Промежуточные таблицы истинности:
B∧C:

(B∧C)∧D:

A∨U:

(A∨U)∨((B∧C)∧D):

Общая таблица истинности:

Логическая схема:

---

Совершенная дизъюнктивная нормальная форма (СДНФ):

По таблице истинности:
F_сднф = ¬A∧¬U∧B∧C∧D ∨ ¬A∧U∧¬B∧¬C∧¬D ∨ ¬A∧U∧¬B∧¬C∧D ∨ ¬A∧U∧¬B∧C∧¬D ∨ ¬A∧U∧¬B∧C∧D ∨ ¬A∧U∧B∧¬C∧¬D ∨ ¬A∧U∧B∧¬C∧D ∨ ¬A∧U∧B∧C∧¬D ∨ ¬A∧U∧B∧C∧D ∨ A∧¬U∧¬B∧¬C∧¬D ∨ A∧¬U∧¬B∧¬C∧D ∨ A∧¬U∧¬B∧C∧¬D ∨ A∧¬U∧¬B∧C∧D ∨ A∧¬U∧B∧¬C∧¬D ∨ A∧¬U∧B∧¬C∧D ∨ A∧¬U∧B∧C∧¬D ∨ A∧¬U∧B∧C∧D ∨ A∧U∧¬B∧¬C∧¬D ∨ A∧U∧¬B∧¬C∧D ∨ A∧U∧¬B∧C∧¬D ∨ A∧U∧¬B∧C∧D ∨ A∧U∧B∧¬C∧¬D ∨ A∧U∧B∧¬C∧D ∨ A∧U∧B∧C∧¬D ∨ A∧U∧B∧C∧D
Логическая cхема:

---

Совершенная конъюнктивная нормальная форма (СКНФ):

По таблице истинности:
F_скнф = (A∨U∨B∨C∨D) ∧ (A∨U∨B∨C∨¬D) ∧ (A∨U∨B∨¬C∨D) ∧ (A∨U∨B∨¬C∨¬D) ∧ (A∨U∨¬B∨C∨D) ∧ (A∨U∨¬B∨C∨¬D) ∧ (A∨U∨¬B∨¬C∨D)
Логическая cхема:

---

Построение полинома Жегалкина:

По таблице истинности функции
Построим полином Жегалкина:
F_ж = C₀₀₀₀₀ ⊕ C₁₀₀₀₀∧A ⊕ C₀₁₀₀₀∧U ⊕ C₀₀₁₀₀∧B ⊕ C₀₀₀₁₀∧C ⊕ C₀₀₀₀₁∧D ⊕ C₁₁₀₀₀∧A∧U ⊕ C₁₀₁₀₀∧A∧B ⊕ C₁₀₀₁₀∧A∧C ⊕ C₁₀₀₀₁∧A∧D ⊕ C₀₁₁₀₀∧U∧B ⊕ C₀₁₀₁₀∧U∧C ⊕ C₀₁₀₀₁∧U∧D ⊕ C₀₀₁₁₀∧B∧C ⊕ C₀₀₁₀₁∧B∧D ⊕ C₀₀₀₁₁∧C∧D ⊕ C₁₁₁₀₀∧A∧U∧B ⊕ C₁₁₀₁₀∧A∧U∧C ⊕ C₁₁₀₀₁∧A∧U∧D ⊕ C₁₀₁₁₀∧A∧B∧C ⊕ C₁₀₁₀₁∧A∧B∧D ⊕ C₁₀₀₁₁∧A∧C∧D ⊕ C₀₁₁₁₀∧U∧B∧C ⊕ C₀₁₁₀₁∧U∧B∧D ⊕ C₀₁₀₁₁∧U∧C∧D ⊕ C₀₀₁₁₁∧B∧C∧D ⊕ C₁₁₁₁₀∧A∧U∧B∧C ⊕ C₁₁₁₀₁∧A∧U∧B∧D ⊕ C₁₁₀₁₁∧A∧U∧C∧D ⊕ C₁₀₁₁₁∧A∧B∧C∧D ⊕ C₀₁₁₁₁∧U∧B∧C∧D ⊕ C₁₁₁₁₁∧A∧U∧B∧C∧D

Так как F_ж(00000) = 0, то С₀₀₀₀₀ = 0.

