Quantifier free logic

[Zuhair]

Dear Sirs

I'm writing here especially about Tony Orlow's idea of replacing the
quantifiers
using membership e and a set U which informally stands for the
universe
of discourse of what is within it.

Tony Orlow suggested the following:

(x e U -> phi(x)) to stand for Ax.phi(x)

~(x e U ->~phi(x)) to stand for Ex.phi(x)

I personally prefer the following:

(x e U -> phi(x)) to stand for AxeU(phi(x))

~(x e U -> ~phi(x)) to stand for ExeU(phi(x))

However there is a lot of matters needs to be understood
and stipulated in order for this to work.

first we need to know that statements like x e U
are not propositions, they are propositional functions!
a proposition is a statement that is either true or false
like "Plato is a man", however a statement like
"x is a man" is a statement the truth of which
depends on the value x is substituted for, so
"x is a man" is neither true nor false in the sense
of how Plato is a man is true and how
Margret is a man is false. Since for every
individual value of x "x is a man" is either true
or false, then it is "x is a man" is called as
propositional function. Now the stipulation that
x is a variable that may range over many values
and x e U is a propositional function by itself
bares the seed of quantification, the next
property that we need to stipulate is the
scope of quantification. We need to understand that
x e U -> phi(x), which is a formula that has two open
formulas where x is free in each, we need to understand
that whatever is substituted for x in formula x e U
will be simultaneously substituted for phi(x)
because phi(x) is the total string of symbols over
which quantification by x e U range over.
So Socrates e U -> Socrates is Greek
we cannot wright Plato e U -> Socrates is Greek,
of course this peculiarity is confined to the the use
of x e U, and it doesn't pass over to other situations.
so the above is first order logic but under disguise of
propositional logic.

Now K_h has presented the following proof

first) ~(x e U ->~phi(x)) -> x e U & phi(x)
which is correct!

second) x e U & phi(x) -> (x e U -> phi(x))
The idea is that
x e U & phi(x) -> phi(x) -> (~xeU or phi(x))
->(xeU ->phi(x))

thus concluding that this method would lead to

ExeU(phi(x)) -> AxeU(phi(x))

which is so weak since it is only true if U is
empty or contains only one element.

This proof of K_h presented above is FALSE.

The reason is it violates the scope of quantification
lets take it a step by step

we have: x e U & phi(x) (which stands for ExeU(phi(x)))
then he concluded that

x e U & phi(x) -> phi(x)

and here is the error, the scope of quantification of x e U & phi(x)
don't go beyond that phi(x) mentioned in that statement, the
next phi(x) has x free that is unbound by x present in x e U, so
we can have

Plato e U & Plato is Greek -> Mohammad is Greek

which is False.

So we don't have ((xeU) & (phi(x))) -> (phi(x))
since the conclusion is out of the quantification of the antecedent!

So the proof fails.

The message is: You cannot use all rules of propositional logic freely
you must appreciate bounding.

Now let's see the other direction

((xeU) -> (phi(x))) -> ((xeU)^(phi(x)))

it seems reasonable since we maintain
that x e U must hold for some x.

So to wright some axioms of ZF.

Extensionality:

xeU->(yeU->((zeU->(zex<->zey))->x=y))

this is reduced to

xeU & y e U -> ((zeU->(zex<->zey))->x=y)

Empty set:

x e U ^ (y e U -> ~yex)

Of course those axioms stipulate U which is
is not written in ZF axioms however we can
add it as a primitive to the language of ZF
and write all axioms with all quantifiers bounded
in U. And we'll get U as a model of ZF.

However we need to stipulate further ways
of eliminating quantifiers, like
(xeU ->phi(x)) -> phi(x)
and like
phi(x) -> xeU^phi

etc..

Regards

Zuhair

[William Elliot]

On Wed, 11 Jan 2012, Zuhair wrote:

> I'm writing here especially about Tony Orlow's idea of replacing the
> quantifiers using membership e and a set U which informally stands for
> the universe of discourse of what is within it.
>
> Tony Orlow suggested the following:
>
> (x e U -> phi(x)) to stand for Ax.phi(x)
>

U subset phi

> ~(x e U ->~phi(x)) to stand for Ex.phi(x)
>

not (U subset not.phi)

{x} subset A for x in A

> I personally prefer the following:
>
> (x e U -> phi(x)) to stand for AxeU(phi(x))
>

WhatdoesAxeUmean?

[Zuhair]

For all x in U.

