Well, always-terminating implies non-Turing-complete, so that's a way to guarantee it.
The proof is easy: just set for all n, f(n) to be the nth program in your toy language in lexicographical order and then set for all n, g(n):=f(n)(n) + 1. This g is clearly Turing-computable (you can modify an interpreter to compute it) and not in your language (a la Cantor).
The proof is easy: just set for all n, f(n) to be the nth program in your toy language in lexicographical order and then set for all n, g(n):=f(n)(n) + 1. This g is clearly Turing-computable (you can modify an interpreter to compute it) and not in your language (a la Cantor).