logic formula: transform to cnf / dnfb

Question

I try to transform some formulas into CNF and DNF but I am not sure how to use distributivity Law here.

Given: $$ ((C \lor D) \land (A\rightarrow D)) <-> (C \rightarrow A ) $$

I applied the rules for Implication and Biimplication and transformed it into NNF:

$$ (((\lnot C \land \lnot D) \land (A\land \lnot D)) \lor \lnot C \lor A) \land ((C \land \lnot A) \lor (C \lor D) \land (\lnot A \lor D)) $$

I knew that I have now to use Distributivity to get CNF and DNF, but I am not sure how to do it properly. I began with it and are already on the 2nd page, since I always get more Clauses :O.

Answer 1

For an expression with just three variables, the easiest way is probably to write down the truth table. This directly represents the (non-optimized) DNF.

CNF: d & (a+!c)

DNF: ad + !cd

Karnaugh map:

             ac
       00  01  11  10
      +---+---+---+---+
   0  | 0 | 0 | 0 | 0 |
d     +---+---+---+---+
   1  | 1 | 0 | 1 | 1 |
      +---+---+---+---+

You get the CNF by looking at all terms with output value 0. Inverting all input literals gives you the CNF clauses.