Proving if Boolean Equations are valid I need to prove algebraically that:
$$ab + abc'd + abde' + abc'e + a'b = b$$
$$(wxyz)(wxyz' + wx'yz + w'xyz + wxy'z) = 0$$
 A: 
$$ab + abc'd + abde' + abc'e + a'b = b\tag{1}$$

\begin{align} ab + abc'd +abde' + abc'e + a'b 
& = b(\color{blue}{\bf a} + ac'd + ade' + ac'e + \color{blue}{\bf a'}) \quad\quad\quad\quad\quad\tag{distributive rule}\\ \\
& = b(\color{blue}{\bf 1} + a(c'd + de' + c'e)) \tag{$\color{blue}{\bf a + a' = 1}$}\\ \\
& = b\cdot 1\tag{why?}\\ \\
& = b \tag{why?}
\end{align}

$$(wxyz) * (wxyz' + wx'yz + w'xyz + wxy'z) = 0\tag{2}$$

\begin{align} & (wxyz) * (wxyz' + wx'yz + w'xyz + wxy'z) \\ \\
& = wxyzwxyz' + wxyzwx'yz + wxyzw'xyz + wxyzwxy'z \\ \\
& = wxy(zz') + w(xx')yz + (ww')xyz + wx(yy')z \tag{why?}\\ \\
& = wxy\cdot 0 + wyz\cdot 0 + 0 \cdot xyz + wxz\cdot 0 \tag{why?}\\ \\
& = 0 
\end{align}

Here we need 


*

*the distributive rule, 

*the identities

*

*$\;0\cdot a = 0$, 

*$1 + a = 1$, and 


*absorption: 

*

*$aa = a$, 

*$a+a = a$.



These are crucial to know, on the spot, as is DeMorgan's, which was not needed in either problem.
A: Hint for the first one: Note that $$ab+abc'd+abde'+abc'e+a'b = ab(1+c'd+de'+c'e)+a'b$$ and that $$ab+a'b=(a+a')b=1*b=b.$$ What is $1+x$ for any $x$ in the boolean algebra?

Hint for the second one: Note that, for example, $$wxyz*wxyz'=wwxxyyzz'=wxyzz'=wxy*0=0,$$ and use distributivity.
