We define: F as the set of all computable functions. P is a subgroup of F. And we say P is trivial if P = emptyset or P=F. prove: 1-if P is trivial the language L={(M) | M is a TM that computes a function that belongs to P} is recursive. 2- if P is non trivial then L is not in R.
Im having a hard time proving the case when P=F. any suggestions ?
In order for the problem to make sense, $F$ has to be the set of all partial computable functions. Then when $P=F$, the language $L$ is by definition the set of all possible $\langle M\rangle$.
The result is false if $P$ is the set of all total computable functions. In that case $L$, far from being recursive, is the $\Pi_2$-complete set known as $\it Tot$.