I have a formula

$(L\Leftrightarrow (A\vee J))$

and I am to turn it into DNF and CNF. When I use de Morgan rules and so on, the formula looks like

$(L\Rightarrow (A\vee J))\wedge ((A\vee J)\Rightarrow L)$

$(\lnot L\vee (A\vee J))\wedge ((\lnot A\wedge \lnot J)\vee L)$

the CNF is pretty easy its

$(\lnot L \vee A\vee J)\wedge (L\vee \lnot A)\wedge (L\vee \lnot J)$

But how can I make DNF out of this formula?

  • $\begingroup$ Do you mean "KNF" (as in the title) or "CNF" (Conjunctive Normal Form, as in the text)? $\endgroup$ – Rory Daulton Jan 1 '16 at 14:35
  • $\begingroup$ sorry its CNF ( but in my national language its KNF :) ) $\endgroup$ – trolkura Jan 1 '16 at 14:43
  • 1
    $\begingroup$ Do a Karnaugh map. Collect fullest lines/squares first. $\endgroup$ – Alex Jan 1 '16 at 15:02
  • 1
    $\begingroup$ See maths.usyd.edu.au/u/don/courses/math1904/lect11.pdf $\endgroup$ – Alex Jan 1 '16 at 15:03
  • $\begingroup$ i have tried it with Karnaugh maps , but didnt get the right result , could you elaborate? I havent found anything usefull in that pdf $\endgroup$ – trolkura Jan 1 '16 at 18:49

A simple way to calculate a DNF logically equivalent to the formula $L \Leftrightarrow (A \lor J)$ is to look at its truth table and "build" the formula corresponding to the "disjunction" of the rows for which $L \Leftrightarrow (A \lor J)$ is true.

$\begin{array}{ccccc} A & J & L & A \lor J & L \Leftrightarrow (A \lor J) \\ \mathtt{t} & \mathtt{t} & \mathtt{t} & \mathtt{t} & \mathtt{t} \\ \mathtt{t} & \mathtt{t} & \mathtt{f} & \mathtt{t} & \mathtt{f} \\ \mathtt{t} & \mathtt{f} & \mathtt{t} & \mathtt{t} & \mathtt{t} \\ \mathtt{t} & \mathtt{f} & \mathtt{f} & \mathtt{t} & \mathtt{f} \\ \mathtt{f} & \mathtt{t} & \mathtt{t} & \mathtt{t} & \mathtt{t} \\ \mathtt{f} & \mathtt{t} & \mathtt{f} & \mathtt{t} & \mathtt{f} \\ \mathtt{f} & \mathtt{f} & \mathtt{t} & \mathtt{f} & \mathtt{f} \\ \mathtt{f} & \mathtt{f} & \mathtt{f} & \mathtt{f} & \mathtt{t} \\ \end{array}$

The formula $L \Leftrightarrow (A \lor J)$ is true at rows 1 (which corresponds to the formula $A \land J \land L$), row 3 (which corresponds to the formula $A \land \lnot J \land L$), row 5 (which corresponds to the formula $\lnot A \land J \land L$), and row 8 (which corresponds to the formula $\lnot A \land \lnot J \land \lnot L$). Hence, a DNF logically equivalent to $L \Leftrightarrow (A \lor J)$ is \begin{equation*} (A \land J \land L) \lor (A \land \lnot J \land L) \lor (\lnot A \land J \land L) \lor (\lnot A \land \lnot J \land \lnot L). \end{equation*}


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