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For any two events A and B with $Pr(B) > 0$, prove that $Pr(A^c|B) = 1− Pr(A|B)$.

Is there a valid way to finish a proof if a step moves a term across the equality as follows?

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$Pr(A^c|B) = 1− Pr(A|B)$

$Pr(A^c|B) + Pr(A|B) = 1 - Pr(A|B) + Pr(A|B)$

$1 = (Pr(A\cap B) + Pr(A^C \cap B))/Pr(B) $

$= Pr(B \cap (A \cup A^C)) / P(B) $

$= Pr(B)/Pr(B) = 1$

Therefore $Pr(A^c|B) + Pr(A|B) = 1$, and $Pr(A^c|B) = 1− Pr(A|B)$

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    If you start at $1=P(B)/P(B)$ and work backwards, it might be easier to see that you've already proved it.2012-08-05

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Why not just drop the first line after "Suppose"? Everything after that is correct and proves the proposition. Good job!