Count the number of literals in the following expression :

F = AB' + BC' + CD' + DE'

According to me, the answer should be 8. But my solution suggests that the answer should be 6. Can anyone help me out. What am I doing wrong?

Thanks in advance!

  • 2
    $\begingroup$ I also count 8. Your solution is wrong. Switch a solution manual. $\endgroup$ Commented Dec 28, 2018 at 12:56

1 Answer 1


Does your material use the (strange) convention of notating $\neg A$ as $A'$?

If so, there are 6 literals (5 without the LHS). If $A'$ is distinct from $A$, there are 9 (8 without LHS) and the solution book is wrong.

  • $\begingroup$ Yes you are right @orlp. This book demotes complements by " ' ". But when I looked up on the web, its almost mentioned everywhere that a variable and it's complement should be counted as two. Also, LHS shouldn't be considered, as far as I know... Can the variable and its implement be counted as one? $\endgroup$ Commented Dec 28, 2018 at 16:59
  • $\begingroup$ @AbhilashMishra It's all just a bit ambiguous. It seems that this exercise counts it as such. $\endgroup$
    – orlp
    Commented Dec 28, 2018 at 19:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.