'When Are Relations Neither True Nor False? [Logic]

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From: Jesse F. Hughes on 30 Mar 2010 09:04

Newberry <newberryxy(a)gmail.com> writes:

> Indeed. But if we leave out all the vacuous sentences we can still do
> all the useful arithmetic as we know it. Although all the people on
> this board believe that such sentences are true nobody argued that
> they were useful. Aatu even said that they did not belong in ordinary
> mathematical reasoning. Furthermore there is a reason to think that
> they are neither true nor false. I cannot think of any good reason for
> claiming that 1 + 1 = 2 is not true.

You seem to have misrepresented Aatu's claims. Moreover, you're just
wrong. I've argued repeatedly that some sentences of the form

~(Ex)(P & Q)

occur in ordinary mathematical reasoning (and hence are useful), even
when (Ex)P is false. An example occurred in sci.math recently.

Simon C. Roberts gave a purported proof of FLT[1], by arguing:

~(En)(Ea,b,c)(a^n + b^n = c^n & a, b, c are pairwise coprime).

Let's denote a^n + b^n = c^n by P(a,b,c,n) and a, b, c are pairwise
coprime by Q(a,b,c), so that Simon's argument attempts to show that

~(En)(Ea,b,c)( P(a,b,c,n) & Q(a,b,c) ). (1)

Of course, I am *not* claiming that he proved what he claims. That's
beside my point. A poster named bill replied that (1) is not Fermat's
last theorem[2], which has the form

~(En)(Ea,b,c) P(a,b,c,n). (2)

Arturo responded[3] by proving

(En)(Ea,b,c) P(a,b,c,n) -> (En)(Ea,b,c)( P(a,b,c,n) & Q(a,b,c) ). (3)

Hence, a proof of (1) yields a proof of (2) by modus tollens.

According to you, however, if (2) is true (and I assume we all know
that (2) was proved by Wiles), then (1) is meaningless. Yet, no one
here balked at the claim that (1) could be used to prove (2) (once (3)
was proved). No one here had any trouble understanding what (1)
means. Everyone in the thread accepted this form of mathematical
argument as beyond suspicion -- although the claim that Simon actually
proved (1) is regarded as doubtful.

So, you're just plain wrong. These statements that you call
meaningless occur in ordinary mathematical reasoning all the time.

Footnotes:
[1] Message id
<1917288606.455209.1269716329839.JavaMail.root(a)gallium.mathforum.org>,
in the thread "Another Proof of Fermats Last Theorem".

[2] Message id
<50f09d88-a96b-464c-aec5-be000f0be40d(a)x23g2000prd.googlegroups.com>.

[3] Message id
<e6768d43-7706-41f4-bff8-8e666d693c3a(a)j21g2000yqh.googlegroups.com>.

--
"There's lots of things in this old world to take a poor boy down.
If you leave them be, you can save yourself some pain.
You don't have to live in fear, but you best have some respect,
For rattlesnakes, painted ladies and cocaine." -- Bob Childers

From: Transfer Principle on 30 Mar 2010 15:42

On Mar 26, 9:55 am, MoeBlee <jazzm...(a)hotmail.com> wrote:
> On Mar 26, 12:20 am, Transfer Principle <lwal...(a)lausd.net> wrote:
> > proof of 0.999...=1 to which I linked explicitly lists the
> > Replacement Schema as one of the axioms that is used
> > in the proof.
> Replacement isn't needed to prove .999... = 1. With David Ullrich's
> help, I posted a rigorous proof (in Z-regularity) of .999... = 1 in a
> thread for that purpose a few years ago.
> That 1 is the multiplicative identity does not require replacement.

As usual, I was unable to find Ullrich's proof via a Google search,
but I
did go back to the Metamath proof and find where the Replacement
Schema was used.

Let's begin by trying to proof that 1 is the multiplicative identity
of N,
in other words:

AneN (n*1 = n)

So we begin by noting that we are considering N to be the set of
nonzero finite ordinals, omega\{0}. So we use the recursive definition
of multiplication of ordinals to obtain:

AneN (n*1 = n*0 + n)
AneN (n*1 = 0 + n)

But hold on a minute. Multiplication of ordinals is defined using
transfinite recursion -- which requires the Replacement Schema!

At this point, one might point out that we only need to define the
operations on the _finite_ ordinals, and so we ought to use finite,
rather than _trans_finite, recursion. But finite recursion -- though
it
apparently avoids Replacement Schema -- requires the Axiom of
_Infinity_ in its proof. It may seem ironic that _finite_ recursion
requires Infinity, yet _trans_finite recursion doesn't, but this is
confimed at the following two links:

http://us.metamath.org/mpegif/inf3.html
http://us.metamath.org/mpegif/mmnotes2004.txt

Thus, by using transfinite recursion rather than finite recursion, the
Metamath proof of 1 as the multiplicative identity avoids the Axiom
of Infinity.

