Промежуточные таблицы истинности:
X1∨X2:

(X1∨X2)∨X3:

((X1∨X2)∨X3)∨X4:

¬(((X1∨X2)∨X3)∨X4):

Общая таблица истинности:

Логическая схема:

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Совершенная дизъюнктивная нормальная форма (СДНФ):

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

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Совершенная конъюнктивная нормальная форма (СКНФ):

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

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Построение полинома Жегалкина:

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

Так как F_ж(0000) = 1, то С₀₀₀₀ = 1.

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

Таким образом, полином Жегалкина будет равен:
F_ж = 1 ⊕ X1 ⊕ X2 ⊕ X3 ⊕ X4 ⊕ X1∧X2 ⊕ X1∧X3 ⊕ X1∧X4 ⊕ X2∧X3 ⊕ X2∧X4 ⊕ X3∧X4 ⊕ X1∧X2∧X3 ⊕ X1∧X2∧X4 ⊕ X1∧X3∧X4 ⊕ X2∧X3∧X4 ⊕ X1∧X2∧X3∧X4
Логическая схема, соответствующая полиному Жегалкина: