I have a question about the last step in the proof of Mason's Theorem. I will write it.
Mason's Theorem. If $a, b$ and $c$ are relatively prime polynomials such that $a + b = c$, then $$\max \{ \deg(a) , \deg(b) , \deg(c)\} \leq N_0(abc) -1,$$ where $N_0(f)$ is the number of distinct factors of $f$.
Proof: Let $\displaystyle a(t) = \prod_{i=1}^{s_a} (t-\alpha_i)^{u_i},~ b(t) = \prod_{i=1}^{s_b}(t-\beta_i)^{v_i}, ~c(t) = \prod_{i=1}^{s_c}(t-\gamma_i)^{w_i} $
$N_0(abc) = s_a + s_b + s_c$. Let $\displaystyle f = \frac{a}{c},~ g= \frac{b}{c}$. Hence $f + g = 1 \Rightarrow f' + g' = 0 \Rightarrow f' = - g'$.
$\displaystyle\frac{f'}{f} = \sum_{i=1}^{s_a} \frac{u_i}{t-\alpha_i} - \sum_{i=1}^{s_c} \frac{w_i}{t-\gamma_i}, ~~ \frac{g'}{g} =\sum_{i=1}^{s_b} \frac{v_i}{t-\beta_i} - \sum_{i=1}^{s_c} \frac{w_i}{t-\gamma_i}$
$\displaystyle \frac{b}{a} = \frac{g}{f} =-\frac{f'/f}{g'/g}=-\frac{M f'/f}{M g'/g},\tag{1} $
where $M =\prod_{i=1}^{s_a}(t-\alpha_i) \prod_{i=1}^{s_b}(t-\beta_i)\prod_{i=1}^{s_c}(t-\gamma_i)$.
So the denominator and nominator in $(1)$ has the degree at most $s_a + s_b + s_c -1 $. Then we conclude the inequality of the theorem.
I understand all the proof except the last thing.