In fact, the implication looks like an equivalence too.

All satisfying solutions to a boolean formula

are valid answers by programs for NP

is equivalent to saying

the class #P equals the class NP.

Any agreement? Looks trivial. However, complexity classes are untrivial, so… negating both sides of the equivalence also produces an equivalence: Not all solutions are valid is equivalent to #P!=NP.

I have a program for counting solutions to boolean formulas, and after some years of experience, I observed that SAT coNP and #P are roughly equivalent in time complexity. I have published previously a related result called #P=#Q, where the number of satisfying assignments equals the number of valid quantifications of any original boolean formula (Satisfiability 2002, Introduction to Qspace). My program runs fast on small to moderate sizes, and when #P is linear, I can decide all qbfs of the original formula.

My theoretical position is:

coNP=NP=#P=#Q=qbf=allQBFs=?Exp

with #P=#Q being the central thesis.

source code for program bob is available, send me email at gmail, pehoushek1. Sincerely, Daniel Pehoushek, 1983 GREs: 2380/2400