I have tried simplify some expressions to DNF and CNF but I'm stuck in one step and I can't find some rule or law what can I apply on it. I used Wolfram and found that my expression is not in DNF/CNF yet.

On DNF I'm stuck here: My expression but the DNF is:

final DNF

On CNF I'm stuck here:

my CNF expresison but the CNF from calculators is:

founded CNF

Please can someone help me with some tip, rule or law what can by applied to solve the problem? I spent hours with it but with no progress.








Applied to your first expression:

$$C'D'+CD+B'D'+BD \overset{Adjacency \ x \ 2}{=}$$

$$BC'D'+B'C'D'+CD+B'D'+BCD+BC'D \overset{Absorption \ x \ 2}{=}$$

$$BC'D'+CD+B'D'+BC'D \overset{Adjacency}{=}$$


  • $\begingroup$ Thats gr8, i try something similar but with some mistakes :) Thanks a lot, save my day. Havent you any other super advice for that CNF expression? :) $\endgroup$ – jcop Sep 28 '18 at 6:04
  • $\begingroup$ @kampus For the CNF: The $A + B'+C+D)$ term gets absorbed by the $A + C+D$ term, and the $A+B'+C'+D'$ term gets absorbed by the $A + C'+D'$ term. Also, I note that there are two $B+C'+D'$ terms at the end .. one can be eliminated by Idempotence. $\endgroup$ – Bram28 Sep 28 '18 at 17:18
  • $\begingroup$ Thank you so much, i have not notice that i can do such absorbtion. Cookie for both answers, you have save me :) $\endgroup$ – jcop Oct 1 '18 at 8:50

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