Prove i. A’D + BD + B’C + AB’D = B’C + D
ii. (x'y' + z)' + z + xy + wz = x + y + z

I'm new to Boolean algebra and I've tried all the rules for Boolean algebra without success. Help anyone?

 

Consider the following identities:

A' = A'B + A'B'   and
B  = BA' + BA
 
Now
A'D + BD + B'C + AB'D  =
(A' + B + AB')D + B'C   =
(A'B + A'B' + A'B + AB + AB')D + B'C =
(1)D + B'C

Show us what you got for the other one - eh?
I shouldn't have to do all the heavy lifting.

 

a'd + bd + b'c + ab'd
= a'd + d(b+ab') + b'c
= a'd + d(a + b) + b'c
= a'd + ad + db + b'c
= d(a' + a) + db + b'c
= d + db + b'c
= d(1 + b) + b'c
= d + b'c

 

Prove i. A’D + BD + B’C + AB’D = B’C + D
ii. (x'y' + z)' + z + xy + wz = x + y + z

I'm new to Boolean algebra and I've tried all the rules for Boolean algebra without success. Help anyone?

The 1st question is solved by 2 gentlemen
Now the 2nd question is wrong as the lhs is 1
the actual question must be

(x'y' + z')' + z + xy + wz = x + y + z