 Solution

Context-free languages are closed under union: use a new start symbol with alternatives for two renamed context-free grammars. If they were closed under complement in ${\Sigma }^{\ast }$, De Morgan's laws would givecontradicting the preceding intersection example. Therefore some CFL has a non-context-free complement. This is an existence argument; it does not assert that both complements of the particular $L_1,L_2$ are non-context-free.

A concrete choice is $L=\{a,b,c{\}}^{\ast }\setminus \{a^n{b}^n{c}^n:n\ge 0\}$. Words outside $a^{\ast }{b}^{\ast }{c}^{\ast }$ form a regular language. Among words inside it, membership in $L$ means the first two counts or the last two counts differ. Each unequal-count comparison has a context-free grammar: for $a^i{b}^j$ with $i>j$, use $T\to aTb\mid A$, $A\to aA\mid a$; for $i<j$, replace $A$ by $B\to bB\mid b$. Append any number of $c$'s, or prefix any number of $a$'s for the analogous $b,c$ comparison. Taking these unions gives a context-free language whose complement is the non-context-free equal-three-count language. These productions are definitions of the example, not a restriction to strict Chomsky normal form.