How to show the function is not Turing computable? Having the function:

$$f(y) = \begin{cases}
         \ 1 &amp;\text{if }\forall n \Phi_y(n)=n\lor \Phi_y(n) \!\uparrow\\
         \ 0 &amp;\text{otherwise.}
\end{cases}$$

By the rule of thumb it should not be computable. How to prove it formally?
If it's computable, I assumed a function like this exists:
$$g(y) = \begin{cases}
         \ 1 &amp;\text{if $f$(y) = 1 }\\
         \ 0 &amp;\text{if $f$(y) = 0}
\end{cases}$$

I assumed $g(y)$ is computable. Now there exists a function $\Phi_{x_0}(x_0)$ which simulates $g(y)$. 
This means: $\Phi_{x_0}(x_0) \downarrow$ iff $\Phi_{x_0} (x_0) \uparrow$, which is a contradiction. Is this sufficient as a proof?
Since I assume this is a homework question, I won't give a fully detailed proof but rather a sketch with some gaps to fill.

Consider the function 
$$g(x)=
\begin{cases} 
  x+1 &amp; \text{ if } \varphi_x(x)\downarrow\\
  \uparrow &amp; \text{ otherwise}.\end{cases}$$
This function is computable (convince yourself of that!), so it has an index $e$. 

Now consider the function $f$ on $e$, then computing $f(e)$ comes down to verifying whether $\forall n\  \varphi_e(n)\uparrow$. Can you see why this contradicts that the halting problem is not computable?