Notes on computational linguistics.pdf - UCLA Department of ...

Stabler - Lx 185/209 2003
L0 and L1 is L2. These two axioms entail that there is a list L which is the result of appending [the,cat,is]
with [on,the,mat]. 9 Prolog can prove this fact:
| ?- append([the,cat,is],[on,the,mat],L).
L = [the,cat,is,on,the,mat] ? ;
no
The proof of the goal,
append([the,cat,is],[on,the,mat],[the,cat,is,on,the,mat])
can be depicted by the following proof tree:
append([the,cat,is],[on,the,mat],[the,cat,is,on,the,mat])
append([cat,is],[on,the,mat],[cat,is,on,the,mat])
append([is],[on,the,mat],[is,on,the,mat])
append([],[on,the,mat],[on,the,mat])
This axiomatization of append behaves nicely on a wide range of problems. It correctly rejects
| ?- append([the,cat,is],[on,the,mat],[]).
no
| ?- append([the,cat,is],[on,the,mat],[the,cat,is,on,the]).
no
We can also use it to split a list:
| ?- append(L0,L1,[the,cat,is,on,the,mat]).
L0 = []
L1 = [the,cat,is,on,the,mat] ? ;
L0 = [the]
L1 = [cat,is,on,the,mat] ? ;
L0 = [the,cat]
L1 = [is,on,the,mat] ? ;
L0 = [the,cat,is]
L1 = [on,the,mat] ? ;
L0 = [the,cat,is,on]
L1 = [the,mat] ? ;
L0 = [the,cat,is,on,the]
L1 = [mat] ? ;
L0 = [the,cat,is,on,the,mat]
L1 = [] ? ;
no
9 Using the successor function s and 0 to represent the numbers, so that s(0)=1, s(s(0))=2,…, notice how similar the definition of
append is to the following formulation of Peano’s axioms for the sum relation:
sum(0,N,N).
sum(s(N0),N1,s(N2)) :- sum(N0,N1,N2).
20