Let $(\Omega, \mathcal A,P)$ be a probability space and $A,B,T\in \mathcal A$. I want to verify the identity $$P(A\mid B)=P(A\mid T)P(T\mid B)+P(A\mid T^c)P(T^c\mid B)$$
If I draw a probability tree then I can clearly see where the formula comes from by following the event $A$ from the top.
However, I would much rather like to prove the identity above using explicit formulas. I tried the following: $$P(A\mid T)P(T\mid B)+P(A\mid T^c)P(T^c\mid B)$$ $$= \frac{P(A\cap T)}{P(T)}\cdot\frac{P(T\cap B)}{P(B)} + \frac{P(A\cap T^c)}{P(T^c)}\cdot\frac{P(T^c\cap B)}{P(B)}$$ $$= \cdots$$
But here is where I am stuck. How can I show that this is equal to $P(A\mid B)$ ?
Your identity ought to say: $$ P(A\mid B)=P(A\mid T\cap B)P(T\mid B)+P(A\mid T^c\cap B)P(T^c\mid B). $$
I suspect there was at least a tacit assumption that $T\subseteq B$ and $T^c\subseteq B$, so that $P(T)=P(T\cap B)$. That would imply $P(B)=1$. In that case, changing $P(T)$ to $P(T\cap B)$ you your denominators, or changing $P(T\cap B)$ to $P(T)$ in your numerators, would be correct and results in a cancellation.