Промежуточные таблицы истинности:
¬X4:

X4∨(¬X4):

X2∨X3:

¬X1:

(¬X1)∨X3:

(X4∨(¬X4))∧(X2∨X3):

((X4∨(¬X4))∧(X2∨X3))∧((¬X1)∨X3):

Общая таблица истинности:

Логическая схема:

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Совершенная дизъюнктивная нормальная форма (СДНФ):

По таблице истинности:
F_сднф = ¬X4∧¬X2∧X3∧¬X1 ∨ ¬X4∧¬X2∧X3∧X1 ∨ ¬X4∧X2∧¬X3∧¬X1 ∨ ¬X4∧X2∧X3∧¬X1 ∨ ¬X4∧X2∧X3∧X1 ∨ X4∧¬X2∧X3∧¬X1 ∨ X4∧¬X2∧X3∧X1 ∨ X4∧X2∧¬X3∧¬X1 ∨ X4∧X2∧X3∧¬X1 ∨ X4∧X2∧X3∧X1
Логическая cхема:

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Совершенная конъюнктивная нормальная форма (СКНФ):

По таблице истинности:
F_скнф = (X4∨X2∨X3∨X1) ∧ (X4∨X2∨X3∨¬X1) ∧ (X4∨¬X2∨X3∨¬X1) ∧ (¬X4∨X2∨X3∨X1) ∧ (¬X4∨X2∨X3∨¬X1) ∧ (¬X4∨¬X2∨X3∨¬X1)
Логическая cхема:

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Построение полинома Жегалкина:

По таблице истинности функции
Построим полином Жегалкина:
F_ж = C₀₀₀₀ ⊕ C₁₀₀₀∧X4 ⊕ C₀₁₀₀∧X2 ⊕ C₀₀₁₀∧X3 ⊕ C₀₀₀₁∧X1 ⊕ C₁₁₀₀∧X4∧X2 ⊕ C₁₀₁₀∧X4∧X3 ⊕ C₁₀₀₁∧X4∧X1 ⊕ C₀₁₁₀∧X2∧X3 ⊕ C₀₁₀₁∧X2∧X1 ⊕ C₀₀₁₁∧X3∧X1 ⊕ C₁₁₁₀∧X4∧X2∧X3 ⊕ C₁₁₀₁∧X4∧X2∧X1 ⊕ C₁₀₁₁∧X4∧X3∧X1 ⊕ C₀₁₁₁∧X2∧X3∧X1 ⊕ C₁₁₁₁∧X4∧X2∧X3∧X1

Так как F_ж(0000) = 0, то С₀₀₀₀ = 0.

Далее подставляем все остальные наборы в порядке возрастания числа единиц, подставляя вновь полученные значения в следующие формулы:
F_ж(1000) = С₀₀₀₀ ⊕ С₁₀₀₀ = 0 => С₁₀₀₀ = 0 ⊕ 0 = 0
F_ж(0100) = С₀₀₀₀ ⊕ С₀₁₀₀ = 1 => С₀₁₀₀ = 0 ⊕ 1 = 1
F_ж(0010) = С₀₀₀₀ ⊕ С₀₀₁₀ = 1 => С₀₀₁₀ = 0 ⊕ 1 = 1
F_ж(0001) = С₀₀₀₀ ⊕ С₀₀₀₁ = 0 => С₀₀₀₁ = 0 ⊕ 0 = 0
F_ж(1100) = С₀₀₀₀ ⊕ С₁₀₀₀ ⊕ С₀₁₀₀ ⊕ С₁₁₀₀ = 1 => С₁₁₀₀ = 0 ⊕ 0 ⊕ 1 ⊕ 1 = 0
F_ж(1010) = С₀₀₀₀ ⊕ С₁₀₀₀ ⊕ С₀₀₁₀ ⊕ С₁₀₁₀ = 1 => С₁₀₁₀ = 0 ⊕ 0 ⊕ 1 ⊕ 1 = 0
F_ж(1001) = С₀₀₀₀ ⊕ С₁₀₀₀ ⊕ С₀₀₀₁ ⊕ С₁₀₀₁ = 0 => С₁₀₀₁ = 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
F_ж(0110) = С₀₀₀₀ ⊕ С₀₁₀₀ ⊕ С₀₀₁₀ ⊕ С₀₁₁₀ = 1 => С₀₁₁₀ = 0 ⊕ 1 ⊕ 1 ⊕ 1 = 1
F_ж(0101) = С₀₀₀₀ ⊕ С₀₁₀₀ ⊕ С₀₀₀₁ ⊕ С₀₁₀₁ = 0 => С₀₁₀₁ = 0 ⊕ 1 ⊕ 0 ⊕ 0 = 1
F_ж(0011) = С₀₀₀₀ ⊕ С₀₀₁₀ ⊕ С₀₀₀₁ ⊕ С₀₀₁₁ = 1 => С₀₀₁₁ = 0 ⊕ 1 ⊕ 0 ⊕ 1 = 0
F_ж(1110) = С₀₀₀₀ ⊕ С₁₀₀₀ ⊕ С₀₁₀₀ ⊕ С₀₀₁₀ ⊕ С₁₁₀₀ ⊕ С₁₀₁₀ ⊕ С₀₁₁₀ ⊕ С₁₁₁₀ = 1 => С₁₁₁₀ = 0 ⊕ 0 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 1 = 0
F_ж(1101) = С₀₀₀₀ ⊕ С₁₀₀₀ ⊕ С₀₁₀₀ ⊕ С₀₀₀₁ ⊕ С₁₁₀₀ ⊕ С₁₀₀₁ ⊕ С₀₁₀₁ ⊕ С₁₁₀₁ = 0 => С₁₁₀₁ = 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 = 0
F_ж(1011) = С₀₀₀₀ ⊕ С₁₀₀₀ ⊕ С₀₀₁₀ ⊕ С₀₀₀₁ ⊕ С₁₀₁₀ ⊕ С₁₀₀₁ ⊕ С₀₀₁₁ ⊕ С₁₀₁₁ = 1 => С₁₀₁₁ = 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 = 0
F_ж(0111) = С₀₀₀₀ ⊕ С₀₁₀₀ ⊕ С₀₀₁₀ ⊕ С₀₀₀₁ ⊕ С₀₁₁₀ ⊕ С₀₁₀₁ ⊕ С₀₀₁₁ ⊕ С₀₁₁₁ = 1 => С₀₁₁₁ = 0 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 1 = 1
F_ж(1111) = С₀₀₀₀ ⊕ С₁₀₀₀ ⊕ С₀₁₀₀ ⊕ С₀₀₁₀ ⊕ С₀₀₀₁ ⊕ С₁₁₀₀ ⊕ С₁₀₁₀ ⊕ С₁₀₀₁ ⊕ С₀₁₁₀ ⊕ С₀₁₀₁ ⊕ С₀₀₁₁ ⊕ С₁₁₁₀ ⊕ С₁₁₀₁ ⊕ С₁₀₁₁ ⊕ С₀₁₁₁ ⊕ С₁₁₁₁ = 1 => С₁₁₁₁ = 0 ⊕ 0 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 1 = 0

Таким образом, полином Жегалкина будет равен:
F_ж = X2 ⊕ X3 ⊕ X2∧X3 ⊕ X2∧X1 ⊕ X2∧X3∧X1
Логическая схема, соответствующая полиному Жегалкина: