Young, independent Markov, looking for a chain. I'm having trouble doing the "$\Longrightarrow$" part of the first proof, which is Lemma 1 from page 6, from these lecture notes. In the lecture notes the proof is a one-liner, from which I can't infer, how an explicit proof goes.
What is to be shown is that if $(X_m)_{m\geq 0}$ is a Markov process [ which is defined as meaning 
$$ (X_m)_{m>n} | X_n = i_n$$(which is a conditional random variable, taking values in $S\times S \times \cdots $, the set of infinitely long tuples from $S$, where $S$ denotes the state space) is independent of $$ (X_m)_{m<n} | X_n = i_n,$$(which similarly is  a conditional random variable, taking values in $S^n$) for any $n$ and $i_0,\ldots ,i_{n+1} \in S$], then $$P(X_{n+1}=i_{n+1} \mid X_n=i_n,\ldots, X_0=i_0)=P(X_{n+1}=i_{n+1} \mid X_n=i_n).$$
How can I do this ?
 A: Fix $i_n$ and define $P^{i_n}(B):=P(B|X_n=i_n)$ for any event $B$. Let $A$ be the event $\{X_{n-1}=i_{n-1},\ldots,X_0=i_0\}$. Then conditional independence implies
$$P^{i_n}(\{X_{n+1}=i_{n+1}\}\cap A)=P^{i_n}(X_{n+1}=i_{n+1})P^{i_n}(A).$$
Using the definition of conditional probability, we have
$$P(X_{n+1}=i_{n+1}\,|\,X_n=i_n,\ldots,X_0=i_0)=P^{i_n}(X_{n+1}=i_{n+1}\,|\,A)=\frac{P^{i_n}(\{X_{n+1}=i_{n+1}\}\cap A)}{P^{i_n}(A)}\\=\frac{P^{i_n}(X_{n+1}=i_{n+1})P^{i_n}(A)}{P^{i_n}(A)}=P(X_{n+1}=i_{n+1}\,|\,X_n=i_n).$$
EDIT: It seems like your main concern is notational, so I'll give some more details. If $B$ is some event, we might write $P^B(C):=P(C|B)$. The reason for this is that $P^B$ is just another probability measure (i.e. $P^B(\emptyset)=0$ and $P^B$ is countably additive), so we can do everything with $P^B$ that we can with $P$. Your main confusion seems to be how to make sense of $P^B(A|C)$, so let's unpack it using the definitions:
$$P^B(C|A):=\frac{P^B(C\cap A)}{P^B(A)}=\frac{P(C\cap A|B)}{P(A|B)}=\frac{\frac{P(C\cap A\cap B)}{P(B)}}{\frac{P(A\cap B}{P(B)}}=\frac{P(C\cap A\cap B)}{P(A\cap B)}=P(C|A\cap B).$$
This agrees with intuition: the conditional probability given $B$, of $C$, given $A$, is the same as the probability of $C$ given both $A$ and $B$. I did not explain that to you, so I apologize for that.
With that in mind, hopefully what you referred to as "the third line from the bottom" is now clear: in our case, $B$ is the event $\{X_n=i_n\}$, $C$ is the event $\{X_{n+1}=i_{n+1}\}$, and $A$ is as defined up top. Then what I have written is precisely $P(C|B\cap A)=P^B(C|A)$. The next equality is just definition of conditional probability: $P^B(C|A)=\frac{P^B(C\cap A)}{P^B(A)}$. And the next line is what uses our assumption of conditional independence: $C$ ("the future") and $A$ ("the past") are independent conditioned on $B$ ("the present"), which precisely means that $P^B(C\cap A)=P^B(C)P^B(A)$.
Perhaps there is also confusion where I write $\{X_n=i_n,X_{n-1}=i_{n-1}\}$ instead of $\{X_n=i_n\}\cap\{X_{n-1}=i_{n-1}\}$, but I would encourage you to get used to such abbreviations as they are abundant in probability theory. Likewise, the use of $P^{i_n}$ is particularly common in Markov chain analysis. For homogeneous Markov chains in particular, the initial distribution (that is, the distribution of $X_0$) is largely irrelevant and it is both easier and more useful to consider probabilities of the form $P^x:=P(\cdot|X_0=x)$ for each $x\in S$.
As a final note, I will ask you to please watch your tone in comments. This probably wasn't intentional, but you came across quite hostile at times when I was only trying to help.