We know that mathematicians prefer parsimony with regards to the
axioms used in a proof. So if T is a theory and phi is a axiom (that
is undecidable in T), then given two proofs of a statement, one in T
and the other in T+phi, the proof in T is usually preferable. If the
proof in T is only slightly longer than the proof in T+phi, then the
proof in T is still preferable. Only if the proof in T is
significantly
longer (say 10, or 100 times as long) than the proof in T+phi will
mathematicians consider assuming phi to make the proof shorter.

But what if there's a statement that's not provable in T, but we have
a choice between two axioms (or schemata) to add to T in order to
produce a proof? For "1 is the multiplicative identity" appears not to
be provable in Z-Infinity (assuming it's consistent), but we can find
a
proof in either Z (using finite induction) or ZF-Infinity (using
transfinite
induction instead). Which proof is considered preferable?

On one hand, the goal isn't to define N, but _R_. One can avoid
Infinity when trying to prove "n*1 = n for all natural numbers," but
good luck trying to prove "r*1 = r for all _real_ numbers" (or even
anything else about _R_) in ZF-Infinity. Since we're going to use
Infinity to define R anyway, there's no real point in avoiding the
axiom
when discussing the properties of N. Thus, it's better to give the
proof avoiding Replacement Schema instead -- and this is exactly
what MoeBlee and Ullrich did -- work in Z(-Regularity -- we already
know how to avoid Regularity, so I won't keep writing "-Regularity"
each time I refer to a theory).

On the other hand, we already know that some so-called "cranks"
and finitists don't work in ZF(C) -- and we already know that the
majority of them find _Infinity_, not _Replacement_, objectionable. So
in deference to the finitist, we ought to avoid Infinity in as many
proofs
as possible, even if it requires using other axioms and schemata such
as Replacement. Thus, to such posters, the proof in ZF-Infinity would
be preferable to the proof in Z. But of course, a(n ultra)finitist has
no
business even referring to 0.999... in the first place, unless to
denote
only _finitely_ many nines after the decimal point.

A similar debate often occurs when discussing Q -- can a finitist or
"crank" work with rational numbers? On one hand, rationals are often
defined as equivalence classes of ordered pairs, so 1/2 would be
defined as {(1,2), (2,4), (3,6), (4,8), ...} -- this is evidently an
infinite
set, so Infinity is necessary. But on the other hand, we can define
rationals to be only a single ordered pair representing the lowest-
terms
fraction, so 1/2 would be defined as (1,2), which can be defined
without
the Axiom of Infinity. Metamath uses the former definition, for why
should one have to keep dividing by the gcd of the numerator and
denominator to get a fraction into lowest terms just to avoid
Infinity,
when the ultimate goal is to define not Q, but _R_, where we'd use
Infinity anyway? But the finitist/"crank" trying to avoid Infinity at
all
costs will use the latter definition of rational number.

Returning to N, since Metamath has already defined via transfinite
recursion using Replacement, it might as well use it in its proof of
"1 is the multiplicative identity" and all that. But MoeBlee and
Ullrich
were explicitly working in Z(-Regularity), and so they're going to use
the construction of R that doesn't require transfinite induction or
the
Replacement Schema.

I'm not sure whether MoeBlee and Ullrich wish to avoid Replacement
in the same way that a finitist/"crank" wishes to avoid Infinity. It
may
be that MoeBlee and Ullrich are open to using Replacement when it
is absolutely inevitable, but as long as they can avoid the schema --
and one can construct R without it -- they'll do so. (And in order to
reduce my habit of lumping posters together, it could be that MoeBlee
and Ullrich are divided on this issue -- perhaps one of them will use
Replacement when it's necessary, and the other will avoid it like the
plague, just as a finitist/"crank" avoids Infinity.)

And therefore, even without finding MoeBlee's or Ullrich's proof, I
can
figure out what their proof is. We basically take the Metamath proof
of "0.999...=1" and replace all instances of transfinite induction
with
finite induction. Then voila -- we'll have a proof of "0.999...=1"
which
doesn't require Replacement and thus works in Z(-Regularity). QED

From: Transfer Principle on 30 Mar 2010 15:57

On Mar 26, 10:04 am, MoeBlee <jazzm...(a)hotmail.com> wrote:
> On Mar 26, 12:42 am, Transfer Principle <lwal...(a)lausd.net> wrote:
> > as powerful
> This cries for a definition or explanation of what is meant by 'as
> powerful'. In mathematical logic we have various notions such as
> 'intepretability' and 'conservative extension'. But some of your
> comments seem not to use 'as powerful' in such technical senses
> (otherwise some of your questions in this regard would be non-
> starters).

Marshall Spight was the first to use the phrase in this thread. I only
use the phrase in reponse to Spight's post.

Here is a quote from the post from earlier in this thread, back on the
2nd of March, at 4:11PM Greenwich time:

"One thing that the cranks and crankophiles never understand
is that the systems they come up with add a lot of complexity
while actually removing functionality or utility. To do so
merely to avoid some counterintuitive but harmless
property (such as vacuous truth, in Newberry's case) is
a huge waste of time.
Less powerful; more work to use: that's a crank theory for you."

