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Consider some probabilistic events indicated by capital letters. Let $P$ stay for probability, $\perp$ for independence, $|$ for conditional. Suppose $$ (1) \text{ }A_1\perp B_1 | C \text{, i.e., $P(A_1,B_1|C)=P(A_1|C)\times P(B_1|C)$} $$ $$ (2) \text{ }A_2\perp B_2| C \text{, i.e., $P(A_2,B_2|C)=P(A_2|C)\times P(B_2|C)$} $$ $$ (3) \text{ }A_3\perp B_3| C \text{, i.e., $P(A_3,B_3|C)=P(A_3|C)\times P(B_3|C)$} $$ $$ (4) \text{ }A_1,A_2, A_3 \text{ mutually independent conditional on $C$, i.e.} $$ $$ P(A_1,A_2,A_3|C)=P(A_1|C)\times P(A_2|C)\times P(A_3|C) $$ $$ (5) \text{ }B_1,B_2,B_3 \text{ mutually independent conditional on $C$, i.e.,} $$ $$ P(B_1, B_2, B_3| C)=P(B_1| C)\times P(B_2| C)\times P(B_3| C) $$ Can we show that $$ A_1,A_2, A_3 \text{ mutually independent conditional on $B_1,B_2,B_3,C$, i.e. } $$ $$ P(A_1,A_2,A_3|B_1,B_2,B_3,C)=P(A_1|B_1,B_2,B_3,C)\times P(A_2|B_1,B_2,B_3,C)\times P(A_3|B_1,B_2,B_3,C) $$

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No, we cannot show that since this is not true in general under this conditions. Let us provide counterexample.

Consider tossing fair coin twice and set $C=\Omega=\{HH, TT, HT, TH\}$. For this case conditional independence is the same as unconditional.

Define events $$A_1=\{HT,HH\},\, A_2=\{TH, HH\},\, B_1=\{TT, HH\}, \, B_2=A_3=B_3=\Omega.$$

Check that $A_1$, $A_2$ and $A_3$ are mutually independent, $B_1$, $B_2$ and $B_3$ are mutually independent.

Note that the equality $$\mathbb P(A_1\cap A_2\cap A_3)=\mathbb P(A_1)\mathbb P(A_2)\mathbb P(A_3)$$ is not sufficient in general for independence: for three events pairwise independence should also hold. The same is true for conditional independence, so the last phrase in your post does not define mutual conditional independence.

Next, $A_1$ and $B_1$ are independent since $$ \mathbb P(A_1\cap B_1)=\mathbb P(HH)=\frac14 = \mathbb P(A_1)\mathbb P(B_1) $$ $A_2$ and $B_2$ are independent too, as well as $A_3$ and $B_3$.

Calculate conditional probabilities: $$ \mathbb P(A_1\cap A_2\cap A_3\mid B_1\cap B_2\cap B_3) = \mathbb P(HH\mid HH \text{ or } TT) = \frac12, $$ $$ \mathbb P(A_1\mid B_1\cap B_2\cap B_3) = \mathbb P(HT \text{ or } HH\mid HH \text{ or } TT) = \frac12, $$ $$ \mathbb P(A_2\mid B_1\cap B_2\cap B_3) = \mathbb P(TH \text{ or } HH\mid HH \text{ or } TT) = \frac12, $$ $$ \mathbb P(A_3\mid B_1\cap B_2\cap B_3) = 1. $$ And $$\frac12\neq \frac12 \cdot \frac12 \cdot 1.$$

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