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Given: $(A+B) \cdot (B+C) \cdot(C+A)$

Associative
$abc=(ab)c$
$\biggl (\underbrace{(A+B) \cdot (B+C)} \biggr ) \cdot (C+A)$
Distributive
$x(y+z)=xy+xz$

 
$\Bigl (AB+AC+\underbrace{BB}+BC) \Bigr ) \cdot (C+A)$
Idempotent
$xx=x$

 
$\Bigl (AB+AC+ \underbrace{B+BC} \Bigr ) \cdot (C+A)$

Associative,
Absorption
$x+xy=x$

$\underbrace{(AB+AC+B) \cdot (C+A)}$
Distributive $ABC+\underbrace{AA}B+A\underbrace{CC}+\underbrace{AA}C+BC+BA$
Idempotent $\underbrace{ABC+AB}+\overbrace{AC+AC}^{AC}+BC+BA$
Associative,
Absorption
$\underbrace{AB}+AC+BC+\underbrace{BA}$
Idempotent $AB+AC+BC$

Hence, the simplified expression is:
$$AB + AC + BC$$

1 votes
1 votes

Note: This is a classic Example of Self Dual function.

For any logical Expression, two times dual gives the same expression.

In Self dual Expression one time dual gives the same expression.

$f = (A+B).(B+C).(C+A) $

$f^d = AB + BC + CA$

The dual does not look like the original function but simple boolean simplification leads to the original function.

$f^d = AB + CA + BC$

CA = AC                                                     (Commutative)

$f^d = AB + AC + BC$

now taking A common

$f^d = A (B + C) + BC$

now OR is distributive over AND

$f^d = (A + BC)(B + C + BC)$

here B + BC = B                                          (absorption law)

         A + BC = (A + B)(A + C)                      (+ is distributive over . )

$f^d = (A + B)(A + C)(B + C)$

now after applying commutativity, we get 

$f^d = (A+B).(B+C).(C+A) $

Therefore

$f^d = f$

The special thing about the given boolean expression is that it is a self dual function.

 

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