In the road to Godel's incompleteness theorem, Problem saying:
Show that Frame For $G:\omega\times\omega\times\omega^{n}\rightarrow\omega$ a recursive function, the function $F:\omega\times\omega^{n}\rightarrow\omega$ , given by
$ F(a,\bar{b})=G(\bar{F}(a,\bar{b}),a,\bar{b})$ is recursive.
And solution begins with
$H(a,\bar{b})=\mu x(Seq(x)\wedge lh(x)=a\wedge\forall i<a((x)_{i+1}=G(Init(x,i),i,\bar{b}))$
I tried to show it by using definitions. So,
$F(a,\bar b)=G(\bar F(a,\bar b),a,\bar b)=G(<F(0,\bar b),\cdots ,F(a-1,\bar b)>,a,\bar b)=G(\mu x(lh(x)=a\wedge (x)_1 =F(0,\bar b)\wedge \cdots \wedge(x)_a=F(a-1,\bar b),a,b)$
I'm totally lost. How do I get that equation?
*) Seq(x) is a relation x is a sequence number for some sequence and Init is a initial sequence function
**) Definition of $\bar F$ is given a function $F:\omega\times \omega^k\rightarrow \omega$. a function from $\omega \times \omega^k$ to $\omega$ given by $\bar F(a,\bar b)=<F(0,\bar b),\dots , F(a-1),\bar b)>$, that is, $\mu x(lh(x)=a\wedge\forall i<a((x)_{i+1} =F(i,\bar b))).$