Distribution function (CDF) of the sum of two random variables + law of iterated expectations I'm taking my first probability class, and we're studying sums of independent random variables. We're using Ross's First Course in Probability. It states the definition of a convolution, but doesn't give much more depth into the subject. I checked out statlect.com's page on the sums of independent random variables, but I'm struggling with the first proof on the page.

It states the following theorem: Let $X$ and $Y$ be two independent random variables and denote $F_{X}(x)$ and $F_{Y}(y)$ their distribution functions. Let: $Z=X+Y$ and denote the distribution of $Z$ by $F_{Z}(z)$. The following holds: $F_{Z}(z) = E[F_{X}(z-Y)] = E[F_{Y}(z-X)]$

Proof:

$F_{Z}(z)=P(Z\leq z)\\\;\;\;\;\;\;\;\;\;=P(X+Y\leq z)\\\;\;\;\;\;\;\;\;\;=P(X\leq z-Y)\\\;\;\;\;\;\;\;\;\;=*E[P(X\leq z-Y|Y=y)]* \text{by the law of iterated sums}\\\;\;\;\;\;\;\;\;\;=E[F_{X}(z-Y)]$

The law of iterated sums states: $E[E[X|Y]]=E[X]$. I follow the proof of that theorem just fine. However, I'm struggling to see how it allows you to make the jump from $P(X\leq z-Y)$ to $E[P(X\leq z-Y|Y=y)]$.

In Ross's A First Course in Probability, the proof for $F_{Z}(z)=E[F_{X}(z-Y)]$ makes more sense. But I can't link the steps in his proof to see where iterated expectations come in to play. Any help would be greatly appreciated!
I will denote $P(X\le z-Y)$ as $F_X(z-Y)$, the pdf as $f()$, and integrals using $\int_{lb}^{ub}$:

First, write CDF of $z-Y$ as conditional probability on the value of $Y$.

$$F_X(z-Y) = \int_{y=-\infty}^{\infty} F_X(z-y|Y=y)f_Y(Y=y)dy   \qquad (1)$$

In other words, you are finding the cumulative probability that $X$ is less than $z-Y$ by conditioning on the value of the r.v. $Y$ and then integrating over all possible values of $Y$.

Recall the definition of "expectation"

$$E_Y[Y] = \int_{y=-\infty}^{y=\infty} yf_{Y}(y)dy     \qquad (2)$$

You will see that you can rewrite eqn $(1)$ as 

$$E_Y[F_X(z-y|Y=y)]$$

In general, when working with conditional probability/expectation proofs I find it helpful to write things out in terms of integrals.
I will denote $P(X\le z-Y)$ as $F_X(z-Y)$, the pdf as $f()$, and integrals using $\int_{lb}^{ub}$:

First, write CDF of $z-Y$ as conditional probability on the value of $Y$.

$$F_X(z-Y) = \int_{y=-\infty}^{\infty} F_X(z-y|Y=y)f_Y(Y=y)dy   \qquad (1)$$

In other words, you are finding the cumulative probability that $X$ is less than $z-Y$ by conditioning on the value of the r.v. $Y$ and then integrating over all possible values of $Y$.

Recall the definition of "expectation"

$$E_Y[Y] = \int_{y=-\infty}^{y=\infty} yf_{Y}(y)dy     \qquad (2)$$

You will see that you can rewrite eqn $(1)$ as 

$$E_Y[F_X(z-y|Y=y)]$$

In general, when working with conditional probability/expectation proofs I find it helpful to write things out in terms of integrals.