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

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From: Newberry on 25 Mar 2010 00:28

On Mar 24, 9:34 am, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote:
> Newberry <newberr...(a)gmail.com> writes:
> > On Mar 24, 3:58 am, stevendaryl3...(a)yahoo.com (Daryl McCullough)
> > wrote:
> >> Newberry says...
>
> >> >On Mar 23, 5:42=A0am, stevendaryl3...(a)yahoo.com (Daryl McCullough)
> >> >wrote:
> >> >> The problem with this talk about statements being "meaningless" is
> >> >> that often the only way that you *know* that ~(Ex)Px is necessarly true
> >> >> is by proving a pair of statements:
>
> >> >> 1. (Ax) Px -> Qx
>
> >> >> 2. (Ax) Px -> ~Qx
>
> >> >> together, these imply
>
> >> >> 3. ~(Ex) Px
>
> >> >> which you claim implies that 1&2 are meaningless.
>
> >> >Maybe it is like a proof by contradiction. You assume (Ex)Px and
> >> >derive ~(Ex)Px.
>
> >> >How do you know that there are no alternative proofs?
>
> >> I don't know that there are no alternative proofs, but I
> >> know that the proof using 1-3 works perfectly fine. I can't
> >> see any *advantage* to calling some statements meaningless,
> >> especially when you can derive true consequences from them.
>
> >> >> I think it makes more sense to say that every statement
> >> >> of arithmetic is meaningful.
>
> >> >No, system with gaps has better properties than a system without gaps..
>
> >> Better in what sense?
>
> > It can solve the Liar paradox.
>
> Perhaps you can remind us how that goes? How do you represent the
> Liar's paradox in your logic?
>
> > Tarki's theorem does not apply to it.
>
> >> >Plus
>
> >> > (x)((x = x + 1) -> (x = x + 2))
>
> >> >does not look particularly meaningful to me.
>
> >> The classical meaning of an implication
>
> > The classical "meaning of an implication is known to be paradixical.
>
> Paradoxical in what sense?
>
> >> "A implies B"
> >> is that it is *not* the case that A is true and B is false.
> >> So for A implies B to be true, there are three possibilities:
>
> >> 1. A is true and B is true.
> >> 2. A is false and B is true.
> >> 3. A is false and B is false.
>
> >> In the case x=x+1 -> x=x+2, both are false, so we are
> >> in case 3. There is nothing meaningless about this. It's
> >> not a particularly *interesting* observation. But who
> >> cares?
>
> > Right on! No harm will be done to arithmetic if such sentences are not
> > provable.
>
> This particular sentence does not appear useful. The fact that I can
> prove
>
> ~(E a b)( a^2/b^2 = 2 and gcd(a,b) = 1 )
>
> is historically useful. I use it to prove that sqrt(2) is
> irrational.
>
> You really need to stop pretending that if you show us a single
> "useless" sentence of the form ~(E x)(Px & Qx), then all such
> sentences are useless (when ~(Ex)Px is true). We actually use some of
> these to prove interesting facts -- including ~(E x)Px.
>
> Moreover, you said something stronger than that it is not provable.
> You said that it's meaningless.
>
> Let me give you another example. I suppose you agree that the
> following is a true statement about sets:
>
> (Ax)( (Ay)(y in x -> s(y) in x) -> (x = {} or x is infinite) ).
>
> That is, any set closed under the successor operation is either
> infinite or empty.
>
> Suppose we could prove that a particular finite set is closed under
> the successor operation. That is, that
>
> (Ay)(y in x -> s(y) in x) and x is not infinite.
>
> We would conclude that x is empty, right? Except, we couldn't
> conclude that, since if x is empty, (Ay)(y in x -> s(y) in x) is
> neither true nor false. Oops!
>
> Evidently, the empty set is not closed under the successor operation
> (according to you), since
>
> (Ay)( y in {} -> s(y) in {} )
>
> is neither true nor false. So my question is this: since the empty
> set is not closed under the successor operation, would you say that
>
> (Ax)( (Ay)(y in x -> s(y) in x) -> (x is infinite) )
>
> is a true statement?

I think it is true.

> (I hope not.)

Why?

