booleanboolean-algebra

How to proof (x+y)(x’+z)(y+z) = (x+y)(x’+z)

I made OR in the end, but I cannot solve that. I just got the final answer xy+xz+yz without a x, if I have a x, that will equal.

Solution

  • I think your final answer should be x'y + xz + yz, if you introduce xx' which evaluates to zero, you will be able to get to the final answer. Nevertheless, please find the complete solution:

    (x + y)(x' + z)(y + z)
(xx' + yx' + zx + zy)(y + z)
(x'y + xyz + yz + x'yz + xz + yz)
(x'y + yz + xz)

introduce xx'
(x' + z)(y + x)

    Expanding the above: x'y + yz + x'x + xz => x'y + yz + xz which is equal to above

    Hope this has answered your question