How is it that $P(RL^C)$ is equal to $P(R)$?

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I have a formula which says:

$$P(R | L^C) = \cfrac{P(RL^C)}{P(L^C)} = \cfrac{P(R)}{1-P(L)}$$ I understand how $P(L^C) = 1-P(L)$, but can't find the formula which would allow $P(RL^C)$ to equal $P(R)$

The context behind the problem is: Joe is 80 percent certain that his missing key is in one of the two pockets of his hanging jacket, being 40 percent certain it is in the left-handed pocket and 40 percent certain it is the right-hand pocket. If a search of the left-hand pocket does not find the key, what is the conditional probability that it is in the other pocket?

R -> the event that it is in the right hand pocket L -> the event that it is in the left hand pocket

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$R$ and $L$ are disjoint events, so $R \subset L^C$. Hence, $RL^C=R$.