>>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        |   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