🚀 go-pugleaf

Arity Prolog, given the call

findall(X,member(X,[Y,Z]),S).

responds

X = _0034
Y = _0048
Z = _005C
S = [_00FC,00EC]

This is the same result you find surprising on the TI.  What you
seem to be surprised by is the different variables in the list 
formed in S.

Arity says, "The findall predicate is similar to bagof, except
that any free variable is assumed to be existentially quantified."
Given that, the above response strikes me as perfectly logical.

Think of what the query is.  It asks: What would I find if I
looked for anything that was a member of a list which contained
something followed by something?
	The answer: You would get something and something.

The fact that these are different variables in the answer points
out the fact that bound variables in a different scope are not
to be identified even if they are typographically the same.  In
logic classes it is important to explain that (Ex)policeman(x)
and (Ex)murderer(x) does not given the conclusion that some
policeman is a murderer.  It is safer, therefore, to write
(Ex)policeman(x) and (Ey)murderer(y).

If, therefore, you had a list containing some policeman followed
by some policeman (Ex)policeman(x),(Ey)policeman(y), and then
you asked what you would find in that list, the answer:
(Ev)policeman(v), (Ew)policeman(w) would be correct.  Since
Prolog does not include the quantifiers indicating scope, the
use of divergent variables is almost required.

bob wengert		wengert@clio.las.uiuc.edu