## Think you can prove this for me?

Sequences, counting (including probability), logic and truth tables, algorithms, number theory, set theory, etc.
pbj
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### Think you can prove this for me?

Hey guys, my professor showed me this equation and I sort of understand the logic of it. However, he didn't supply a proof for it and I am not waiting till next class to see what it looks like. Think you can show me?

You have to use demorgans theorem, here it is in a circuit, but I want to see it written out, no kmaps or truth tables please:

stapel_eliz
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Think you can show me? You have to use demorgans theorem...I want to see it written out, no kmaps or truth tables please:

pbj
Posts: 3
Joined: Tue Feb 11, 2014 6:06 pm
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### Re: Think you can prove this for me?

I'd do that, but I'm not really sure what the code the upper bar looks like, you know unbroken over multiple variables, how can I represent that?

I hope I didn't look like I was looking for homework answers. This isn't in my homework and I rather understand the content rather then cheat.

If not, I'll take a picture.

pbj
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### Re: Think you can prove this for me?

Okay so I want to express the following in only ors and complements. Is this correct?

AB+A'B'+A'C
(AB)''+(A+B)'+(B'+C)''
(A'+B')'+(A+B)"+(B'+C)''

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Joined: Sun Feb 22, 2009 11:12 pm

### Re: Think you can prove this for me?

I don't know this way of doing these. I think this and this are what they're talking about in the picture. I think it means:

X = NOT[(NOT-A AND NOT-B) AND (NOT-C AND NOT-D)]

And you want to see if you can get this down to (A AND B) OR (C AND D). Is that right? The (NOT-A AND NOT-B) goes to NOT(A OR B) so:

X = NOT[NOT(A OR B) AND NOT(C OR D)]

And NOT(P) AND NOT(Q) goes to NOT(P OR Q) so:

X = NOT[NOT((A OR B) OR (C OR D))]

The two NOTs cancel so:

X = (A OR B) OR (C OR D)