prolog - Convert list into functor parameter -

i got stuck implement logic. @ instance in program have list named list. length of list variable , don't know in advance. have pass list in functor create fact , unable implement it. eg:

if list [first] should add fact functor(first).
if list [first,second] should add fact functor(first,second).
if list [first,second,third] should add fact functor(first,second,third).
, on...

i trying =.. here unable map variable length constraint. fixed length able perform don't know in advance how many elements there in list.

any suggestions implement logic. thanks.

i don't quite understand problem =.. worked me:

 assert_list(list) :-             term =.. [my_functor|list],             assert(term). 

note use my_functor instead of functor because functor/3 built-in predicate cannot assert ternary functor facts (functor(first, second, third)).

calling it:

 ?- assert_list([first,second,third]).  true. 

checking works:

 ?- listing(my_functor).  :- dynamic user:my_functor/3.   user:my_functor(first, second, third).   true. 

note technically, different n-ary my_functor/n predicates not same predicates. must use different queries in program each n. circumvent this, assert list 1 , argument of my_functor:

 ?- list = [first, second, third],  assert(my_functor(list)).  true.   ?- listing(my_functor).  :- dynamic user:my_functor/3.   user:my_functor([first, second, third]).   true. 

my swi-prolog version 5.7.5.