Under the assumption that a PSPACE-hard problem
A can be solved in polynomial time, is the following argumentation valid?
Since every problem in PSPACE can be reduced to
A in polynomial time, every problem in NP which is a subset of PSPACE is included. That means that each Problem in NP is solvable in polynomial time and therefore P = NP.
Since coNP is a subset of PSPACE as well, we can argue that every problem in coNP can be reduced to
A and therefore be solved in polynomial time as well, which means that P = coNP.
Therefore we can say that P = NP = coNP.
I don't see a problem with this argumentation at the moment, but it seems a bit too simplistic to me, so is this a valid way of arguing for the equality above?