I'm having trouble converting the below formula to disjunctive normal form using logical laws. I found the DNF using truth tables but I am having issues using just logical laws. Here is the formula:

$(A \to (A \land \lnot B)) \land (A \to (B \land \lnot A)))$

The DNF I found using truth tables:

$(\lnot A \land \lnot B) \lor (\lnot A \land B)$

Using the logical laws to get the DNF this was one of my attempts:

$(((\lnot A \lor ( A \land \lnot B)) \land (\lnot A \lor (B \land \lnot A)))$

few more steps..

arrived at : $((\lnot A \lor (A \land \lnot B))$ which is wrong

  • $\begingroup$ The first step is to replace implications with disjunctions. Do you know how to do that? $\endgroup$
    – MJD
    Jun 6 '15 at 7:14
  • $\begingroup$ What trouble did you face? Follow MJD's suggestion and then use De Morgan's or distributivity laws to expand $\endgroup$
    – user21820
    Jun 6 '15 at 7:15
  • $\begingroup$ Yeah I know the procedure to get the DNF I'm just not getting the correct equivalences. I'll update my question with one of my attempts $\endgroup$
    – jn025
    Jun 6 '15 at 7:16
  • $\begingroup$ I basically did implication->distribution->negation->distribution not sure where I went wrong or where I need extra steps $\endgroup$
    – jn025
    Jun 6 '15 at 7:23

(1) $A \to (A \land \lnot B)=\lnot A \lor(A\land \lnot B)=\lnot A\lor \lnot B$

(2) $A \to (B \land \lnot A)=\lnot A \lor (B \land \lnot A)=\lnot A \lor B\land \lnot A=\lnot A \lor \lnot A \land B$

(3) $(\lnot A\lor \lnot B)\land (\lnot A \lor \lnot A \land B)=\lnot A \lor \lnot A\land B \lor \lnot B \land \lnot A=\lnot A$

The final result is $\lnot A$.


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.