> If not, why not?
>
> Normally, I'd think that the following reasoning is valid:
>
> If (A -> (B or C)) and B -> "A is not true", then A -> C.
>
> Am I wrong?
>
> --
> Jesse F. Hughes
>
> "For a gentle introduction to set theory, see Bourbaki (1970)."
> -- Footnote from "Transgressing the Boundaries", Alan Sokal- Hide quoted text -
>
> - Show quoted text -

From: Newberry on 25 Mar 2010 00:41

On Mar 24, 9:34 am, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote:
> Newberry <newberr...(a)gmail.com> writes:
> > On Mar 24, 3:58 am, stevendaryl3...(a)yahoo.com (Daryl McCullough)
> > wrote:
> >> Newberry says...
>
> >> >On Mar 23, 5:42=A0am, stevendaryl3...(a)yahoo.com (Daryl McCullough)
> >> >wrote:
> >> >> The problem with this talk about statements being "meaningless" is
> >> >> that often the only way that you *know* that ~(Ex)Px is necessarly true
> >> >> is by proving a pair of statements:
>
> >> >> 1. (Ax) Px -> Qx
>
> >> >> 2. (Ax) Px -> ~Qx
>
> >> >> together, these imply
>
> >> >> 3. ~(Ex) Px
>
> >> >> which you claim implies that 1&2 are meaningless.
>
> >> >Maybe it is like a proof by contradiction. You assume (Ex)Px and
> >> >derive ~(Ex)Px.
>
> >> >How do you know that there are no alternative proofs?
>
> >> I don't know that there are no alternative proofs, but I
> >> know that the proof using 1-3 works perfectly fine. I can't
> >> see any *advantage* to calling some statements meaningless,
> >> especially when you can derive true consequences from them.
>
> >> >> I think it makes more sense to say that every statement
> >> >> of arithmetic is meaningful.
>
> >> >No, system with gaps has better properties than a system without gaps..
>
> >> Better in what sense?
>
> > It can solve the Liar paradox.
>
> Perhaps you can remind us how that goes? How do you represent the
> Liar's paradox in your logic?

First of all I had in mind the Liar paradox in the natural language.
If you take the position that there are truth value gaps then you can
solve it relatively easily. (As long as you relize that "this sentence
is not true" is indeed not true BUT NOT because it succeeded in
attributing non-truth to itself.)

In my logic the Liar paradox can be expressed as follows.

~(Ex)(Ey)(Pxy & Qy) (L)

where Pxy means that x is a proof of y, Q is satisfied by only one y =
m, and m is Goedel number of (L). It is ~(T v F) i.e. ~T. (I would go
as far as saying that it is meaningless but it does not matter for our
present purpose.)

~(Ex)Pxm (L')

is true.

Let Y = "This sentence is not true", then

~(T v F): Y, ~(Ex)(Ey)(Pxy & Qy)
T: ~T(Y), ~(Ex)Pxm

>
> > Tarki's theorem does not apply to it.
>
> >> >Plus
>
> >> > (x)((x = x + 1) -> (x = x + 2))
>
> >> >does not look particularly meaningful to me.
>
> >> The classical meaning of an implication
>
> > The classical "meaning of an implication is known to be paradixical.
>
> Paradoxical in what sense?

Does not everybody know what the paradox of material implication is?

>
> >> "A implies B"
> >> is that it is *not* the case that A is true and B is false.
> >> So for A implies B to be true, there are three possibilities:
>
> >> 1. A is true and B is true.
> >> 2. A is false and B is true.
> >> 3. A is false and B is false.
>
> >> In the case x=x+1 -> x=x+2, both are false, so we are
> >> in case 3. There is nothing meaningless about this. It's
> >> not a particularly *interesting* observation. But who
> >> cares?
>
> > Right on! No harm will be done to arithmetic if such sentences are not
> > provable.
>
> This particular sentence does not appear useful. The fact that I can
> prove
>
> ~(E a b)( a^2/b^2 = 2 and gcd(a,b) = 1 )
>
> is historically useful. I use it to prove that sqrt(2) is
> irrational.
>
> You really need to stop pretending that if you show us a single
> "useless" sentence of the form ~(E x)(Px & Qx), then all such
> sentences are useless (when ~(Ex)Px is true). We actually use some of
> these to prove interesting facts -- including ~(E x)Px.
>
> Moreover, you said something stronger than that it is not provable.
> You said that it's meaningless.
>
> Let me give you another example. I suppose you agree that the
> following is a true statement about sets:
>
> (Ax)( (Ay)(y in x -> s(y) in x) -> (x = {} or x is infinite) ).
>
> That is, any set closed under the successor operation is either
> infinite or empty.
>
> Suppose we could prove that a particular finite set is closed under
> the successor operation. That is, that
>
> (Ay)(y in x -> s(y) in x) and x is not infinite.
>
> We would conclude that x is empty, right? Except, we couldn't
> conclude that, since if x is empty, (Ay)(y in x -> s(y) in x) is
> neither true nor false. Oops!
>
> Evidently, the empty set is not closed under the successor operation
> (according to you), since
>
> (Ay)( y in {} -> s(y) in {} )
>
> is neither true nor false. So my question is this: since the empty
> set is not closed under the successor operation, would you say that
>
> (Ax)( (Ay)(y in x -> s(y) in x) -> (x is infinite) )
>
> is a true statement? (I hope not.) If not, why not?
>
> Normally, I'd think that the following reasoning is valid:
>
> If (A -> (B or C)) and B -> "A is not true", then A -> C.
>
> Am I wrong?
>
> --
> Jesse F. Hughes
>
> "For a gentle introduction to set theory, see Bourbaki (1970)."
> -- Footnote from "Transgressing the Boundaries", Alan Sokal- Hide quoted text -
>
> - Show quoted text -

From: Newberry on 25 Mar 2010 01:05

On Mar 24, 3:32 am, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote:
> Newberry <newberr...(a)gmail.com> writes:
> > Plus
>
> > (x)((x = x + 1) -> (x = x + 2))
>
> > does not look particularly meaningful to me.
>
> I don't believe you.

