[sniper079]

Hi,

Can somebody help me with this question please, since I can't figure it out how to work it.

Assume that the following rules and facts have been given and that they have been given in this order:
1.	   Bigger(cat,apple).
2.	   Bigger(house,truck).
3.	   Bigger(cat,cat).
4.	   Bigger(lion,cat).
5.	   Bigger(house,lion).
6.	   Bigger(truck,car).
7.	   Bigger_than(A,B):-bigger(A,B).
8.	   Bigger_than(A,B):-bigger(A,C),bigger_than(C,B).

What is the result of executing the query bigger_than(truck,lion)?

This post has been edited by sniper079: 12 April 2009 - 02:19 PM

[mostyfriedman]

this will give you an error, since all the predicates start with a capital letter, in prolog if a word starts with a capital letter then its interpreted as a variable and the name of a predicate cant be a variable.

[sniper079]

sniper079, on 12 Apr, 2009 - 09:11 AM, said:

Hi,

Can somebody help me with this question please, since I can't figure it out how to work it.

Assume that the following rules and facts have been given and that they have been given in this order:
1.	   Bigger(cat,apple).
2.	   Bigger(house,truck).
3.	   Bigger(cat,cat).
4.	   Bigger(lion,cat).
5.	   Bigger(house,lion).
6.	   Bigger(truck,car).
7.	   Bigger_than(A,B):-bigger(A,B).
8.	   Bigger_than(A,B):-bigger(A,C),bigger_than(C,B).

What is the result of executing the query bigger_than(truck,lion)?

but what happens the the query bigger_than(truck,lion)?
I've tried to work it with a prolog programm and this is the result that I'm having, and I'm not sure what I'm doing wrong.
 ?- bigger_than(truck,lion).
ERROR: toplevel: Undefined procedure: bigger_than/2 (DWIM could not correct goal)

[mostyfriedman]

bigger_than is not defined in your knowledge base, you have to define the predicate first

This post has been edited by mostyfriedman: 14 April 2009 - 08:50 AM