Fibonacci sequence implemented in Prolog

gistfile1.prolog

fib(0, 1) :- !.
fib(1, 1) :- !.
fib(N, Result) :- N1 is N - 1, N2 is N - 2, fib(N1, Result1), fib(N2, Result2), Result is Result1 + Result2.

Hi, @Jorgebv02

Brief explanation

You create base cases, and now DEFINE the problem in logical terms.Let's dig into that.... commas (,) represent ANDs.

Remember that the fibonacci sequence is defined recursively, S(n) = S(n-1) + S(n-2) where S(0)=1 and S(1)=1. What we are doing is stating by using predicates that ...

fib(Index, Value) is fib(Index-1, Value1) and fib(Index-2, Value2) and Value = Value1 + Value2.

Obviously at this point, Value1 + Value2 = Value is like saying S(n-1)+S(n-2)=S(n)

How to use it?

Just type on your terminal fib(7,21). or if you want to know the value given an index type the first letter of the param in uppercase fib(7,V).

I think
fib(0, 1) :- !.
should be
fib(0, 0) :- !.
because the fibonnaci sequence starts with 0 so, if you think about it, fibonacci at index 0 is 0.
For example:
0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
0 | 1 | 1 | 2 | 3 | 5 | 8 | 13 |
Therefore, fibonacci at index 7 is actually 13.

when am I supposed to write simple expressions(non recursive) like "N1 is N - 1"and " Result is Result1 + Result2".on the right of the expression and when should it be on the left

What will be the modification required to print the whole series up to the given index?

fibonacci(0,1):-!. % cut down the backtracking
fibonacci(1,1):-!.

fibonacci(N,X):-
  N > 0, % the constraint to make N always greater than 0; put the integer less than 0 doesn't make sense
  N2 is N - 2, % it doesn't work to pass N-2 into the fibonacci predicate. The "is" is the arithmetic operation, not "=" in other languages 
  N1 is N - 1,
  fibonacci(N2,X2), % get the X2 value
  fibonacci(N1,X1), % get the X1 value
  X is X1 + X2. % calculate the Fibonacci number