Zuhair

[Zuhair]

On Jan 11, 2:53 pm, William Elliot <ma...@panix.com> wrote:

> On Wed, 11 Jan 2012, Zuhair wrote:
> > I'm writing here especially about Tony Orlow's idea of replacing the
> > quantifiers using membership e and a set U which informally stands for
> > the universe of discourse of what is within it.
>
> > Tony Orlow suggested the following:
>
> > (x e U -> phi(x)) to stand for Ax.phi(x)
>
> U subset phi

No

>
> > ~(x e U ->~phi(x)) to stand for Ex.phi(x)
>
> not (U subset not.phi)
>
> {x} subset A for x in A

NO

[MoeBlee]

On Jan 11, 5:53 am, William Elliot <ma...@panix.com> wrote:

> WhatdoesAxeUmean?

Ax(xeU ->

So AxeU P

stands for

Ax(xeU -> P)

MoeBlee

[MoeBlee]

On Jan 11, 5:02 am, Zuhair <zaljo...@gmail.com> wrote:
> the scope of quantification

There are no quantifiers in Orlow's languages, so there is no scope of
quantification.

After interviewing Orlow, as best I can pin down now, he gives
languages as having

variables, possibly predicate symbols (including '='), possibly
function symbols, connectives, also, in every language, the special 2-
place predicate symbol 'e' and the special variable 'U', and no
quantifiers

And we are to apply ordinary syntax and semantics to the above, except
'e' maps to the membership relation on the universe of the
interpretation (he says something different, but it doesn't make
sense, so, unless he comes up with something that does make sense, I
take it that 'e' maps to the membership relation), and 'U' maps to the
universe itself (so, I would recommend he say that 'U' is a constant
with a fixed interpretation).

But if 'U' maps to the universe itself, then, in ordinary semantics,
that would violate regularity, since variables and constants map to
members of the universe, thus 'U' is mapped to an object that is a
member of itself. Moreover, it would require that EVERY universe be a
member of itself, thus that every set is a member of itself, which is
even stronger than merely violating regularity.

Moreover, it's not been shown that this semantics provides for a
formula that expresses that, say P, holds for all members of the
universe.

And according to Orlow, xeU is always satisfied. So, for any formula
P(x), we have:

xeU -> P(x) is satisfied iff P(x) is satisfied.

So xeU -> P(x) is semantically equivalent with P(x).

But, in ordinary semantics, P(x) is NOT equivalent with AxP(x).
So, xeU -> P(x) does NOT capture universal quantification.

MoeBlee

[Zuhair]

On Jan 11, 7:42 pm, MoeBlee <modem...@gmail.com> wrote:
> On Jan 11, 5:02 am, Zuhair <zaljo...@gmail.com> wrote:
>
> > the scope of quantification
>
> There are no quantifiers in Orlow's languages, so there is no scope of
> quantification.
>

Possibly there is explicitly non in Orlow's, but he uses x e U and
phi(x)
and all of those are *propositional functions* so there must
be some form of subtle quantification!, when he writes
((xeU)->(phi(x))) this is actually a propositional function
of two propositional functions and obviously each substitution
of x must be the same in both xeU and in phi(x) in the above
statement, but this simultaneous substitution do not necessarily
extend to what is beyond it and this is what I mean by the scope
of quantification! and all of this is present in a subtle manner in
Orlow's definitions and in his arguments. At the end we will reach
at a first order logic in which quantification is expressed using
different
symbols and with some stipulations of special logical contexts which
is generally in my opinion a complex approach that if it survive
contradictions
and weakness.

I can understand for example: ((xeU)^((yeU)->(~yex))) to stand for
Ex(xeU & Ay(yeU->~yex)) which simply mean that there exist
an element of U that is empty of elements of U.

Now:
((xeU)^((yeU)->(~yex))) uses less characters than
Ex(xeU & Ay(yeU->~yex)) .

Not only that it can be further simplified as to reduce
brackets to

xeU ^ (yeU ->~yex) to represent
Ex(xeU&Ay(yeU->~yex))

lets take maximal simplification of both statements

xeU^(yeU->~yex) while
ExeU(AyeU(~yex))

It seems that the first is more elegant really. That's why
he say "superfluous" in his threads. Anyhow.

