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我想在prolog中做一個謂詞，它將我給出的值替換爲多項式的變量，然後計算結果。這裏是我的代碼：爲什麼不統一？ Prolog問題
as_monomial(X, m(X, 0, [])) :- number(X), !.
as_monomial(^(Y, Z), m(1, Z, [v(Z, Y)])) :- !.
as_monomial(*(X, ^(Y, Z)), m(G, K, Q)) :- as_monomial(X, m(G, TD, Vars)), K is (TD + Z), compress_monomial([v(Z, Y)| Vars], A), ordina_m(A, Q), !.
as_monomial(*(X, Y), m(G, K, Q)) :- as_monomial(X, m(G, TD, Vars)), K is (TD + 1), compress_monomial([v(1, Y)| Vars], A), ordina_m(A, Q), !.
as_monomial(-(X), m(-A, Y, L)) :- as_monomial(X, m(A, Y, L)).
as_monomial(X, m(1, 1, [v(1, X)])).
ordina_m(List, Sorted) :- sort(2, @=<, List, Sorted).
ordina_var(List, Sorted) :- sort(0, @=<, List, Sorted).
compress_monomial([], []) :- !.
compress_monomial([X| Xs], A2) :- compress_monomial(Xs, A), compress_monomial2(X, A, A2), !.
is_monomial(m(_C, TD, VPs)) :- integer(TD), TD >= 0, is_list(VPs).
is_polynomial(poly(M)) :- is_list(M), foreach(member(Monomio, M), is_monomial(Monomio)).
variables(Poly1, Result) :- is_polynomial(Poly1), variabili(Poly1, Result), !.
variables(Poly1, Result) :- as_polynomial(Poly1, Result1), variabili(Result1, Result), !.
variabili(poly([]), []) :- !.
variabili(poly([m(_, _, [])| Xs]), Ys) :- variabili(poly(Xs), Ys), !.
variabili(poly([m(X, Y, [v(_, A)| Vs])| Xs]), Z) :- variabili(poly([m(X, Y, Vs)| Xs]), Ys), ordina_var([A| Ys], R), compressV(R, Z), !.
compressV([], []).
compressV([X|T],[X|T1]):- member(X,T),!,canc(X,T,R), compressV(R,T1).
compressV([X|T],[X|T1]) :- compressV(T,T1).
canc(_L, [], []).
canc(L, [L|S], Z) :- canc(L, S, Z).
canc(L, [H|S], [H|Z]):- canc(L, S, Z), !.
as_polynomial(+(X, Y), poly(C)) :- as_monomial(Y, G), as_polynomial(X, poly(Gs)), compress_polynomial([G| Gs], C), !.
as_polynomial(-(X, Y), poly(C)) :- as_monomial(-Y, G), as_polynomial(X, poly(Gs)), compress_polynomial([G| Gs], C), !.
as_polynomial(X, poly([X])) :- is_monomial(X), !.
as_polynomial(X, poly([Q])) :- as_monomial(X, Q), !.
compress_polynomial([], []) :- !.
compress_polynomial([X| Xs], A2) :- compress_polynomial(Xs, A), compress_polynomial2(X, A, A2), !.
compress_polynomial2(m(X, Y, Z), [], [m(X, Y, Z)]) :- !.
compress_polynomial2(m(X, Y, Z), [m(X1, Y, Z)| Xs], [m(X2, Y, Z)| Xs]) :- X2 is (X + X1), !.
compress_polynomial2(X, [Y| Ys], [Y| Z]) :- compress_polynomial2(X, Ys, Z), !.
polyval(Poly1, V, Result) :- is_polynomial(Poly1), variables(Poly1, Vars), poly_val(Poly1, Vars, V, Result), !.
polyval(Poly1, V, Result) :- as_polynomial(Poly1, P1), variables(P1, Vars), poly_val(P1, Vars, V, Result), !.
poly_val(poly([]), , , poly([])) :- !.
poly_val(poly([m(X, Y, Z)| Xs]), Vars, V, poly([R| Ys])) :- poly_val(poly(Xs), Vars, V, poly(Ys)), print(m(X, Y, Z)), mon_val(m(X, Y, Z), Vars, V, R), !.
mon_val(m(X, Y, []), [_], [_], m(X, Y, [])) :- !.
mon_val(m(X, Y, [v(W, Z)| Vs]), [Z| Vs2], [Val| Vvs], m(X2, Y2, Z2)) :- integer(Val), mon_val(m(X, Y, Vs), Vs2, Vvs, m(X3, Y2, Z2)), X2 is (X3 * (Val^W)), !.
mon_val(m(X, Y, [v(W, Z)| Vs]), [_| Vs2], [_| Vvs], m(X, Y2, Z2)) :- mon_val(m(X, Y, [v(W, Z)| Vs]), Vs2, Vvs, m(X, Y3, Z2)), Y2 is (Y3 + W), !.
我希望我把你需要證明的所有代碼，以防萬一請告訴我，我爲此道歉。我知道切割，但目前，這只是一個試驗。我的問題是在mon_val，因爲它看起來不想統一。我使用的查詢的一個例子是polyval（x + x + y，[1,3]，Q）。其中輸出爲「false」，它應該返回poly（m（1,0）[m]，m（1,0），m（3,0，[]））。你能幫我做嗎？我只是想解決這個問題，後來我還會實現這些數字之間的和數，這些數字相當簡單，其餘的代碼我都有。謝謝你們