory
premises is not a valid proof in these logics. E.g. see Appendix A of
http://arxiv.org/abs/math.LO/0506475
which reproduces the argument given in my published paper in IJQI, Vol.
3, No. 1 (2005) pp. 263-267 ("The quantum superposition principle
justified in new non-Aristotelian finitary logic").
Regards, RS
In article <1134713007.253164.75…@z14g2000cwz.googlegroups.com>,
"sradhakr" <sradh…@in.ibm.com> wrote:
>Barb Knox wrote:
>> If you want provability and truth to be the same, you can dispense with
>> all the modal machinery and just use some existing well-thought-out
>> constructive logic (e.g. Intuitionistic).
>False. Truth and provability are *not* the same in
>intuitionistic/constructive logics. If you claim otherwise, show me a
>valid proof of the law of non-contradiction i.e., of ~(P&~P), in these
>logics.
OK, here’s a Fitch-style Intuitionistic ND proof:
1. | P ^ ~P A
|——-
2. | P 1 ^E
3. | ~P 1 ^E
4. ~(P ^ ~P) 1,2,3 RAA
>Any "proof" of ~(P&~P) that you produce from contradictory
>premises is not a valid proof in these logics.
Eh? I’ve given a perfectly valid Intuitionistic proof. On what grounds
do you object to it (if you do)?
[snip]
–
—————————
| BBB b \ Barbara at LivingHistory stop co stop uk
| B B aa rrr b |
| BBB a a r bbb | Quidquid latine dictum sit,
| B B a a r b b | altum viditur.
| BBB aa a r bbb |
—————————–
- Hide quoted text — Show quoted text -
Barb Knox wrote:
> In article <1134713007.253164.75…@z14g2000cwz.googlegroups.com>,
> "sradhakr" <sradh…@in.ibm.com> wrote:
> >Barb Knox wrote:
> >> If you want provability and truth to be the same, you can dispense with
> >> all the modal machinery and just use some existing well-thought-out
> >> constructive logic (e.g. Intuitionistic).
> >False. Truth and provability are *not* the same in
> >intuitionistic/constructive logics. If you claim otherwise, show me a
> >valid proof of the law of non-contradiction i.e., of ~(P&~P), in these
> >logics.
> OK, here’s a Fitch-style Intuitionistic ND proof:
> 1. | P ^ ~P A
> |——-
> 2. | P 1 ^E
> 3. | ~P 1 ^E
> 4. ~(P ^ ~P) 1,2,3 RAA
> >Any "proof" of ~(P&~P) that you produce from contradictory
> >premises is not a valid proof in these logics.
> Eh? I’ve given a perfectly valid Intuitionistic proof. On what grounds
> do you object to it (if you do)?
*Any* proposition can be proven in intiuitionistic logic if you start
with the premise P&~P. The above "proof" is fundamentally flawed and
*should not* be accepted as a valid proof of ~(P&~P). The above "proof"
is indistinguishable from the case where you deduce an *arbitrary*
proposition (say, denoted by the "absurdity symbol") from P&~P and then
*substituted* ~(P&~P) for the absurdity symbol and claimed that as a
proof of ~(P&~P). It is obvious that any proposition can be proved that
way.
Indeed, if you think carefully, you will see that the very concept of
"proof" requires the law of non-contradiction in the first place and
you have already *presumed* ~(P&~P) in your so-called proof– it is
totally absurd to claim an RAA proof of ~(P&~P). What you could do is
to make a straightforward, bald assertion of ~(P&~P) without claiming
to prove it. But that would violate the intuitionistic philosophy of
truth as provability. See the reference that I have cited in my earlier
post for my objections to ~(P&~P) as a theorem of a theory T in which P
is undecidable (i.e., neither provable nor refutable in T).
Regards, RS
On 15 Dec 2005 12:16:35 -0800, "Charlie-Boo" <ch…@aol.com> wrote:
- Hide quoted text — Show quoted text -
>David C. Ullrich wrote:
>> On 14 Dec 2005 18:10:58 -0800, "Charlie-Boo" <ch…@aol.com> wrote:
>> >Prove ( P = |- Q ) => |- ( P = Q )
>> Various people have asked what the heck this means.
>> You should _say_ what it means, instead of replying
>> with questions.
>[...]
>> Hint: My best attempt at deciphering it leads to
>> something obviously false:
>> "If P equals ‘Q has a proof’ then there is a
>> proof of ‘P equals Q’."
>That’s it. What was so hard about that?
Very curious – a minute ago I saw you say that it
meant something else.
>> Obviously false, since if P equals ‘Q has a proof’
>> then P does not equal Q.
>Why is that false? (Once again, what you are so adamantly convinced of
>is just not so.)
If this is not clearly false you must be assuming that I’m
being as sloppy with the language as you are. I’m not –
when I say "equals" I mean "equals", not "is logically
equivalent to". For example, P is not equal to P&P,
although they’re certainly equivalent.
