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Predicate nth0/4
Availability::- use_module(library(lists)).(can be autoloaded)
Source[det]nth0(?N, ?List, ?Elem, ?Rest)
Select/insert element at index. True when Elem is the N’th (0-based) element of List and Rest is the remainder (as in by select/3) of List. For example: 
?- nth0(I, [a,b,c], E, R).
I = 0, E = a, R = [b, c] ;
I = 1, E = b, R = [a, c] ;
I = 2, E = c, R = [a, b] ;
false.
?- nth0(1, L, a1, [a,b]).
L = [a, a1, b].
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LogicalCaptain said (2020-04-04T16:19:29):0 upvotes 0 0 downvotes
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Replacing an element in a list (with 0-based indexing)

from https://stackoverflow.com/questions/61028457/an-easier-way-to-split-a-list-at-the-n-th-place-in-prolog/61029588

replace_nth0(List, Index, OldElem, NewElem, NewList) :-
   % predicate works forward: Index,List -> OldElem, Transfer
   nth0(Index,List,OldElem,Transfer),
   % predicate works backwards: Index,NewElem,Transfer -> NewList
   nth0(Index,NewList,NewElem,Transfer).
?- replace_nth0([a,b,c,d], 0, Oe, replaced(Oe), NewList).
Oe = a,
NewList = [replaced(a), b, c, d].

?- replace_nth0([a,b,c,d], 1, Oe, replaced(Oe), NewList).
Oe = b,
NewList = [a, replaced(b), c, d].

?- replace_nth0([a,b,c,d], 2, Oe, replaced(Oe), NewList).
Oe = c,
NewList = [a, b, replaced(c), d].

?- replace_nth0([a,b,c,d], 3, Oe, replaced(Oe), NewList).
Oe = d,
NewList = [a, b, c, replaced(d)].