On Mon, 30 Mar 2009, Mark J. Reed wrote:
> > ( $a <= any(-1,+1) <= $b ) == ( $a <= any(-1,+1) && any(-1,+1) <= $b
> > )
>
> Clearly, the RHS is true for $a == $b == 0, but I'm not sure the LHS
> shouldn't also be. Isn't it just syntactic sugar for the RHS?
I suspect not. Rather I think th
On Mon, Mar 30, 2009 at 9:44 PM, Martin D Kealey
wrote:
> This would certainly be false:
>
> ( $a <= any(-1,+1) <= $b ) == ( $a <= any(-1,+1) && any(-1,+1) <= $b )
Clearly, the RHS is true for $a == $b == 0, but I'm not sure the LHS
shouldn't also be. Isn't it just syntactic sugar for the
On Mon, 30 Mar 2009, Jon Lang wrote:
> Here's another useful one:
>
> any($x) eqv all($x) eqv one($x) eqv $x
>
> but:
>
> none($x) !eqv $x
>
> That is, applying any, all, or one to a one-item list produces the
> equivalent to a single item. For an empty list: any() eqv all() eqv
> (). But
Here's another useful one:
any($x) eqv all($x) eqv one($x) eqv $x
but:
none($x) !eqv $x
That is, applying any, all, or one to a one-item list produces the
equivalent to a single item. For an empty list: any() eqv all() eqv
(). But what about one() and none()?
--
Jonathan "Dataweaver
Richard Hainsworth conjectured:
> 1) Is the following true for an any junction?
> any( ... , any('foo','bar')) === any(...,'foo','bar')
>
> If yes, then
> if an 'any' junction is contained in an outer 'any', the inner 'any' can be
> factored out?
Yes. More precisely, an 'any' that is directly nes
On Sat, Mar 28, 2009 at 10:19:31AM -0500, Patrick R. Michaud wrote:
> On Sat, Mar 28, 2009 at 02:08:22PM +0300, Richard Hainsworth wrote:
> > 3) Conjecture: The following is true of all junction types, eg.,
> > junc(..., junc(...)) === junc(..., ...)
>
> The conjecture is false for one/none juncti
On Sat, Mar 28, 2009 at 02:08:22PM +0300, Richard Hainsworth wrote:
> 3) Conjecture: The following is true of all junction types, eg.,
> junc(..., junc(...)) === junc(..., ...)
The conjecture is false for one/none junctions:
one(0, one(1, 1)) # true
one(0, 1, 1) # fal
