How do i do this Boolean Algebra Proof?

Question

How do i reduce Y = (((AB)' * B)' * ((AB)' *A)')'
to Y = A'B + AB'
' Means not
* means times as well.

Answer 1

Y = (((AB)'*B)')*((AB)'*A)')'
= ((A'*B'*B)')*(A'*B'*A)')'
= ((A+B+B')*(A+B+A')'
= (A*B)'
= A'B+AB'

Answer 2

( ( (AB)' * B)' * ( (AB)' *A )' )'

Your first step is to use DeMorgan's law on the inner negations. Remember that (AB)' = A' + B', and (A + B)' = A'B'.

( ( (AB)'' + B') * ( (AB)'' +A' ) )'

Note that, by double negation, C'' = C. Applying these to (AB)'', we get

( ( (AB) + B') * ( (AB) +A' ) )'

Now, we apply the outside negation to the whole thing, since we have a product of two expressions.

( (AB) + B')' + ( (AB) +A' )'

Use DeMorgan's law once again

(AB)'(B'') + ( (AB)' A'' )

By double negation,

(AB)'(B) + (AB)'(A)

By DeMorgan's law, again,

(A' + B')B + (A' + B')A

By the distributive law,

(A'B + B'B) + (A'A + AB')

But, a negation ANDed with itself is 0, so

(A'B + 0) + (0 + AB')

Which we can now eliminate the 0, as per the OR rules.

A'B + AB'

Answer 3

Reduce (AB)' to A' + B'
AB + B` to A + B'
(A + B)' to A'B'



Y = (((AB)' * B)' * ((AB)' *A)')' = ((AB + B') * (AB + A'))' =
= ((A + B') * (A' + B))' = (A + B')' + (A' +B)' = A'B + AB'

Answer 4

Y = ( ( (A * B)' * B)' * ( (A * B)' * A)' )'

Y = (((A*B' + A'*B + A'*B')*B)' * ((A'*B + A*B' + A'*B') * A)' )'

Y = ( ( A' * B )' * (A * B')' )'

Y = ( (A*B' + A'*B' + A*B) * (A'*B' + A'*B + A*B) )'

Y = ( A*B + A'*B' )'

Y = A'*B + A*B'