Describe the error in the following fallacious “proof” that P ≠ NP. Assume that P = NP and obtain a contradiction. If P = NP, then SAT ∈ P and so for some k, SAT ∈ TIME(n^k). Because every language in NP is polynomial time reducible to SAT, you have NP ⊆ TIME(n^k). Therefore, P ⊆ TIME(n^k). But by the time hierarchy theorem, TIME(nk+1) contains a language that isn’t in TIME(n^k), which contradicts P ⊆ TIME(n^k). Therefore, P ≠ NP.