Evaluating expressions with indexicals in SICStus Prolog - prolog

Question

I'm trying to write a predicate in SICStus Prolog such that given an expression, I can evaluate it (possible several times). The following works as expected:
?- A is 1, H = A+2+2, C is H.
C = 5 ?
yes
And so does this more elaborate code:
testing(Variables, Updates, Values, Result):-
assert(temp(Variables, Updates)),
temp(Values, Result),
abolish(temp/2).
evaluate([],[]).
evaluate([Term|T1],[H|T2]):-
H is Term,
evaluate(T1,T2).
Now, if instead A is an indexical, say A in 1..3, it doesn't work anymore. Any ideas about how to fix it?
The longer code is supposed to be used as follows:
?- testing([A,B,C], [A+1,B+C,max(A,B)], [0,0,0], Result), evaluate(Result, R).
Result = [0+1,0+0,max(0,0)],
R = [1,0,0] ?
yes
But is suffers from the same problem as the small example: I can't provide ranges of values in this way:
?- Val1 in 1..2, Val2 in 3..10, testing([A,B], [A+1,B+A], [Val1,Val2], Result), evaluate(Result, R).
Any suggestions?

Answer 1

My current solution requires replacing is/2 with val_of/2. It works, but I still believe there should be a better/faster implementation.
testing(Variables, Updates, Values, Result):-
assert(temp(Variables, Updates)),
temp(Values, Result),
abolish(temp/2).
evaluate([],[]).
evaluate([Term|T1],[H|T2]):-
val_of(H,Term),
evaluate(T1,T2).
val_of(E,E):- number(E),!.
val_of(E,E):- var(E),!.
val_of(V,E1+E2):- !, val_of(V1,E1), val_of(V2,E2), V #= V1+V2.
val_of(V,E1-E2):- !, val_of(V1,E1), val_of(V2,E2), V #= V1-V2.
val_of(V,max(E1,E2)):- !, val_of(V1,E1), val_of(V2,E2), V #= max(V1,V2).
val_of(V,min(E1,E2)):- !, val_of(V1,E1), val_of(V2,E2), V #= min(V1,V2).
val_of(V,abs(E1,E2)):- !, val_of(V1,E1), val_of(V2,E2), V #= abs(V1,V2).
A test example:
| ?- X in 1..3, testing([A,B], [A+1,B], [X,0], R), evaluate(R,R1).
R = [X+1,0],
R1 = [_A,0],
X in 1..3,
_A in 2..4 ?
yes

Answer 2

I think that all you need is just
evaluate([],[]).
evaluate([Term|T1],[H|T2]):-
H #= Term,
evaluate(T1,T2).
but the temp/2 relation is unnecessary, so a real simplification could be:
testing(Variables, Updates, Values, Result):-
maplist(#=, Updates, Values), Result=Variables.
results in
?- testing([A,B,C], [A+1,B+C,max(A,B)], [0,0,0], Result).
A = -1,
B = C, C = 0,
Result = [-1, 0, 0].
(note: tested in SWI-Prolog, after ?- [library(clpfd)].)

Answer 3

My final solution is a modified version of my original code based on the useful answers and comments provided by #CapelliC and #false:
testing(Variables, Updates, Values, Result):-
copy_term(Variables-Updates, Values-Result).
evaluate([],[]).
evaluate([Term|T1],[H|T2]):-
call(H #= Term),
evaluate(T1,T2).
The main issue in my original code was the missing call/1 in evaluate/2.
A test example in SICStus Prolog looks like this:
?- A in 1..3, testing([C,D,R],[C+1,max(D,5),R],[A,0,0],Res), evaluate(Res,T).
Res = [A+1,max(0,5),0],
T = [_A,5,0],
A in 1..3,
_A in 2..4 ?
yes