We have AB+AB'+A'C+AC
We can rewrite above equation as
AB(C+C')+AB'(C+C')+A'(B+B')C+A(B+B')C ....(As (C+C')=0, (B+B')=0)
We get
ABC+ABC'+AB'C+AB'C'+A'BC+A'B'C+ABC+AB'C
Eliminating Common Terms and rewrite equation as.
ABC+ABC'+AB'C+AB'C'+A'BC+A'B'C
expression in terms of minterms is
m7+m6+m5+m4+m3+m1
K-Map will be
|
|
C'B' |
C'B |
CB |
CB' |
|
|
00 |
01 |
11 |
10 |
A' |
0 |
0 |
1 |
1 |
0 |
A |
1 |
1
|
1
|
1
|
1
|
Thus we can write Reduced equation as
A+C