Diagonalization out of partial recursive functions

Question

So generally partial recursive functions don't diagonalize. But isn't this function an exception?

$\phi(x)=\lambda_{x}(x)+1 $ if $\lambda_{x}(x)$ halts and $0$ else.

Completely no clue... It seems this function is diagonalizable??

Answer 1

It is not clear what you mean when you say that a function "diagonalize". I assume that the $\lambda_e$ you used is an enumeration of the partial recursive functions.

Then the function $\phi$ you defined is not recursive, and you can prove it by diagonalization, because if $\phi$ was recursive then there would be a code $e$ so that $\phi=\lambda_e$ and then so that $\phi(e)=\lambda_e(e)$. But by definition of $\phi$ this is not possible, as $\phi(e)$ is always distinct from $\lambda_e(e)$.

What is recursive is the function $\phi(x)=\lambda_x(x) + 1$ if $\lambda_x(x)$ halts, and $\phi(x)$ does not halt otherwise.