Is it try what SAT for a boolean formula in CNF in NP-full?
And for DNF it is trivial?
But can we just transform any CNF to DNF?
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Yes, for DNF it is trivial ... but note that converting a CNF into DNF is very costly: you have to do a general distribution of all terms over all terms. For example, suppose you have a CNF with $5$ conjuncts, and each conjunct has $5$ literals. That doesn't look so bad, right? Well, the DNF would have $5^5=3125$ disjuncts! So it may take a while before you run into a term that is satisfiable ... if there is one at all.
Indeed, converting from CNF to DNF is in effect a brute force way to see if you can find a way to make at least one literal in each conjunct in the CNF true, because in the transformation you are systematically creating all such combinations of literals ... so no, there is no paradox here.
In sum, the problem in your thinking is that we can 'just' transform any CNF to DNF in order to solve the SAT problem. No, that conversion takes exponential time.
Bram28
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