CODEWITHSUNDEEP
logical-operatorsboolean-expressionkarnaugh-map
Tharindi
Tharindi

Reputation: 11

Stuck in simplifying logic expression using Boolean algebra also simplified by karnaugh maps

I also done this with a karnaugh map. My answers for this logical expression are different. I know the output can't be 1 for this question but I can't get the right one using Boolean laws.

A'B'C'D'+A'B'CD'+A'BC'D+AB'C'D'+AB'CD'+ABCD

My answers for this logical expression are different. I know the output can't be 1 for this question but I can't get the right one using Boolean laws.

A'B'C'D'+A'B'CD'+A'BC'D+AB'C'D'+AB'CD'+ABCD

Upvotes: 1

Views: 868

Answers (1)

Kit Ostrihon
Kit Ostrihon

Reputation: 834

I suspect that the problem is in assuming that ¬A¬C + AC is equal to 1. (lines 2 and 3 in your expression.)

If you look at that closely in a truth table, it becomes clear that line 2 and line 3 are not equal to 1 for those inputs.

input | output
A  C  | ¬A¬C    AC     ¬A¬C+AC
------|-----------------------
0  0  | 1*1=1   0*0=0   1+0=1
0  1  | 1*0=0   0*1=0   0+0=0 (!)
1  0  | 0*1=0   1*0=0   0+0=0 (!)
1  1  | 0*0=0   1*1=1   0+1=1

You will get the same result as with the Karnaugh map following the laws of Boolean algebra.

¬A¬B¬C¬D + ¬A¬BC¬D + ¬AB¬CD + A¬B¬C¬D + A¬BC¬D + ABCD
¬A¬B¬D*(¬C+C)      + ¬AB¬CD + A¬B¬D*(¬C+C)     + ABCD
¬A¬B¬D*1           + ¬AB¬CD + A¬B¬D*1          + ABCD
¬A¬B¬D             + ¬AB¬CD + A¬B¬D            + ABCD
  ¬B¬D*(¬A+A)      + ¬AB¬CD +                    ABCD
  ¬B¬D*1           + ¬AB¬CD +                    ABCD
  ¬B¬D             + ¬AB¬CD +                    ABCD

Upvotes: 0

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