Is there a relativized form of Rice Theorem?
You can prove your idea via the usual recursion theorem proof of Rice's theorem. 

proof:

Suppose that $M_{yes}$ is a machine with both property $P_1$ and $P_2$ and that $M_{no}$ is a machine with property $P_1$ but without property $P_2$. Such machines must exist by the nontrivial assumption.

Then suppose there exists a decider $D$ which accepts any input $&lt;X&gt;$ such that machine $X$ satisfies both property $P_1$ and $P_2$, rejects any input $&lt;X&gt;$ such that machine $X$ satisfies property $P_1$ but not $P_2$, and has any behavior on all other inputs.

Consider the following machine:

$M =$ on input $w$:


Obtain the description $&lt;M&gt;$ via the recursion theorem.
Run $D$ on $&lt;M&gt;$
If $D$ accepts $&lt;M&gt;$, simulate $M_{no}$ on $w$ and output the same answer
If $D$ rejects $&lt;M&gt;$, simulate $M_{yes}$ on $w$ and output the same answer


This language $L(M)$ of this machine is either $L(M_{yes})$ or $L(M_{no})$ depending on whether $D$ rejects or accepts $&lt;M&gt;$. Notice that since $P_1$ is a semantic property shared by both $M_{yes}$ and $M_{no}$, we know that in either case, $M$ has property $P_1$. Then by definition of $D$, $D$ accepts $&lt;M&gt;$ if and only if $M$ has property $P_2$. Thus, we have that $L(M) = L(M_{no})$ if $M$ has property $P_2$ (i.e., $D$ accepts $&lt;M&gt;$) and $L(M) = L(M_{yes})$ if $M$ does not have property $P_2$ (i.e., $D$ rejects $&lt;M&gt;$). This, however, contradicts the fact that $M_{yes}$ has property $P_2$, $M_{no}$ does not have property $P_2$, and $P_2$ is semantic. 

By contradiction, we can conclude that decider $D$ as described cannot exist. In other words, $P_2$ given $P_1$ is undecidable.