How to show Boolean identity : $(ab + c + d)(c' + d)(c' + d + e) = abc' + d$

Question

How to show Boolean identity : $(ab + c + d)(c' + d)(c' + d + e) = abc' + d$

Answer 1

First note that by the absorption identity $x(x+y)=x$ (because "$x+y\ge x$"), we have $(c' + d)(c' + d + e)=(c' + d)$, then the left side goes $$(ab + c + d)(c' + d) = abc' + cc' + dc'+abd+cd+d = \\ abc' + d(c+c')+abd+d = abc' +abd+d =\\ abc'+d$$ The last equality is because of the dual absorption identity $xy+x=x$ (because "$xy\le x$").

Answer 2

$$(ab + c + d)(c' + d)(c' + d + e) = abc' + d$$

Start by using the distributive rule: E.g., $(p + q)(r + s) = pr + ps + qr + qs$.

You will also need the reverse application of the distributive law.

E.g., $$\begin{align} (ab + c + d)(c' + d)(c' + d + e) & = (abc' + abc + cc' + cd + ce + dc' + dd + de)(c' + d + e) \\ \\ & = (abc' + abc + cd + ce + dc' + d + de)(c' + d + e) \\ \\ & = \quad \vdots \\ \\ & = abc' + d \end{align} $$

$ + cc' + = ...+ 0 + ...$, which can be omitted.

$dd = d$.

You'll need to apply the distributive rules many times...it gets tedious, but it can be done. Whenever you get a term $+ pp' +$, you can omit it. Whenever you get $pp$ it reduces to $p$.

You need to use the boolean identities you're learning; and they are best learned (and remembered) by using them to prove identities, to notice patterns in which an identity can be applied, etc.

If you have any questions as you take a stab at this, I'll be happy to clarify, correct, or confirm your work, if you post questions/progress in a comment below.

Answer 3

I suggest that you start by reading What are a list of helpful boolean identities for solving boolean functions?. You can do this proof by applying the identities listed there. Please post more questions when you run into problems along the way.

Answer 4

(ab + c + d)(c' + d)(c' + d + e) = (abc' + cc' + dc' + abd + cd + dd)(c' + d + e) =

since cc' = F, and dd = d:

(abc' + dc' + abd + cd + d)(c' + d + e) = (abc' + d(c' + ab + c + T))(c' + d + e) =

since x(a + b + ... + T) = x:

(abc' + d)(c' + d + e) = abc'c' + dc' + abc'd + dd + abc'e + de =

abc' + dc' + abc'd + d + abc'e + de = abc'(T + d + e) + d(c' + T + e) = abc' + d