Zuhair


Zuhair

[MoeBlee]

On Jan 11, 12:05 pm, Zuhair <zaljo...@gmail.com> wrote:
> On Jan 11, 7:42 pm, MoeBlee <modem...@gmail.com> wrote:> On Jan 11, 5:02 am, Zuhair <zaljo...@gmail.com> wrote:
>
> > > the scope of quantification
>
> > There are no quantifiers in Orlow's languages, so there is no scope of
> > quantification.
>
> Possibly there is explicitly non in Orlow's, but he uses x e U and
> phi(x)
> and all of those are *propositional functions* so there must
> be some form of subtle quantification!

He's proposing a system without quantifiers. Without quantifiers,
there's no scope of quantifiers. If he wishes to define an alternative
notion of scope then he'd need to tell us his definition and his
revised proof rules regarding it.

> each substitution
> of x must be the same in both xeU and in phi(x)

There is no substitution going on at this point. In any case, an
assignment for the variables will assign only one object to each
variable. That is already a done deal.

> I can understand for example: ((xeU)^((yeU)->(~yex))) to stand for
> Ex(xeU & Ay(yeU->~yex))

Not without an alternative semantics and alternative proof syntax.

Just thinking that one thing kinda seems like another and then waving
the hand and saying "Voila, this stands for this" does not itself
ensure the desired properties and equivalences.

MoeBlee

[Zuhair]

On Jan 11, 7:42 pm, MoeBlee <modem...@gmail.com> wrote:

Yes I agree with you saying that if xeU is always satisfied, or
in other words xeU is satisfied for all values of x, then
xeU ->P(x) is nothing but saying P(x). But I think Orlow
failed to express himself here, it is easier to stipulate
that U is just a constant referring to some specific set
and that xeU is not necessarily always satisfied, we
only need to say that when xeU is satisfied for some
value of x then P(x) for the same value of x will be
satisfied, and this really implies universal quantification
over U. By the way U is indeed an object in the universe
of discourse D which is not U, but what we can say
is that U will represent the universe of discourse for
objects in D that are in U something similar to how
V in Morse-Kelley can be interpreted as a model of
ZF, i.e. a class in which all statements of ZF are
satisfied. So U can represent the universe(or a model)
of a subtheory of the theory in which U is written. In this
way we will avoid conflicts with Regularity.

Zuhair

[Zuhair]

On Jan 11, 9:12 pm, MoeBlee <modem...@gmail.com> wrote:
> On Jan 11, 12:05 pm, Zuhair <zaljo...@gmail.com> wrote:
>
> > On Jan 11, 7:42 pm, MoeBlee <modem...@gmail.com> wrote:> On Jan 11, 5:02 am, Zuhair <zaljo...@gmail.com> wrote:
>
> > > > the scope of quantification
>
> > > There are no quantifiers in Orlow's languages, so there is no scope of
> > > quantification.
>
> > Possibly there is explicitly non in Orlow's, but he uses x e U and
> > phi(x)
> > and all of those are *propositional functions* so there must
> > be some form of subtle quantification!
>
> He's proposing a system without quantifiers. Without quantifiers,
> there's no scope of quantifiers. If he wishes to define an alternative
> notion of scope then he'd need to tell us his definition and his
> revised proof rules regarding it.
>
> > each substitution
> > of x must be the same in both xeU and in phi(x)
>
> There is no substitution going on at this point. In any case, an
> assignment for the variables will assign only one object to each
> variable. That is already a done deal.
>
> > I can understand for example: ((xeU)^((yeU)->(~yex))) to stand for
> > Ex(xeU & Ay(yeU->~yex))
>
> Not without an alternative semantics and alternative proof syntax.

Of course, which is what I was saying, it can be done I think, but the
result would be complex even if for some statements it might look
elegant. This needs lots of work really, and I do not say that I
already
lied down all the required steps, but I do think it is possible.
Possibly
my understanding of Orlow's is mistaken, anyhow.

[Zuhair]

On Jan 11, 9:12 pm, MoeBlee <modem...@gmail.com> wrote:
> On Jan 11, 12:05 pm, Zuhair <zaljo...@gmail.com> wrote:
>
> > On Jan 11, 7:42 pm, MoeBlee <modem...@gmail.com> wrote:> On Jan 11, 5:02 am, Zuhair <zaljo...@gmail.com> wrote:
>
> > > > the scope of quantification
>
> > > There are no quantifiers in Orlow's languages, so there is no scope of
> > > quantification.
>
> > Possibly there is explicitly non in Orlow's, but he uses x e U and
> > phi(x)
> > and all of those are *propositional functions* so there must
> > be some form of subtle quantification!
>
> He's proposing a system without quantifiers. Without quantifiers,
> there's no scope of quantifiers. If he wishes to define an alternative
> notion of scope then he'd need to tell us his definition and his
> revised proof rules regarding it.
>

It doesn't matter, I'm speaking about my take on that matter
in which quantification occurs but it is expressed in different
ways using e and U symbols instead of A and E quantifiers
so in my take on this subject, there is quantification, there
is a scope of it. I don't know if Orlow's agrees with me on that
possibly he didn't express his ideas in a clear manner, possibly
he meant another thing, I don't know, what matters here is
my take on that subject which is apparently different from
Orlow's.