Далее подставляем все остальные наборы в порядке возрастания числа единиц, подставляя вновь полученные значения в следующие формулы:
F_ж(10000) = С₀₀₀₀₀ ⊕ С₁₀₀₀₀ = 1 => С₁₀₀₀₀ = 0 ⊕ 1 = 1
F_ж(01000) = С₀₀₀₀₀ ⊕ С₀₁₀₀₀ = 1 => С₀₁₀₀₀ = 0 ⊕ 1 = 1
F_ж(00100) = С₀₀₀₀₀ ⊕ С₀₀₁₀₀ = 0 => С₀₀₁₀₀ = 0 ⊕ 0 = 0
F_ж(00010) = С₀₀₀₀₀ ⊕ С₀₀₀₁₀ = 0 => С₀₀₀₁₀ = 0 ⊕ 0 = 0
F_ж(00001) = С₀₀₀₀₀ ⊕ С₀₀₀₀₁ = 0 => С₀₀₀₀₁ = 0 ⊕ 0 = 0
F_ж(11000) = С₀₀₀₀₀ ⊕ С₁₀₀₀₀ ⊕ С₀₁₀₀₀ ⊕ С₁₁₀₀₀ = 1 => С₁₁₀₀₀ = 0 ⊕ 1 ⊕ 1 ⊕ 1 = 1
F_ж(10100) = С₀₀₀₀₀ ⊕ С₁₀₀₀₀ ⊕ С₀₀₁₀₀ ⊕ С₁₀₁₀₀ = 1 => С₁₀₁₀₀ = 0 ⊕ 1 ⊕ 0 ⊕ 1 = 0
F_ж(10010) = С₀₀₀₀₀ ⊕ С₁₀₀₀₀ ⊕ С₀₀₀₁₀ ⊕ С₁₀₀₁₀ = 1 => С₁₀₀₁₀ = 0 ⊕ 1 ⊕ 0 ⊕ 1 = 0
F_ж(10001) = С₀₀₀₀₀ ⊕ С₁₀₀₀₀ ⊕ С₀₀₀₀₁ ⊕ С₁₀₀₀₁ = 1 => С₁₀₀₀₁ = 0 ⊕ 1 ⊕ 0 ⊕ 1 = 0
F_ж(01100) = С₀₀₀₀₀ ⊕ С₀₁₀₀₀ ⊕ С₀₀₁₀₀ ⊕ С₀₁₁₀₀ = 1 => С₀₁₁₀₀ = 0 ⊕ 1 ⊕ 0 ⊕ 1 = 0
F_ж(01010) = С₀₀₀₀₀ ⊕ С₀₁₀₀₀ ⊕ С₀₀₀₁₀ ⊕ С₀₁₀₁₀ = 1 => С₀₁₀₁₀ = 0 ⊕ 1 ⊕ 0 ⊕ 1 = 0
F_ж(01001) = С₀₀₀₀₀ ⊕ С₀₁₀₀₀ ⊕ С₀₀₀₀₁ ⊕ С₀₁₀₀₁ = 1 => С₀₁₀₀₁ = 0 ⊕ 1 ⊕ 0 ⊕ 1 = 0
F_ж(00110) = С₀₀₀₀₀ ⊕ С₀₀₁₀₀ ⊕ С₀₀₀₁₀ ⊕ С₀₀₁₁₀ = 0 => С₀₀₁₁₀ = 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
F_ж(00101) = С₀₀₀₀₀ ⊕ С₀₀₁₀₀ ⊕ С₀₀₀₀₁ ⊕ С₀₀₁₀₁ = 0 => С₀₀₁₀₁ = 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
F_ж(00011) = С₀₀₀₀₀ ⊕ С₀₀₀₁₀ ⊕ С₀₀₀₀₁ ⊕ С₀₀₀₁₁ = 0 => С₀₀₀₁₁ = 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
F_ж(11100) = С₀₀₀₀₀ ⊕ С₁₀₀₀₀ ⊕ С₀₁₀₀₀ ⊕ С₀₀₁₀₀ ⊕ С₁₁₀₀₀ ⊕ С₁₀₁₀₀ ⊕ С₀₁₁₀₀ ⊕ С₁₁₁₀₀ = 1 => С₁₁₁₀₀ = 0 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 = 0
