R: Boolean Identity AB+AB=A o A sum of 2 logically adjacent m minterms can be simplified by eliminating one variable, which is the different literal in the two minterms o The resulting product term has m-1 literals
Boolean Identity AB+AB’=A A sum of 2 logically adjacent mminterms can be simplified by eliminating one variable , which is the different literal in the two minterms . The resulting product term has m-1 literals
R: Boolean Identity AB+ABA o All minterm groups must occur in a power of 2, 2n, n is the number of variables to be eliminated by the rou The resulting product term have m-n literals, which are the common variables to all minterms
Boolean Identity AB+AB’=A All minterm groups must occur in a power of 2, 2n , n is the number of variables to be eliminated by the group. The resulting product term have m-n literals, which are the common variables to all minterms
3mFA'BCD, m3=ABCD R Common literals CD00011110 0412 00 2 BD 5139 0 A'B'D 3A/715 410 10 MSB=A: LSB=D
m1=A’B’C’D, m3=A’B’CD 00 01 11 10 AB CD 00 01 11 10 0 4 12 8 1 5 13 9 3 7 15 11 2 6 14 10 MSB=A ; LSB=D 1 1 Common literals A’ ,B’ ,D m1+m2=A’B’ D m=4, n=1 m-1=3
=ABCD, m9=ABCD CD200011110 2 Comon Literals. 04128 00 B C D 5139 01 5m1+mg=BCD 371511 26140 图=4,n=1 MSB=A. LSB=D
m1=A’B’C’D, m9=AB’C’D 00 01 11 10 AB CD 00 01 11 10 0 4 12 8 1 5 13 9 3 7 15 11 2 6 14 10 MSB=A ; LSB=D 1 1 Common literals: B’, C’, D m1+m9=B’C’D m=4, n=1 m-1=3
2 mOFABCD, MFAB'CD R m2=ABCD, 3=A'B'CD CD200011110 原mo+m1=ABC 0△4128 00 原m2+m3=ABC 01/1/5 139 15 5 Common literals 61410 圆A,B MSB=A. LSB=D R mo+m1+m2+m3=A'B m=4,=2 m==2
m0=A’B’C’D’, m1=A’B’C’D m2=A’B’CD’, m3=A’B’CD 00 01 11 10 AB CD 00 01 11 10 0 4 12 8 1 5 13 9 3 7 15 11 2 6 14 10 MSB=A ; LSB=D 1 1 1 1 m0+m1=A’B’C’ m2+m3=A’B’C Common literals A’, B’ m0+m1+m2+m3=A’B’ m=4, n=2 m -n =2