Help Please! Cant do this Boolean Algebra problem, brain not good enough : (

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Jaaames
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Hi, would really appreciate it if someone could help me with this problem:

Show with boolean algebra that:

NotA . NotB XOR NotC . NotD = (A + B) XOR (C + D)

Thanks very much in advance
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RogerOxon
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(Original post by Jaaames)
Hi, would really appreciate it if someone could help me with this problem:

Show with boolean algebra that:

NotA . NotB XOR NotC . NotD = (A + B) XOR (C + D)

Thanks very much in advance
De Morgan's gets you part of the way, then you need to use a property of XOR.
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Jaaames
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(Original post by RogerOxon)
De Morgan's gets you part of the way, then you need to use a property of XOR.
Cheers Roger,

I thought so, this is what I've got so far:

Name:  Question 6 workings.png
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What do you mean by 'need to use a property of XOR'?

Thanks for your help.
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RogerOxon
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First apply De Morgan's to each side of your XOR:

\overline{x}.\overline{y}= \overline{x+y}

You should have something that differs from what you want by having a not on each side of the XOR. Look at the truth table for XOR to see the next step.
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Jaaames
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(Original post by RogerOxon)
First apply De Morgan's to each side of your XOR:

\overline{x}.\overline{y}= \overline{x+y}

You should have something that differs from what you want by having a not on each side of the XOR. Look at the truth table for XOR to see the next step.
Thanks Roger,

After a bit of head scratching I managed to do it:
Name:  Question 6 workings.png
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Size:  430.3 KBThanks so much for your help though
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Jaaames
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(Original post by Jaaames)
Thanks Roger,

After a bit of head scratching I managed to do it:
Name:  Question 6 workings.png
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Size:  430.3 KBThanks so much for your help though
Name:  Question 6 workings 2.jpg
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RogerOxon
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(Original post by Jaaames)
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Good. Here's how I would do it:

\overline{x}.\overline{y}= \overline{x+y}

\therefore (\overline{A}.\overline{B}) \; XOR \; (\overline{C}.\overline{D}) = \overline{A+B} \; XOR \; \overline{C+D}

x \; XOR \; y=\overline{x} \; XOR \; \overline{y} (Easiest to show this with a Karnaugh map)

\therefore \overline{A+B} \; XOR \; \overline{C+D} = (A+B) \; XOR \; (C+D)

\therefore (\overline{A}.\overline{B}) \; XOR \; (\overline{C}.\overline{D}) = (A+B) \; XOR \; (C+D)
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