Why is repeat/0 gross? It implements a failure-driven loop. If you want to force your code to backtrack over all choice points, that's what you need to do. You

Why is repeat/0 gross? It implements a failure-driven loop. If you want to force your code to backtrack over all choice points, that's what you need to do. You can't get the same result with recursion. For instance if you have a nondeterministic predicate (definition) and want to loop over all its results, if you try to do that with recursion you'll just generate the same result at every step of the recursion. Instead you want to iterate over all the choice points. To demonstrate its behaviour you can try that on the SWI-Prolog command line: ?- repeat, between(1,3,I). I = 1 ; I = 2 ; I = 3 ; I = 1 ; I = 2 ; I = 3 ; I = 1 . The ";" are where I ask for more results, the "." is where I said "stop". If you tried to recurse around between(1,3,I), you'd get I bound to "1" every step through the recursion- and still leave two choice points behind for every "1". It's been a while since I used repeat/0, but in general it's used to write generators. It usually makes more sense to use findall/3 and friends, which simply iterate over choice points without adding new ones. For example: ?- findall(I, between(1,3,I), Is). Is = [1, 2, 3]. People complain about dirty imperative constructs in Prolog like that all the time, but they are very useful to a pragmatic programmer and everytime anyone tries to "fix" Prolog by removing the impurities, they break it.