Question

I have the next function from the book:

% Signature: select (X,HasXs,OneLessXs)/3 
% purpose: The list OneLessXs is the result of removing 
 one occurrence of X from the list HasXs. 
select(X,[X|Xs],Xs).  *rule number 1* 
select(X,[Y|Ys],[Y|Zs]) :- select(X,Ys,Zs).  * rule number 2*

?- select(4,[2,3,2,4,5,2,4],X].
X=[2,3,2,5,2,4]

But I didn’t understand how it finds the correct answer. After it removes all the Y=!X, It come to a rule number 1, with: Xs=4,5,2,4, then it return true. and what then? If it continue to rule 2, then he also remove the next “4”. If it doesn’t continue to rule 1, then How Zs is [2,3,2,5,2,4]? I think Im missing a basic rule.

Answer 1

You can understand why this is happening by thinking of how it’s going to be executed.
So you have:

% I use K here instead of X, so there is no
% confusion with the X on the rules
select(4,[2,3,2,4,5,2,4], K).

First, it will check rule 1, but 4 != 2 so it will continue to rule 2. At rule 2, you have these “bindings”:

X = 2
[Y | Ys] = [2, 3, 2, 4, 5, 2, 4] (which means Y = 2, Ys = [3, 2, 4, 5, 2, 4])
[Y | Zs] = [2 | Zs] (because Y was binded to 2)

And K is binded to [2 | Zs] and not just Zs. The next select call which will be select(4, [3, 2, 4, 5, 2, 4], Zs) finds Zs and so on. That’s why the returned result is:
K=[2,3,2,5,2,4] and not K=[5, 2, 4] as you was expecting.