> > each substitution
> > of x must be the same in both xeU and in phi(x)
>
> There is no substitution going on at this point. In any case, an
> assignment for the variables will assign only one object to each
> variable. That is already a done deal.

Possibly in Orlow's system or in your understanding of his system yes,
but not in my take on that subject, I don't agree with this fixed
assignment
for the variables, in my take this will lead to problems.

[MoeBlee]

On Jan 11, 12:22 pm, Zuhair <zaljo...@gmail.com> wrote:
> This needs lots of work

Presently it's incoherent. I have not precluded that it can be revised
to make it coherent and working.

MoeBlee

[MoeBlee]

On Jan 11, 12:17 pm, Zuhair <zaljo...@gmail.com> wrote:
> xeU is not necessarily always satisfied,

At one point he said that xeU is always satisfied (or 'true'). I might
have missed that he revised that. In any case, he just needs to say
how it should work. He said to use ordinary semantics, so if 'U' does
not have a fixed intepretation as the universe so that xeU is not
always satisfied, and 'U' itself can be interpreted as some other
member of the domain, then xeU has no special import toward
quantification. And 'U' can't be interpreted as a subset of the domain
(unless that subset is also a member of the domain) since 'U' is not a
predicate symbol but rather it is a variable (actually, I would
suggest it be a constant). Or, he has to specify his exact alternative
semantics regarding 'U'.

MoeBlee

[K_h]

"Zuhair" <zalj...@gmail.com> wrote in message
news:183b87dc-0e05-4f1f...@w4g2000vbc.googlegroups.com...

>
> This proof of K_h presented above is FALSE.
>
> The reason is it violates the scope of quantification
> lets take it a step by step
>
> we have: x e U & phi(x) (which stands for ExeU(phi(x)))
> then he concluded that
>
> x e U & phi(x) -> phi(x)
>
> and here is the error, the scope of quantification of x e U & phi(x)
> don't go beyond that phi(x) mentioned in that statement, the

Whoa, you were attempting to replace quantifiers with non-quantified
expressions and so there will be no scope of quantification. The
simplification rule of inference, p^q-->p, can easily be shown to be true with
a simple truth table.


Best

[Zuhair]

If you look at his arguments you see him saying that U is like a
universe
of discourse for arguments satisfied within U, and this is true, I
think
he confused it with the universe of discourse of the theory in which
U is written. I already gave you an nice example, take a theory To
which is a set of sentences including the symbol U, now we can
get a subtheory of To which is T1 where all its non U statements are
satisfied within U, like saying in ZF we can construct the model
of Z. Something like this sort can be work out.

Zuhair

[Zuhair]

On Jan 12, 5:05 am, "K_h" <KHol...@SX729.com> wrote:
> "Zuhair" <zaljo...@gmail.com> wrote in message

No it cannot be shown here, because some of those non-quantified
expression
act themselves as quantifiers, so the logic I'm speaking about is not
really
quantifier free logic. If ((xeU)->(phi(x))) act as Ax.(xeU->phi(x))
then x variables
in phi(x) are bound by x in (xeU), so you cannot infer ((xeU)-
>(phi(x)))->(phi(x))
because assignments to occurrences of x in the first phi(x) is not
necessary
the same as those in the next phi(x), so your proof fails. However
your proof
will stand the course if we say all occurrences of a variable will be
assigned
the same value each time, which is also a form of subtle
quantification.

Zuhair

[MoeBlee]

On Jan 12, 1:11 am, Zuhair <zaljo...@gmail.com> wrote:

> so the logic I'm speaking about is not
> really
> quantifier free logic.

Then it's not the logic Orlow is proposing.

MoeBlee

[MoeBlee]

P.S

On Jan 12, 1:11 am, Zuhair <zaljo...@gmail.com> wrote:

> so the logic I'm speaking about is not
> really
> quantifier free logic.

Then it's not Orlow's proposal.

Orlow says xeU is always satisifed and that there are no quantifiers
in his language.

