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Problem 12

# Describe the error in the following fallacious "proof" that $\mathrm{P} \neq \mathrm{NP}$$. Assume that$$\mathrm{P}=\mathrm{NP}$ and obtain a contradiction. If $$\mathrm{P}=\mathrm{NP}$$, then $$S A T \in \mathrm{P}$$ and so for some $$k$$, $$S A T \in T I M E\left(n^{k}\right)$$. Because every language in NP is polynomial time reducible to $$S A T$$, you have NP $\subseteq \operatorname{TIME}\left(n^{k}\right) .$$Therefore,$$\mathrm{P} \subseteq \operatorname{TIME}\left(n^{k}\right) .$ But by the time hierarchy theorem, $$\operatorname{TIME}\left(n^{k+1}\right)$$ contains a language that isn't in $$\operatorname{TIME}\left(n^{k}\right)$$, which contradicts $\mathrm{P} \subseteq \operatorname{TIME}\left(n^{k}\right)$$. Therefore,$$\mathrm{P} \neq \mathrm{NP}$.

Expert verified
The error in the fallacious "proof" is the incorrect implication that NP ⊆ TIME(n^k) because SAT ∈ TIME(n^k). If P = NP, it means that there exists a polynomial-time algorithm for each problem in NP, not necessarily within a specific n^k complexity. Also, the "proof" misuses the Time Hierarchy Theorem to reach a contradiction, which does not hold. Consequently, this "proof" does not prove that P ≠ NP.
See the step by step solution

## Step 1: Understand the Assertion

First, understand that the assertion is that P != NP, and the "proof" is trying to reach a contradiction by assuming that P = NP.

## Step 2: Analyze the Assumption and Reasoning

The assumption that the "proof" starts with is that P = NP. It then claims that since SAT ∈ TIME(n^k) and every language in NP is polynomial time reducible to SAT, NP ⊆ TIME(n^k). This is the first place to look for an error in reasoning.

## Step 3: Identify the Error in the Implication of NP ⊆ TIME(n^k)

The error in the reasoning is the implication that NP ⊆ TIME(n^k) just because SAT ∈ TIME(n^k). Remember that just because SAT has a polynomial-time solution, it doesn't mean that all problems in NP can be solved in that specific polynomial time. If P = NP, it means that there exists a polynomial-time algorithm for each problem in NP, not necessarily within a specific n^k complexity.

## Step 4: Understand the Misuse of Time Hierarchy Theorem

In the "proof", the Time Hierarchy Theorem is used to point out that TIME(n^(k+1)) contains a language that isn't in TIME(n^k). While this statement is true, it cannot be concluded that P ⊆ TIME(n^k) since P is the class of all decision problems that can be solved by a deterministic Turing machine in polynomial time, which could include any polynomial and not just n^k.

## Step 5: Realize the Contradiction Does Not Hold

The contradiction in the "proof" that P ⊆ TIME(n^k), but there is a language not in TIME(n^k) does not hold. As explained in the previous steps, the assumption that NP ⊆ TIME(n^k) from the fact that SAT ∈ TIME(n^k) is incorrect, and therefore the contradiction does not prove that P != NP. In conclusion, the error in the fallacious "proof" lies in the implication that NP ⊆ TIME(n^k) based on SAT ∈ TIME(n^k), and the misuse of the Time Hierarchy Theorem to reach a contradiction. Therefore, this "proof" does not provide evidence that P != NP.

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