# If AA is independent of (B,C)(B,C) then A−B−CA-B-C is a Markov chain. [on hold]

I need to proof the following statement:

If AA is independent of (B,C)(B,C) then A−B−CA-B-C is a Markov chain.

My current approach is to write the probability mass function:

p(a,b,c)=p(a)p(b|a)p(c|a,b)p(a,b,c) = p(a)p(b|a)p(c|a,b)

so that my goal is to proof that p(c|a,b)=p(c|b)p(c|a,b) = p(c|b). But apparently I need stronger assumptions to do so as for example that AA is independent of BB.

Please could you provide a proof of the statement above or a counterexample?

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