So, by Orlow, we have

|= xeU

So, for any formula P:

|= (xeu -> P) <-> P

and

|= (xeU & P) <-> P

so

|= (xeU & P) <-> (xeU -> P)

So, If "xeu & P" is supposed to have a semantic interpretation as
"there exists an x such that P", and "xeU -> P" is supposed to have a
semantic interpretation as "for all x, P", then "there exists and x
such that P" and "for all x, P" turn out equivalent.

So, this is yet another demonstration that Orlow's proposal, as he's
given it, is nonsense.

And YOUR arguments about this and that don't vindicate his proposal if
your arguments use quantifiers in the language, since Orlow does not
have quantifiers in his language.

MoeBlee

[Zuhair]

My take on this subject is possibly different from Orlow's. What I
want to do is to express quantification in another way that somehow
bears some superficial resemblance to what Orlow stated or possibly
that what Orlow wanted to say.

It is not difficult to express quantification in some other way than
the customary way

for example lets stipulate that whenever x appears in an expression
then it is assigned the same value. So all occurrences of x in phi(x)
would
can only be bound by the same quantifier.
Of course this is a restriction on ordinary way of writing FOL.
Now lets stipulate that whenever a variable is written in upper case
then it is universally quantified, and if written in lower case then
it
is existential quantified, now the order of quantification will not be
specified if it is the same order of the appearance of those variables
in the formula, if it is the reverse order then we wright * over the
first
variable, Now the order is neither the same nor the reverse then
we need to specify them by a number after the first occurrence
of each variable. | will be used to delineate the end of
quantification
the last variable.
This will let us get rid of the use of A and E altogether.

Examples: *ZeX<->ZeY| ->X=Y this will represent
Extensionality.

Also: ~*Y e x, this will stand for ExAy(~y e x)

you see this way is a more abbreviated way of writing FOL
but there is not difference in principle. It is just first order
logic written in another way with the benefit of abbreviation
only.

Now can we do a similar approach to the above but
instead of using difference in case we use
membership of some set U and use some logical
contexts to differentiate universal from existential quantification,
I think we can and this is my aim here (which might not be
Orlow's original aim but that would be the most that we
can make of the general notion of using e and U to get
rid of logical quantifiers in FOL, i think).

A simple trial is to stipulate that U is a primitive constant
e is a primitive binary relation, No symbols for quantifiers
variables are only assigned for objects, rules of formation
of formulas in first order logic are all there (except for
quantification symbols since they are absent).
we maintain that in any formula phi(x) , the formula
x e U only occur once and it will be the first subformula
of phi(x) in which x occur, and all occurrences of x whenever
it appears are assigned the same value. if phi is a sentence
then every variable xi in phi must have its first occurrence
in the sub- formula xi e U of phi.

Now we take (x e U -> phi(x)) to mean Ax(xeU ->phi(x))
So for example we cannot have

(x e U ->phi(x)) ->phi(x)

because x in the conclusion is not bound by x e U (the scope
of quantification that began by x e U ends after at the bracket
closing the bracket immediately before xeU.

we can have (x e U ->(phi(x) ->phi(x))) yes but not the above
we can have of course
(x e U ->phi(x)) -> phi(y)

Now given the restrictions and stipulations above we may
be able to represent universal and existential quantification
in the following manner:

(x e U -> phi(x)) to represent Ax(xeU ->phi(x))
(x e U ^ phi(x)) to represent Ex(xeU->phi(x))

I think this can work, I'm not sure though.

so for example:

x e U & y e U ->((z e U->(zex<->zey)) ->x=y) to represent

AxeU(AyeU(AzeU(zex<->zey) ->x=y)

x e U ^ (y e U -> ~y e x) to represent

Ex(Ay(~yex))

Regards

Zuhair

[Zuhair]

Yes there is a big difference between my take on this subject and
Orlow's, I've said that almost ten times here, and I said many times
that what
I am concerned about is MY take on this subject and not Orlow's
specifics, I only look at Orlow's contribution of the "bare" original
idea only, the general idea only is Orlow's, my stipulations are of
course different, and i think
it is the way one can make most of the "bare" idea of Orlow's, and the
bare idea
I am referring to is to represent universal and existential
quantification
using symbols of e and U only, and thus getting rid of the symbols A
and E
It is only this part of Orlow's argument that I am taking and I must
name it
after him because he is the one who first present it, even if he meant
another
thing, he still remain the first one who had presented it, of course
my further
on specifics and details may greatly depart from Orlow's, possibly
Orlow's wanted
to say that first order logic can be reduced to propositional logic
using e and U
which is of course not my concern here, those details I'm giving here
represent
MY take on this subject. I don't know if this is clear.

