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?
Show:
$Pr(A^c|B) = 1− Pr(A|B)$
$Pr(A^c|B) + Pr(A|B) = 1 - Pr(A|B) + Pr(A|B)$
$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 $Pr(A^c|B) = 1− Pr(A|B)$