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Boolean algebra
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How do I get from
`X'Y'+X'Y+XY'`
to
`(XY)'`
using Boolean algebra, where ' denotes inverse?
>>
>>52271006
>X'Y'+X
Is that easier to understand
>>
>>52271052
Fortunately not.
>>
>>52271006
you can split (xy)' to x' + y'
>>
`¬X¬Y+X¬Y+X¬Y`

Easier on the eyes.
`¬(XY)=(¬X+¬Y)`

After DeMorgans Law.
>>
>>52271342
`¬X¬Y + ¬XY + X¬Y`

Whoops
>>
>>52271353
Now apply the Resolution Rule or whatever it's called in English to this and you should end up with
`¬X+¬Y`
>>
>>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.
>>
`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)'`
>>
>>52271006
Thanks for reminding the horror of using some logic program from the 80s in a VM as part of the boolean algebra course.
>>
>>52271501
Thanks!

Over and out.
>>
>>52271006
`X | Y | X' | Y' | XY | XY' | X'Y | X'Y' | X'Y' + X'Y + XY' | (XY)'0 | 0 | 1  | 1  | 0  |  0  |  0  |  1   |         1        |   10 | 1 | 1  | 0  | 0  |  0  |  1  |  0   |         1        |   11 | 0 | 0  | 1  | 0  |  1  |  0  |  0   |         1        |   11 | 1 | 0  | 0  | 1  |  0  |  0  |  0   |         0        |   0`