Zuhair

[Zuhair]

On Jan 12, 10:59 am, MoeBlee <modem...@gmail.com> wrote:

Yes, I agree with you about your conclusions about Orlow's arguments,
truly if x e U is always satisfied then matters would go in the way
you are pointing to, that is correct no doubt. Of course this doesn't
touch
my take on this subject. But it is no doubt correct argumentation on
your
behalf.

Zuhair

[Zuhair]

its better to use brackets to delineate the beginning
and end of quantification.

Zuhair

[MoeBlee]

On Jan 12, 5:03 am, Zuhair <zaljo...@gmail.com> wrote:
> whenever a variable is written in upper case
> then it is universally quantified, and if written in lower case then
> it
> is existential quantified,

(1) Please give recursive definitions to determine the scope of the
quantification.

For example

PX -> GX

How do I know whether that is to mean

AXPX -> AXGX

or

AX(PX -> GX) ?

(But I guess you attempt to address this later in your post?)

(2) What about variables that are not bound by either universal or
existential quantification?

> now the order of quantification will not be
> specified if it is the same order of the appearance of those variables
> in the formula, if it is the reverse order then we wright * over the
> first
> variable, Now the order is neither the same nor the reverse then
> we need to specify them by a number after the first occurrence
> of each variable. | will be used to delineate the end of
> quantification
> the last variable.
> This will let us get rid of the use of A and E altogether.

That's not easy to follow.

Would you please give recursive definitions for your syntax and also
an inductive formulation for satisfiability of formulas?

Also, for context, would you please say whether your proposal is one
that you believe is presently formulated in a fully satisfactory
manner or whether you're just sketching some ideas now of which the
details would need to be, if possible, worked out later?

MoeBlee

[Zuhair]

>
> Also, for context, would you please say whether your proposal is one
> that you believe is presently formulated in a fully satisfactory
> manner or whether you're just sketching some ideas now of which the
> details would need to be, if possible, worked out later?
>
> MoeBlee

The later, it is only a sketch the details of which if possible would
be
worked out later.

Zuhair

[K_h]

"Zuhair" <zalj...@gmail.com> wrote in message

news:6a75145f-7b61-4f27...@i26g2000vbt.googlegroups.com...

On Jan 12, 5:05 am, "K_h" <KHol...@SX729.com> wrote:
> > "Zuhair" <zaljo...@gmail.com> wrote in message
> >

> > Whoa, you were attempting to replace quantifiers with non-quantified
> > expressions and so there will be no scope of quantification. The
> > simplification rule of inference, p^q-->p, can easily be shown to be true
> > with
> > a simple truth table.
> >
> > Best
>
> No it cannot be shown here, because some of those non-quantified
> expression
> act themselves as quantifiers, so the logic I'm speaking about is not
> really
> quantifier free logic.

That is a contradiction. If the non-quantified expressions act as quantifiers
then they really are expressions for quantification. In other words, it is
wrong to call them non-quantified in the first place.

> If ((xeU)->(phi(x))) act as Ax.((xeU)->(phi(x)))
> then x variables

But the only difference here is that you are dropping the Ax symbol. The
problem with ((xeU)->(phi(x))) then becomes how does one determine if it is a
propositional function or just a proposition? Ordinarily, ((xeU)->(phi(x))) is
a propositional function whereas Ax.((xeU)->(phi(x)))
is just a proposition.

> in phi(x) are bound by x in (xeU), so you cannot infer ((xeU)-
> >(phi(x)))->(phi(x))
> because assignments to occurrences of x in the first phi(x) is not
> necessary
> the same as those in the next phi(x), so your proof fails.

p ^ q ---> p

proof:

1. TRUE
2. TRUE v ~q, 1, add
3. (pv~p) v ~q, 2, neg
4. p v (~pv~q), 3, assoc
5. (~pv~q) v p, 4, comm
6. ~(p^q) v p, 5, DeM
7. p^q --> p, 6, M.I.

[Zuhair]

I know that but it is not applicable to the case I am speaking about.
And yes this logic has quantification, I said quantifier free logic
I didn't say quantification free logic, I just want to get rid of the
KNOWN quantifiers, but definitely there is quantification. Now

(xeU -> phi(x)) is a proposition iff every variable xi in phi(x)
occur in a subformula of phi(x) of the form (xi e U ->Q(xi))
or (xi e U ^ Q(xi)) in other word (xeU ->phi(x)) is a proposition
iff it is a sentence.

