We know that the language AiBiCi $=\{$a$^$ib$^$ic$^$i $|$ i$>=0\}$ is not context$-$free. Show that the complement of AiBiCi is context$-$free. We can break it down into four cases:

$(1)$ w is in a$*$b$*$c$*,$ but the number of a$\text{'}$s and the number of b$\text{'}$s in w are not equal,

$(2)$ w is in a$*$b$*$c$*,$ but the number of b$\text{'}$s and the number of c$\text{'}$s in w are not equal,

$(3)$ w is in a$*$b$*$c$*,$ but the number of a$\text{'}$s and the number of c$\text{'}$s in w are not equal,

$(4)$ w is not in a$*$b$*$c$*.$

Show that each of the four cases are context$-$free, then since the context$-$free languages are closed under union, the complement of AiBiCi must be context$-$free.