I've been thinking about conversion from CNF to DNF. Assume a worst case CNF formula with $k$ disjunctions, each containing exactly $l$ elements and no variable is used twice. Example with $k=3$ and $l=2$:$(a \lor b) \land (c \lor d) \land (e \lor f)$Obviously, this will result in a DNF of length $k^l$, as all possible conjunction combinations need to be listed and due to all variables being unique, nothing will vanish.Given that, computation time complexity will at least be exponential and lay in $o(k^l)$.Now, two questions:Can we say that time complexity will also be within $o(exp(n))$, where n denotes formula length?If 1. is true, wouldn't that prove that this problem isn't part of $P$, and, given CNF -> DNF is NP-hard NP-complete, that $P \neq NP$?Edit: I'm aware CNF -> DNF is NP-hard, but is it NP-complete too? 
What's wrong here, or, is CNF to DNF conversion in o(exp(n))?
complexity theory;complexity classes
Can we say that time complexity will also be within $o(exp(n))$, where $n$ denotes formula length?No, because no one has proven that distribution of terms is the only way to convert CNF to DNF.  Obviously simple distribution of terms produces an exponential blowup in the formula size, but there may be other ways to compress the formula so that the same truth table is produced.  We already known that given a truth table clever algorithms can produce smaller formulas than full disjunction over the distributed terms.If 1. is true, wouldn't that prove that this problem isn't part of $P$, and, given CNF -> DNF is NP-hard NP-complete, that $P \neq NP$?No.  If CNF to DNF conversion could be accomplished in polynomial time, its output would necessarily fit into polynomial space.  Therefore you could scan this output in polynomial time and decide satisfiability.  Thus P would be equivalent to NP.  But if CNF to DNF conversion takes exponential time it says nothing about whether some other algorithm could efficiently solve SAT and bring P and NP together.