Returning false if item is not in list (PROLOG)-CodePudding

Below is the code to remove a key (A) from a list if it is there. If it isn't there, it currently returns the entire list. I would like it instead to return 'false.' Sample outputs will be below as well.

mySelect(_, [], []).
mySelect(X, [Y|K], [Y|M]):- mySelect(X, K, M), (X \= Y).
mySelect(X, [X|K], R) :- mySelect(X, K, R).

Currently this will output:

?- my_delete(c,[a,b,c,d],R). 
R = [a, b, d] .

?- my_delete(e,[a,b,c,d],R). 
R = [a, b, c, d] .

I would like it to output:

?- my_delete(c,[a,b,c,d],R). 
R = [a, b, d] .

?- my_delete(e,[a,b,c,d],R). 
false .

Any pointers would be greatly appreciated!

CodePudding user response：

You can add another rule which first ensures that X is a member of L, before applying mySelect.

delete(X, L, R) :-  member(X, L), mySelect(X, L, R).

CodePudding user response：

Either partition the list into matches and non-matches:

filter( X , Ys, Zs ) :- filter(X,Ys,[_|_],Zs).

filter( _ ,    []  ,    []  ,    []  ) .
filter( X , [X|Ys] , [X|Xs] ,    Zs  ) :- !, filter(X,Ys,Xs,Zs) .
filter( X , [Y|Ys] ,    Xs  , [Y|Zs] ) :-    filter(X,Ys,Xs,Zs) .

or count the number of items and test the count for being greater than zero:

filter( X , Ys, Zs ) :- filter(X,Ys,0,N,Zs), N > 0 .

filter( _ ,    []  , N , N ,    []  ) .
filter( X , [X|Ys] , T , N ,    Zs  ) :- T1 is T 1, !, filter(X,Ys,T1,N, Zs) .
filter( X , [Y|Ys] , T , N , [Y|Zs] ) :-            !, filter(X,Ys,T,N,Zs) .