Q:

J K Flip Flop and Boolean Algebra

Y=K'J'Q+K'J+KJQ'
The output Y should be that of a JK flip-flop. That is:   Y=JQ'+K'Q

I tried to solve the following way:
1. Y=K'J'Q+K'J+KJQ'
2. =K'(J'Q+J)+KJQ'
3. =JK'+K'Q+JKQ'
4. =J(K'+KQ')+K'Q
5. =J(K'+Q')+K'Q
6. =JK'+JQ'+K'Q
It appears as if the term JK' is 0. Please help.

A:

Start with the original equation:
Y = K'J'Q + K'J + KJQ'
The trick is to create two terms that are equivalent to the middle term, using the fact that X + X' = 1:
Y = J'K'Q + JK'(Q + Q') + JKQ'
Y = J'K'Q + JK'Q + JK'Q' + JKQ'
Now factor the pairs of terms:
Y = K'Q(J + J') + JQ'(K + K')
Simplify, again using the fact that X + X' = 1:
Y = K'Q + JQ'
This is really obvious if you draw the Karnaugh map for the function. The JK' term is redundant, given the other two terms.

