[m-users.] Confusion with inferencing for determinism

[artem.kozlov at posteo.net]

Hi Everyone,

I'm trying to learn Mercury and along the way decided to work on "99 
Prolog Problems". And immediately get confused with inferencing for the 
determinism. I'm trying to write my_last predicate which should return 
last element of a list.
The first immediate solution that I tried was:
my_last([X], X) .
my_last([_ | XS], X) :- my_last(XS, X) .

Which works OK but Mercury infers that my_last is nondet, but it should 
be semidet because it doesn't cover empty list case, but otherwise 
produce only one result. I looked into Mercurly standard lib where last 
predicate indeed has semidet determinism. I tried to ask in IRC and 
people suggested that I should have only one clause for predicate, but 
apparently that is not the main issue with my version. Even if I rewrite 
my function to:

my_last(YS, X) :- (YS = [X] ; YS = [_ | XS], my_last(XS, X)) .

it still nondet. In comparison to built in one:

my_last([H | T], Last) :- ( T = [], Last = H ; T = [_ | _], my_last(T, 
Last)).

which is semidet.

Maybe that is a silly question, but could someone explain how these 
versions of my_last are different and why only the last one is inferred 
as semidet.

Thanks,

Artem