Simplifying the expression using Boolean Algebra Simplifying the expression using Boolean Algebra into sum-of-products (SOP) expressions
. refers to AND
+ refers to OR
(y' + x) ∙ (z + z') ∙ (y' + x') + (z + x'∙z) ∙ (x + y)
This is what I have so far.
(y' + x) ∙ (z + z') ∙ (y' + x') + (z + x'∙z) ∙ (x + y)
= (y' + x) ∙ 1 ∙ (y' + x') + (z + x'∙z) ∙ (x + y) (Using Complement Element, x+x' = 1)
= (y' + x) ∙ (y' + x') + (z + x'∙z) ∙ (x + y) (Using Identity Element, 1 . x = x . 1 = x)
Updated with new question
( (a + b) ∙ (a' + c') )' + (b + c')' + a∙b'∙c
= ( (a + b) ∙ (a' + c') )' + (b' . c) + a∙b'∙c
= (( a ∙ (a' + c') + b ∙ (a' + c') )' + (b' . c) + a∙b'∙c
= (( a ∙ a' + a ∙ c') + (a' . b + b ∙ c') )' + (b' . c) + a∙b'∙c
= (( a ∙ c') + (a' . b  + b ∙ c') )' + (b' . c) + a∙b'∙c
= ( a ∙ c')' . (a'. b + b ∙ c')'  + (b' . c) + a∙b'∙c
= ( a' + c) . (a + b' ∙ b' + c)  + (b' . c) + a∙b'∙c
I am stuck here.
How do I continue?
Any help please?

Here’s a start:
$$\begin{align*}
(y'+x)\cdot(y'+x')&=y'\cdot(y'+x')+x\cdot(y'+x')\\
&=y'\cdot y'+y'\cdot x'+x\cdot y'+x\cdot x'\\
&=y'+x'\cdot y'+x\cdot y'\\
&=y'+(x'+x)\cdot y'\\
&=y'+y'\\
&=y'
\end{align*}$$
Note also that $z+x'\cdot z=(1+x')\cdot z=z$.
