Let's suppose we have 4 random variables $X,Y,Z$ and $T$ and that the following equations hold about the entropy: $$H(T|X)=H(T)$$ $$H(T|X,Y)=0$$ $$H(T|Y)=H(T)$$ $$H(Y|Z)=0$$ $$H(T|Z)=0$$
Also, the following statements are easy to prove and hold: $$H(T|X,Y,Z)=I(Z;T|X,Y)=0$$ $$I(X;T|Y,Z)=I(Y;T|X,Z)=0$$ $$H(Y|X,Z,T)=I(X;Y|Z,T)=0$$ $$H(T|X,Z)=0$$ $$H(T|Y,Z)=0$$
Now, I want to prove that: $$H(T)=0$$
I have tried many different things, but I got nothing. Now, I am thinking if there is a way to decompose the entropy like \begin{eqnarray*} H(T)&=&\sum\limits_tp_T(t)\log_2p_T(t)\\ &=&\sum\limits_t\Big\{\sum\limits_{x,y,z}p_{TXYZ}(t,x,y,z)\log_2p_{TXYZ}(t,x,y,z)\Big\}\\ &=&\sum\limits_t\Big\{\sum\limits_{x,y,z}p_{TXYZ}(t|x,y,z)p_{XYZ}(x,y,z)\log_2\big\{p_{T|XYZ}(t|x,y,z)p_{XYZ}(x,y,z)\big\}\Big\}\\ \end{eqnarray*}
You cannot prove that because it's not true. Counterexample:
Let $X,Y$ be iid fair Bernoulli $(p=1/2)$, let $T= X \oplus Y$ (XOR, or sum in modulus 2). Then $T$ is independent from $X$ and from $Y$ (each taken alone), but it's a deterministic function of the pair $X,Y$. Hence the first three equations, hold. But $H(T)=1$
The last two equations involve an extra variable $Z$, it's easy to construct an example such that $Z$ determines the values of $X$ and $Y$ (and hence $Z$), but still $X$ and $Y$ are iid. Say, $Z$ uniform on $\{0,1,2,3\}$ and $X=Z \mod 2$ , $Y=\lfloor Z/2\rfloor $