Question

Simplify using boolean logic and then apply DeMorgan's Law to convert to negative logic: T = A!B + B!C + (C!D)(A + C)

Answer 1

Given

T = A!B + B!C + (C!D)(A + C)

T = A'B + B'C + (C'D)(A + C)

T = A'B + B'C + (AC'D + CC'D)  { By Distributive law P(Q+R) = PQ+PR }

T = A'B + B'C + (AC'D + (0)D) { We know that PP'= 0 }

T = A'B + B'C + (AC'D + 0) { We know that P(0)= 0 }

​​​​​​​​​​​​​​T = A'B + B'C + AC'D { We know that P+0= P }

T = !AB + !BC + A!CD

Which is Required Simplified Boolean Expression

Given T = A'B + B'C + AC'D

Above Function in K-map as follows

K-map B'C CD 00 01 11 AB A'B AC'D .

From the above K-map F (A, B, C, D) = ​ m (2, 3, 4, 5, 6, 7, 9, 10, 11, 13)

Truth Table: F (A, B, C, D) = ​ m (2, 3, 4, 5, 6, 7, 9, 10, 11, 13)

A	B	C	D	F
0	0	0	0	0
0	0	0	1	0
0	0	1	0	1
0	0	1	1	1
0	1	0	0	1
0	1	0	1	1
0	1	1	0	1
0	1	1	1	1
1	0	0	0	0
1	0	0	1	1
1	0	1	0	1
1	0	1	1	1
1	1	0	0	0
1	1	0	1	1
1	1	1	0	0
1	1	1	1	0

Truth Table: From the above Truth Table Negative Logic is Convert 0’s to 1’s and 1’s to 0’s

A	B	C	D	F
1	1	1	1	1
1	1	1	0	1
1	1	0	1	0
1	1	0	0	0
1	0	1	1	0
1	0	1	0	0
1	0	0	1	0
1	0	0	0	0
0	1	1	1	1
0	1	1	0	0
0	1	0	1	0
0	1	0	0	0
0	0	1	1	1
0	0	1	0	0
0	0	0	1	1
0	0	0	0	1

Truth Table Negative Logic F (A, B, C, D) = ​ m (0, 1, 8, 12, 14, 15)

K-map CD 00 01 B+C 11 A+B' A'+C+D' 11

Negative Logic is T = (B+C’)(A+B’)(A’+C+D’)