In reality one can regard the formula xeU as a quantifying
formula over x in phi(x) in the expression (xeU -> phi(x))
so this logic do contain quantifiers but in forms of formulas.

To see where you argument fails

(xeU ->phi(x)) ->phi(x)

this was your intermediate step which is not applicable here
because simultaneous assignments for x in xeU and phi(x)
ends by the bracket closing the opening bracket before xeU
so x variables in conclusion phi(x) are not necessarily
assigned the same values as x variables in (xeU ->phi(x))
so your argument fails because it crosses the borders
of quantification.

Mind you I'm not speaking of Orlow's arguments, I'm speaking
about my take on this subject which only shares with Orlow
the general intention to use xeU-> and xeU^ instead
of universal and existential quantification respectively. My
approach mainly differ from Orlow's in addressing quantification
and scope of quantification, so my approach is just another
way of writing FOL, possibly Orlow's is different it seems
as if we want his logic to be totally free from quantification
and as if wants to melt FOL into propositional logic, I don't
know but if that is his aim, then this differs from the approach
I'm speaking about here.

Zuhair

[K_h]

"Zuhair" <zalj...@gmail.com> wrote in message

news:89524b91-a8a7-482a...@cs7g2000vbb.googlegroups.com...

On Jan 12, 11:13 pm, "K_h" <KHol...@SX729.com> wrote:
> > "Zuhair" <zaljo...@gmail.com> wrote in message
> >
> > news:6a75145f-7b61-4f27...@i26g2000vbt.googlegroups.com...
> > On Jan 12, 5:05 am, "K_h" <KHol...@SX729.com> wrote:
> >
> I know that but it is not applicable to the case I am speaking about.
> And yes this logic has quantification, I said quantifier free logic
> I didn't say quantification free logic, I just want to get rid of the
> KNOWN quantifiers, but definitely there is quantification. Now

You wanted to quantify using expressions like xeU so you did not get rid of
quantifiers. You introduced ambiguity, then, because there is no way to tell
the difference between a proposition and a propositional function.

> To see where you argument fails
>
> (xeU ->phi(x)) ->phi(x)
>
> this was your intermediate step which is not applicable here

This was never a step of mine. What I wrote was (xeU)^phi(x)-->phi(x)!

> because simultaneous assignments for x in xeU and phi(x)
> ends by the bracket closing the opening bracket before xeU
> so x variables in conclusion phi(x) are not necessarily
> assigned the same values as x variables in (xeU ->phi(x))
> so your argument fails because it crosses the borders
> of quantification.

So you want to use brackets to denote scope of quantification without
quantification symbols? Well, brackets are used in propositional functions
too! So how to tell them apart? I was correct in everything I posted. Now
you are changing things piece-meal, e.g. by adding "bracket quantification",
etc. I suggest you work up your whole theory and then post or publish it.

[Zuhair]

Yes, I need to make a full work up and present it.

Zuhair

[Graham Cooper]

This problem was solved 50 years ago by replacing existential
variables with functions.

Ex ...x...

==> x()

Ay Ex ...x...

==> x(y)

**x DEPENDS ON y**

This ties in with the definition of predicates and functions.

INTRODUCTION TO ARTIFICIAL INTELLIGENCE
PHILLIP C JACKSON JR

P218

An INTERPRETATION for a set of variable-symbols is given by specifying
the universe of discourse; that is, the set of values they can
assume. The universe of discourse for a set of variable symbols is
denoted by the letter D. For example, D might be the set of numbers
{-1, 0, +1}. If D denotes a set k then D^n denotes the set of all n-
tuples of D. Thus,

{(-1,0,+1)}^2 = {(-1,-1), (-1,0), (0,-1) ... (+1,+1)}

If D contains m elements, then D^n contains m^n elements. An
INTERPRETATION of an n-ary predicate symbol P associates each element
of D^n with exactly one element of the set {true,false}.
....

P 219

For n=0 we define D^n to be the set containing (), the zero tuple. By
our definition, the INTERPRETATION of a zero-ary function must be a
CONSTANT; rather than write expressions like "f()" we can usually
denote constants by letters a, b, c.

*************

EXTENDING zero-ary function interpretation further, f() can represent
an EXISTENTIALLY QUANTIFIED VARIABLE... a unspecified constant.

*************

e.g.

