Possible typos in the proof of Recursion Theorem Below is the proof I took from textbook A Course in Mathematical Analysis by Proof D. J. H. Garling.
I think there are some typos, which I highlighted in the attached picture.



*

*$g(s(n))= f(n)$


Since the theorem states $a_{s(n)}=f(a_{n})$, I think the right version should be: $g(s(n))=f(g(n))$, not $g(s(n))= f(n)$.
2.$(n,a)\in P$ and $(n,a')\in P$
I think the right version should be: $(n,a)\in g$ and $(n,a')\in g$
3.$(s(m),f(b))\in g$
Since the author is trying to prove $g'\in S$, I think the right version should be: $(s(m),f(b))\in g'$, not $(s(m),f(b))\in g$.
4.$(s(m),f(b))\notin g$
As in 3., I think the right version should be: $(s(m),f(b))\notin g'$, not $(s(m),f(b))\notin g$.
Please check whether I am right or wrong about these typos!
 A: Ad 1. Indeed, $n\in P$, not $n\in A$, hence $f(n)$ would not even make sense. But the correct formula  should be
$$ g(s(n))=f(g(n)).$$It is only after finding this nice $g$ that we can set $a_n=g(n)$ an have our recursively defined sequence.
Ad 2. Yes. Again, $\in P$ would not even make sense. If you follow the argument, you will notice that $\in g$ is used and shown, so at least the typo is no show-stopper for the proof.
Ad 3 and 4. No and then Yes. That step of the argument should read 

Then $(s(m),f(b))\in g$. Thus if $(s(m),f(b))\notin g'$ then $(s(m),f(b))=(s(n),a')$.


There seem to be more typos, e.g., there is a superfluous $S$ in what should read

Since $(0,\bar a)\in R$ for all $R\in S$, $(0,\bar a)\in g$.

A: 3 is no type and 4 should read: "Thus if $(s(m),b)\notin g'$.."
The author is indeed proving that $g'\in S$. This by induction. 
From
$g'=g-\left\{ \left(s\left(n\right),a'\right)\right\} $ and $\left(0,\overline{a}\right)\in g$
it follows that $\left(0,\overline{a}\right)\in g'$. 
Then it remains
to prove that $\left(m,b\right)\in g'\implies\left(s\left(m\right),f\left(b\right)\right)\in g'$.
Suppose that this is not true for some $m\in P$ so that $\left(m,b\right)\in g'\wedge\left(s\left(m\right),f\left(b\right)\right)\notin g'$.
This route is taken by the author but unfortunately a typo was made ($g$ instead of $g'$).
It will be shown now that this assumption leads to contradiction.
From $\left(m,b\right)\in g'$ it follows that $\left(m,b\right)\in g$
so that $\left(s\left(m\right),f\left(b\right)\right)\in g$. 
Then
from $\left(s\left(m\right),f\left(b\right)\right)\notin g'$ it follows
that $\left(s\left(m\right),f\left(b\right)\right)=\left(s\left(n\right),a'\right)$
and consequently $m=n$ and $f\left(b\right)=a'$. 
However it was
assumed that $n\in U$ and that for a unique $a\in A$ we had $\left(n,a\right)\in g$.
So $\left(n,b\right)=\left(m,b\right)\in g$ tells us that $b=a$.
Then $a'=f\left(b\right)=f\left(a\right)$ and a contradiction is
found. 
Proved is now that $g'\in S$. 
[This leads to a contradiction again, which allows the conclusion that $n\in U\implies s(n)\in U$.]
