CODEWITHSUNDEEP
boolean-logicboolean-expression
user3225981
user3225981

Reputation: 107

Simplify boolean expression example

i have the following boolean to simplify 

(A'C'D')+(A'B'D')+(A'BC')+(A'BD)

the furthest i got was ,

A'C'(D'+B)+A', which was from factoring out a A' from A'B'D'+A'BD. Is there anything else that can be done? I've been trying different stuff and I cant simplify it anymore.

Upvotes: 1

Views: 441

Answers (1)

Kit Ostrihon
Kit Ostrihon

Reputation: 834

Actually you should be able to simplify the original expression 

¬a·¬c·¬d + ¬a·¬b·¬d + ¬a·b·¬c + ¬a·b·d

to either one of these minimal forms:

¬a·(b + ¬d)·(¬b + ¬c + d)
¬a·¬b·¬d + ¬a·b·¬c + ¬a·b·d

The given expression next to it's minimal DNF and minimal CNF in Karnaugh maps (generated using latex):

Three equivalent expressions in Karnaugh maps

You can also check it by applying the laws of Boolean algebra:

¬a·¬c·¬d + ¬a·¬b·¬d     + ¬a·b·¬c + ¬a·b·d
¬a·¬b·¬d + ¬a·¬c·¬d     + ¬a·b·¬c + ¬a·b·d        //just permuting
¬a·¬b·¬d + ¬a·¬c·(¬d    +      b) + ¬a·b·d        //distributivity
¬a·¬b·¬d + ¬a·¬c·(¬b·¬d +      b) + ¬a·b·d        //distributivity
¬a·¬b·¬d + ¬a·¬c·¬b·¬d  + ¬a·¬c·b + ¬a·b·d        //distributivity
¬a·¬b·¬d                + ¬a·¬c·b + ¬a·b·d        //absorption

¬a·¬b·¬d + ¬a·b·¬c + ¬a·b·d                       //minimal DNF

¬a·¬b·¬d  + ¬a·b·d                      + ¬a·b·¬c //just permuting
¬a·(¬b·¬d +    b·d                      + b·¬c)   //distributivity
¬a·((¬b + b)·(¬b + d)·(¬d + b)·(¬d + d) + b·¬c)   //distributivity
¬a·(     (1)·(¬b + d)·(¬d + b)·(1)      + b·¬c)   //complementation
¬a·(         (¬b + d)·(¬d + b)          + b·¬c)   //identity for ·
¬a·(  (¬b + d + b·¬c)·(¬d + b + b·¬c))            //distributivity
¬a·(  (¬b + d +   ¬c)·(¬d + b + b·¬c))            //distributivity
¬a·(  (¬b + d +   ¬c)·(¬d + b))                   //absorption

¬a·(¬b + ¬c + d)·(b + ¬d)                         //minimal CNF

Upvotes: 1

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