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im struggling with this Boolean expression.

If anyone could help me solve it that would be great, I've been struggling with this for over a week now and would really like someones help.

https://i.stack.imgur.com/DjLwy.jpg

DarkArmy
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  • the image does not open on my computer – gt6989b Nov 02 '17 at 16:26
  • sorry, now try it should work. – DarkArmy Nov 02 '17 at 16:27
  • same problem, but it could be a domain restriction on my side. Want to rewrite this in MathJax? – gt6989b Nov 02 '17 at 16:34
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    Over a week? That's dedication, kudos to you! But yes, after some point you need to realize you could use some help. Now, did you generate any thoughts or attempts in that week+'s worth of effort and thinking? – Bram28 Nov 02 '17 at 16:35
  • Well the result im only getting is that the output is ABCD, but i know that isn't right i'm just stuck and need a answer please. – DarkArmy Nov 02 '17 at 16:44
  • Is this for a class? If so, I would suggest you talk to your professor or TA. If the best you can do after a week's struggle is ABCD, then an answer is the last thing you need; you first need some elementary instruction and understanding in how these circuits work. Sorry to be harsh but it needs to be said. – Bram28 Nov 02 '17 at 16:49
  • It's fine man i understand thanks anyway. – DarkArmy Nov 02 '17 at 16:58

1 Answers1

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$X=A\wedge (B\vee \neg C) \wedge (C \vee D)$

$X=A\wedge (C \implies B) \wedge (C \vee D)$

$X=A\wedge (D \vee B)$

$\wedge$ is and

$\vee$ is or

$\neg$ is not

J. Sadek
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  • Im not not sure if this is right because in msot of the exampels that were taught to us in lectures didnt involve the ^ -> symbols. If you could try make it a bit more specific for me that would be great buddy! – DarkArmy Nov 02 '17 at 16:59
  • can you tell me exactly what part is problematic? – J. Sadek Nov 02 '17 at 17:02
  • A.(B+C ̅).(C+D) = (AB+A C ̅).(C+D) = ABC + ABD + AC ̅C + ACD = ABC + ABD + AC ̅D = X

    is this right?

     – DarkArmy Nov 02 '17 at 18:26
  • Yes but you can continue simplifying at the end you should find $X=A(B+D)$ – J. Sadek Nov 02 '17 at 18:52
  • I am really sorry I didnt understand you were not understating my notations you should ov told me you were using AB for A and B and A+B for A or B – J. Sadek Nov 02 '17 at 18:54