A simple FD example

so.prolog

subject(t1, math).
subject(t2, math).
subject(t2, english).
subject(t3, sports).
subject(t3, art).
subject(t3, english).
subject(t4, math).
subject(t4, art).
subject(t5, sports).
subject(t5, math).
subject(t5, english).

hours(math, 12).
hours(english, 8).
hours(art, 4).
hours(sport, 2).

p(A, B, A-B).

filterS(_, [], []).
filterS(S, [_-S-X|Xs], [X|Ys]) :- !, filterS(S, Xs, Ys).
filterS(S, [_|Xs], Ys) :- filterS(S, Xs, Ys).


% findall(T-S, subject(T, S), X), length(X, N), length(Y, N), fd_domain(Y, 2, 3), maplist(pair, X, Y, Z).

%% constraint_subject_hours(Subject, L) :-
%%     hours(Subject, H),
%%     filterS(Subject, L, Vars),
%%     fd_domain(Vars,0,3).

% A direct modelling of the specific problem  
soln(X) :-
    X = [T1M, T2M, T2E, T3S, T3A, T3E, T4M, T4A, T5S, T5M, T5E],
    fd_domain(X, 0, 10),
    %% subject constraints
    T1M + T2M + T4M + T5M #= 12,
    T2E + T3E + T5E #= 8,
    T4A + T3A #= 4,
    T3S + T5S #= 2,

    fd_domain(T, 0, 10),
    T #=< T1M, T1M #=< T+1,
    T #=< T2M + T2E, T2M + T2E #=< T+1,
    T #=< T3S + T3E + T3A, T3S + T3E + T3A #=< T+1,
    T #=< T4M + T4A, T4M + T4A #=< T+1,
    T #=< T5S + T5M + T5E, T5S + T5M + T5E #=< T+1,
    fd_labeling(X).