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### Think you can prove this for me?

Posted: Tue Feb 11, 2014 6:11 pm
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:

Posted: Tue Feb 11, 2014 6:32 pm
Think you can show me? You have to use demorgans theorem...I want to see it written out, no kmaps or truth tables please:

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

Posted: Tue Feb 11, 2014 6:44 pm
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.

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

Posted: Wed Feb 12, 2014 8:39 pm
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)''

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

Posted: Wed Feb 12, 2014 11:56 pm
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)