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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!

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    $\begingroup$ I also count 8. Your solution is wrong. Switch a solution manual. $\endgroup$ – Yuval Filmus Dec 28 '18 at 12:56
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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.

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  • $\begingroup$ 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? $\endgroup$ – Abhilash Mishra Dec 28 '18 at 16:59
  • $\begingroup$ @AbhilashMishra It's all just a bit ambiguous. It seems that this exercise counts it as such. $\endgroup$ – orlp Dec 28 '18 at 19:19

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