Question

I’ve just starting working with Prolog and I don’t understand how to work with multiple predicates.
For example I have to solve the following problem: Substitute in a list a value with all the elements of another list.
This is the code that I managed to write so far:

domains
    elem=integer
    list=elem*

predicates
    %append to a list already created another list. There are 3 list parameters 
    %because I don't know other method 
    append (list,list,list)
    %This will search the list (the first one) for the searched element and   
    %it is it will replace it with the list(the second one). The result will be
    %kept in the third list.
    add(list,list,elem,list)

goal
    add([1,2,3,2,1],[4,5],2,L),
    write (L).  
clauses
    add ([],[_],_,[]).
    add ([A|L],L1,E,[A|L2]):-
        add(L,L1,E,L2).
    add ([E|L],L1,E,L2):-
        add(L,L1,E,L2).
    append([],[],L2).
    append([],[X|L1],[X|L2]):-
        append([],L1,L2).

Answer 1

Does your append definition is working? I think should be

append([], L, L).
append([X|Xs], Ys, [X|Zs]):-
        append(Xs, Ys, Zs).

The append predicate it’s one of the most basic tools in Prolog programming, better to keep the usual behaviour, or to change name…

Instead of add, a better name could be replace_elem_with_list.
To implement it you should iterate, inspecting each element, and when you find a match to what’s required to replace append the list instead of copying the element.

Something like

% replace_elem_with_list(ToSearch, Replacement, SoughtElem, Result)
replace_elem_with_list([E|Es], Replacement, E, Result) :-
  !, replace_elem_with_list(Es, Replacement, E, Rs),
  append(Replacement, Rs, Result).

I’ll leave you the other 2 cases that you will need to cover (when element doesn’t match and recursion base, which are similar to append)

the result:

?- replace_elem_with_list([1,2,3,4,2,3,4],[a,b],2,L).
L = [1, a, b, 3, 4, a, b, 3, 4].