Let $A,B$ be two events in the probability space $(\Omega,\mathcal F,P)$ and suppose $P(B)>0$. Let $\mathcal G=\{\varnothing,\Omega,B,B^c\}$. Then $\mathcal G$ is a $\sigma$-field. The following holds $$P(A|\mathcal G)=P(A|B)1_B+P(A|B^c)1_{B^c}$$
Knowing the result, it is easy to prove it. But I am wondering if I don't know the RHS beforehand, how would I obtain it from the left? In other words, what is the intuition behind this equality?
$\mathbb{E}\left[\mathsf{1}_{A}\mid\mathcal{G}\right]$ is measurable wrt $\mathcal{G}$ which means that constants $c,d\in\mathbb R$ exists with: $$\mathbb{E}\left[\mathsf{1}_{A}\mid\mathcal{G}\right]=c\mathsf{1}_{B}+d\mathsf{1}_{B^{\complement}}$$
Next to that by definition: $$\mathbb{E}\mathsf{1}_{A}\mathsf{1}_{B}=\mathbb{E}\left[\mathbb{E}\left[\mathsf{1}_{A}\mid\mathcal{G}\right]\mathsf{1}_{B}\right]\text{ and }\mathbb{E}\mathsf{1}_{A}\mathsf{1}_{B^{\complement}}=\mathbb{E}\left[\mathbb{E}\left[\mathsf{1}_{A}\mid\mathcal{G}\right]\mathsf{1}_{B^{\complement}}\right]$$
Working this out we find: $$P\left(A\cap B\right)=cP\left(B\right)\text{ and }P\left(A\cap B^{\complement}\right)=dP\left(B^{\complement}\right)$$
Then $c$ can be recognized as $P\left(A\mid B\right)$ and $d$ as $P\left(A\mid B^{\complement}\right)$.