At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
|dw:1327917442864:dw| X'Y'Z' + YZ no other solution for sure ... you are welcome
2nd one = X'Y'+X(Y'Z+YZ') = X'Y'+Y'Z+YZ' = X'Y' + (Y xor Z)
Chris ... we have not x'y' ... please check your answer . I'm sure its not right answer
What do you mean by "we have not x'y'"? The (2nd) given formula was X'Y'Z' + X'Y'Z + XY'Z + XYZ' and there you can easily see, that the first two terms lead to X'Y'(Z'+Z) = X'Y'. Or maybe I misunderstood you?
look to original Boolean equ X'Y'Z' + X'YZ + XYZ X'Y'Z + XY'Z + XYZ' yours do you have any difference before simplification
I'm all the time talking about the 2nd (second, the lower, the latter, the last) of the two expressions that were mentioned in the original post. (At least, as far as I can understand, the two lines with two expressions, one consisting of three, the other of four terms, are two independent expressions, each of he has to reduce.) Maybe I see a different posting, but for me the 2nd expressions reads: \[X'Y'Z' + X'Y'Z + XY'Z + XYZ'\].
in second term you consider y' while its not ..
any way .. logic is simple .. and simplification is easy . its just need to take it easy and step by step .. I'm sure that you can solve harder than this, I like your way of deffence because that is mean ... You are completely understand logic ... respect for you buddy .. see u around