Abbreviating FOL with identity and membership2
in [Math]
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From: zuhair on 1 May 2010 14:12 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
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
From: Jesse F. Hughes on 1 May 2010 14:58
zuhair <zaljohar(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 |