Evidently from what you say elsewhere in the thread
what you really mean, and what you thought I meant,
is this:
"If P is equivalent to ‘Q has a proof’ then there is a
proof of ‘P is equivalent to Q’."
This is also false, as has been pointed out, although
it’s not quite so _obviously_ false as my first interpretation,
where I was assuming that "=" referred to equality.
Why is it false? Consider the special case where P literally
equals "Q is provable". Then your theorem implies this,
for any Q:
(*) "It is provable that ‘Q is provable’ is equivalent to Q."
Or, in extremely bad notation:
(*) |- (|-Q == Q)
(bad notation because the first |- means |-, while
the second |- refers to some encoding of provability…)
This is false. It contradicts the incompleteness theorem,
for example, which says that in any system satisfying this
and that there exists Q which is true but not provable.
- Hide quoted text — Show quoted text -
>> So that must not be what you mean.
>Thanks for the vote of confidence.
>> >(P and Q have the same sets of free variables.) This simple theorem (I
>> >created 12/1/05) provides the link between Theory of Computation and
>> >Proof Theory (Incompleteness in Logic) that theoreticians such as
>> >myself have been looking for since the 1930′s. (Each Theory of
>> >Computation theorem becomes an Incompleteness theorem in Logic,
>> >providing almost trivial formal derivation of the exact results of
>> >Godel, Rosser and Smullyan.)
>> Awesome.
>Would it be awesome (interesting, new, useful) if it were true?
>C-B
>> >Hint: Consider the equivalent ( P = |-Q ) => ( P = |- P )
>> ************************
>> David C. Ullrich
************************
David C. Ullrich
sradhakr wrote:
> Barb Knox wrote:
>>In article <1134713007.253164.75…@z14g2000cwz.googlegroups.com>,
>> "sradhakr" <sradh…@in.ibm.com> wrote:
>>>Any "proof" of ~(P&~P) that you produce from contradictory
>>>premises is not a valid proof in these logics.
>>Eh? I’ve given a perfectly valid Intuitionistic proof. On what grounds
>>do you object to it (if you do)?
> *Any* proposition can be proven in intiuitionistic logic if you start
> with the premise P&~P.
Ok, please produce, in the same way, a proof of the
proposition P&~P, then.
(I mean "P&~P", *not* "(P&~P) -> (P&~P)".)
[snip more nonsense]
groente
— Sander
Charlie-Boo wrote:
> Barb Knox wrote:
>>OK then, using some modal logic for provability and a more-common
>>logical notation, you’re claiming:
>>(*) (P <-> []Q) -> [](P <-> Q).
> Yes, but I use the standard symbol |- for provable instead of box [].
Actually, [] is the standard symbol, not |-.
[snip]
> This seems valid. I really need only the "hint": that P = |-P.
> Since that means (|-P)=(|-Q) I thought that would imply |-(P=Q) but
> maybe not. Do you agree that P = |-P ?
NO! Provability is *not* the same as truth!
[snip]
groente
— Sander
H. J. Sander Bruggink wrote:
- Hide quoted text — Show quoted text -
> sradhakr wrote:
> > Barb Knox wrote:
> >>In article <1134713007.253164.75…@z14g2000cwz.googlegroups.com>,
> >> "sradhakr" <sradh…@in.ibm.com> wrote:
> >>>Any "proof" of ~(P&~P) that you produce from contradictory
> >>>premises is not a valid proof in these logics.
> >>Eh? I’ve given a perfectly valid Intuitionistic proof. On what grounds
> >>do you object to it (if you do)?
> > *Any* proposition can be proven in intiuitionistic logic if you start
> > with the premise P&~P.
> Ok, please produce, in the same way, a proof of the
> proposition P&~P, then.
> (I mean "P&~P", *not* "(P&~P) -> (P&~P)".)
That is precisely the point of my objection. If the claimed "proof" of
~(P&~P) is allowed to go through, then just about any proposition,
including P&~P should also be provable.
In other words, the claimed proof shows that from the hypothesis P&~P
one can conclude ~(P&~P). But from the hypothesis P&~P one can conclude
any proposition, so what is the basis for the claimed proof?
> [snip more nonsense]
??????
If you have something meaningful to say, say it. Otherwiise just keep
shut.
Regards, RS
- Hide quoted text — Show quoted text -
sradhakr wrote:
> H. J. Sander Bruggink wrote:
>>sradhakr wrote:
>>>Barb Knox wrote:
>>>>In article <1134713007.253164.75…@z14g2000cwz.googlegroups.com>,
>>>>"sradhakr" <sradh…@in.ibm.com> wrote:
>>>>>Any "proof" of ~(P&~P) that you produce from contradictory
>>>>>premises is not a valid proof in these logics.
>>>>Eh? I’ve given a perfectly valid Intuitionistic proof. On what grounds
>>>>do you object to it (if you do)?
