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 2 Added "important" technical side-condition. edited Jun 20 at 9:19 David Richerby 74.7k1616 gold badges117117 silver badges208208 bronze badges [...] the structure formerly known as the Polynomial Hierarchy collapses to the level above $$\text{P}=\text{NP}$$. This claim makes no sense. If $$\text{P}=\text{NP}$$, then the whole polynomial hierarchy is equal to $$\text{P}$$ and there is no level above that. That is, we show that $$\text{co-NP}\subseteq\text{NP}\setminus{P}$$. $$\text{co-NP}\subseteq\text{NP}\setminus{P}$$ is unconditionally false, since $$\text{P}\subseteq\text{co-NP}$$ (and $$\mathrm{P}\neq\emptyset)$$. I wouldn't recommend spending any time on this paper. [...] the structure formerly known as the Polynomial Hierarchy collapses to the level above $$\text{P}=\text{NP}$$. This claim makes no sense. If $$\text{P}=\text{NP}$$, then the whole polynomial hierarchy is equal to $$\text{P}$$ and there is no level above that. That is, we show that $$\text{co-NP}\subseteq\text{NP}\setminus{P}$$. $$\text{co-NP}\subseteq\text{NP}\setminus{P}$$ is unconditionally false, since $$\text{P}\subseteq\text{co-NP}$$. I wouldn't recommend spending any time on this paper. [...] the structure formerly known as the Polynomial Hierarchy collapses to the level above $$\text{P}=\text{NP}$$. This claim makes no sense. If $$\text{P}=\text{NP}$$, then the whole polynomial hierarchy is equal to $$\text{P}$$ and there is no level above that. That is, we show that $$\text{co-NP}\subseteq\text{NP}\setminus{P}$$. $$\text{co-NP}\subseteq\text{NP}\setminus{P}$$ is unconditionally false, since $$\text{P}\subseteq\text{co-NP}$$ (and $$\mathrm{P}\neq\emptyset)$$. I wouldn't recommend spending any time on this paper. 1 answered Jun 19 at 9:05 David Richerby 74.7k1616 gold badges117117 silver badges208208 bronze badges [...] the structure formerly known as the Polynomial Hierarchy collapses to the level above $$\text{P}=\text{NP}$$. This claim makes no sense. If $$\text{P}=\text{NP}$$, then the whole polynomial hierarchy is equal to $$\text{P}$$ and there is no level above that. That is, we show that $$\text{co-NP}\subseteq\text{NP}\setminus{P}$$. $$\text{co-NP}\subseteq\text{NP}\setminus{P}$$ is unconditionally false, since $$\text{P}\subseteq\text{co-NP}$$. I wouldn't recommend spending any time on this paper.