I am studying Ebbinghaus book "Mathematical Logic". I am in the section concerning the undecidability of first-order logic, trying to solve the following problem:
Show that the set of satisfiable $S_{\infty}$-sentences is not R-enumerable.
Can anyone tell me if what I did is correct? Here's what I did:
Suppose, by contradiction, that the set of satisfiable $S_{\infty}$-sentences is $R$-enumerable. Since the set of non satisfiable $S_{\infty}$-sentences is equal to the set of $S_{\infty}$-sentences whose negation is valid and the set of valid $S_{\infty}$-sentences is $R$-enumerable, we obtain that the set of non satisfiable $S_{\infty}$-sentences is also $R$-enumerable. If both a set and it's complement are $R$-enumerable, we can conclude that the set is $R$-decidable. So, we can conclude that the set of satisfiable $S_{\infty}$-sentences is $R$-decidable.
If the set of satisfiable $S_{\infty}$-sentences were $R$-decidable, we would be able to $R$-decide the set of valid $S_{\infty}$-sentences, because a sentence is valid iff its negation is not satisfiable. This contradicts the fact (this fact is the main result of the section) that the set of valid $S_{\infty}$-sentences is not $R$-decidable. So, we conclude that the set of satisfiable $S_{\infty}$-sentences is not R-enumerable.
Other paths for solving the question are also welcomed. Thanks in advance.
Yes, that's correct: the set of satisfiable sentences is a priori co-r.e., so it can't be r.e. without being recursive. But we already know that it's not recursive - or rather, we know something not recursive which is reducible to it, so it can't be recursive.