Question

a) Simplify the Boolean function F = AB + (AC)′ + AB ′C(AB + C).

Answer 1

Answer on Question #78937 – Math – Discrete Mathematics

Question

a) Simplify the Boolean function $F = AB + (AC)' + AB'C(AB + C)$.

Solution

$AB + (AC)' + AB'C(AB + C) = AB + A' + C' + ABB'C + AB'C$ (according to formulas 1, 2 and 3)

$AB + A' + C' + ABB'C + AB'C = AB + A' + C' + AB'C$ (according to formulas 4 and 5)

$AB + A' + C' + AB'C = A' + B + C' + AB'C$ (according to formula 8)

$A' + B + C' + AB'C = A' + C' + (B + AB'C)$

$A' + C' + (B + AB'C) = A' + C' + (B + AC)$ (according to formula 8)

$A' + C' + (B + AC) = A' + B + C' + AC$

$A' + B + C' + AC = A' + B + C' + C$ (according to formula 8)

$A' + B + C' + C = A' + B' + 1$ (according to formula 6)

$A' + B + 1 = 1$ (according to formula 7)

Formula 1 (OR Distributive law): $A(B + C) = AB + AC$

Formula 2: $(AB)' = A' + B'$

Formula 3: $AA = A$

Formula 4: $AA' = 0$

Formula 5: $A + 0 = A$

Formula 6: $A + A' = 1$

Formula 7: $A + 1 = A' + 1 = 1$

Formula 8: $AB + A' = A' + B \Leftrightarrow A'B + A = A + B$

Proof of formula 8:

Formula 8.1 (AND Distributive law): $A + (BC) = (A + B)(A + C)$

$A'B + A = (A + A')(A + B)$ (according to formula 8.1)

$(A + A')(A + B) = 1(A + B)$ (according to formula 6)

$1(A + B) = A + B$

$AB + A' = (A' + A)(A' + B)$ (according to formula 8.1)

$(A' + A)(A' + B) = 1(A' + B)$ (according to formula 6)

$1(A' + B) = A' + B$

Answer: $F = 1$.

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