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

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 $$\textbf{P} = \textbf{NP}$$?

• If an NP-complete problem is solvable in polynomial time, then by definition NP=P. Check other questions on the same topic. Jan 24 '19 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. Jan 24 '19 at 10:31

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).