Verify two probability equations Suppose a new case has inputs $X = x$ and has target features $Y$; the aim is to compute $P(Y \mid X = x \land e)$, where $e$ is the set
of training examples. This is the probability distribution of
the target variables given the particular inputs and the examples.
The role of a model is to be the assumed generator of
the examples. If we let $M$ be a set of disjoint and covering
models, then reasoning by cases and the chain rule give:
\begin{equation}
P(Y \mid x \land e) = \sum_{m\in M} P(Y \land m\mid x \land e) 
\end{equation}
and
\begin{equation}
\sum_{m\in M} P(Y \land m\mid x \land e) = \sum_{m\in M} P(Y \mid m \land x\land e) \times P(m\mid x\land e)
\end{equation}
I want to verify these two equations with the theorems from the definition of probability.
http://artint.info/html/ArtInt_196.html 
I was thinking using maybe the chain rule but then I got stuck what kind of theorems can help me here? Also I got some what confused with $e$ the training examples and $M$ the set of disjoint models.
Thanks!
 A: The link you provided actually contains a link where some necessary hints can be found. For the first equation, let's assume that $M=\{m_i\}$ is countable, then: $$Y\wedge M=Y\wedge (m_1\vee m_2\vee\dotsb)=(Y\wedge m_1)\vee(Y\wedge m_2)\vee\dotsb$$ since $m_i$s are disjoint: \begin{align*}P(Y\wedge M)&=P\bigl((Y\wedge m_1)\vee(Y\wedge m_2)\vee\dotsb\bigr)\\&=P(Y\wedge m_1)+P(Y\wedge m_2)+\dotsb\\&=\sum_{m\in M}P(Y\wedge m).\end{align*} Using the basic addition rule: $$P(Y\vee M\mid x\wedge e)=P(Y\mid x\wedge e)+P(M\mid x\wedge e)-P(Y\wedge M\mid x\wedge e).$$ Here $P(Y\vee M\mid x\wedge e)=P(M\mid x\wedge e)=1$ for $M$ is covering, so $$P(Y\mid x\wedge e)=P(Y\wedge M\mid x\wedge e)=\sum_{m\in M}P(Y\wedge m\mid x\wedge e).$$ The second equation is easier to derive, basically \begin{align}\frac{P(Y \land m\mid x \land e)}{P(m\mid x \land e)}&=\frac{P(Y \land m\wedge x \land e)/P(x\wedge e)}{P(m\wedge x \land e)/P(x\wedge e)}\\&=\frac{P(Y \land m\wedge x \land e)}{P(m\wedge x \land e)}\\&=P(Y\mid m\wedge x\wedge e).\end{align} So $$P(Y \land m\mid x \land e)=P(Y\mid m\wedge x\wedge e)P(m\mid x \land e).$$
