True when all three arguments are lists, and the members of List3 are the members of List1 followed by the members of List2.
Appends lists List1 and List2 to form List3:
| ?- append([a,b], [a,d], X). X = [a,b,a,d] | ?- append([a], [a], [a]). no | ?- append(2, [a], X). no
Takes List3 apart:
| ?- append(X, [e], [b,e,e]). X = [b,e] | ?- append([b|X], [e,r], [b,o,r,e,r]). X = [o,r] | ?- append(X, Y, [h,i]). X = , Y = [h,i] ; X = [h], Y = [i] ; X = [h,i], Y =  ; no
Suppose L is bound to a proper list (see lib-lis-prl). That is, it has the form [T1,...,Tn] for some n. In that instance, the following things apply:
)has at most one solution, whatever X and Y are, and cannot backtrack at all.
)has at most n+1 solutions, whatever X and Y are, and though it can backtrack over these it cannot run away without finding a solution.
), however, can backtrack indefinitely if X and Y are variables.
The following examples are
perfectly ordinary uses of
To enumerate adjacent pairs of elements from a list:
next_to(X, Y, /*in*/ List3) :- append(_, [X,Y|_], List3).
To check whether Word1 and Word2 are the same except for
a single transposition. (
library(lists) would be
better for this task.)
one_transposition(Word1, Word2) :- append(Prefix, [X,Y|Suffix], Word1), append(Prefix, [Y,X|Suffix], Word2). | ?- one_transposition("fred", X). X = "rfed" ; X = "ferd" ; X = "frde" ; no
Given a list of words and commas, to backtrack through the phrases delimited by commas:
comma_phrase(List3, Phrase) :- append(F, [','|Rest], List3), !, ( Phrase = F ; comma_phrase(Rest, Phrase) ). comma_phrase(List3, List3). | ?- comma_phrase([this,is,',',um,',',an, example], X). X = [this,is] ; X = [um] ; X = [an,example] ; no