The proof that P = NP implies EXP = NEXP uses "padding".

${\displaystyle \mathrm {EXP} \subseteq \mathrm {NEXP} }$  by definition, so it suffices to show ${\displaystyle \mathrm {NEXP} \subseteq \mathrm {EXP} }$ .

Let L be a language in NEXP. Since L is in NEXP, there is a non-deterministic Turing machine M that decides L in time ${\displaystyle 2^{n^{c}}}$  for some constant c. Let

${\displaystyle L'=\{x1^{2^{|x|^{c}}}\mid x\in L\},}$ 

where '1' is a symbol not occurring in L. First we show that ${\displaystyle L'}$  is in NP, then we will use the deterministic polynomial time machine given by P = NP to show that L is in EXP.

${\displaystyle L'}$  can be decided in non-deterministic polynomial time as follows. Given input ${\displaystyle x'}$ , verify that it has the form ${\displaystyle x'=x1^{2^{|x|^{c}}}}$  and reject if it does not. If it has the correct form, simulate M(x). The simulation takes non-deterministic ${\displaystyle 2^{|x|^{c}}}$  time, which is polynomial in the size of the input, ${\displaystyle x'}$ . So, ${\displaystyle L'}$  is in NP. By the assumption P = NP, there is also a deterministic machine DM that decides ${\displaystyle L'}$  in polynomial time. We can then decide L in deterministic exponential time as follows. Given input ${\displaystyle x}$ , simulate ${\displaystyle DM(x1^{2^{|x|^{c}}})}$ . This takes only exponential time in the size of the input, ${\displaystyle x}$ .

The ${\displaystyle 1^{d}}$  is called the "padding" of the language L. This type of argument is also sometimes used for space complexity classes, alternating classes, and bounded alternating classes.

• Arora, Sanjeev; Barak, Boaz (2009), Computational Complexity: A Modern Approach, Cambridge, p. 57, ISBN 978-0-521-42426-4