Boolean Algebra Simplification of (x'y'+z)'+z+xy+wz

Question

How would you simplify the following? I'm having a bit of trouble with the first part with negation. How would DeMorgan’s Theorem be applied here?

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

Please provide answer in detail.

Update:

The complete question that I got was to prove that

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

equals

x+y+z

Answer 1

Initial expression:

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

Apply DeMorgan's Theorem:

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

Simplify (a'b + a = b + a):

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

Apply DeMorgan's Theorem:

x + y + z + xy + wz

Rearrange (commutativity/associativity):

x + xy + y + z + wz

Factor:

x(1 + y) + y + z(1 + w)

Simplify (1 + a = 1):

x + y + y + z

Simplify (a + a = a):

x + y + z