你的代码有几个问题:

• filter([],[],0). 在处理任何不以 0 作为最小值的列表时不会统一，这不是您想要的.无论结束递归的最小值如何，您都希望它统一.
• 您编写 filter([H1|T1],[H2|T2],Z) 的方式及其主体将使两个列表始终具有相同数量的元素，当事实上，第二个应该至少少一个.

• filter([],[],0). will not unify when working with any list that does not have 0 as its minimum value, which is not what you want. You want it to unify regardless of the minimum value to end your recursion.
• The way you wrote filter([H1|T1],[H2|T2],Z) and its body will make it so that the two lists always have the same number of elements, when in fact the second one should have at least one less.

filter/3 的正确实现如下:

filter([],[],_).
filter([H1|T1],L2,Z):-
    + number(H1),
    write("ERROR: List parameter contains a non-number element"),
    !;
    H1 = Z -> filter(T1,T2,Z), L2 = [H1|T2];
    filter(T1,L2,Z).