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In other words, if P(A | C) = P(A) and P(B | C) = P(B), can it be demonstrated that P(A & B | C) = P(A & B)? If yes, how?

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No.

Take two independent Bernoulli random variables $X_1$ and $X_2$, equidistributed on $\{-1, 1\}$. Set $X_3 = X_1X_2$. Then you can check that $X_3$ is independent of $X_1$ and $X_2$ (but not of the pair $(X_1,X_2)$, of course), and has the same distribution as $X_1$ and $X_2$.

Take $A=\{X_1=1\}$, $B=\{X_2=1\}$, and $C=\{X_3=-1\}$. Then your conditions are fulfilled, however $\Pr(A \cap B \mid C) = 0$ but $\Pr(A\cap B) = \frac{1}{4}$.

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