The pipe operator in prolog returns one or more atomic Heads and a Tail list.
?- [a,b,c] = [a,b|[c]].
true.
Nesting multiple pipes in a single match can be done similar to this:
?- [a,b,c] = [a|[b|[c]]].
true.
What does the statement [a|b|c] infer about a, b and c?
EDIT
So far, all I can deduce is:
?- [a,b,c] = [a|b|c].
false.
I am more interested in any techniques to find the answer rather than to answer this borderline useless question.
EDIT2
I'm clearly not too familiar with prolog, a simple assignment answered my question...
?- R = [a|b|c].
R = [a| (b'|'c)].
What exactly is going on with (b'|'c)?
Since I'm your lecturer I, here's my answer.
(Oh, and I can confirm that this isn't homework, it's related to the practice exam).
The syntax [a|b|c] actually doesn't seem to be standard Prolog, and some implementations interpret it differently. (Had I known this I might not have used it.)
Some interpret it as [a|[b|c]]. (As I intended.)
But with SWI Prolog (and probably others):
?- [a|b|c] = [a|[b|c]].
false.
The (b '|' c) is actually constructed using '|' rather than '.' as a list would be. So, the the second | is not interpreted as being part of constructing a list at all.
To confirm this, the following succeeds:
?- X=(b|c), [a|b|c] = [a|X]. X = (b'|'c) .
The '|' here seems to be a another binary operator on terms just like '.'.
Instead of [a|b|c] the standard in Prolog is to use [a,b|c].
(I only decided to use [a|b|c] in Programming Paradigms because it more directly relates to the notation a::b::c from F#, and we only saw a tiny glimpse of Prolog. In future I guess I'll relate to [a|[b|c]] and then give [a,b|c] as an abbreviation.)