Boolean algebra

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Anonymous

Boolean algebra 2016-01-06 16:32:29 Post No. 52271006

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Boolean algebra 2016-01-06 16:32:29 Post No. 52271006

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How do I get fromX'Y'+X'Y+XY'to(XY)'using Boolean algebra, where ' denotes inverse?

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>>52271006

>X'Y'+X

Is that easier to understand

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>>52271052

Fortunately not.

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>>52271006

you can split (xy)' to x' + y'

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¬X¬Y+X¬Y+X¬Y

Easier on the eyes.¬(XY)=(¬X+¬Y)

After DeMorgans Law.

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>>52271342¬X¬Y + ¬XY + X¬Y

Whoops

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>>52271353

Now apply the Resolution Rule or whatever it's called in English to this and you should end up with¬X+¬Y

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>>52271378

I appreciate the help, but it's the last step that's confusing me. Could you please explain the "Resolution rule"? I think that's what I don't get.

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x'y' + x'y + xy'

x'(y' + y) + xy'

x'(1) + xy'

x' + xy'

absorption rule reversed: (x' = x'+x'y')

(x' + x'y') + xy'

x' + x'y' + xy'

x' + (x' + x)y'

x' + (1)y'

x' + y'

(xy)'

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>>52271006

Thanks for reminding the horror of using some logic program from the 80s in a VM as part of the boolean algebra course.

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>>52271501

Thanks!

Over and out.

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>>52271006X | Y | X' | Y' | XY | XY' | X'Y | X'Y' | X'Y' + X'Y + XY' | (XY)'

0 | 0 | 1 | 1 | 0 | 0 | 0 | 1 | 1 | 1

0 | 1 | 1 | 0 | 0 | 0 | 1 | 0 | 1 | 1

1 | 0 | 0 | 1 | 0 | 1 | 0 | 0 | 1 | 1

1 | 1 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 0

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