Simplify the following functions using Boolean algebra:
Y_1=ABC + AB'C'+AB'C.
Y_2=A'BC' + AB'C + AB'C'.
For Y_1:
ABC + AB'C' + AB'C
= AB(C + C') + AB'C
= AB + AB'C
= AB(1 + C)
= AB
For Y_2:
A'BC' + AB'C + AB'C'
= AB'C' + AB'C + A'BC'
= AB'C'(1 + C) + A'BC'
= AB'C' + A'BC'
= BC'(A' + A)
= BC'
To simplify the given functions using Boolean algebra, we will apply the following Boolean identities and theorems:
1. Identity Law:
- A + 0 = A
- A * 1 = A
2. Domination Law:
- A + A' = 1
- A * A' = 0
3. Idempotent Law:
- A + A = A
- A * A = A
4. De Morgan's Laws:
- (A + B)' = A' * B'
- (A * B)' = A' + B'
Using these laws, let's simplify the given functions step-by-step:
Function Y_1 = ABC + AB'C' + AB'C:
1. Apply the Idempotent Law to remove the duplicate terms:
- Y_1 = ABC + ABC' + AB'C
2. Apply Distribution Law:
- Y_1 = AC(B + B') + AB'C
3. Apply the Domination Law to simplify (B + B') to 1:
- Y_1 = AC(1) + AB'C
4. Apply the Identity Law to simplify AC(1) to AC:
- Y_1 = AC + AB'C
The simplified form of Y_1 is AC + AB'C.
Function Y_2 = A'BC' + AB'C + AB'C':
1. Apply the De Morgan's Law to each term:
- Y_2 = (A'BC')' * (AB'C)' * (AB'C')'
2. Apply the De Morgan's Law again:
- Y_2 = (A' + B' + C) * (A' + B + C') * (A + B' + C')
3. Apply Distribution Law:
- Y_2 = (A' + B' + C)(A'A + A'B + A'C' + AB' + BB' + BC + CA + CB' + CC')
4. Apply the Domination Law to simplify terms like A'A, B'B, and C'C to 0:
- Y_2 = (A' + B' + C)(A'B + A'C' + AB' + BC + CA + CB')
5. Apply the Identity Law to remove redundant terms like A'B, AB', and BC:
- Y_2 = A'B + A'C' + CA + CB'
The simplified form of Y_2 is A'B + A'C' + CA + CB'.