fund with findall

Given
foo(1, 2).
foo(3, 2).
foo(1, 10).

funky(Src, Dest, Matches) :-
	findall(x(Src, Dest), foo(Src, Dest), Matches).

Then funky(1, B, M) produces

M = [x(1, 2), x(1, 10)].

I find this slightly disturbing because it means that Dest got bound to
2 separate values, while remaining unbound at the end.

If anyone could help me understand a bit better what's going on, I would
appreciate it.  The behavior is very useful since it works with Src or
Dest is instantiated, but I'm a little surprised it works.

Thanks.
Ross Boylan