CNF Satisfiability Problem is NP-Complete

Theorem
The Conjunctive Normal Form Boolean Satisfiability Problem (Abbreviated as CNF SAT) is NP-complete.

CNF SAT is NP
Given a CNF SAT problem a potential solution can be verified in polynomial  time by checking every clause in $L$ to see if they all have at least  one true un-negated variable or one false negated variable.

Because a solution can be verified or rejected in polynomial time the problem is NP

CNF SAT is NP-hard
We will show this by reducing the Boolean Satisfiability(SAT) problem to CNF SAT.

The algorithm to convert the SAT problem to CNF SAT is recursive. Wherever $A$, $B$,and $C$ are seen in the output it is understood that the  algorithm would call itself on those formulas and convert them into CNF.

iff and xor
The first step will be to remove every instance mutual implication and exclusive or from the problem. In every clause with one of these functions there must be at least one that does not have any instance of mutual implication or exclusive or in its arguments. Introduce a new variable, $x$, and replace $A \implies B$ with $x$ or replace $\neg (A \implies B)$ with $\neg x$ as appropriate and add the following four clauses to $L$.
 * $A \lor B \lor x$
 * $A \lor \neg B \lor \neg x$
 * $\neg A \lor B \lor \neg x$
 * $\neg A \lor \neg B \lor x$

Repeat this procedure until there are no more mutual implications or exclusive ors in the list of clauses. It is important to note that the amount this procedure adds to the size of the final problem is bounded. Specifically, every instance of mutual implication or exclusive or increases the number of variables by one, increases the number of "new" symbols in $L$ by a fixed number and reproduces every "old" symbol in $L$ at most four times. In total, this step increases the size of the problem at most linearly.

The remaining functions that need to be considered can be converted to CNF with only a constant amount of increase to the size by a constant amount per production.

Given a logical expression it will either be a variable, a unary function, or a binary function.

Single Variable
If $A$ is a single variable then it is in CNF form and we are done

Unary functions
With the two constant functions can simply replace the replace $f_F \left( A  \right)$ and $f_T \left( A \right)$ with $F$ and $T$ respectively. These can be removed from the final expression rather easily.

The identity function is also fairly trivial. Simply replace $f_I \left( A \right)$ with the CNF conversion of $A$.

With the $\neg$ function the replacement will depend on the argument. If the argument is a single variable ($x$) then the output $\neg x$ is in  CNF. If the argument is one of the constant functions then replace $\neg f_F \left( A \right)$ and $\neg f_T \left( A \right)$ with $T$  and $F$ respectively. If the argument is a binary function of the form $f \left({A, B}\right)$ then replace the function with the appropriate  one, following the list below.


 * $\neg f_F \left({A, B}\right) \to f_T \left({A, B}\right)$
 * $\neg f_T \left({A, B}\right) \to f_F \left({A, B}\right)$


 * $\neg A \land B \to A \uparrow B$
 * $\neg A \uparrow B \to A \land B$


 * $\neg (A \implies B) \to \neg (A \implies B)$
 * $\neg \neg (A \implies B) \to (A \implies B)$


 * $\neg \operatorname{pr}_1 \left({A, B}\right) \to \overline {\operatorname{pr}_1} \left({A, B}\right)$
 * $\neg \overline {\operatorname{pr}_1} \left({A, B}\right) \to \operatorname{pr}_1 \left({A, B}\right)$


 * $\neg (B \implies A) \to \neg (B \implies A)$
 * $\neg \neg (B \implies A) \to (B \implies A)$


 * $\neg \operatorname{pr}_2 \to \overline {\operatorname{pr}_2}$
 * $\neg \overline {\operatorname{pr}_2} \to \operatorname{pr}_2$


 * $\neg A \lor B \to A \downarrow B$
 * $\neg A \downarrow B \to A \lor B$

Binary Functions
Replace $\neg f_T \left({A, B}\right)$ with $T$.

$A \lor B$ is already in CNF.

Replace $A \implies B$ with $\neg A \lor B$.

Replace $\operatorname{pr}_1 \left({A, B}\right)$ with $A$.

Replace $B \implies A$ with $A \lor \neg B$.

Replace $\operatorname{pr}_2 \left({A, B}\right)$ with $B$.

In the case of $A \land B$ the clause either takes the form $A \land B$ or  $(A \land B) \lor C$. In the first case simply replace the clause with the two clauses $A$ and $B$. In the second case introduce the new variable $x$ and replace the clause with the three clauses
 * $x \lor C$
 * $\neg x \lor A$
 * $\neg x \lor B$

A simple case analysis will show that the above clauses have a solution iff $(A \land B) \lor C$.

Replace $A \uparrow B$ with $\neg A \lor \neg B$ Replace $\overline {\operatorname{pr}_2} \left( {A, B} \right)$ with $\neg B$.

Replace $\neg (A \implies B)$ with $A \land \neg B$ and use the rule for $A \land B$.

Replace $\overline {\operatorname{pr}_2} \left( {A, B} \right)$ with $\neg A$.

Replace $\neg (B \implies A)$ with $\neg A \land B$ and use the rule for $A \land B$.

Replace $A \downarrow B$ with $\neg A \land \neg B$ and use the rule for $A \land B$.

Replace $f_F \left({A, B}\right)$ with $F$.

At this point there may be some instances of $T$ and $F$ in the clauses. They can be removed by deleting any clause from $L$ that contains $T$ in any of its conjunctions and removing $F$ from any clause that contains it. (Note that if any clause simplifies to $F$ the problem is trivially unsatisfiable and we are done.)

In the end the size and number of variables in the CNF problem is $O(n)$ where $n$ is the number of symbols required to write down the original SAT problem and the conversion can be done in polynomial time. SAT (an NP-hard problem) is reducible to CNF SAT. CNF is NP-hard.

CNF SAT is NP-complete.