I have a problem with this assignment:

The 26th one:

enter image description here

I am supposed to minimize this, and I get to

" XY + ZY + ( Z*(inverted X) )"

But solution to this problem is enter image description here

I don't know how to minimize more the solution I got up there :/

Does anyone know?


See if this makes sense... z+y goes away because the OR function is handled by the other two that have those inputs. Best way to do this is draw a K-map. I used xy on the vertical axis, and z on the horizontal and the equation (x+z)(!x+y) was obvious.

Then, because you expand that equation to (!xx) + (!xz) + (zy) + (xy) and notice that (!xx) has no contribution, and you create another K-map for the remaining terms (excluding (!xx)) the equation (xy) + (!xz) becomes obvious.

  • \$\begingroup\$ Oh... But when you are left with only XY+Z*(! X) + ZY, how do you minimize that without K map, what rules of boolean algebra could we use to minimize it? \$\endgroup\$ – user197072 Aug 28 '18 at 14:26
  • \$\begingroup\$ How about... !xz + !!z + !!y + xy = !x(z+z) + x(y+y) = !xz+xy Assuming, I applied that DeMorgan's identity correctly for the zy term... I dunno, could be crap-math :/ \$\endgroup\$ – CapnJJ Aug 28 '18 at 18:40
  • \$\begingroup\$ To what identity did you apply deMorgan? How did you get the. !xz +!!z +!!y + xy \$\endgroup\$ – user197072 Aug 30 '18 at 10:40
  • \$\begingroup\$ zy = (!!z+!!y) \$\endgroup\$ – CapnJJ Aug 30 '18 at 15:56

Given that (x+z)(/x+y), there are four conditions in which it can be true. We can eliminate one of them, x/x, as it is always false. That leaves xy+/xz+yz. The third term is redundant, since if yz is true, one of the preceding terms will be true depending on the value of x.


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