Clifford Theorem as an easy corollary of Riemann-Roch Theorem I'm studying Fulton's algebraic curves book and on page 109 he proves the Clifford's theorem:

I have these doubts:

1.Why does he consider only the divisors $D\ge 0$ and $W-D\ge 0$?
2.What does Fulton mean by "since otherwise we work with $D-P$ and get a better inequality"?

Any help is very welcome and appreciated.
I really need help.
Thanks a lot.  
 A: First note that if the theorem is true for some $D$, it is also true for any linearly equivalent divisor. Moreover if $l(D)>0$, then $D$ is linearly equivalent to an effective divisor. So you can safely assume that $D$ is effective.
Assume that $D$ is effective. You can also assume that $W-D$ is effective by replacing $W$ by an other one : if $l(W-D)>0$ then there exist $f$ such that $W-D+\operatorname {div}(f)\geq 0$, hence put $W'=W+\operatorname{div}(f)$. It is linearly equivalent to $W$ (hence canonical) and $W'-D\geq 0$.
For your second question : this is some kind of argument by induction. So you can do as follows for more explanations on this step.
You want to prove Clifford theorem by induction on the degree of $D$. (Note that the hypotheses implies $\deg D\geq 0$). The case $\deg D=0$ is trivial. Assume the conclusion of Clifford theorem holds for $\deg D<n$ and let $D$ be a divisor of degree $n$ that satisfies the hypotheses. If $l(D-P)=l(D)$ for some point $P$, then $D-P$ satisfies the hypotheses : $l(D-P)=l(D)>0$ and $l(W-(D-P))\geq l(W-D)>0$. Since it is of degree strictly less that $D$, by induction
$$l(D-P)\leq \frac{1}{2}\deg(D-P) + 1$$
but since $l(D-P)=l(D)$ and $\deg(D-P)=\deg P-1$, replacing these equalities in the inequality above gives an inequality that is better that the conclusion, namely :
$$ l(D)\leq \frac{1}{2}\deg D + \frac{1}{2} < \frac{1}{2}\deg D+1.$$
Hence you can assume that $l(D-P)\neq l(D)$ for all $P$.
A this point, change $D$ and $W$ so that $D$ and $W-D$ are effective as in your first question, and finish the proof.