>>>*Any* proposition can be proven in intiuitionistic logic if you start
>>>with the premise P&~P.
>>Ok, please produce, in the same way, a proof of the
>>proposition P&~P, then.
>>(I mean "P&~P", *not* "(P&~P) -> (P&~P)".)
> That is precisely the point of my objection. If the claimed "proof" of
> ~(P&~P) is allowed to go through, then just about any proposition,
> including P&~P should also be provable.
Fine, show me that proof of P&~P, then.
> In other words, the claimed proof shows that from the hypothesis P&~P
> one can conclude ~(P&~P). But from the hypothesis P&~P one can conclude
> any proposition, so what is the basis for the claimed proof?
Yes, for any proposition Q, you can prove (P&~P) -> Q.
What’s your point?
>>[snip more nonsense]
> ??????
> If you have something meaningful to say, say it. Otherwiise just keep
> shut.
I apologize. It was not nice of me to call you nonsense
"nonsense".
groente
— Sander
H. J. Sander Bruggink wrote:
- Hide quoted text — Show quoted text -
> sradhakr wrote:
> > H. J. Sander Bruggink wrote:
> >>sradhakr wrote:
> >>>Barb Knox wrote:
> >>>>In article <1134713007.253164.75…@z14g2000cwz.googlegroups.com>,
> >>>>"sradhakr" <sradh…@in.ibm.com> wrote:
> >>>>>Any "proof" of ~(P&~P) that you produce from contradictory
> >>>>>premises is not a valid proof in these logics.
> >>>>Eh? I’ve given a perfectly valid Intuitionistic proof. On what grounds
> >>>>do you object to it (if you do)?
> >>>*Any* proposition can be proven in intiuitionistic logic if you start
> >>>with the premise P&~P.
> >>Ok, please produce, in the same way, a proof of the
> >>proposition P&~P, then.
> >>(I mean "P&~P", *not* "(P&~P) -> (P&~P)".)
> > That is precisely the point of my objection. If the claimed "proof" of
> > ~(P&~P) is allowed to go through, then just about any proposition,
> > including P&~P should also be provable.
> Fine, show me that proof of P&~P, then.
> > In other words, the claimed proof shows that from the hypothesis P&~P
> > one can conclude ~(P&~P). But from the hypothesis P&~P one can conclude
> > any proposition, so what is the basis for the claimed proof?
> Yes, for any proposition Q, you can prove (P&~P) -> Q.
> What’s your point?
That therefore no valid proof of ~(P&~P) should start with the
hypothesis P&~P. Understand this point before you hit the keyboard
again.
> >>[snip more nonsense]
> > ??????
> > If you have something meaningful to say, say it. Otherwiise just keep
> > shut.
> I apologize. It was not nice of me to call you nonsense
> "nonsense".
????
Read what I wrote above.
Regards, RS
"H. J. Sander Bruggink" <brugg…@phil.uu.nl> writes:
> In other words, the claimed proof shows that from the hypothesis P&~P
> one can conclude ~(P&~P). But from the hypothesis P&~P one can conclude
> any proposition, so what is the basis for the claimed proof?
Your comments are based on a confusion. The assumption P&~P is
discharged in the derivation. Ex falso quodlibet is not used.
"sradhakr" <sradh…@in.ibm.com> writes:
> In other words, the claimed proof shows that from the hypothesis P&~P
> one can conclude ~(P&~P). But from the hypothesis P&~P one can conclude
> any proposition, so what is the basis for the claimed proof?
Your comments are based on a confusion. The assumption P&~P is
discharged in the derivation. Ex falso quodlibet is not used.
(Apologies for earlier misattributed version.)
Torkel Franzen wrote:
> "sradhakr" <sradh…@in.ibm.com> writes:
> > In other words, the claimed proof shows that from the hypothesis P&~P
> > one can conclude ~(P&~P). But from the hypothesis P&~P one can conclude
> > any proposition, so what is the basis for the claimed proof?
> Your comments are based on a confusion. The assumption P&~P is
> discharged in the derivation. Ex falso quodlibet is not used.
> (Apologies for earlier misattributed version.)
Ex falso quodlibet is not directly used, I agree. But EFQ *could* be
used to deduce just about whatever we want from P&~P. So an assertion
of P&~P as a hypothesis is in principle the same as asserting an
arbitrary proposition Q. So is it surprising that Q could in
particular, be ~(P&~P)? You might argue that the actual proof doesn’t
run this way. But what, precisely, is the "absurdity" that you deduce
from P&~P, in order to conclude ~(P&~P) in the claimed proof? The fact
that you can deduce an arbitrary proposition from P&~P? If so, that is
a tacit use of EFQ, and invalidates the conclusion of ~(P&~P) from the
same hypothesis P&~P, or so I claim. Let me know what you think.
Regards, RS