序言 - 试图解决文本谜题

Question

我目前正在尝试学习一点Prolog。作为一个练习我试图解决以下几个谜语：

这些规则给出：

*Every person that has neither a car nor a plane, has a bike. 
*Every person that doesn't have a plane but has a bike, has a car 
*Every person that doesn't have a plane but has a car, has a truck 
*Every person that doesn't have a truck but has a boat, doesn't have a plane 
*Every person that doesn't have a boat but has a plane, doesn't have a car 

现在有四个人：

*Person1 doesn't have a car but has a boat 
*Person2 doesn't have a boat but has a plane 
*Person3 doesn't have a plane but has a bike 
*Person4 doesn't have a bike but has a car 

哪个人没有一辆卡车？

我想出来的，到目前为止是这样的：

doesnthave(car,pa). 
has(boat,pa). 
doesnthave(boat,pb). 
has(plane,pb). 
doesnthave(plane,pc). 
has(bike,pc). 
doesnthave(bike,pd). 
has(car,pd). 

has(bike,X) :- doesnthave(car,X),doesnthave(plane,X). 
has(car,X) :- doesnthave(plane,X),has(bike,X). 
has(truck,X) :- doesnthave(plane,X),has(car,X). 
doesnthave(plane,X) :- doesnthave(truck,X),has(boat,X). 
doesnthave(car,X) :- doesnthave(boat,X),has(plane,X). 

现在，这似乎并不足够。或者这不是在prolog中解决像这样的难题的方法吗？

编辑：看起来前两个陈述是矛盾的。它们一起产生：每个既没有汽车也没有飞机的人拥有一辆汽车。我不确定是否有明智的解决方案。

Answer 1

不知道这个解决方案，但知识的简单的表示一定帮助：

has(car, pa, n). has(boat, pa, y). 
has(boat, pb, n). has(plane, pb, y). 
has(plane, pc, n). has(bike, pc, y). 
has(bike, pd, n). has(car, pd, y). 

has(bike, X, y) :- has(car, X, n), has(plane, X, n). 
has(car, X, y) :- has(plane, X, n), has(bike, X, y). 
has(truck, X, y) :- has(plane, X, n), has(car, X, y). 
has(plane, X, n) :- has(truck, X, n), has(boat, X, y). 
has(car, X, n) :- has(boat, X, n), has(plane, X, y). 

现在我们可以查询有什么所有制（即注意这是一个一对多的关系）

?- setof((P,T,R), has(T,P,R), L), maplist(writeln, L). 
pa,boat,y 
pa,car,n 
pb,boat,n 
pb,car,n 
pb,plane,y 
pc,bike,y 
pc,car,y 
pc,plane,n 
pc,truck,y 
pd,bike,n 
pd,car,y 
L = [ (pa, boat, y), (pa, car, n), (pb, boat, n), (pb, car, n), (pb, plane, y), (pc, bike, y), (pc, car, y), (pc, ..., ...), (..., ...)|...]. 

注意我放置P erson之前T运输，然后我们可以目测检查结果...

似乎t他的解决方案是一个三重...