# Closure properties of a non-regular language under complement? [duplicate]

• Assume I have L1 which is a regular language, so we know since regular language is closed under complement, the complement of L1 is also a regular language.
• But let's say if the complement of L1 is a non-regular language, is it safe to conclude that L1 is a non-regular language as well?

Since I'm trying to prove a language L1 is not a regular language, and the pumping lemma doesn't work well with this case. But I can easily prove the complement of L1 is not regular, I'm wonder if that option is possible.

## marked as duplicate by Apass.Jack, xskxzr, Evil, Yuval Filmus formal-languages StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); May 21 at 16:33

Yes, non-regular languages are closed under complement as well.

Suppose the complement of L1 is a non-regular language. If L1 is regular, then "the complement of L1 is also a regular language", which is not true. Hence L1 cannot be regular.

More generally, suppose we have defined a collection of languages as myLanguages. Then

myLanguages are closed under complement $$\Longleftrightarrow$$ non-myLanguages are closed under complement

For example, we have

• non-context-free languages are not closed under complement.
• non-context-sensitive languages are closed under complement.
• non-deterministic-context-free languages are closed under complement.