1
目前正在進行一些PROLOG練習 - 對此很新，所有這些都與我相關。我有以下的知識基礎：prolog - 提取信息
/* The structure of a subject teaching team takes the form:
team(Subject, Leader, Non_management_staff, Deputy).
Non_management_staff is a (possibly empty) list of teacher
structures and excludes the teacher structures for Leader and
Deputy.
teacher structures take the form:
teacher(Surname, Initial,
profile(Years_teaching,Second_subject,Club_supervision)).
Assume that each teacher has his or her team's Subject as their
main subject. */
team(computer_science,teacher(may,j,profile(20,ict,model_railways)),
[teacher(clarke,j,profile(32,ict,car_maintenance))],
teacher(hamm,p,profile(11,ict,science_club))).
team(maths,teacher(vorderly,c,profile(25,computer_science,chess)),
[teacher(o_connell,d,profile(10,music,orchestra)),
teacher(brankin,p,profile(20,home_economics,cookery_club))],
teacher(lynas,d,profile(10,pe,football))).
team(english,teacher(brewster,f,profile(30,french,french_society)),
[ ],
teacher(flaxman,j,profile(35,drama,debating_society))).
team(art,teacher(lawless,m,profile(20,english,film_club)),
[teacher(walker,k,profile(25,english,debating_society)),
teacher(brankin,i,profile(20,home_economics,writing)),
teacher(boyson,r,profile(30,english,writing))],
teacher(carthy,m,profile(20,music,orchestra))).
subject(X):- team(X,_,_,_).
leader(X) :- team(_,X,_,_).
deputy(X) :- team(_,_,_,X).
non_management(X) :-
team(_,_,Non_management,_),
member(X,Non_management).
exists(X) :-
subject(X);
leader(X);
deputy(X);
non_management(X).
我現在已經寫了一個規則，（Q）老師和教師B的縮寫，其中教師A和教師B是不同學科小組，每個小組有家政作爲他們的第二個 科目，並且每個都有姓布蘭金。
我被困在如何比較知識庫中的所有實體。在此之前，我只提取單個實體的值（在這個例子中是單個教師）。例如：
question1(Initial,Surname) :-
exists(teacher(Surname,Initial,profile(_,english,_))).
任何幫助非常感謝。