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There are various Boolean identities some of which are as follows:

  • $A + A = A$

  • $A . A = A$

  • $A + 1 = 1$

  • $AB = BA$

  • $A + AB = A$

and the list goes on

But why doesn't there exist the following Boolean identities which to me seems correct ?

1) $A + B = 1$

2) $A \cdot B = 0$

Proof of above identity

thanks for your precious time.

skyking
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Ahtisham
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1 Answers1

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Because if $A=B=0$ then $A+B=0$ which makes the first invalid. And if $A=B=1$ it makes $A\cdot B=1$ which makes the second invalid.

Your "proof" is flawed since the truth table doesn't include all combination of values for $A$ and $B$. A proper truth table is:

$$\begin{matrix} A & B & A+B & A\cdot B \\ \hline 0 & 0 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 1 & 0 & 1 & 0 \\ 1 & 1 & 1 & 1 \end{matrix}$$

skyking
  • 16,654
  • But A and B are two different variable how can they have same value if you assign 1 to A then you can't assign same to B ?????????? – Ahtisham Nov 06 '17 at 07:39
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    Just because they are different doesn't require them to be different. How do you interpret distributive laws ($A(B+C)= AB+AC$)? If you require $A$, $B$ and $C$ to have different values you'd run into serious problems. – skyking Nov 06 '17 at 07:41
  • thanks got you. I thought maybe this would be interesting question. – Ahtisham Nov 06 '17 at 07:50