I'm still in the process of grokking computational complexity.
However, I came across a statement like the above in an old midterm paper I'm reviewing, and I'm not sure I completely follow its logic.
$NP$ is the class of solveable decision problems that can be verified in polynomial time.
So when we say that $A$ is a NP Complete problem, it is understandably contained within the $NP$ class.
However, if $\bar{A}$ is included in the class $NP$, wouldn't these two postulates essentially indicate that ALL problems are $NP$?
Any insight into how to look at this would be much appreciated!