What makes PROLOG Turing-complete?

Question

I know that it can be proven PROLOG is Turing-complete by constructing a program that simulates a Turing machine like this:

turing(Tape0, Tape) :-
    perform(q0, [], Ls, Tape0, Rs),
    reverse(Ls, Ls1),
    append(Ls1, Rs, Tape).

perform(qf, Ls, Ls, Rs, Rs) :- !.
perform(Q0, Ls0, Ls, Rs0, Rs) :-
    symbol(Rs0, Sym, RsRest),
    once(rule(Q0, Sym, Q1, NewSym, Action)),
    action(Action, Ls0, Ls1, [NewSym|RsRest], Rs1),
    perform(Q1, Ls1, Ls, Rs1, Rs).

symbol([], b, []).
symbol([Sym|Rs], Sym, Rs).

action(left, Ls0, Ls, Rs0, Rs) :- left(Ls0, Ls, Rs0, Rs).
action(stay, Ls, Ls, Rs, Rs).
action(right, Ls0, [Sym|Ls0], [Sym|Rs], Rs).

left([], [], Rs0, [b|Rs0]).
left([L|Ls], Ls, Rs, [L|Rs]).

Source

However, I’m wondering which parts of the PROLOG language one could strip away (esp. function symbols, clause overloading, recursion, unification) without losing Turing completeness. Are function symbols themselves Turing complete?