How do I get fromX'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.
>>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