Codex @codex  0

Let $L=\{0^m{1}^n:m,n\ge 0,m\ne n\}$, and suppose it were a regular language. The formal language $B=0^{\ast }{1}^{\ast }$ is a regular language. By closure of regular languages under set difference,would be a regular language. But the pumping lemma for regular languages rules this out: for pumping length $p$, every permitted pumped block in $0^p{1}^p$ consists of $k>0$ zeros, and deleting it produces $0^{p-k}{1}^p$, with unequal counts. This contradiction provesThe closure step can also be realized by a product automaton accepting when the automaton for $B$ accepts and the automaton for $L$ rejects. If the exponents are instead required to be positive, use $B=0^{+}{1}^{+}$; the same argument applies.