Let $A$ and $B$ be events with $P(A) > 0$ and $P(B) > 0$. We say that an event $B$ suggests an event $A$ if $P(A|B) > P(A)$, and does not suggest event A if $P(A|B) < P(A)$.
1.25(b). Assume that $P(B^C) > 0$. Show that $B$ suggests $A$ if and only if $B^C$ does not suggest $A$.
Solution: $\color{green}{P(B)}P(A) + \color{green}{P(B^C)}P(A) \color{green}{= 1} P(A)$
(by the Total Probability Theorem) $\qquad = P(B)P(A | B) + P(B^C)P(A | B^C)$
$ \iff P(B^C) [ P(A) − P(A|B^C) ] = P(B) [ P(A|B) − P(A) ] \quad \blacksquare$
How and why would one discern or foreknow to write $P(A)$ in two ways as above? This "trick" feels guileful; is there an easier/more natural proof? Or could someone please demystify it?
I tried to prove via a string of equivalences: $P(A|B) > P(A) \mathop{\iff}^{?} P(A|B^C) < P(A)$.
$RHS \iff \dfrac{P(A \cap B^C)}{P(B^C)} < P(A) \iff \dfrac{P(A \cap B^C)}{1 - P(B)} < P(A)$.
but I don't savvy how to simplify the latter.