Prolog 1st exercise

[_,_,[_|X]|Tai]=[[],0,[1,2,3],4,5,6]
Tai = [4, 5, 6],
X = [2, 3]

add(X,L,N).
add(X,[],[X]).
add(X,[Y],[Y,X|[]]).
add(X,[H|T],[H|N]):-add(X,T,N).

memberr(X,[X|T]).
memberr(X,[H|T]):-
    memberr(X,T).

member(X,[23,50,16,14,30,14]), Y is X*X, Y<400.
X = 16,
Y = 256
X = 14,
Y = 196
X = 14,
Y = 196



first(X,[X|L]).
first(X,[[5,2],3,4]).
X = [5, 2]

snd(X,[_,X|L]).
X = 3

lastt(X,[X]).//IF i ONLY WIRTE E.G A(X,[X]). and a(X,[2]). it will give me 2
lastt(X,[_|L]):-
    lastt(X,L).
lastt(X,[[5,2],3,4]).
X=4


append([],B,B).
append([H|A],B,[H|AB]):-
    append(A,B,AB).
append([1,2],[3,4],X).
//opisano v tetradkata-stiga do dunoto,koeto e ([],B,B). 2roto B nqma stoinost tui kato AB nqma => 2roto B priema stoinostta na 1voto.
//B=[3,4]->AB=[3,4] i se vrushtame nazad 2|AB->[2,3,4], 1|AB->[1,2,3,4]
https://www.youtube.com/watch?v=0LR_N_EMjwk

2nd exercise
1member using append
append([],B,B).
append([H|A],B,[H|AB]):-
    append(A,B,AB).

member(X,L):-
    append(A,[X],C),
    append(C,B,L),
    append(_,[X|_],L).
member(2,[1,2,3,4])
true


2last using append
last(X,L):-append(_,[X],L).

PREFIX
prefix(P,L):-append(P,_,L).
This says that list P is a prefix of list L when there is some list such that L is the result of concatenating P with that list. (We use the anonymous variable since we don’t care what that other list is: we only care that there is some such list or other.) This predicate successfully finds prefixes of lists, and moreover, via backtracking, finds them all:
refix(X,[1,2,3,4]).
X = []
X = [1]
X = [1, 2]
X = [1, 2, 3]
X = [1, 2, 3, 4]

suffix(S,L):-append(_,S,L).
That is, list S is a suffix of list L if there is some list such that L is the result of concatenating that list with S . This predicate successfully finds suffixes of lists, and moreover, via backtracking, finds them all:
suffix(X,[1,2,3])
X = [1, 2, 3]
X = [2, 3]
X = [3]
X = []



??
infix(A,L):-append(P,_,L),
    append(_,A,P).

//infix(A,[1,2,3])
A = []
A = [1]
A = []
A = [1, 2]
A = [2]
A = []
A = [1, 2, 3]
A = [2, 3]
A = [3]
A = []


%dali S e pod-vo na L

subset([],[]).
        subset([X|S],[X|L]) :-
            subset(S,L).
        subset(S, [_|L]) :-
            subset(S,L).
subset(P,[1,2,3])
P = [1, 2, 3]
P = [1, 2]
P = [1, 3]
P = [1]
P = [2, 3]
P = [2]
P = [3]
P = []


del(X,[X|Tail],Tail).

del(X,[Head|Tail],[Head|NewList]):-
    del(X,Tail,NewList).

remove(X,L,N):-append(A,[X|B],L),
    append(A,B,N).
remove(X,[X|L],L).
remove(X,[H|L],[H|B]):-remove(X,L,B).

add(X,L,N):-remove(X,N,L).