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I am honestly confused over here. I understand how bit manipulation works. But when it comes to Boolean Algebra I seem to be a bit stumped.

So if A = 1, then ~A = 0.

But in this example (~A * ~B) + (~C * ~D) = 0 + 0 or 0 | 0 = 0. But in this example it equals 1. Karnaugh Map

Seven Segment Display

Kendrick Lamar
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  • If $A=1$, then $\sim A = 0$, sure. But also, if $A=0$, then $\sim A=1$. The $1$ in the K-map says that when all four inputs are false, you want the e segment to lit up. By the way, that box of the K-map corresponds to $\sim A * \sim B * \sim C * \sim D$, which is true when all four inputs are false. – Fabio Somenzi May 02 '17 at 04:21
  • Ah I was thinking that might be that case because in programming if you do if (A && B && C && D) which means its a true statement. I get it now, I was a bit confused on why the ~ was in there I thought A was 1 by default. But it turns out those are the cases to get the output segment of e. – Kendrick Lamar May 02 '17 at 04:34
  • How do I upvote you and close this? – Kendrick Lamar May 02 '17 at 04:40
  • I posted my comment as an answer. – Fabio Somenzi May 02 '17 at 04:48

1 Answers1

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If $A=1$, then $\sim\! A=0$, sure. But also, if $A=0$, then $\sim\! A=1$. The $1$ in the K-map says that when all four inputs are false, you want the e segment of the display to lit up.

By the way, that box of the K-map corresponds to the minterm

$$ \sim\! A \,∗ \sim\! B\, ∗ \sim\! C\, ∗ \sim\! D \enspace,$$

which is true when all four inputs are false.

Fabio Somenzi
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