3
$\begingroup$

Is this calucation of DNF and CNF for the formula $A \land (A \lor C) \implies (C \lor B)$ correct? $$ \begin{array}{|c|c|} \hline \text{Given:} & A \land (A \lor C) \implies (C \lor B) \\ \hline {} & (\neg(A \land (A \lor C)) \lor (C \lor B)) \\ \hline {} & (\neg A \lor \neg(A \lor C) \lor (C \lor B)) \\ \hline {} & (\neg A \lor (\neg A \land \neg C) \lor (C \vee B)) \\ \hline \text{DNF:} & \neg A \lor (\neg A \land \neg C) \lor C \lor B \\ \hline {} & \neg A \lor (C \lor B) \\ \hline \text{DNF and CNF:} & \neg A \lor C \lor B \\ \hline \end{array} $$ (Original picture of the calculation here.)

When I got the DNF, I applied the absorption rule in order to get DNF/CNF.

$\endgroup$
  • $\begingroup$ Thanks for your correction @Jendrik Stelzner - did you also check the calculations? $\endgroup$ – user3352632 Aug 26 '18 at 20:41
  • $\begingroup$ It has been some time since I’ve dealt with DNF/CNF, but your calculations seem correct to me. But I would prefer it if someone with more expertise would post an answer. $\endgroup$ – Jendrik Stelzner Aug 26 '18 at 20:55
1
$\begingroup$

The calculus is correct.   You can also check the final answer to be sure it is an equivalence.

If $A$ is false, $A\wedge(A\vee C)\to(C\vee B)$ is true, as is the case when either $C$ or $B$ is true.   So if and only if $\neg A\vee B\vee C$ do we have the implication.


Note also an easier route would have been to apply absorption equivalence first, then implication equivalence: $A\wedge(A\vee C)\to(C\vee B)\\\equiv A\to (C\vee B)\\\equiv \neg A\vee C\vee B$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.