Obtain CNF of (p -> q) ^ (¬q <-> ¬p) and obtain DNF of (p v q) -> (¬(p v q))

Steps to Solve

1. Identify the logical expressions
   The expressions given for transformation are:

   1. For CNF: $(\neg P \rightarrow \neg X) \land (\neg Q \leftrightarrow \neg P)$
   2. For DNF: $P \lor (Q \land P)$

2. Convert implications and biconditionals for CNF
   Recall that:

   • $A \rightarrow B$ is equivalent to $\neg A \lor B$
   • $A \leftrightarrow B$ is equivalent to $(A \rightarrow B) \land (B \rightarrow A)$

   For the expression $(\neg P \rightarrow \neg X) \land (\neg Q \leftrightarrow \neg P)$, we can rewrite it:

   • $\neg P \rightarrow \neg X$ becomes $\neg(\neg P) \lor \neg X \equiv P \lor \neg X$
   • $\neg Q \leftrightarrow \neg P$ becomes $(\neg Q \rightarrow \neg P) \land (\neg P \rightarrow \neg Q)$, which simplifies as follows:
     • $\neg Q \rightarrow \neg P$ becomes $Q \lor \neg P$
     • $\neg P \rightarrow \neg Q$ becomes $P \lor \neg Q$

   Putting it all together: $$(P \lor \neg X) \land ((Q \lor \neg P) \land (P \lor \neg Q))$$

3. Distribute to achieve CNF
   The CNF can be simplified to a conjunction of disjunctions:
   $$ (P \lor \neg X) \land (Q \lor \neg P) \land (P \lor \neg Q) $$

4. Simplifying DNF
   For the expression $P \lor (Q \land P)$:

   • Use the distributive property:
     $$ P \lor (Q \land P) = P $$
     (since if $P$ is true, the entire expression is true regardless of $Q$)