Trying exercise 2.6 in Hoff's Bayesian Statistics book.
Given $A\perp B \space \vert C$
Prove $A^c \perp B^c \space \vert C$
My attempt:
$$ \text{Definition}: A\perp B \space \vert C \iff P(A \cap B \space \vert C) = P(A \vert C)P(B \vert C) $$ Using the RHS: $$ RHS = (1-P(A^C \vert C))(1-P(B^C \vert C)) $$ $$ RHS = 1 - P(A^c \vert C) - P(B^c \vert C) + P(A^c \vert C)P(B^c \vert C) $$ Setting LHS to this new RHS: $$ P(A \cap B \vert C) = 1 - P(A^c \vert C) - P(B^c \vert C) + P(A^c \vert C)P(B^c \vert C) $$ $$ \implies P(A \cap B \vert C) + P(A^c \vert C) + P(B^c \vert C) - 1 = P(A^c \vert C)P(B^c \vert C) $$
This is where I am stuck. My RHS is now what I need per the definition. Unsure how to get $P(A^c \cap B^c \vert C)$ on the LHS though.
I tried something like this...
Note: $$ P(A^c \cap B) = P(A^c \cap B \vert C) + P(A^c \cap B^c \vert C) $$ $$ P(B^c \cap B) = P(B^c \cap A \vert C) + P(B^c \cap A^c \vert C) $$ Now, the LHS becomes: $$ P(A^c \cap B \vert C) + 2P(A^c \cap B^c \vert C) + P(B^c \cap A \vert C) + P(A \cap B \vert C) - 1 $$ Not sure if this is the right way to go though. Would appreciate some help!
To conclude, note that $$P(A^c \cap B \mid C) + P(A^c \cap B^c \mid C) + P(B^c \cap A \mid C) + P(A \cap B \mid C) = 1.$$