Does the uncomputability of Kolmogorov complexity follows from Rice Theorem? I was reading the proof of uncomputability of Kolmogorov complexity by Li and Vitányi book and thinking if there isn't another way to do this simpler using Rice theorem. I came up with this argument. Can you say to me if it is correct?
Definition: The set $A ⊆ N$ is an index set if it has the property that if $x ∈ A$ and $\phi_x = \phi_y$ , then $y ∈ A$.
Rice theorem states:
Theorem: Suppose that $A$ is an index set not equal to $∅$ or $\mathbb{N}$. Then A is not computable.
The conditional absolute complexity can be defined as (by Li and Vitányi book's definition):
$$C(x|y) = \min \{ |p| : \phi(\langle y,p \rangle) = x \}$$
We have: $\phi(\langle y,p \rangle) = \phi_y(p)$ by the enumeration theorem and property of the universal additively optimal $\phi$:
$$C(x|y) = \min \{ |p| : \phi_y(p) = x \}$$
But $A = \{ p : \phi_y(p) = x \}$ is a index set. Because if any  $p \in A$ and  $\phi_p= \phi_q$, then (this is the step I have doubts) $\phi_y(p) = \phi_y(q)$, because  $\phi_y$  can be interpreted as a program that receives other programs $p$ and $q$, and if the programs are equal so is $\phi_y(p)$ and $\phi_y(q)$. Then $q \in A$.
So $A$ is a index set and also is not $\emptyset$ or $\mathbb{N}$ (because  there is a  program $p$ such that $\phi_y(p) \not = x$). So, by Rice Theorem, $A$ is not computable, and $C(x|y)$ also. And so $C(x) = C(x|\varepsilon)$ is also uncomputable.

Your argument that $A$ is an index set is not correct: you haven't used anything particular about $y$, so your argument would imply that $$\{p: \phi_y(p)=x\}$$ would be an index set for every choice of $y$. But that's clearly incorrect: for example, take $x=1$ and let $\phi_y(p)=1$ if $p=0$ and $0$ otherwise. By the padding lemma, this is not an index set.
While we may want to think of $\phi_y$ as acting on computable functions in a well-defined way rather than natural numbers, that doesn't mean that it actually does.
