这个问题完全不清楚：是关于合并排序列表吗？排序的数字列表，也许？它是关于一种只保留第二个列表的第一个列表/前缀的共享后缀的副本的附加吗？是不是更一般意义上的删除重复项？
这是一个删除共享后缀/前缀的解决方案。它类似于 slago 的解决方案，除了它使用两个谓词来表示计算可能处于的两种不同状态：merge“复制”从第一个参数到第三个参数的元素；在某些时候它切换到mergerest，它“继续复制”但要求它的第一个参数是第二个参数的非空前缀。
merge(Xs, Ys, Zs) :-
mergerest(Xs, Ys, Zs).
merge([X|Xs], [Y|Ys], [X|Zs]) :-
merge(Xs, [Y|Ys], Zs).
mergerest([X], [X|Ys], [X|Ys]).
mergerest([X|Xs], [X|Ys], [X|Zs]) :-
mergerest(Xs, Ys, Zs).
使用 false 的动物定义：
?- animal(A), animal(B), dif(A, Mutation), merge(A, B, Mutation).
A = [a,l,l,i,g,a,t,o,r],
B = [t,o,r,t,u,e],
Mutation = [a,l,l,i,g,a,t,o,r,t,u,e] ;
A = [c,a,r,i,b,o,u],
B = [o,u,r,s],
Mutation = [c,a,r,i,b,o,u,r,s] ;
A = [c,h,e,v,a,l],
B = [a,l,l,i,g,a,t,o,r],
Mutation = [c,h,e,v,a,l,l,i,g,a,t,o,r] ;
A = [c,h,e,v,a,l],
B = [l,a,p,i,n],
Mutation = [c,h,e,v,a,l,a,p,i,n] ;
A = [v,a,c,h,e],
B = [c,h,e,v,a,l],
Mutation = [v,a,c,h,e,v,a,l] ;
false.
仅绑定第三个参数的行为，例如：
?- merge(Xs, Ys, [1, 2, 3]).
Xs = [1],
Ys = [1, 2, 3] ;
Xs = [1, 2],
Ys = [1, 2, 3] ;
Xs = Ys, Ys = [1, 2, 3] ;
Xs = [1, 2],
Ys = [2, 3] ;
Xs = [1, 2, 3],
Ys = [2, 3] ;
Xs = [1, 2, 3],
Ys = [3] ;
false.
更笼统地说：
?- length(Zs,_), merge(Xs, Ys, Zs).
Zs = Xs, Xs = Ys, Ys = [_4262] ;
Zs = Ys, Ys = [_4262, _4268],
Xs = [_4262] ;
Zs = Xs, Xs = Ys, Ys = [_4262, _4268] ;
Zs = Xs, Xs = [_4262, _4268],
Ys = [_4268] ;
Zs = Ys, Ys = [_4262, _4268, _4274],
Xs = [_4262] ;
Zs = Ys, Ys = [_4262, _4268, _4274],
Xs = [_4262, _4268] ;
Zs = Xs, Xs = Ys, Ys = [_4262, _4268, _4274] ;
Zs = [_4262, _4268, _4274],
Xs = [_4262, _4268],
Ys = [_4268, _4274] ;
Zs = Xs, Xs = [_4262, _4268, _4274],
Ys = [_4268, _4274] ;
Zs = Xs, Xs = [_4262, _4268, _4274],
Ys = [_4274] ;
Zs = Ys, Ys = [_4262, _4268, _4274, _4280],
Xs = [_4262] ;
Zs = Ys, Ys = [_4262, _4268, _4274, _4280],
Xs = [_4262, _4268] . % ad nauseam
快速失败：
?- merge([a|_],_,[b|_]).
false.