# Does $X \rightarrow y$ being CDH imply that, given $X$ distinguishing between $y, r$ is DDH $\forall r\in Z_{p}^{*}$

In one proof I show that, given a cyclic group $$Z_{p}^{*}$$ where $$p$$ is prime, and a set of information $$X$$ computing $$y\in Z_{p}^{*}$$ is as difficult as solving Computational Diffie Hellman (CDH) problem.

In another proof can I make the argument that as transformation $$X \rightarrow y$$ is proven to be CDH, given $$X$$, distinguishing between any random $$r\in Z_{p}^{*}$$ from $$y$$ is as difficult as Decisional Diffie Hellman (DDH).

First, it might be that CDH is hard, yet the corresponding DDH is easy. In fact, this is generally the case in $$\mathbb{Z}_p^*$$, so you should be careful about that. See the footnote.
Second, there is a trivial counter example to your question: let $$g$$ be a generator of the subgroup of squares modulo $$p$$ (assume for simplicity that $$p$$ is a safe prime). Then, computing $$g^{ab} \bmod p$$ given $$X = {g^a, g^b}$$, for random $$a,b$$, is infeasible assuming the hardness of CDH over the subgroup of squares (which is implied by the hardness of CDH over $$\mathbb{Z}_p^*$$). Now, define $$y$$ as the value obtained by replacing the least significant bit of $$g^{ab}\bmod p$$ with zero. Computing $$y$$ from $$X = {g^a, g^b}$$ is as hard as CDH over the subgroup of squares (the reduction looses a factor 2). Yet, distinguishing $$y$$ from a random element is obviously not implied by DDH over this group (I focused on the subgroup of squares because DDH is broken over $$\mathbb{Z}_p^*$$, see below).
It is easy to distinguish $$g^{ab} = y \bmod p$$ from a random element $$r$$ of $$\mathbb{Z}_p^*$$: with probability 1/2, $$r$$ is not a square, and this is easy to check by computing the Legendre symbol (in contrast, $$y$$ is always a square).