Proving if Boolean Equations are valid

Question

I need to prove algebraically that:

$$ab + abc'd + abde' + abc'e + a'b = b$$

$$(wxyz)(wxyz' + wx'yz + w'xyz + wxy'z) = 0$$

Answer 1

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?

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

Answer 2

$$ab + abc'd + abde' + abc'e + a'b = b\tag{1}$$

\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}

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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.