Information content of two bits Let $X \in\{0,1\}$ and $Y \in\{0,1\} $ be two uniformly distributed bits. Let $B$ be an arbitrary random variable such that $I(X:B)=0$, $I(Y:B)=0$, and $I(X \oplus Y:B)=0$, then is it true that $I(X,Y:B)=0$? 
($I(X:Y)$ is Shannon’s mutual information.)
 A: Zero mutual information can be interpreted as independence as well. The set of assumptions conclude that:
\begin{align*}
P((X,Y)\in\{(0,0),(0,1)\}\big| B=0)&= P((X,Y)\in\{(0,0),(0,1)\}\big| B=1)\\
P((X,Y)\in\{(1,0),(1,1)\}\big| B=0)&= P((X,Y)\in\{(1,0),(1,1)\}\big| B=1)\\
P((X,Y)\in\{(0,0),(1,0)\}\big| B=0)&= P((X,Y)\in\{(0,0),(1,0)\}\big| B=1)\\
P((X,Y)\in\{(0,1),(1,1)\}\big| B=0)&= P((X,Y)\in\{(0,1),(1,1)\}\big| B=1)\\
P((X,Y)\in\{(0,0),(1,1)\}\big| B=0)&= P((X,Y)\in\{(0,0),(1,1)\}\big| B=1)\\
P((X,Y)\in\{(0,1),(1,0)\}\big| B=0)&= P((X,Y)\in\{(0,1),(1,0)\}\big| B=1).
\end{align*}
Let $$U_i=[P((X,Y)=(0,0)|B=i),P((X,Y)=(0,1)|B=i),P((X,Y)=(1,0)|B=i),P((X,Y)=(1,1)|B=i)]^T.$$
Then, by removing redundant equations, we get
\begin{align*}
\begin{bmatrix}
1& 1&0&0\\
0& 0&1&1\\
0& 1&0&1\\
0& 1&1&0
\end{bmatrix}\cdot U_0=
\begin{bmatrix}
1& 1&0&0\\
0& 0&1&1\\
0& 1&0&1\\
0& 1&1&0
\end{bmatrix}\cdot U_1,
\end{align*}
which results in $U_0=U_1$. Or in other words
$$\forall b,x,y:~P((X,Y)=(x,y)\big| B=0)=P((X,Y)=(x,y)\big| B=1),$$
which concludes independence and $I(X,Y;B)=0$.