F_ж(11010) = С₀₀₀₀₀ ⊕ С₁₀₀₀₀ ⊕ С₀₁₀₀₀ ⊕ С₀₀₀₁₀ ⊕ С₁₁₀₀₀ ⊕ С₁₀₀₁₀ ⊕ С₀₁₀₁₀ ⊕ С₁₁₀₁₀ = 1 => С₁₁₀₁₀ = 0 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 = 0
F_ж(11001) = С₀₀₀₀₀ ⊕ С₁₀₀₀₀ ⊕ С₀₁₀₀₀ ⊕ С₀₀₀₀₁ ⊕ С₁₁₀₀₀ ⊕ С₁₀₀₀₁ ⊕ С₀₁₀₀₁ ⊕ С₁₁₀₀₁ = 1 => С₁₁₀₀₁ = 0 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 = 0
F_ж(10110) = С₀₀₀₀₀ ⊕ С₁₀₀₀₀ ⊕ С₀₀₁₀₀ ⊕ С₀₀₀₁₀ ⊕ С₁₀₁₀₀ ⊕ С₁₀₀₁₀ ⊕ С₀₀₁₁₀ ⊕ С₁₀₁₁₀ = 1 => С₁₀₁₁₀ = 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 = 0
F_ж(10101) = С₀₀₀₀₀ ⊕ С₁₀₀₀₀ ⊕ С₀₀₁₀₀ ⊕ С₀₀₀₀₁ ⊕ С₁₀₁₀₀ ⊕ С₁₀₀₀₁ ⊕ С₀₀₁₀₁ ⊕ С₁₀₁₀₁ = 1 => С₁₀₁₀₁ = 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 = 0
F_ж(10011) = С₀₀₀₀₀ ⊕ С₁₀₀₀₀ ⊕ С₀₀₀₁₀ ⊕ С₀₀₀₀₁ ⊕ С₁₀₀₁₀ ⊕ С₁₀₀₀₁ ⊕ С₀₀₀₁₁ ⊕ С₁₀₀₁₁ = 1 => С₁₀₀₁₁ = 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 = 0
F_ж(01110) = С₀₀₀₀₀ ⊕ С₀₁₀₀₀ ⊕ С₀₀₁₀₀ ⊕ С₀₀₀₁₀ ⊕ С₀₁₁₀₀ ⊕ С₀₁₀₁₀ ⊕ С₀₀₁₁₀ ⊕ С₀₁₁₁₀ = 1 => С₀₁₁₁₀ = 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 = 0
F_ж(01101) = С₀₀₀₀₀ ⊕ С₀₁₀₀₀ ⊕ С₀₀₁₀₀ ⊕ С₀₀₀₀₁ ⊕ С₀₁₁₀₀ ⊕ С₀₁₀₀₁ ⊕ С₀₀₁₀₁ ⊕ С₀₁₁₀₁ = 1 => С₀₁₁₀₁ = 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 = 0
F_ж(01011) = С₀₀₀₀₀ ⊕ С₀₁₀₀₀ ⊕ С₀₀₀₁₀ ⊕ С₀₀₀₀₁ ⊕ С₀₁₀₁₀ ⊕ С₀₁₀₀₁ ⊕ С₀₀₀₁₁ ⊕ С₀₁₀₁₁ = 1 => С₀₁₀₁₁ = 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 = 0
F_ж(00111) = С₀₀₀₀₀ ⊕ С₀₀₁₀₀ ⊕ С₀₀₀₁₀ ⊕ С₀₀₀₀₁ ⊕ С₀₀₁₁₀ ⊕ С₀₀₁₀₁ ⊕ С₀₀₀₁₁ ⊕ С₀₀₁₁₁ = 1 => С₀₀₁₁₁ = 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 = 1
F_ж(11110) = С₀₀₀₀₀ ⊕ С₁₀₀₀₀ ⊕ С₀₁₀₀₀ ⊕ С₀₀₁₀₀ ⊕ С₀₀₀₁₀ ⊕ С₁₁₀₀₀ ⊕ С₁₀₁₀₀ ⊕ С₁₀₀₁₀ ⊕ С₀₁₁₀₀ ⊕ С₀₁₀₁₀ ⊕ С₀₀₁₁₀ ⊕ С₁₁₁₀₀ ⊕ С₁₁₀₁₀ ⊕ С₁₀₁₁₀ ⊕ С₀₁₁₁₀ ⊕ С₁₁₁₁₀ = 1 => С₁₁₁₁₀ = 0 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 = 0
