-2

Gentlemen

I need a hint to simply this expression since I'm quite rusty in my boolean algebra.

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

I however have made thus far 

A'B'C'+ B(A'C+AC')
A'B'C'+ B( AA' + CC' + A'C +AC') //SUBSTITUTE FOR 2 ZEROS
A'B'C'+ B( A(A'+C')+ C(A'+C')) 
A'B'C'+ B( A(AC)' + C(AC)') //DE MORGAN'S LAW
A'B'C'+ B(AC)'(A+C)

Any help will be much appreciated.

Desper
  • 1

3 Answers3

1

Why do you think this can be made any simpler?

$A'B'C' + A'BC + ABC'\;$ is the shortest disjunctive normal form you can get for this expression.

Graham Kemp
  • 129,094
0

If you consider this a simplification, then:

$$[\neg({A}\vee{B}\vee{C})]\vee[B\wedge({A}\oplus{C})]$$

If you consider this a simplification, then:

$${A}\vee{B}\vee{C} \implies B\wedge({A}\oplus{C})$$

barak manos
  • 43,109
0
A'B'C' + A'BC + ABC' 
(A+B+C)' + B(A'C + AC')
(A+B+C)' + B(A XOR C)

Focusing on the first SOP term we can use De-Morgans theorem to replace the AND gate with 3 inverted inputs to a NOR gate.

The second thing to note is when factoring the final 2 SOP terms, you are left with A'C + AC' which can be recognised as the XOR function.

adam3141
  • 1