# Conditional Independence involving four events

I am trying to verify if $$A \perp C | H$$ and $$B \perp C | H$$ implies $$A\cap B \perp C | H$$.

So far, I did the below -- but feel I'm skipping something in the step indicated (*) below. Any help would be greatly appreciated (and this is not any homework).

$$P(A,B |C,H) = \frac{P(A,B,C,H)}{P(C,H)}$$

$$= \frac{P(A|B,C,H)P(B|C,H)P(C,H)}{P(C,H)}$$

$$= P(A|B,C,H)P(B|C,H)$$

$$= P(A|B,C,H)P(B|H) \hspace{2em} \because B \perp C | H$$

$$= P(A|B,H)P(B|H) \hspace{1em} ... step (*)$$

$$= P(A,B|H)$$

When $$A, B, C, H$$ are events, this property seems wrong. For example, when $$H = \Omega$$, then it amounts to say "if $$A$$ and $$C$$ are independent, $$B$$ and $$C$$ are independent, then $$A \cap B$$ and $$C$$ are independent." But this is clearly not the case.
As a counterexample, set $$\Omega = \{(u_1, u_2, u_3): u_i \in \{0, 1\}\}$$ such that $$P(\{u_1, u_2, u_3\}) = \frac{1}{8}$$ for any $$(u_1, u_2, u_3) \in \Omega$$ (that is, $$\Omega$$ is the sample space of outcomes of tossing a coin for three times). Let
\begin{align*} & A = \{(u_1, u_2, u_3): u_1 = u_2\}, \\ & B = \{(u_1, u_2, u_3): u_1 = u_3\}, \\ & C = \{(u_1, u_2, u_3): u_2 = u_3\}. \end{align*} Then it is easy to verify that $$P(A \cap C) = P(B \cap C) = P(A)P(C) = P(B)P(C) = \frac{1}{4}$$, i.e., $$A$$ and $$C$$ are independent, $$B$$ and $$C$$ are independent. However, $$P(A \cap B \cap C) = \frac{1}{4} \neq P(A \cap B)P(C) = \frac{1}{4} \times \frac{1}{2} = \frac{1}{8}.$$