目前正在做一些 PROLOG 练习——对这一切都很陌生，所以请多多包涵。我有以下知识库：
/* 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).
我现在要写一条规则，(q) 老师 A 和老师 B 的首字母，其中老师 A 和老师 B 在不同的学科组，每个人都以家政学作为第二个学科，每个人都姓 Brankin。
我被困在如何比较知识库中的所有实体上。在此之前，我只从单个实体（在本例中为单个教师）提取值。例如：
question1(Initial,Surname) :-
exists(teacher(Surname,Initial,profile(_,english,_))).
非常感谢任何帮助。