Af Er MISSING(f, r) << CANTORS THEOREM

==>

MISSING(f, r(f))

where r(f) = ANTIDIAG(f)

MISSING(f, ANTIDIAG(f)) << CANTORS THEOREM WITHOUT QUANTIFIERS

MISSING(x,y) <=> An x(n)=/= y

Herc

[Tony Orlow]

Hi Zuhair -

I haven't had time for a few days to spend on this on the computer,
but thanks to you and the rest of the discussion, I think I've gained
some insight into what I was leaving out. I'm not sure what happened
to the thread I started, but I see you've started this thread to
continue the discussion, and I appreciate that. I have had some
moments to reflect and come to these conclusions.

In a sense I was correct, on a per-element basis, at the most basic
logical level, but that truth tables allowed for an empty universe,
and for elements outside the universe. Certainly, with FOL requiring a
non-empty universe containing all objects under discourse, Ax P(x) ->
Ex P(x) is true and not the reverse. So, you and others are correct in
that regard, and what I am leaving out is the connection between
element truth and set truth.

ExeS P(x) is shorthand for saying, for x_n in S, x_1 v x_2 v x_3 v
x_4..... = T, false otherwise
AxeS P(x) is shorthand for saying, for x_n in S, x_1 ^ x_2 ^ x_3 ^
x_4..... = T, false otherwise

The null universe may be off limits to FOL, but it is not in general
logic, and perhaps can be used as the basis for a recursive definition
of the quantifiers, as follows:

AP Exe{} P(x) = (xe{} ^ P(x)) = F
AP Axe{} P(x) = (xe{} -> P(x)) = T

For element-wise addition of elements:
ExeS P(x) v P(y)) <-> EzeSu{y} P(z) ('u' is union)
(AxeS P(x) v P(y)) <-> AzeSu{y} P(z)

or, for set-wise combination of elements:

ExeS P(x) v EyeT P(x) <-> Eze(SuT) P(x)
AxeS P(x) ^ AyeT P(x) <-> Aze(SuT) P(x)

If we replace ExeS P(x) with xeS ^ P(x) in set-wise existential
quantification we get
(xeS ^ P(x)) v (yeT ^ P(y)) -> (ze(SuT) ^ P(z))

And, if we replace AxeS P(x), with xeS -> P(x) in set-wise universal
quantification we get
(xeS ->P(x)) ^ (yeT -> P(y)) -> ((ze(SuT)) -> P(z))

Now, I should point out a semantic aspect of my grammar. Any object,
function or collection named in the statement is assumed to exist. Any
object in the premise of an implication (x in "x->y", y in "x<-y", and
both x and y in "x<->y") are assumed to be universally quantified
unless specifically declared as a constant using an inclusion
statement ("xeS" or "yeT"). Also, the same symbol in a statement
always refers to the same object; 'x' cannot represent two different
objects in the same statement. So, the above statements read as
follows:

For all x, y, S and T, if x in S and P(X), or y in T and P(Y), then
exists z in (SuT) and P(z)
For all x, y, S and T, if x in S implies P(X), and y in T implies
P(Y), then for all z, z in (SuT) implies P(z)

I think there are a few more steps to be filled in here, and I'll have
to mull it over a bit, but I hope this addresses some of the questions
about my proposal and indicates a direction that might lead to a
proper statement of my intentions.

Thanks to all for your input.

Peace,

Tony

[Graham Cooper]

On Jan 16, 2:28 am, Tony Orlow <bonyto...@gmail.com> wrote:
>
> Hi Zuhair -
>
> I haven't had time for a few days to spend on this on the computer,
> but thanks to you and the rest of the discussion, I think I've gained
> some insight into what I was leaving out.

Ew I got a mention this time!

Herc

[Tony Orlow]

Herc -

Sometimes you make some sense, but I'm not sure how I mentioned you
this time. What do you think of the ideas I put forth? I rather think
the rules for element-wise addition to S cover any countable sets, and
uncountable sets are covered by the set-wise combinations, since they
are generally defined using some P(x) anyway. I'm surprised I'm not
getting more response to this, yea or nay.

Peace,

Tony

[Zuhair]

The problem with that is that it goes to infinitary logic which is not
FOL.

Zuhair

[Tony Orlow]

Isn't the axiom of infinity expressible in FOL? You may assume that
axiom has been stated so N exists, since it can be stated in my more
primitive logic as follows:

1. {}eN ^ (xeN -> {x}eN)

Is there really a problem?

Peace,

Tony