Abbreviating FOL with identity and membership2 [Logic]

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From: zuhair on 1 May 2010 14:02

From: zuhair on 1 May 2010 14:12

From: zuhair on 1 May 2010 14:14

On May 1, 1:12 pm, zuhair <zaljo...(a)gmail.com> wrote:
> On May 1, 12:22 pm, MoeBlee <jazzm...(a)hotmail.com> wrote:
>
>
>
> > On May 1, 9:42 am, zuhair <zaljo...(a)gmail.com> wrote:
>
> > > The quantifiers need not be actually written if writing them
> > > makes no difference to the meaning of the formula
>
> > > Example: writing the Extensionality axiom (see below)
>
> > [...]
>
> > > Extensionality: zx<>zy > x=y
>
> > WRONG
>
> > Az(zex <-> zey) -> x=y
>
> > is NOT equivalent with
>
> > Az((zex <-> zey) -> x=y)
>
> > Please get a book on the basics of this subject.
>
> > MoeBlee
>
> Still I need to make sure of this issue, since what I had in my mind
> was a kind of * exhaustive* quantification notation, which means that
> the quantification ends one the quantified variable seize to appear
> any more,
>
> so for example the variable z do not appear on the right of the
> implication
> so my aim was to say that z quantification would be restricted
> to zx<>zy.
>
> so when I wrote zx<>zy >y=x
>
> I actually meant
>
> for all z ( z e x <-> z e y ) -> y=x
>
> But I'm not sure if that exhaustive principle would work.
>
> Zuhair

I admit that this exhaustive principle would not work, so we need to
extend the dot notation to cover quantification also.

Zuhair

From: zuhair on 1 May 2010 14:40

On May 1, 12:22 pm, MoeBlee <jazzm...(a)hotmail.com> wrote:
> On May 1, 9:42 am, zuhair <zaljo...(a)gmail.com> wrote:
>
> > The quantifiers need not be actually written if writing them
> > makes no difference to the meaning of the formula
>
> > Example: writing the Extensionality axiom (see below)
>
> [...]
>
> > Extensionality: zx<>zy > x=y
>
> WRONG
>
> Az(zex <-> zey) -> x=y
>
> is NOT equivalent with
>
> Az((zex <-> zey) -> x=y)
>
> Please get a book on the basics of this subject.
>
> MoeBlee

Ok Moe, sorry for the above replies, the following is the correct one:

All right, my way of writing it was correct, actually, but I myself
had forgotten the rational behind it.

zx<>zy>y=x

this is read as for all z ( z in x iff z in y ) -> x=y, which is
Extensionality.

there is nothing wrong with the notation.

The reason here is because I am using quantification itself as a flow
controller.

To make matters clearer zx<>zy>y=x CANNOT be used to symbolize

for all z (( z in x iff z in y) -> x=y) as you thought!

this is actually written as zx<>zy.>x=y

the reason is as you see there is no dot notation whatsoever in
zx<>zy>y=x

so if we say that this is taken to represent
for all z (( z in x iff z in y) -> x=y), then the question wold
be:which connective we work first? is it the bi-conditional or the
implication?
as you see this would be "undetermined" because there is no dot
notation
on the left of any connective!

The only way zx<>zy>y=x would make sense is if the quantification by z
would end before the implication, since by then zx<>zy would be
considered
as one block, therefore obviating the need for dot notation.

I agree that this is complex somehow though.

Zuhair

From: zuhair on 1 May 2010 14:51

On May 1, 1:40 pm, zuhair <zaljo...(a)gmail.com> wrote:
> On May 1, 12:22 pm, MoeBlee <jazzm...(a)hotmail.com> wrote:
>
>
>
>
>
> > On May 1, 9:42 am, zuhair <zaljo...(a)gmail.com> wrote:
>
> > > The quantifiers need not be actually written if writing them
> > > makes no difference to the meaning of the formula
>
> > > Example: writing the Extensionality axiom (see below)
>
> > [...]
>
> > > Extensionality: zx<>zy > x=y
>
> > WRONG
>
> > Az(zex <-> zey) -> x=y
>
> > is NOT equivalent with
>
> > Az((zex <-> zey) -> x=y)
>
> > Please get a book on the basics of this subject.
>
> > MoeBlee
>
> Ok Moe, sorry for the above replies, the following is the correct one:
>
> All right, my way of writing it was correct, actually, but I myself
> had forgotten the rational behind it.
>
> zx<>zy>y=x
>
> this is read as for all z ( z in x iff z in y ) -> x=y, which is
> Extensionality.
>
> there is nothing wrong with the notation.
>
> The reason here is because I am using quantification itself as a flow
> controller.
>
> To make matters clearer zx<>zy>y=x CANNOT be used to symbolize
>
> for all z (( z in x iff z in y) -> x=y) as you thought!
>
> this is actually written as zx<>zy.>x=y
>
> the reason is as you see there is no dot notation whatsoever in
> zx<>zy>y=x
>
> so if we say that this is taken to represent
> for all z (( z in x iff z in y) -> x=y), then the question wold
> be:which connective we work first? is it the bi-conditional or the
> implication?
> as you see this would be "undetermined" because there is no dot
> notation
> on the left of any connective!
>
> The only way zx<>zy>y=x would make sense is if the quantification by z
> would end before the implication, since by then zx<>zy would be
> considered
> as one block, therefore obviating the need for dot notation.
>
> I agree that this is complex somehow though.
>
> Zuhair

To add to that, still the idea of Exhaustive quantification holds
also.
I thought it want work but actually it works.

To correct some earlier errors:

Example:

t _x yt::.<> w _k uw::<> u is a wiener ordered pair
i _s,r isru > i subset k:
j _p,q jpqu 0eq. > j=x:.
> yw

is the abbreviation of

for all t Exist x for all y ( y e t <-> for all w
( Exist k for all u ( u e w <->
(u is a wiener ordered pair &
for all i (Exist sr iesereu -> i subset k)&
for all j (Exist pq (jepeqeu & 0eq) -> j=x)))
-> yew)).

so we have 5 dot notations, abbreviating 14 bracket!

Zuhair

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