# RE: Any and Every... (was Re: Eval)

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From: "KELLEHER,KEVIN (Non-HP-Roseville,ex1)" <kevin_kelleher@non.hp.com>
To: ll1-discuss@ai.mit.edu
Subject: RE: Any and Every... (was Re: Eval)
Date: Tue, 8 Jan 2002 15:06:52 -0800

"Any" is not the union of all types,
but one member from the group of all types.
"Every" is all members.

If you are looking for "any", you are looking for one.

If you are looking for "every", you are looking for many.

The phrases "the union of all types" and "the intersection
of all types" do not make sense.  You have to have more
than one group to have unions and intersections,
but there is only one group: types.

Are you using "group" in its mathematical sense of
a set of elements plus a binary operation that maps
every ordered pair of elements to an element in the set,
such that the operation is associative, there is some
element that is an identity for the binary operation,
and for every element there is some element (calle`d the
inverse of the first element) such that the operation
maps either pair of these elements to the identity?

Or are you using the word "group" as a loose (and confusing)
synonym for "set"?

Or what?

--Guy

```