F_ж(11101) = С₀₀₀₀₀ ⊕ С₁₀₀₀₀ ⊕ С₀₁₀₀₀ ⊕ С₀₀₁₀₀ ⊕ С₀₀₀₀₁ ⊕ С₁₁₀₀₀ ⊕ С₁₀₁₀₀ ⊕ С₁₀₀₀₁ ⊕ С₀₁₁₀₀ ⊕ С₀₁₀₀₁ ⊕ С₀₀₁₀₁ ⊕ С₁₁₁₀₀ ⊕ С₁₁₀₀₁ ⊕ С₁₀₁₀₁ ⊕ С₀₁₁₀₁ ⊕ С₁₁₁₀₁ = 1 => С₁₁₁₀₁ = 0 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 = 0
F_ж(11011) = С₀₀₀₀₀ ⊕ С₁₀₀₀₀ ⊕ С₀₁₀₀₀ ⊕ С₀₀₀₁₀ ⊕ С₀₀₀₀₁ ⊕ С₁₁₀₀₀ ⊕ С₁₀₀₁₀ ⊕ С₁₀₀₀₁ ⊕ С₀₁₀₁₀ ⊕ С₀₁₀₀₁ ⊕ С₀₀₀₁₁ ⊕ С₁₁₀₁₀ ⊕ С₁₁₀₀₁ ⊕ С₁₀₀₁₁ ⊕ С₀₁₀₁₁ ⊕ С₁₁₀₁₁ = 1 => С₁₁₀₁₁ = 0 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 = 0
F_ж(10111) = С₀₀₀₀₀ ⊕ С₁₀₀₀₀ ⊕ С₀₀₁₀₀ ⊕ С₀₀₀₁₀ ⊕ С₀₀₀₀₁ ⊕ С₁₀₁₀₀ ⊕ С₁₀₀₁₀ ⊕ С₁₀₀₀₁ ⊕ С₀₀₁₁₀ ⊕ С₀₀₁₀₁ ⊕ С₀₀₀₁₁ ⊕ С₁₀₁₁₀ ⊕ С₁₀₁₀₁ ⊕ С₁₀₀₁₁ ⊕ С₀₀₁₁₁ ⊕ С₁₀₁₁₁ = 1 => С₁₀₁₁₁ = 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 1 = 1
F_ж(01111) = С₀₀₀₀₀ ⊕ С₀₁₀₀₀ ⊕ С₀₀₁₀₀ ⊕ С₀₀₀₁₀ ⊕ С₀₀₀₀₁ ⊕ С₀₁₁₀₀ ⊕ С₀₁₀₁₀ ⊕ С₀₁₀₀₁ ⊕ С₀₀₁₁₀ ⊕ С₀₀₁₀₁ ⊕ С₀₀₀₁₁ ⊕ С₀₁₁₁₀ ⊕ С₀₁₁₀₁ ⊕ С₀₁₀₁₁ ⊕ С₀₀₁₁₁ ⊕ С₀₁₁₁₁ = 1 => С₀₁₁₁₁ = 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 1 = 1
F_ж(11111) = С₀₀₀₀₀ ⊕ С₁₀₀₀₀ ⊕ С₀₁₀₀₀ ⊕ С₀₀₁₀₀ ⊕ С₀₀₀₁₀ ⊕ С₀₀₀₀₁ ⊕ С₁₁₀₀₀ ⊕ С₁₀₁₀₀ ⊕ С₁₀₀₁₀ ⊕ С₁₀₀₀₁ ⊕ С₀₁₁₀₀ ⊕ С₀₁₀₁₀ ⊕ С₀₁₀₀₁ ⊕ С₀₀₁₁₀ ⊕ С₀₀₁₀₁ ⊕ С₀₀₀₁₁ ⊕ С₁₁₁₀₀ ⊕ С₁₁₀₁₀ ⊕ С₁₁₀₀₁ ⊕ С₁₀₁₁₀ ⊕ С₁₀₁₀₁ ⊕ С₁₀₀₁₁ ⊕ С₀₁₁₁₀ ⊕ С₀₁₁₀₁ ⊕ С₀₁₀₁₁ ⊕ С₀₀₁₁₁ ⊕ С₁₁₁₁₀ ⊕ С₁₁₁₀₁ ⊕ С₁₁₀₁₁ ⊕ С₁₀₁₁₁ ⊕ С₀₁₁₁₁ ⊕ С₁₁₁₁₁ = 1 => С₁₁₁₁₁ = 0 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 1 ⊕ 1 = 1

Таким образом, полином Жегалкина будет равен:
F_ж = A ⊕ U ⊕ A∧U ⊕ B∧C∧D ⊕ A∧B∧C∧D ⊕ U∧B∧C∧D ⊕ A∧U∧B∧C∧D
Логическая схема, соответствующая полиному Жегалкина: