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Proposition

I have a collection of independent events B, C and D. I am trying to check if for another event $A$, $P(A|B,C,D) = P(A|B,D) \cdot P(A|C,D)$

Argument $P(A|B,C,D) = \dfrac{P(A, B, C, D)}{P(B, C, D)} = \dfrac{P(A, B, C, D)}{P(B,C|D) \cdot P(D)} = \dfrac{P(A, B, C, D)}{P(B|D) \cdot P(C|D) \cdot P(D)} = \dfrac{P(A, B, C, D)}{P(B|D) \cdot P(C, D)}$

However, the numerator seems intractable. Any suggestions regarding the truthiness of the proposition and guidance regarding how to go about proving or disproving this?

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    Strange to try to prove this since, when $(A,B,C,D)$ is independent, the LHS is $P(A)$ while the RHS is $P(A)^2$. – 2017-02-08
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    Yes, your comment makes sense. Intuitively, I expected the probability to increase with increased evidence, and it would be devastating if my original proposition turned true. I am using this equation to make a proposition $ P(A|B,C,D) = \lambda P(A|D)$, for $\lambda > 1$. – 2017-02-08
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    But why should such a comparison hold, at all? Try $A=C$ then the LHS is $1$, then $A=D\setminus B$ then the LHS is $0$. – 2017-02-08
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    hmm.. this gets me back to the drawing board... – 2017-02-08

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