Weird law of total probability-like identity

Question

Is this true:

$$ P({+}c\mid {+}s) = P({+}c\mid{+}s,{+}r) P({+}r) + P({+}c\mid{+}s,{-}r)P({-}r) $$

, where ${+}x$ means $x = $true and $-x$ means $x= $false ($s,c,r$ are all boolean variables) and the commas$(,)$ mean "AND". To me, it looks like the Law of total probability: $p(a) = p(a\mid x)\cdot p(x) + p(a\mid \lnot x)\cdot p(\lnot x)$, with $a$ somehow replaced (???) by ${+}c\mid{+}s$ and $x$ by ${+}r$

Answer 1

Idiosyncratic notation aside, and assuming that $S$ and $R$ are independent, then this is indeed the Law of Total Probability for a conditional probability expression.

$$\def\P{\operatorname{\sf P}}\begin{align} \P(C\mid S) ~&=~ \dfrac{\P(C,S)}{\P(S)}&&\text{conditional probability}\\[1ex]&=~\dfrac{\mathsf P(C,S,R)+\P(C,S,R^\complement)}{\P(S)} && \text{Law of Total Probability}\\[1ex]&=~\dfrac{\mathsf P(C\mid S,R)\P(S,R)+\P(C\mid S,R^\complement)\P(S,R^\complement)}{\P(S)} &&\text{conditional probability} \\[1ex]&=~ \P(C\mid S, R) \P(R\mid S) + \P(C\mid S,R^\complement)\P(R^\complement\mid S) &&\text{conditional probability}\\[1ex] &=~ \P(C\mid S, R) \P(R) + \P(C\mid S,R^\complement)\P(R^\complement) &~\star~& \text{independence assumed}\\[3ex] \P({+}c\mid{+}s)~&=~\P({+}c\mid {+}s,{+}r)\P({+}r)+\P(+c\mid {+}s,{-}r)\P({-}r) && \text{notational equivalence}\end{align} $$

Answer 2

The equation is correct when $S$ and $R$ are independent of each other.

Hint

$$P(X|Y,Z) = \frac{P(X,Y,Z)}{P(Y,Z)}$$

If $Y$ and $Z$ are independent,

$$P(Y,Z) = P(Y)P(Z)$$

And finally,

$$\sum_{z}P(X,Y,Z=z) = P(X,Y)$$

Using above equations and the assumption, it is easy to derive your identity.