I'm reading a book "Diagonalization and Self-Reference" by Raymond Smullyan. There's an exercise in the Part 1 of Chapter 4. In this exercise the set $\mathit P$ is the set of all $\it provable$ sentences, and the set $\mathit R$ is the set of all $\it refutable$ sentences.
Exercise 3 . Let as say that $\mathit H$ $\it almost$ represents $\mathit A$ if $\mathit H$ is provable for every $\mathit n$ in $\mathit A$ and not refutable for any $\mathit n$ not in $\mathit A$ (i.e., $n \in A \Rightarrow H(n) \in P$, and $n \notin A \Rightarrow H(n) \notin R $. Prove that for a consistent system $\mathit Z$, if $\mathit H$ represents $\mathit A$ then $\mathit H$ almost represents $\mathit A$. Then prove the following strenghtening of Thereom 3.1: if $\mathit Z$ is consistent and $\mathit R^*$ is almost representable in $\mathit Z$, then $\mathit Z$ is incomplete.
Notions of $\it representation$ and $\it consistency$ were defined previously in the book:
We recall that for a predicate $\mathit H$ of degree 1 and any set $\mathit A$ of numbers, $\mathit H$ is said to represent $\mathit A$ if for all numbers $\mathit n$, $\mathit n \in \mathit A \equiv \mathit H(n) \in P.$
The system is called $\it consistent$ if no sense is both provable and refutable.
The system studied in that chapter is the $\it representation$ system as defined in the book. The exercise deals with the $\it representation$ system, but I believe that the definitions I provided are sufficient to solve the exercise without reference to the remaining properties of the system using only the set theory. It must be simple, but I'm missing something. I can show that if $\mathit H$ represents $\mathit A$ then $n \in A \Rightarrow H(n) \in P$, but I'm not able to prove the second property $n \notin A \Rightarrow H(n) \notin R $. I think that I have to use the fact that the system is $\it consistent$, but I do not know how. If we assume that $n \notin A$ holds, but $H(n) \notin R $ does not (i.e $H(n) \in R $), then $H(n) \notin P $ since $\mathit H$ represents $\mathit A$ and we can't derive a contradiction from consistency property(i.e. that there's no sense that belongs to the set $\mathit P$ and $\mathit R$ simultaneously). Please, help me with that.