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