SAT can be verified in polynomial time, by a conversion to CNF, then verifying the SAT of the CNF in polynomial. What is wrong with this argument?

Question

we know that any expression can be converted to a CNF in linear time where the CNF result is not 'logically equivalent' to the original expression but the 'satisfiability' is invariant meaning that the CNF is satisfiable if and only if the original expression is satisfiable (this can be done for example by calculating the 'Tseitin derivative'). Given that the satisfiability of the CNF can be checked in polynomial time, then the satisfiability of the original expression can be checked in polynomial time. Therefore, the satisfiability of any expression can be checked in polynomial time. However, the SAT problem is NP-complete so this is not true. But where am I going wrong here? I assume that the explanation comes from "CNF is not logically equivalent to the original expression" but then again this is not important in checking the satisfiability (?). Thank you in advance for any help.

Answer 1

CNF sat cannot be checked in linear time. In fact even the restricted 3-CNF SAT is NP-complete. See https://en.wikipedia.org/wiki/Boolean_satisfiability_problem#3-satisfiability.

Answer 2

Given that the satisfiability of the CNF can be checked in polynomial time

There is no known deterministic polynomial time algorithm for CNF formulas. The assumption you are making above is not known to be valid. This assumption does indeed imply P=NP so that not a problem in your reasoning. The problem is instead that you are 'begging the question', i.e. you have assumed the conclusion with this statement.