Question

How can I simplify the following Boolean expression? A'C+A'.B+AB'C+BC

Answer 1

One can use a 3 variable Karnaugh map

I have added simplification by Bollean algebra.

Explanation:

Converting the primes to bars:

`f = barAC+barAB+AbarBC+BC`

Let's start with an empty 3 variable Karnaugh map:

#bar(ul(| (" ",barBbarC,barBC, BC,BbarC), (barA,0,0,0,0), (A,0,0,0,0) |))#

Enter the 1s corresponding to the first term into the map:

`f = color(red)(barAC)+barAB+AbarBC+BC`

#bar(ul(| (" ",barBbarC,barBC, BC,BbarC), (barA,0,color(red)(1),color(red)(1),0), (A,0,0,0,0) |))#

Enter the 1s corresponding to the second term into the map:

`f = barAC+color(red)(barAB)+AbarBC+BC`

#bar(ul(| (" ",barBbarC,barBC, BC,BbarC), (barA,0,1,color(red)(1),color(red)(1)), (A,0,0,0,0) |))#

Enter the 1s corresponding to the third term into the map:

`f = barAC+barAB+color(red)(AbarBC)+BC`

#bar(ul(| (" ",barBbarC,barBC, BC,BbarC), (barA,0,1,1,1), (A,0,color(red)(1),0,0) |))#

Enter the 1s corresponding to the fourth term into the map:

`f = barAC+barAB+AbarBC+color(red)(BC)`

#bar(ul(| (" ",barBbarC,barBC, BC,BbarC), (barA,0,1,color(red)(1),1), (A,0,1,color(red)(1),0) |))#

We are ready to begin minimization with the following map:

#bar(ul(| (" ",barBbarC,barBC, BC,BbarC), (barA,0,1,1,1), (A,0,1,1,0) |))#

Please observe that the 4 1s in the center tell us that the function they are only dependent on C:

#bar(ul(| (" ",barBbarC,barBC, BC,BbarC), (barA,0,color(red)(1),color(red)(1),1), (A,0,color(red)(1),color(red)(1),0) |))#

`f_"mimimized" = C`

Add the term `barAB` for the pair of 1s in the upper right:

#bar(ul(| (" ",barBbarC,barBC, BC,BbarC), (barA,0,1,color(red)(1),color(red)(1)), (A,0,1,1,0) |))#

`f_"mimimized" = C+barAB`

Starting with the function with notation converted from primes to bars:

`f = barAC+barAB+AbarBC+BC`

Group all of the terms containing C together:

`f = barAB+barAC+AbarBC+BC`

Remove a factor of C:

`f = barAB+C(barA+AbarB+B)`

AND the `barA` term with 1 in the form of `(B + barB)`

`f = barAB+C(barA(B + barB)+AbarB+B)`

Distribute the `barA`

`f = barAB+C(barAB + barAbarB+AbarB+B)`

AND the B term with 1 in the form of `(A+barA)`

`f = barAB+C(barAB + barAbarB+AbarB+(A+barA)B)`

Distribute the B:

`f = barAB+C(barAB + barAbarB+AbarB+AB+barAB)`

The two terms in red are duplicated, therefore one can be eliminated:

`f = barAB+C(color(red)(barAB) + barAbarB+AbarB+AB+color(red)(barAB))`

`f = barAB+C(barAbarB+AbarB+AB+barAB)`

The 4 terms are all of the possible case of A AND B, therefore, it is 1:

`f = barAB+C`