Intuition behind why NP could be equals to co-NP , if we "assume" P not equals NP

Question

if we "assume" P not equals NP , this and this suggests NP could be equals to co-NP .

I searched a lot , and went through a lot of materials . I can't seem to quite intuitively understand why this could be true . It would be really helpful if someone can lead me to any resource or help me intuitively understand why this could be true .

Answer 1

Here's one possible intuition, which could be totally misguided given that we haven't yet resolved P versus NP.

The complexity class P can be thought of as all problems for which the following statement is true: there's a boolean function solveProblem such that

• solveProblem runs in polynomial time in the size of its input,
• for any input x where the answer is "yes," solveProblem(x) returns true, and
• for any input x where the answer is "no," solveProblem(x) returns false.

Now, let's look at the definition of NP. It's the set of problems for which the following statement is true: there's a boolean function checkYesAnswer such that

• checkYesAnswer runs in polynomial time,
• for any input x where the answer is "yes", there is some y where checkYesAnswer(x, y) returns true, and
• for any input x where the answer is "no," checkYesAnswer(x, y) always returns false.

The difference between P and NP, then, is that when you're looking at the complexity class P, you're given just the input and need to make a yes/no determination, whereas in the class NP there may be some auxiliary information that helps you make a yes/no decision. That makes the distinction between the definitions of P and NP pretty large - in one case you need to solve the problem, and in the other you need to check whether some auxiliary information helps you.

Now, here's another, equivalent definition of co-NP. The class co-NP consists of all problems where there's a boolean function checkNoAnswer such that

• checkNoAnswer runs in polynomial time,
• for any input x where the answer is "yes," checkNoAnswer(x, y) always returns true, and
• for any input x where the answer is "no," there is a choice of y where checkNoAnswer(x, y) returns true.

In other words, the definitions of NP and co-NP are very closely related to one another, with the only difference being that the branches for "yes" and "no" are swapped. So in that sense, even if P ≠ NP, it still could be plausible that NP and co-NP could be equal, since we're basically just flipping the role of the words "yes" and "no."

There's also good reasons to see why NP might not be co-NP. You can think of the distinction between the definitions as follows: with NP, the answer is "yes" if there exists some y with some property, whereas with co-NP the answer is "yes" if for all y something holds. And since the ∃ and ∀ quantifiers work differently, perhaps there's something else going on here.

Answer 2

If P=NP, then NP=co-NP. The proof for this statement critically depends on the assumption P=NP.

Now if P!=NP, the proof does not work, and so the question whether NP=co-NP remains open. It's as simple as that.