$P(X+Y=k | Y=y) $ to $ P(X=k-y)$ 
Let $ X ∼ Geom (p)$ and $Y ∼ U (0, 1, . . . , n)$ be random variables. Find the probability distribution of $Z = X + Y$ .

Don't give me the solution to the question above.
What I did so far:
$$ P(X+Y=k) $$
$$ = \sum_{y=0}^{n} P(X+Y=k | Y=y)*P(Y=y) $$
Now, my only question is, why I'm allowed to conclude the following:
$$ = \sum_{y=0}^{n} P(X=k-y)*P(Y=y) $$


*

*Logically it seems correct but then I can conclude the following as well: $P(X+Y=k \cap Y=y) $ to $ P(X=k-y)$. Is it correct also?

 A: By definition, $$P(X+Y=k)=\sum_{\{(x,y)|x+y=k\}} P(X=x \cap Y=y)$$
You write one of them in terms of $k$ and another one:
$$P(X+Y=k)=\sum_{y=0}^{n} P(X=k-y \cap Y=y)$$
If they're independent, $$P(X+Y=k)=\sum_{y=0}^{n} P(X=k-y)P(Y=y)$$
which is exactly what you want to have. If not,$$\sum_{y=0}^{n} P(X=k-y \cap Y=y)=\sum_{y=0}^{n} P(X=k-y | Y=y)P(Y=y)$$ and that's it. This step comes from conditioning on $Y$ (and total probability if you wish).

However, consider what you propose:$$P(X+Y=k)=\sum_{y=0}^{n}P(X+Y=k \cap Y=y)=\sum_{y=0}^{n}P(X=k-y)$$
The event $\{X+Y=k \cap Y=y\}$ implies $\{X=k-y\} \cap \{Y=y\}$. The reason you can't leave $\{Y=y\}$ is that you're kind of considering two events in one space. $y$ gives you only the iterator if you drop it (which doesn't have any meaning in this case). 
Another way to see this is by conditional probability. When considering $P(X=k-y | Y=y)$, you know that $\{Y=y\}$ has already happened. You're taking some information away in your proposition: 
Independent case: $P(X+Y=k \cap Y=y)=P(X+Y=k)P(Y=y)$. Since you're summing over $y$, $P(X+Y=k \cap Y=y)=P(X=k-y)P(Y=y)$ for each $y$.
Dependent case: $P(X+Y=k \cap Y=y)=P(X+Y=k|Y=y)P(Y=y)=P(X=k-y|Y=y)P(Y=y)$. 
In either way, $P(X+Y=k \cap Y=y)\neq P(X=k-y)$.
