On Mon, Mar 30, 2009 at 9:44 PM, Martin D Kealey <mar...@kurahaupo.gen.nz> 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 == $b == 0, but I'm not sure the LHS shouldn't also be. Isn't it just syntactic sugar for the RHS? Logically, you might want it to mean something like ∃$x: $x == any(-1,+1) && $a <= $x && $x <= $b, but I don't think it does. -- Mark J. Reed <markjr...@gmail.com>
