TITLE: if P=NP, then is E=NE?
QUESTION [6 upvotes]: If the computational complexity class P equals to NP, does the complexity class E equal to the class NE?
E is defined as $DTIME(O(2^{O(n)}))$
NE is defined as $NTIME(O(2^{O(n)}))$
Thank you very much.

REPLY [13 votes]: Yes, by using the padding argument.  The Wikipedia article on the padding argument contains a proof of the implication P=NP ⇒ EXP=NEXP, and the implication P=NP ⇒ E=NE can be proved in the same way.  I leave the details as an exercise.