# Reduction of complement from complexity class co-np and p

Let P $$\neq$$ NP. D is in the complexity class co-NP. B is in the complexity class P. Let $$\bar{D}$$ be the complement of D, then $$\bar{D}$$ $$\leq _ {p}$$ B.

Is this statement true or false? My guess is that it's wrong because D is $$\in$$ co-NP and therefore $$\bar{D}$$ is $$\in$$ NP. From the reduction $$\bar{D}$$ $$\leq_ {p}$$B, it would follow that P = NP, which contradicts to the assumption.

Is that correct or am I having a mistake?

• D and B could just be in P. – Yuval Filmus Feb 28 at 6:59
• Can you explain why D could be in P? Does that mean if D is in co-np, that the complement of D is not in NP? – user101036 Feb 28 at 7:13
• @x_x Because D is simply in coNP, not coNP-complete. – dkaeae Feb 28 at 7:27
• But isn't coNP all problems, that have a complement in NP? – user101036 Feb 28 at 12:06