How does AB(1+C'D) simplify into AB in boolean algebra? I cannot compare their truth tables since literal number of these two expressions is not same. So can anyone explain this to me?

P.S: I found this in solution of a problem of Digital Logic book by Morris Mano and Micheal Ciletti. But since it is a brief solution of a problem, there are no explanations given.

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    $\begingroup$ If I'm not mistaken, $1+X$ for any $X$ is equal to $1$, whereas $X1 = X$ for any $X$ as well. $\endgroup$ – Nagase Oct 4 '15 at 20:34

In Boolean algebra, 1 represents True and + represents OR.

True OR anything = True.

1 + anything = 1.

1 + C'D = 1.

AB (1 + C'D) = AB.

  • $\begingroup$ And then 1 and anything = anything....so, AB(1+C'D)=AB ... Got it, thanks. $\endgroup$ – Asiful Nobel Oct 4 '15 at 20:42

So you have AB(1+C'D)=AB
1+C'D is equal to 1.
We can make a truth table for 1+C'D
So it seems as we have an issue. If C=F and D=T this statement is not true. So, we can say that this isn't a tautology by counterexample.

  • $\begingroup$ Since when does 1+1 = 0? $\endgroup$ – IanF1 Oct 4 '15 at 20:35
  • $\begingroup$ In boolean algebra all addition is modulus 2. Boolean algebra can be also defined as $( \mathbb{Z}_2,+_2) $ $\endgroup$ – Eli Sadoff Oct 4 '15 at 20:36
  • $\begingroup$ @EliSadoff - I think you are confusing a Boolean ring with a Boolean algebra. $\endgroup$ – Nagase Oct 4 '15 at 20:37
  • $\begingroup$ The notation with boolean algebra is also not consistent. When I learned boolean algebra + was the XOR operator whereas ^ was the and operator and V was the or operator $\endgroup$ – Eli Sadoff Oct 4 '15 at 20:39
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    $\begingroup$ Oh sorry. Well my answer makes no sense now because I used notation incorrectly. :( $\endgroup$ – Eli Sadoff Oct 4 '15 at 20:45

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