Abbreviating FOL with identity and membership2
in [Math]
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From: Daryl McCullough on 2 May 2010 10:05 In article <hrk0g901qot(a)drn.newsguy.com>, Daryl McCullough says... > >Jesse F. Hughes says... > >>Yes, number of characters is the most accurate way to indicate >>readability. Good job! > >When I was in college, one of my classmates liked to program in APL, whose claim >to fame was the ability to write incredibly powerful programs in just one line >of code. Some examples are in Wikipedia: >http://en.wikipedia.org/wiki/APL_%28programming_language%29 > >The following expression sorts a word list stored in matrix X according to word >length: > >X[⍋X+.≠' ';] > >This following immediate-mode expression generates a typical set of Pick 6 >lottery numbers: six pseudo-random integers ranging from 1 through 40, >guaranteed non-repeating, and displays them sorted in ascending order: > >x[⍋x←6?40] Hmm. Those lines displayed correctly in the preview. Anyway, go to the Wiki page to see examples of APL. -- Daryl McCullough Ithaca, NY
From: zuhair on 2 May 2010 12:16 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 I admit my weakness about the basics. A question regarding the basics of this subject, is the following a well formed formula in FOL(e,=): For all x ( For all y ( For all z ((z e x iff z e y) -> x=y) ) ). z is only appearing on the left of the implication, and all variables other than z are quantified *before* z, so is that acceptable? I might accept the following formula For all z ( For all x ( For all y ((z e x iff z e y) -> x=y) ) ) . since x and y are quantified *after* z, so quantification over z can extend as far as variables x and y appear in the formula, so it doesn't matter if z seize to appear after the implication. I just thought that quantification over a variable z closes after the last formula z or the last variable quantified *after* z appears in. If the answer is yes, which mean that it is a well formed formula, then in these circumstances we must modify the notation to accommodate for that, so we either extend the dot notation to cover such cases of quantification, or we simply keep the rule of exhaustive quantification (as a notation) and always wright the formula z=z on the other side to indicate that quantification extend over to the other size. Zuhair
From: Marshall on 2 May 2010 13:09 On May 2, 6:59 am, stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: > Jesse F. Hughes says... > > >Yes, number of characters is the most accurate way to indicate > >readability. Good job! > > When I was in college, one of my classmates liked to program in APL, whose claim > to fame was the ability to write incredibly powerful programs in just one line > of code. APL is a fabulous example to study for the power and generality of its operators. It's also a fabulous example to study to learn what to avoid in syntax design. Marshall
From: zuhair on 2 May 2010 17:25 On May 2, 8:21 am, zuhair <zaljo...(a)gmail.com> wrote: > On May 2, 7:51 am, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote: > > > > > > > zuhair <zaljo...(a)gmail.com> writes: > > > On May 1, 1:58 pm, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote: > > >> zuhair <zaljo...(a)gmail.com> writes: > > >> > 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 > > > >> > hmmm...., > > > >> > it seems that we must apply the dot notation to be sometimes > > >> > on the right of quantifiers also. > > > >> > like: > > > >> > z.zx<>zy:> y=x > > > >> Boy, that's *much* more readable than > > > >> Az(zex <-> zey) -> x=y > > > >> Kudos! > > > >> -- > > >> Jesse F. Hughes > > >> "[M]oving towards development meetings for new release class viewer 5.0 > > >> and since [I]'m the only developer, easy to schedule." > > >> --James S. Harris tweets on code development > > > > No Jesse, you came late. > > > > The correct way to write > > > > Az(zex <-> zey) -> x=y > > > > using this system is > > > > zx<>zy > x=y > > > > I was correct really. > > > Oh? Then how do you write > > > Az((zex <-> zey) -> x=y)? > > If you read my reply to Moe, then you'd seen the answer. > > Az((zex <-> zey) -> x=y) is written in the following way: > > zx<>zy.>x=y more precisely if you meant AxyAz((zex<->zey)->x=y) then it is written as: xy zx<>zy.>x=y if you meant Azxy((zex<->zey)->x=y) then it is written as zx<>zy.>x=y if you meant Azyx((zex<->zey)->x=y) then it is written as zy<>zx.