Depending on the task at hand, you may need to program the finite domain constraint logic (CLPFD).
Prolog , X > 10. nat10/1 , :
nat10(X) :- nat(X), ( X > 10 -> !, fail ; true ).
, X > 10, (!), nat(X) ( , 10 ), fail. (true).
| ?- nat10(X).
X = 1 ? ;
X = 2 ? ;
...
X = 9 ? ;
X = 10 ? ;
(3 ms) no
| ?-