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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

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  • $\begingroup$ 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 $\endgroup$ – joe 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 $\endgroup$ – joe Jun 6 '15 at 7:23
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(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 $\lnot A$.

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