So Spight criticizes so-called "crank" theories as less powerful than
some other theory -- presumably the standard theory (ZFC). My
goal, therefore, is to find a theory that's _as_ powerful as ZFC, so
that Spight would have less reason to criticize it.

In order for me to reduce my habit of misinterpreting other posters,
it would be far better for Spight himself to state what exactly he
meant when he wrote "less powerful." That way, we can agree
on what criteria a theory needs to satisfy in order for it to be "as
powerful" as ZFC, without "removing functionality or utility."

From: Transfer Principle on 30 Mar 2010 16:29

On Mar 27, 7:02 am, "Tim Golden BandTech.com" <tttppp...(a)yahoo.com>
wrote:
> On Mar 26, 5:55 pm, Nam Nguyen <namducngu...(a)shaw.ca> wrote:
> > Any rate, enough talk. Do you have even a single absolute truth you
> > could show me so that I'd realize I've been wrong all along? Let's
> > begin with the natural numbers: which formula in the language of
> > arithmetic could _you_ demonstrate as absolutely true?
> There is a fairly straightforward construction that can yield both
> boolean logic and continuous higher forms, and even a lower form that
> I will call universal.
> Constrain the real numbers to those values whose magnitude is unity.
> We see two options
> +1, -1 .
> Using polysign numbers extend this system to P3.

Ah yes, the polysign numbers. I still remember Golden's constuction
of these sets.

> One might initially consider there to be a three verticed logic here,
> but in the general form we see that the unity values now form a
> continuous circle.

There was a discussion of alternate-valued logic back when tommy1729
proposed using three-valued logic (tommy1729 being, of course, one of
Golden's biggest supporters). But we found out that usually, standard
theorists object to these alternate forms of logic.

For one thing, standard theorists obviously accept two-valued Boolean
logic (FOL), and they appear to be open to continuum-valued logic
(also called "fuzzy" logic). But they tend to object to kappa-valued
logic, where kappa is a cardinal that is strictly between two and the
cardinality of the continuum.

> In two dimensions we see that the same procedure yields a continuum of
> values, though there are arguably those three unique positions
> -1, +1, *1 .

But notice that Golden does acknowledge a continuum of values. So
perhaps this could be a form of fuzzy logic that the standard
theorists
might accept as well.

There's a huge difference between fuzzy logic and Golden's though. For
fuzzy logic usually considers the values to lie in the interval [0,1],
with
0 being false and 1 being true. Golden's fuzzy logic is decidedly
_not_
described by the interval [0,1] at all.

Golden regularly points out that multiplication on the set {-1,+1,*1}
in
P3 is isomorphic to addition in the group Z/3Z. In general, Golden
wishes to construct an n-valued logic by considering a subset of Pn
that's isomorphic to addition in the group Z/nZ, which is also
isomorphic to the multiplicative set of nth roots of unity in C. (As
was
discussed in many previous Golden threads, _addition_ in Pn is not
isomorphic to _addition_ in C, but this current subthread only deals
with multiplication, not addition.)

Thus, our continuum-valued logic can be described by the entire unit
circle in C, not the unit interval [0,1].

Now we ask ourselves, is such a logic even possible. Back in the
tommy1729 three-valued logic threads, the standard theorists often
pointed out that what they needed to see were the laws of inference
for any proposed logic. Without laws of inference, one can't really
call
it a logic at all.

I wouldn't mind taking at the laws of inference for fuzzy logic and
modifying them so that they work for Golden's logic. But of course, I
don't own a textbook on fuzzy logic or its laws of inference, nor do I
plan on owning such a book anytime soon.

> By leaving the Euclidean and working the sphere these forms exist
> naturally.

Hmmm, non-Euclidean geometry and the sphere. This reminds me of
AP's work as well. I wonder whether the AP-adics might work better
if we used Golden's logic instead of FOL. (And before Jesse Hughes
or anyone else protests, I'm fully aware that the link between
Golden's
post here and AP's work is even flimsier than that between Newberry
and Clarke. I'm the one who's trying to unify the so-called "cranks"
as
best as I can in order to find a theory that will satisfy at least two
of
them, which saves me work from having to find a different theory for
each and every "crank.")

From: Jesse F. Hughes on 30 Mar 2010 18:02

Transfer Principle <lwalke3(a)lausd.net> writes:

> There was a discussion of alternate-valued logic back when tommy1729
> proposed using three-valued logic (tommy1729 being, of course, one
> of Golden's biggest supporters). But we found out that usually,
> standard theorists object to these alternate forms of logic.

No, you found no such thing.

Multi-valued logic is a perfectly respectable field. (It does not
follow that Tommy's blatherings are perfectly respectable. I don't
read Timothy's posts, so I don't offer an opinion on them. They might
be blatherings, too, for all I know.)

Of course, most mathematicians would bristle at the suggestion that
they ought to do their work in three-valued logic. But that's not the
same as objecting to three-valued logic per se.

--
Jesse F. Hughes
"Radicals are interesting because they were considered 'radical' by
the people who played with them who wrote a lot of math work that
modern mathematics depends on." --Another JSH history lesson