我是prolog的新手，我有一个列出员工及其主管的分层数据库：
employee(employee1, employee4).
employee(employee2, employee4).
employee(employee3, employee4).
employee(employee4, employee8).
employee(employee5, employee7).
employee(employee6, employee7).
employee(employee7, employee8).
employee(employee8, employee9).
employee(employee9, employee10).
我想建立一个名为coworker（A，B）的谓词，该谓词列出某个雇员拥有的所有同事，包括其上司。
coworker(Who, employee1).
Who = employee2 ?;
Who = employee3 ?;
Who = employee4 ?;
Who = employee5 ?;
Who = employee6 ?;
Who = employee7 ?;
Who = employee8 ?;
Who = employee9 ?;
Who = employee10 ?;
到目前为止，我的代码是：
supervisor(A,B):-employee(B,A).
boss(A,B):-supervisor(A,B).
boss(A,B):-supervisor(A,C),boss(C,B).
coworker(A,B):-boss(C,A),boss(C,B).
coworker(A,B):-employee(A,B).
coworker(A,B):-supervisor(A,B).
coworker(A,B):-boss(X,Y).
但是，我很难完全摆脱同事的所有可能重复。我该怎么做才能解决此问题？