Prove associativity (to prove semiring) So we have this semiring $(S, +, *, 0, 1)$ (note that $+,*,0,1$ are for generic binary operations and the unit and zero elements as identities for corresponding operations, not actual plus and times).
If $r$ is a function from $S\to S$ then $r$ is a reduction for $S$ if for all $a$ and $b$ in $S$
\begin{align}
r(a) &= r(r(a)) \\
r(a+b) &= r(r(a) + b) = r(a + r(b))
\end{align}
If $r$ is a reduction then we have $\def\red{\mathop{\mathrm{red}}\nolimits}\red_r(S) = \def\sr\mathbin{+_r}\def\pr\mathbin{*_r}(S_r, \sr, \pr)$ where
\begin{align}
S_r &= \{ s \in S_r | r(s) = s\} \\
x \sr y &= r(x + y) \\
x \pr y &= r(x * y)
\end{align}
So we are asked if $\red_r$ is a semiring. If so prove it otherwise impose some conditions on $r$ that would guarantee that $\red_r(S)$ is a semiring.
The way I went about it is to try and prove that it is a semiring and if I can't see what conditions are needed for it to be.
So to be a semiring the following must be true:


*

*$(S_r, \sr)$ must be associative

*$(S_r, \pr)$ must be associative

*$\sr$ must be commutative

*RD and LD must hold

*$(S_r, \sr, 0)$ must be a commutative monoid (which I assume means that I have to prove there is an identity for the operation and I think it must also be that $r(0) = 0$ - itself, not something new)

*$(S_r, \pr, 1)$ must be a  monoid 

*identify for $\pr$ must be annihilator for $\pr$



3) $a \sr b = r(a + b) = r(b + a)$ because $+$ is operation of the original set which was a semiring and thus commutative and $b,a$ are part of $S_r$ which is a subset of $S = b \pr a$
5) we must have a zero element such that $a \sr 0_r = 0_r \sr a = a$. Using definition
$a \sr 0_r = r(a + 0_r)$ So if we use the zero element of the original semiring we will have
$r(a + 0) = r(a) = a$
6) $a \pr 1_r = a = r(a \pr 1_r)$ again if we use the unit of the original semiring we will have
$r(a * 1) = r(a) = a$
7) $a \pr 0 = 0$
$a \pr 0 = r(a * 0) = r(0) = 0$
So I am not sure about arbitrarily deciding to use the unit and zero from the original semiring and also I am not sure how to go about associativity.
I basically have to prove that $a \sr (b \sr c) = (a \sr b) \sr c$ but when I unfold the definitions I get
$$r(a + r(b + c)) = r(r(a + b) + c)$$
$+$ is closed to $S$ but I can't really say that is also closed to $S_r$ (if I could I could argue that $b+c$ map to $S_r$ so $r(b+c)$ is $b+c$ but it seems very unlikely)...
Any ideas?
BTW this is about a course using algebra to define network routing problems
 A: For the bit about using the same $0$ from the original ring:  you aren't sure that $0$ from the original ring is in $S_r$.  But whether it is or not, $r(0)$ should be the $0$ of $S_r$. (In fact it need not be:  any constant map is a reduction by your definition.)
For the associative part:  you are almost there.  You know by definition that $r(a+(b+c)) = r(a+r(b+c))$; now use associativity of $+$ in $S$ on the left hand side and apply the identity again.
What about $*$? In this case, these tricks don't work, and for good reason:  it's not necessarily true.  Here's an example:  let $S$ be the semiring of nonnegative-integer-coefficient polynomials in a single variable, and let $r$ be the map that removes the constant term.  Then one can check that $r$ is a reduction, but that there is no identity element for $*$ in $S_r$ as defined. 
One idea to try is introducing an additional condition, similar to the second one in your definition, except for multiplication.  You should be able to use that condition, together with the others, to prove distributivity and associativity for $*$.
