Prove complement a^nb^nc^n is contextfree
You're overcomplicating. There is no need for the grammar to be deterministic, so it's OK for cases to overlap.

Say $\omega \in \overline{L_1}$. At least one of the following statements must be true:


$\omega = a^ib^j\nu \text{ where } i \ge 1, j\ge 0 \text{ and } i \ne j, \text{ and } \nu \text{ does not start with } b$
$\omega = a^ib^jc^k \text{ where } i, j, k \ge 1, \text{ and }j \ne k$(I don't require $i = j$, although if $i \ne j$ then the first condition also applies.)
$\omega \text{ does not start with } a$


These all have reasonably simple grammars.



Just in case it's not obvious, there are a few ways to do $a^jb^k | j\ne k$. One simple one:

$$\begin{align}
S &amp;\to A \mid B \\
A &amp;\to aC \mid aA \\
B &amp;\to Cb \mid Bb \\
C &amp;\to \epsilon \mid aCb
\end{align} $$

$C$ is a balanced sequence of $a$ and $b$, $A$ has more $a$s, $B$ has more $b$s, and $S$ has either more $a$s or more $b$s.