Dimostrare le seguenti espressioni logiche utilizzando i teoremi dell’algebra booleana:
A + A⋅ B = A + B
1)
Soluzione
A + A ⋅ B = A(1 + B ) + A ⋅ B =
A + AB + A ⋅ B =
(
)
A+ B A+ A =
A+ B
A⋅ B ⋅C + A⋅ B ⋅C + A⋅ B ⋅C + A⋅ B ⋅C + A⋅ B ⋅C = A⋅ B + C
2)
Soluzione
A⋅ B ⋅C + A⋅ B ⋅C + A⋅ B ⋅C + A⋅ B ⋅C + A⋅ B ⋅C =
A⋅C + A⋅ B ⋅C + A⋅C =
(
)
A(C + B ) + A ⋅ C =
A C + B ⋅C + A⋅C =
A⋅C + A⋅ B + A⋅C =
A⋅ B + C
A⋅ B ⋅C + A⋅ B ⋅C + A⋅ B ⋅C + A⋅ B ⋅C = C
3)
Soluzione
A⋅ B ⋅C + A⋅ B ⋅C + A⋅ B ⋅C + A⋅ B ⋅C =
(
)
(
) (
)
A⋅C B + B + A⋅C B + B = C A + A = C
4)
A⋅ B ⋅C + A⋅C ⋅ D + A⋅C = A⋅ B + A⋅C + C ⋅ D
Soluzione
A⋅ B ⋅C + A⋅C ⋅ D + A⋅C =
(
)
A(B + C ) + A ⋅ C ⋅ D =
A C + B ⋅C + A⋅C ⋅ D =
A⋅ B + A⋅C + A⋅C ⋅ D =
(
)
A⋅ B + C A + A⋅ D =
A ⋅ B + C(A + D) =
A⋅ B + A⋅C + C ⋅ D
B ⋅ C ⋅ D + CD + A ⋅ B ⋅ C ⋅ D + A ⋅ B ⋅ C = C
5)
Soluzione
(
)
C ⋅ D B + A ⋅ B + CD + A ⋅ B ⋅ C =
C ⋅ D(B + A) + CD + A ⋅ B ⋅ C =
(
)
C (B + D + A ⋅ D + A ⋅ B ) =
C (A + B + A + D ) =
C B ⋅ D + A⋅ D + D + A⋅ B =
C (1) =
C
Si applichino i teoremi di De Morgan alle seguenti espressioni logiche:
Y = A⋅ B ⋅C = A⋅ B + C = A⋅ B + C
Y = A + B ⋅C = A⋅ B ⋅C = A⋅ B ⋅C
(
)
(
)
⎛
⎞
Y = A⋅ B + B + C = A⋅ B ⋅⎜ B + C ⎟ = A⋅ B B + C = A⋅ B + A⋅ B ⋅C = A⋅ B 1+ C = A⋅ B
⎝
⎠
