Boolean expression F = x'y + xyz': - boolean

Question

Using DeMorgan's theorem show that:
a. (A + B)'(A' +B)' = 0
b. A + A'B + A'B' = 1
Boolean expression F = x'y + xyz':
Derive an algebraic expression for the complement F'
Show that F·F' = 0
Show that F + F' = 1
Please Help me

Answer 1

Assuming you know how DeMorgan's law works and you understand the basics of AND, OR, NOT operations:
1.a) (A + B)'(A' + B)' = A'B'(A')'B' = A'B'AB' = A'AB'B' = A'AB' = 0 B' = 0.
I used two facts here that hold for any boolean variable A:
AA' = 0 (when A = 0, A' = 1 and when A = 1, A' = 0 so (AA') has to be 0)
0A = 0 (0 AND anything has to be 0)
1.b) A + A'B + A'B' = A + A'(B + B') = A + A' = 1.
I used the following two facts that hold for any boolean variables A, B and C:
AB + AC = A(B + C) - just like you would do with numeric variables and multiplication and addition. Only here we work with boolean variables and AND (multiplication) and OR (addition) operations.
A + A' = 0 (when A = 0, A' = 0 and when A = 1, A' = 0 so (A + A') has to be 1)
2.a) Let's first derive the complement of F:
F' = (x'y + xyz')' = (x'y)'(xyz')' = (x + y')((xy)' + z) = (x + y')(x' + y' + z) = xx' + xy' + xz + x'y' + y'y' + y'z = 0 + xy' + xz + x'y' + y' + y'z = xy' + xz + y'(x + 1) + y'z = xy' + xz + y' + y'z = xy' + xz + y'(z + 1) = xy' + y' + xz = y'(x + 1) = xz + y'.
There is only one additional fact that I used here, that for any boolean variables A and B following holds:
A + AB = A(B + 1) = A - logically, variable A completely determines the output of such an expression and part AB cannot change the output of entire expression (you can check this one with truth tables if it's not clear, but boolean algebra should be enough to understand). And of course, for any boolean variable A + 1 = A.
2.b) FF' = (x'y + xyz')(xz + y') = x'yxz + x'yy' + xyz'xz + xyz'y'.
x'yxz = (xx')yz = 0xz = 0
xyy'= x0 = 0
xyz'xz = xxy(zz') = xy0 = 0
xyz'y' = xz'(yy') = xz'0 = 0
Therefore, FF' = 0.
2.c) F + F' = x'y + xyz' + xz + y'
This one is not so obvious. Let's start with two middle components and see what we can work out:
xyz' + xz = x(yz' + z) = x(yz' + z(y + y')) = x(yz' + yz + y'z) = x(y(z + z') + y'z) = x(y + y'z) = xy + xy'z.
I used the fact that we can write any boolean variable A in the following way:
A = A(B + B') = AB + AB' as (B + B') evaluates to 1 for any boolean variable B, so initial expression is not changed by AND-ing it together with such an expression.
Plugging this back in F + F' expression yields:
x'y + xy + xy'z + y' = y(x + x') + y'(xz + 1) = y + y' = 1.