How to prove probability problem Show that if $P(C) = 1$, then $P(D|C) = P(D)$ using probability rules.
I tried using conditional probability to get from $P(D|C) = P(D \cap C) / P (C) = P(D \cap C)$ but I am not sure how to continue the proof to get to $P(D)$. Can someone explain what the next steps I should do in order to get to $P(D)$?
 A: I would solve it like this: 
$$
\begin{aligned}
P(D|C) &= P(D\cap C) / P(C) \\
&= P(D \cap C) \\ 
&= P(D) + P(C) - P(D \cup C) \\
&= P(D) + 1 - 1 \\ 
&= P(D)
\end{aligned}
$$
Where our third equation is explained by the rule of addition with some simple rearrangement of terms:
$$ P(D \cup C) = P(D) + P(C) - P(D \cap C)$$
$$ \therefore P(D \cap C) =  P(D) + P(C) - P(D \cup C)$$
And our fourth equation is explained by the fact that $P(D \cup C) = 1$, since it is the probability of either $D$ or $C$ occurring, and we know that $P(C) = 1$
A: I would approach it through a frequentist definition of probability, by showing that the set of events that count as D ∩ C has the same count as the set of events that count as D.  That is P(D ∩ C) = N(D ∩ C)/N(total) and P(D) = N(D)/N(total) can be shown to be the same if every event D is also an event D ∩ C (so that N(D ∩ C) is the same as N(D)).  Because C is universally true, this will be the case.
Edit: Okay - a more extended answer based on your comment
P(D|C) = P(D ∩ C)/P(C)
P(D|C) = P(C|D)*P(D)/P(C)
so we have to show that P(C|D) = 1, as does P(C), which was given.
P(C) = P(C|D)P(D) + P(C|!D)(1-P(D))
Let P(C|D) = 1-a and P(C|!D) = 1-b.
a, b must be nonnegative by the rules of probability (0-1 range)
then
1 = (1-a)P(D) + (1-b)(1-P(D))
1 = P(D) - aP(D) + 1 - P(D) - b + b*P(d)
b(1-P(D)) = -aP(D)
which, given the 0-1 constraints on a,b,P(D) is only possible if i) both a and b are zero, ii) both b and P(D) are zero, or iii) a=0 and P(D) = 1
In the first case and third case, we have that P(C|D) = 1 because a=0 and we have proved what was required.
In the second case, if P(D) = 0 then we can show that this can only be satisified when P(D|C) is zero since P(D) = P(D|C)*P(C)+....  in which case we have still proved that P(D) = P(D|C) as required (since they both = 0).
