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.

  • $\begingroup$ 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

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: $A\wedge(A\vee C)\to(C\vee B)\\\equiv A\to (C\vee B)\\\equiv \neg A\vee C\vee B$


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