Промежуточные таблицы истинности:

Общая таблица истинности:

Логическая схема:

---

Совершенная дизъюнктивная нормальная форма (СДНФ):

По таблице истинности:
В таблице истинности нет набора значений переменных при которых функция истинна!

---

Совершенная конъюнктивная нормальная форма (СКНФ):

По таблице истинности:
F_скнф = (X?∨Y?Z∨X?Y∨X?Z) ∧ (X?∨Y?Z∨X?Y∨¬X?Z) ∧ (X?∨Y?Z∨¬X?Y∨X?Z) ∧ (X?∨Y?Z∨¬X?Y∨¬X?Z) ∧ (X?∨¬Y?Z∨X?Y∨X?Z) ∧ (X?∨¬Y?Z∨X?Y∨¬X?Z) ∧ (X?∨¬Y?Z∨¬X?Y∨X?Z) ∧ (X?∨¬Y?Z∨¬X?Y∨¬X?Z) ∧ (¬X?∨Y?Z∨X?Y∨X?Z) ∧ (¬X?∨Y?Z∨X?Y∨¬X?Z) ∧ (¬X?∨Y?Z∨¬X?Y∨X?Z) ∧ (¬X?∨Y?Z∨¬X?Y∨¬X?Z) ∧ (¬X?∨¬Y?Z∨X?Y∨X?Z) ∧ (¬X?∨¬Y?Z∨X?Y∨¬X?Z) ∧ (¬X?∨¬Y?Z∨¬X?Y∨X?Z) ∧ (¬X?∨¬Y?Z∨¬X?Y∨¬X?Z)
Логическая cхема:

---

Построение полинома Жегалкина:

По таблице истинности функции
Построим полином Жегалкина:
F_ж = C₀₀₀₀ ⊕ C₁₀₀₀∧X? ⊕ C₀₁₀₀∧Y?Z ⊕ C₀₀₁₀∧X?Y ⊕ C₀₀₀₁∧X?Z ⊕ C₁₁₀₀∧X?∧Y?Z ⊕ C₁₀₁₀∧X?∧X?Y ⊕ C₁₀₀₁∧X?∧X?Z ⊕ C₀₁₁₀∧Y?Z∧X?Y ⊕ C₀₁₀₁∧Y?Z∧X?Z ⊕ C₀₀₁₁∧X?Y∧X?Z ⊕ C₁₁₁₀∧X?∧Y?Z∧X?Y ⊕ C₁₁₀₁∧X?∧Y?Z∧X?Z ⊕ C₁₀₁₁∧X?∧X?Y∧X?Z ⊕ C₀₁₁₁∧Y?Z∧X?Y∧X?Z ⊕ C₁₁₁₁∧X?∧Y?Z∧X?Y∧X?Z

Так как F_ж(0000) = (, то С₀₀₀₀ = (.

Далее подставляем все остальные наборы в порядке возрастания числа единиц, подставляя вновь полученные значения в следующие формулы:
F_ж(1000) = С₀₀₀₀ ⊕ С₁₀₀₀ = ( => С₁₀₀₀ = ( ⊕ ( = 0
F_ж(0100) = С₀₀₀₀ ⊕ С₀₁₀₀ = ( => С₀₁₀₀ = ( ⊕ ( = 0
F_ж(0010) = С₀₀₀₀ ⊕ С₀₀₁₀ = ( => С₀₀₁₀ = ( ⊕ ( = 0
F_ж(0001) = С₀₀₀₀ ⊕ С₀₀₀₁ = ( => С₀₀₀₁ = ( ⊕ ( = 0
F_ж(1100) = С₀₀₀₀ ⊕ С₁₀₀₀ ⊕ С₀₁₀₀ ⊕ С₁₁₀₀ = ( => С₁₁₀₀ = ( ⊕ 0 ⊕ 0 ⊕ ( = 0
F_ж(1010) = С₀₀₀₀ ⊕ С₁₀₀₀ ⊕ С₀₀₁₀ ⊕ С₁₀₁₀ = ( => С₁₀₁₀ = ( ⊕ 0 ⊕ 0 ⊕ ( = 0
F_ж(1001) = С₀₀₀₀ ⊕ С₁₀₀₀ ⊕ С₀₀₀₁ ⊕ С₁₀₀₁ = ( => С₁₀₀₁ = ( ⊕ 0 ⊕ 0 ⊕ ( = 0
F_ж(0110) = С₀₀₀₀ ⊕ С₀₁₀₀ ⊕ С₀₀₁₀ ⊕ С₀₁₁₀ = ( => С₀₁₁₀ = ( ⊕ 0 ⊕ 0 ⊕ ( = 0
F_ж(0101) = С₀₀₀₀ ⊕ С₀₁₀₀ ⊕ С₀₀₀₁ ⊕ С₀₁₀₁ = ( => С₀₁₀₁ = ( ⊕ 0 ⊕ 0 ⊕ ( = 0
F_ж(0011) = С₀₀₀₀ ⊕ С₀₀₁₀ ⊕ С₀₀₀₁ ⊕ С₀₀₁₁ = ( => С₀₀₁₁ = ( ⊕ 0 ⊕ 0 ⊕ ( = 0
F_ж(1110) = С₀₀₀₀ ⊕ С₁₀₀₀ ⊕ С₀₁₀₀ ⊕ С₀₀₁₀ ⊕ С₁₁₀₀ ⊕ С₁₀₁₀ ⊕ С₀₁₁₀ ⊕ С₁₁₁₀ = ( => С₁₁₁₀ = ( ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ ( = 0
F_ж(1101) = С₀₀₀₀ ⊕ С₁₀₀₀ ⊕ С₀₁₀₀ ⊕ С₀₀₀₁ ⊕ С₁₁₀₀ ⊕ С₁₀₀₁ ⊕ С₀₁₀₁ ⊕ С₁₁₀₁ = ( => С₁₁₀₁ = ( ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ ( = 0
F_ж(1011) = С₀₀₀₀ ⊕ С₁₀₀₀ ⊕ С₀₀₁₀ ⊕ С₀₀₀₁ ⊕ С₁₀₁₀ ⊕ С₁₀₀₁ ⊕ С₀₀₁₁ ⊕ С₁₀₁₁ = ( => С₁₀₁₁ = ( ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ ( = 0
F_ж(0111) = С₀₀₀₀ ⊕ С₀₁₀₀ ⊕ С₀₀₁₀ ⊕ С₀₀₀₁ ⊕ С₀₁₁₀ ⊕ С₀₁₀₁ ⊕ С₀₀₁₁ ⊕ С₀₁₁₁ = ( => С₀₁₁₁ = ( ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ ( = 0
F_ж(1111) = С₀₀₀₀ ⊕ С₁₀₀₀ ⊕ С₀₁₀₀ ⊕ С₀₀₁₀ ⊕ С₀₀₀₁ ⊕ С₁₁₀₀ ⊕ С₁₀₁₀ ⊕ С₁₀₀₁ ⊕ С₀₁₁₀ ⊕ С₀₁₀₁ ⊕ С₀₀₁₁ ⊕ С₁₁₁₀ ⊕ С₁₁₀₁ ⊕ С₁₀₁₁ ⊕ С₀₁₁₁ ⊕ С₁₁₁₁ = ( => С₁₁₁₁ = ( ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ ( = 0

Таким образом, полином Жегалкина будет равен:
F_ж = (