prolog list insert at any position

New to Prolog, trying to write a predicate to give all the options that an element could be inserted in to a list at any position. Ex:

ins(a, [b,c], R). should give:

R = [a,b,c]
R = [b,a,c]
R = [b,c,a]

which it does, but then gives an error 'Out of Global stack'. Is there a way to make this more deterministic, give the results and be done? When it is run in reverse ie. ins(X, Y, [a,b,c]). It gives the expected results then says false indicating it has completed. Code:

app([],L,L).
app([H|T],L2,[H|L]) :- 
    app(T,L2,L).

ins(E, List, R) :-
    R = R1,
    app(R2, R3, R1),
    app([E], R4, R3),
    app(R2, R4, List).

Here is a link to run the code in an online compiler, SWISH (This also has an example of how I hope to use ins, but ins is the problem right now) Any Help would be appreciated!

Solution

Did you notice how this went bad? First, Prolog was very nice and dandy and showed you how smart it is, and only later on it struck you: Buy. More. RAM. Now!

Wouldn't it be better if Prolog would be up front? Before showing any answer?

Well, you can force Prolog to do exactly this. Add a false at the end of your query like so:

?- ins(a, [b,c], R), false.
   resource_error(_). % ERROR: Out of global stack

And the same you can do with your remaining program: Simply add false such that the remaining program still loops or runs out of space. I came up with the following minimal failure-slice

app([],L,L) :- false.
app([H|T],L2,[H|L]) :-
    app(T,L2,L), false.

ins(E, List, R) :-
    R = R1,
    app(R2, R3, R1), false,
    app([E], R4, R3),
    app(R2, R4, List).

?- ins(a, [b,c], R), false.

That means that we have to modify something in the remaining visible part in order to get rid of that looping. In other words: As long as the visible part remains unmodified the error will persist — guaranteed!

For more on this technique to understand reasons for non-termination see failure-slice

The immediate fix would be to put the first app/3-goal last.

But there is something else: You used all kinds of variables that are difficult to fathom. Maybe stick to a more uniform scheme. Also, there is no need for appending [A] using app/3. You actually need only two app/3 goals.