>x=y While Az(zex<->zey)->x=y is written as zx<>zy>x=y > > you must place the dot after zy, otherwise you will end up > with a statement that is not uniquely interpretable (that is > if you take it to represent z being quantified upon over both > sides). > > the only way zx<>zy > x=y, would make sense in this > notation is if quantification over z ends before the implication > since zx<>zy would by then represent a closed sentence, > and not confusion as to its interpret-ability would rise. > > So in short, if quantification over z closes after > x=y, then in order to avoid bi-interpretability, > you must place a dot after zy; while > if quantification over z closes after zy, then you > don't need to place a dot after zy, since the issue > of bi-interpretability disappears by then. > > > > > Is the difference merely in the number of spaces? > > No, the approach here do not use space. > > I think for further reading, its better to refer to my last post on > that matter > it has the same title with number 3 instad of 2. > > > > > > > > Sorry for the confusion. > > > > And I do think that > > > > zx<>zy > x=y > > > > is by far more neat and nicer than the clumsy > > > > Az(zex <-> zey) -> x=y > > > > this have 18 characters > > > while the above have only 10, almost a half size reduction! > > > Yes, number of characters is the most accurate way to indicate > > readability. Good job! > > No, I didn't say that, I said abbreviation is measured by the number > of characters, and it was abbreviation that is the main concern > behind writing this notation. Of course with this extreme > version of abbreviation readability would be somewhat harder, this is > expected, see my reply Elliot in my last post. > > However in addition to abbreviation, I claim that this notation is not > crowded > doesn't contain odd looking symbols like an upside-down A, or > a turnaround E, or and an upside-down large V, etc..., also > it contains less repeatable symbols in them. > > I assume that all these properties would make the formulae written > here > neater, and nicer looking. > > The problem with readability of this syntax, rises from the > abbreviation power of it > and from having many rules in it (please see the last post I referred > to above). > > Zuhair > > > > > -- > > "Mathematicians are rather important in the infrastructures of many > > organizations that protect civilization. I've determined that they > > are a consistent security risk, and seem to have other agendas, other > > loyalties beyond loyalty to their respective nations." -- James Harris
From: zuhair on 2 May 2010 17:26
On May 2, 11:16 am, 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 > > I admit my weakness about the basics. > > A question regarding the basics of this subject, > > is the following a well formed formula in FOL(e,=): > > For all x ( For all y ( For all z ((z e x iff z e y) -> x=y) ) ). > > z is only appearing on the left of the implication, and all variables > other than > z are quantified *before* z, so is that acceptable? > > I might accept the following formula > > For all z ( For all x ( For all y ((z e x iff z e y) -> x=y) ) ) . > > since x and y are quantified *after* z, so quantification over z can > extend as > far as variables x and y appear in the formula, so it doesn't matter > if > z seize to appear after the implication. > > I just thought that quantification over a variable z closes after the > last formula > z or the last variable quantified *after* z appears in. > > If the answer is yes, which mean that it is a well formed formula, > then > in these circumstances we must modify the notation to accommodate for > that, so > we either extend the dot notation to cover such cases of > quantification, or we simply keep the rule of exhaustive > quantification (as a notation) and always wright the formula z=z on > the other side to indicate that quantification extend over to the > other size. To clarify the last statement, we can for example wright Axyz(zex -> y=x) the is written as: xy zx>y=x z=z so putting the formula z=z on the other side, will ensure that quantification is extending over the implication ( this is the consequence of the rule of Exhaustive quantification which is adopted in this notation system, which states that:- "quantification over a variable z closes after the last formula in which z or the last variable quantified *after* z appears in"). however lets take the statement Axyz((zex<->zey)->x=y) now this is written as: xy zx<>zy.>x=y we don't need to write: xy zx<>zy.>x=y z=z since this would be redundant. because from the dot after zy , it is clear that the scope of quantification over z is extending beyond the implication, otherwise we don't need to place a dot after zy. Zuhair |