An easy would be to convert the expression obtained to 0s and 1s. And then, you can notice the fact that there are all possible combinations of odd $\#1$s in each of the terms. Thus it would be the XOR of the inputs.

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+10 votes

The output $F$ of the below multiplexer circuit can be represented by

- $AB+B\bar{C}+\bar{C}A+\bar{B}\bar{C}$
- $A\oplus B\oplus C$
- $A \oplus B$
- $\bar{A} \bar{B} C+ \bar{A} B \bar{C}+A \bar{B} \bar{C}$

+15 votes

Best answer

0

yes, it's A⊕B⊕C, ex-or is an odd function, it means number of boolean variables in un-prime form will be odd.

+1

Thehobo03

A variable has two forms p(non-prime) p'(prime (or) complemented). in exor function there will always be odd number of non-prime variables in each minterm,

A variable has two forms p(non-prime) p'(prime (or) complemented). in exor function there will always be odd number of non-prime variables in each minterm,

0

can you please explain how you got

=(A+B')(A'+B)C+A'BC'+AB'C'

by using

A'B'C+AB'C'+A'BC'+ABC

I am unable to understand this part can uh please explain

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