Can someone explain the solution to this statement? Say C: set of courses
P(x,y): 'x is a prerequisite for course y'
statement: 'some courses have several prerequisites'
symbolically: 
 ∃ x ∈ C, ∃ y ∈ C, ∃ z ∈ C, P(y, x) ∧ P(z, x) ∧ y ≠ z

I don't really understand how you get the symbolic expression from the verbal expression. 
Also, might there be a simpler way of writing this in logical notation?
In addition!
How would you write this:
No course has more than two prerequisites.

Thank you.
 A: The statement
$$\exists x\in C\,\exists y\in C\,\exists z\in C\Big(P(y,x)\land P(z,x)\land y\ne z\Big)$$
says that 

there is a course $x$ such that there are courses $y$ and $z$ that are prerequisites for $x$ and are different courses.

In less convoluted language, this says that there is a course $x$ that has at least two prerequisites, here called $y$ and $z$. The $y\ne z$ clause ensures that $y$ and $z$ aren’t just two names for the same course, i.e., that we really do have two prerequisites here, not one with an alias. The word several in the original statement is being interpreted as at least two. If we defined several to mean at least three, we’d need a more complicated expression:
$$\exists x\in C\,\exists y\in C\,\exists z\in C\,\exists w\in C\Big(P(y,x)\land P(z,x)\land P(w,x)\land y\ne z\land y\ne w\land z\ne w\Big)$$
Here the $y\ne z\land y\ne w\land z\ne w$ part says that no two of the prerequisites $y,z$, and $w$ are really the same course: we really do have three distinct prerequites here for the course $x$.
A: "Some courses have several prerequisites"
$\Rightarrow$ " $\exists x \in C$ such that $x$ has several prerequisites"
$\Rightarrow$ " $\exists x \in C$ such that $x$ has at least two different prerequisites"
$\Rightarrow$ " $\exists x \in C$, $ \exists y, z \in C$ such that $y \neq z$, $P(y, x) \wedge P(z, x)$ "
$\Rightarrow$ $\exists  x, y, z \in C$,  $y \neq z$, $P(y, x) \wedge P(z, x)$ "
For the second part, using similar reasoning we can symbolically write :
$\nexists x, y, z, w \in C$, $P(y,x) \wedge P(z, x) \wedge P(w, x) \wedge y \neq z \neq w $