Trust me.

> You know what it means. It's perfectly clear
> what it means. It means that whenever x = x + 1, then x = x + 2.[1]

The sentence "if it rains then some roads are wet" describes a
possible state of affairs. I can picture to myself what it means. I
can even picture "if it rains then no roads are wet." It is still
conceivable although very unlikely. "If it rains and does not rain
then the roads are wet" does not describe any possible state of
affairs. I cannot picture to myself what it expresses.

The analytic sentences are rather odd. But even then given "all
bachelors are unmarried" if you examine every bachelor you will find
that he is umarried. Given "all married bachelors are unmarried
bachelors" is just like "when it rains and does not rain ..." I cannot
picture anything.

Similaly I cannot picture (x)(x = x+1) -> (x = x+2) any better than I
can picture anything being attributing to married bachelors.

>
> Just as it's perfectly clear what "There are no counterexamples to
> Goldbach's conjecture that are less than 12." You and I both know
> what this means and we won't suddenly be struck stupid if it turns out
> that Goldbach's conjecture is true.
>
> Your claims about the meaninglessness of simple arithmetic statements
> are among the least plausible.
>
> Footnotes:
> [1] In fact, this statement seems obviously true! Suppose
> x = x + 1. Then we may substitute x + 1 for x in the right hand side
> of the equation x = x + 1, thus:
>
> x = x + 1
> = (x + 1) + 1
> = x + 2.
>
> I see nothing the least bit fishy about this reasoning.
>
> --
> "It's one of the easiest tickets to true fame--not this silly stuff
> where people cheer you for a few years and then forget about you--but
> the kind of fame where school kids have to read your biography and do
> reports on you." -- Another reason to support James S. Harris.

From: Newberry on 25 Mar 2010 01:07

On Mar 23, 4:39 am, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote:
> Newberry <newberr...(a)gmail.com> writes:
> > On Mar 22, 6:44 pm, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote:
> >> "Jesse F. Hughes" <je...(a)phiwumbda.org> writes:
>
> >> > "Jesse F. Hughes" <je...(a)phiwumbda.org> writes:
>
> >> >>> Although I don't necessarily agree with the idea of killfiling someone
> >> >>> merely because one disagrees with them, Clarke is only doing the
> >> >>> same thing to Hughes that Daryl McCullough, another poster in this
> >> >>> thread, has done to me. And Clarke has killfiled Hughes for the exact
> >> >>> same reason that McCullough has killfiled me -- because both Hughes
> >> >>> and I have repeatedly posted statements with which Clarke and
> >> >>> McCullough, respectively, disagree. The so-called "cranks" and the
> >> >>> standard theorists have more in common with each other than either
> >> >>> would like to admit.
>
> >> > Two points I failed to make: First, I don't see why you think Clarke
> >> > is a crank.
>
> >> Actually, I should also say that Newberry is not a crank -- at least
> >> not by the standards of sci.math. He's fairly passionate about
> >> replacing FOL with an alternative, which is certainly crank*ish*, but
> >> he's perfectly coherent and relatively competent. (I'm not sure he
> >> has the knowledge necessary to complete his project, but he's still
> >> relatively knowledgeable about the subject.)
>
> > I know enough to cause a lot of damage.
>
> I really wish you wouldn't say things like that when I'm claiming
> you're not a crank. It makes me look foolish.

It is not because of that.

> >> Compare his writings to AP, James S. Harris, Meuckenheim. Newberry is
> >> not in the same league.
>
> >> --
> >> Jesse F. Hughes
>
> >> "Had you told it like it was, it wouldn't be like it is."
> >> -- Albert King
>
> --
> Jesse F. Hughes
> "[I]t's the damndest thing. There's something wrong with every last
> one of you, and I *never* thought that was a possibility. But now I
> feel it's the only reasonable conclusion." --JSH sees some sorta light- Hide quoted text -
>
> - Show quoted text -

From: Nam Nguyen on 25 Mar 2010 02:41

Newberry wrote:

>
> Similaly I cannot picture (x)(x = x+1) -> (x = x+2) any better than I
> can picture anything being attributing to married bachelors.

Of course you can. Let T be the following axiom system:

A1: Ax(x=0)
A2: Ax(Sx=0)
A3: Axy(x+y=0)

Can you now picture (x)(x = x+1) -> (x = x+2)?