# Why does $L\subseteq \textbf{P} \cap \textbf{NP}$ is $\textbf{NP}$-complete imply $\textbf{NP} = \textbf{P}$? [duplicate]

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If I show that a language $$L$$ is contained in $$\textbf{P}$$ and $$\textbf{NP}$$ and I know that the language is $$\textbf{NP}$$-complete, why did I proof that $$\textbf{P} = \textbf{NP}$$?

## marked as duplicate by Pål GD, Derek Elkins, David Richerby complexity-theory 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(); } ); }); }); Jan 24 at 10:51

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• If an NP-complete problem is solvable in polynomial time, then by definition NP=P. Check other questions on the same topic. – Pål GD Jan 24 at 10:10
• I suggest going over the basic definitions of completeness and checking out other related questions on the site, such as the one mentioned here in the comments. – Dean Gurvitz Jan 24 at 10:31

## 1 Answer

Because you have then showed that $$L$$ is an $$\textbf{NP}$$-complete language which, since $$L \in \textbf{P}$$, is decidable in poly-time. Since any other language $$L' \in \textbf{NP}$$ is efficiently reducible to $$L$$ (because of $$\textbf{NP}$$-completeness), $$L' \in \textbf{P}$$ as well. It follows that $$\textbf{NP} \subseteq \textbf{P}$$ (and the other inclusion is trivial).