In 2002, I published a theorem, #P=#Q, that says the number of satisfying assignments equals the number of valid quantifications for any boolean formula. I have done alot of followup work, and am interested in discussion. Fairly recently, I discovered a proof that coNP=NP=#P. I have other results too, that would benefit from discussion. Daniel Pehoushek The Proof of coNP = NP = #P A global notion of validity is introduced. Essentially, the declaration is that any one of all possible solutions can be used to return an answer by the coNP (resp. NP) program. If some solutions are unusable, the program is invalid, because it could then return wrong answers with small effort. With global validity, equality is nearly definitional among the three classes. The dimacs proof is just for fun, showing the proof is easy and uncircular. coNP: Are there zero solutions? NP: Is there at least one polynomially verifiable solution? #P: The problem of counting all solutions. When
Equality in the Hierarchy (coNP=NP=#P=#Q)