# Mutual information decrease with coarse-graining

Let $X,A,Y,B,C,D$ be random binary variables. $D$ is independent from $X,A,C$ and $C$ is independent from $Y,B,D$.

Is it true that:

If $I(Y:B|D=0)\leq \epsilon$ then $I(X\oplus Y:A\oplus B|C=0,D=0)\leq \epsilon \times I(X:A|C=0)$.

This result seems intuitive and I managed to show it for $\epsilon=0$ and $\epsilon=1$, but maybe in the general case we should have a function $f(\epsilon)$ on the right hand side and not simply $\epsilon$. If the result is true, how would you show it using the data processing inequality, the chain rule etc.?

For the case $\epsilon=0$, assume $A=X$ is uniform, and $B=Y$, and $C$ and $D$ are deterministically $0$.
Then you say if $H(B) \le 0$ then $H(A\oplus B)\le 0$.
But $H(A\oplus B) = H(A) = 1 \neq 0$, which contradicts your claim.