I have the following expression:
$$ (((p_3 \lor p_1) \ \land (p_2 \to p_1)) \ \land (p_3 \leftrightarrow p_2)) $$
Work so far:
1) CNF:
$ \begin{align*} &\ (((p_3 \lor p_1) \land (p_2 \to p_1)) \land (p_3 \leftrightarrow p_2)) & \text{(Given)} \\ \equiv &\ (((p_3 \lor p_1) \land (\neg p_2 \lor p_1)) \land (p_3 \leftrightarrow p_2)) & \text{(Simplification of implication)} \\ \equiv &\ (((p_3 \lor p_1) \land (\neg p_2 \lor p_1)) \land ((\neg p_3 \lor p_2 ) \land (p_3 \lor \neg p_2))) & \text{(Simplification of biconditional)} \\ \equiv &\ ((p_3 \lor p_1) \land (\neg p_2 \lor p_1) \land (\neg p_3 \lor p_2 ) \land (p_3 \lor \neg p_2)) & \text{(Associativity of conjunction)} \\ \end{align*} $
Q.E.D. (Whilst the expression could further be simplified, at this point it satisfies the definition of CNF).
2) DNF:
\begin{align*} &\ (((p_3 \lor p_1) \land (p_2 \to p_1)) \land (p_3 \leftrightarrow p_2)) & \text{(Given)} \\ \equiv &\ (((p_3 \lor p_1) \land (\neg p_2 \lor p_1)) \land (p_3 \leftrightarrow p_2)) & \text{(Simplification of implication)} \\ \equiv &\ (((p_3 \lor p_1) \land (\neg p_2 \lor p_1)) \land (( p_3 \land p_2 ) \lor (\neg p_3 \land \neg p_2))) & \text{(Simplification of biconditional)} \\ \equiv &\ ((p_1 \lor (\neg p_2 \land p_3)) \land (( p_3 \land p_2 ) \lor (\neg p_3 \land \neg p_2))) & \text{(Distribution of disjunction over conjunction)} \\ \equiv &\ ((p_1 \land ( p_3 \land p_2 )) \lor (p_1 \land (\neg p_3 \land \neg p_2)) \lor ((\neg p_2 \land p_3) \land ( p_3 \land p_2 )) \lor ((\neg p_2 \land p_3) \land (\neg p_3 \land \neg p_2))) & \text{(Distribution of conjunction over disjunction)} \\ \end{align*}
Q.E.D. (As above, sufficient albeit messy).
Why have I posted this question for verification?
- I am not sure if I have applied the steps correctly.
- I suspect that my approach is too verbose, and since this was taken from a practice exam where it wasn't worth many marks, I want to know if I can simplify my approach/save time anywhere. (So far the only thing I can think of is introducing shorthand terms a,b,c,d for the clauses in the final steps of each calculation, but this assumes my general approach is robust).
Is my solution correct? Can it be improved?