Set cover in which some pairs of sets are forbidden
This problem is way harder than set cover. Here is why...

Intuitively, you can encode independent set as a problem of this type. Indeed, you are given an instance of independent set - a graph $G$ with $n$ vertices, and a number $k$, and the question is whether the graph $G$ has an independent set of size $k$.

We assume that $k$ is large (say, polynomial in $n$). 

It is not hard to form a set system, with $n$ sets, such that any cover of the set system requires at least $\Omega(k/\log n)$ sets, and any specific set of $k$ sets cover the ground set. To this end, have a "large" ground set $X$ (say, of size $t$), and generate $n$ random sets (independently), where you pick every element $x \in X$ to be in the $i$th random set with probability roughly $p = 10(\ln n)/k$. Let $R_1, \ldots, R_n$ be the resulting random sets, and observe that a union of any specific $k$ of them covers the ground set, with high probability. Indeed, the probability that an element of $X$ is not covered, by a specific $k$ sets is bounded by
$$
t (1-p)^k &lt; 1/10,
$$
for $t = n^{O(1)}$.

Similarly, consider $k' = c (k/\log n)$ random sets, where $c$ is a sufficiently small constant. We claim that with high probability their union does not cover $X$. Indeed, the probability a specific element is covered by such $k'$ sets is 
$1- (1-p)^{k'}$ which is a constant, say $\alpha &lt; 1$. As such, the probability that these $k'$ random sets cover $X$ is at most $\alpha^t$. In particular, the probability that any of the possible sets of $k'$ random sets cover $X$ is bounded by 
$$
\alpha^t * \binom{n}{k'} &lt; 1/10,
$$
by taking $t$ to be larger than $20 (k/\alpha) \ln n$ (or some similar quantity).

Now, associate with the $v_i$ vertex in $G$, the set $R_i$. Forbid the pair $\{R_i, R_j\}$ if the edge $v_iv_j$ is in the graph $G$.

Any solution for the resulting instance, would be an independent set in $G$ of size $\Omega(k / \log n)$. The original independent set of size $k$ is a valid solution in the resulting instance (all with good probabilities). But that implies that using your problem, you can get a log approximation to independent set, which is an extremely unlikely event. More generally, a $\beta$ approximation, would imply a $O( \beta \log n)$ approximation for independent set. Since independent set can not be approximated to within $n^{1-\epsilon}$, you essentially get the same result for your problem.

PCP people might be able to derive the above from their known machinery...