Промежуточные таблицы истинности:
A∧B:

¬B:

¬A:

(¬B)→(¬A):

((¬B)→(¬A))∧X:

(((¬B)→(¬A))∧X)∧O:

((((¬B)→(¬A))∧X)∧O)∧R:

(((((¬B)→(¬A))∧X)∧O)∧R)∧A:

(A∧B)→((((((¬B)→(¬A))∧X)∧O)∧R)∧A):

Общая таблица истинности:

Логическая схема:

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Совершенная дизъюнктивная нормальная форма (СДНФ):

По таблице истинности:
F_сднф = ¬A∧¬B∧¬X∧¬O∧¬R ∨ ¬A∧¬B∧¬X∧¬O∧R ∨ ¬A∧¬B∧¬X∧O∧¬R ∨ ¬A∧¬B∧¬X∧O∧R ∨ ¬A∧¬B∧X∧¬O∧¬R ∨ ¬A∧¬B∧X∧¬O∧R ∨ ¬A∧¬B∧X∧O∧¬R ∨ ¬A∧¬B∧X∧O∧R ∨ ¬A∧B∧¬X∧¬O∧¬R ∨ ¬A∧B∧¬X∧¬O∧R ∨ ¬A∧B∧¬X∧O∧¬R ∨ ¬A∧B∧¬X∧O∧R ∨ ¬A∧B∧X∧¬O∧¬R ∨ ¬A∧B∧X∧¬O∧R ∨ ¬A∧B∧X∧O∧¬R ∨ ¬A∧B∧X∧O∧R ∨ A∧¬B∧¬X∧¬O∧¬R ∨ A∧¬B∧¬X∧¬O∧R ∨ A∧¬B∧¬X∧O∧¬R ∨ A∧¬B∧¬X∧O∧R ∨ A∧¬B∧X∧¬O∧¬R ∨ A∧¬B∧X∧¬O∧R ∨ A∧¬B∧X∧O∧¬R ∨ A∧¬B∧X∧O∧R ∨ A∧B∧X∧O∧R
Логическая cхема:

---

Совершенная конъюнктивная нормальная форма (СКНФ):

По таблице истинности:
F_скнф = (¬A∨¬B∨X∨O∨R) ∧ (¬A∨¬B∨X∨O∨¬R) ∧ (¬A∨¬B∨X∨¬O∨R) ∧ (¬A∨¬B∨X∨¬O∨¬R) ∧ (¬A∨¬B∨¬X∨O∨R) ∧ (¬A∨¬B∨¬X∨O∨¬R) ∧ (¬A∨¬B∨¬X∨¬O∨R)
Логическая cхема:

---

Построение полинома Жегалкина:

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

Так как F_ж(00000) = 1, то С₀₀₀₀₀ = 1.

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

Таким образом, полином Жегалкина будет равен:
F_ж = 1 ⊕ A∧B ⊕ A∧B∧X∧O∧R
Логическая схема, соответствующая полиному Жегалкина: