Промежуточные таблицы истинности:
¬X1:

¬X3:

X2∧X4:

X4∧(¬X1):

(X4∧(¬X1))∧(¬X3):

X2∧(¬X1):

(X2∧(¬X1))∧X3:

X4∧X1:

(X4∧X1)∧X3:

(¬X1)∧(¬X3):

((¬X1)∧(¬X3))∧X4:

(X2∧X4)∨((X4∧(¬X1))∧(¬X3)):

((X2∧X4)∨((X4∧(¬X1))∧(¬X3)))∨((X2∧(¬X1))∧X3):

(((X2∧X4)∨((X4∧(¬X1))∧(¬X3)))∨((X2∧(¬X1))∧X3))∨((X4∧X1)∧X3):

((((X2∧X4)∨((X4∧(¬X1))∧(¬X3)))∨((X2∧(¬X1))∧X3))∨((X4∧X1)∧X3))∨(((¬X1)∧(¬X3))∧X4):

Общая таблица истинности:

Логическая схема:

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Совершенная дизъюнктивная нормальная форма (СДНФ):

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

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Совершенная конъюнктивная нормальная форма (СКНФ):

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

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Построение полинома Жегалкина:

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

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 = 0 => С₀₀₁₀ = 0 ⊕ 0 = 0
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 = 0 => С₁₀₁₀ = 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
F_ж(1001) = С₀₀₀₀ ⊕ С₁₀₀₀ ⊕ С₀₀₀₁ ⊕ С₁₀₀₁ = 1 => С₁₀₀₁ = 0 ⊕ 0 ⊕ 0 ⊕ 1 = 1
F_ж(0110) = С₀₀₀₀ ⊕ С₀₁₀₀ ⊕ С₀₀₁₀ ⊕ С₀₁₁₀ = 0 => С₀₁₁₀ = 0 ⊕ 1 ⊕ 0 ⊕ 0 = 1
F_ж(0101) = С₀₀₀₀ ⊕ С₀₁₀₀ ⊕ С₀₀₀₁ ⊕ С₀₁₀₁ = 0 => С₀₁₀₁ = 0 ⊕ 1 ⊕ 0 ⊕ 0 = 1
F_ж(0011) = С₀₀₀₀ ⊕ С₀₀₁₀ ⊕ С₀₀₀₁ ⊕ С₀₀₁₁ = 0 => С₀₀₁₁ = 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
F_ж(1110) = С₀₀₀₀ ⊕ С₁₀₀₀ ⊕ С₀₁₀₀ ⊕ С₀₀₁₀ ⊕ С₁₁₀₀ ⊕ С₁₀₁₀ ⊕ С₀₁₁₀ ⊕ С₁₁₁₀ = 1 => С₁₁₁₀ = 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 1 = 1
F_ж(1101) = С₀₀₀₀ ⊕ С₁₀₀₀ ⊕ С₀₁₀₀ ⊕ С₀₀₀₁ ⊕ С₁₁₀₀ ⊕ С₁₀₀₁ ⊕ С₀₁₀₁ ⊕ С₁₁₀₁ = 1 => С₁₁₀₁ = 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 1 ⊕ 1 = 0
F_ж(1011) = С₀₀₀₀ ⊕ С₁₀₀₀ ⊕ С₀₀₁₀ ⊕ С₀₀₀₁ ⊕ С₁₀₁₀ ⊕ С₁₀₀₁ ⊕ С₀₀₁₁ ⊕ С₁₀₁₁ = 0 => С₁₀₁₁ = 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 = 1
F_ж(0111) = С₀₀₀₀ ⊕ С₀₁₀₀ ⊕ С₀₀₁₀ ⊕ С₀₀₀₁ ⊕ С₀₁₁₀ ⊕ С₀₁₀₁ ⊕ С₀₀₁₁ ⊕ С₀₁₁₁ = 1 => С₀₁₁₁ = 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 1 = 0
F_ж(1111) = С₀₀₀₀ ⊕ С₁₀₀₀ ⊕ С₀₁₀₀ ⊕ С₀₀₁₀ ⊕ С₀₀₀₁ ⊕ С₁₁₀₀ ⊕ С₁₀₁₀ ⊕ С₁₀₀₁ ⊕ С₀₁₁₀ ⊕ С₀₁₀₁ ⊕ С₀₀₁₁ ⊕ С₁₁₁₀ ⊕ С₁₁₀₁ ⊕ С₁₀₁₁ ⊕ С₀₁₁₁ ⊕ С₁₁₁₁ = 1 => С₁₁₁₁ = 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 1 = 1

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