Промежуточные таблицы истинности:
¬X1:

(¬X1)|X3:

X2|X4:

((¬X1)|X3)|(X2|X4):

Y≡(((¬X1)|X3)|(X2|X4)):

Общая таблица истинности:

Логическая схема:

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Совершенная дизъюнктивная нормальная форма (СДНФ):

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

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Совершенная конъюнктивная нормальная форма (СКНФ):

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

---

Построение полинома Жегалкина:

По таблице истинности функции
Построим полином Жегалкина:
F_ж = C₀₀₀₀₀ ⊕ C₁₀₀₀₀∧Y ⊕ C₀₁₀₀₀∧X1 ⊕ C₀₀₁₀₀∧X3 ⊕ C₀₀₀₁₀∧X2 ⊕ C₀₀₀₀₁∧X4 ⊕ C₁₁₀₀₀∧Y∧X1 ⊕ C₁₀₁₀₀∧Y∧X3 ⊕ C₁₀₀₁₀∧Y∧X2 ⊕ C₁₀₀₀₁∧Y∧X4 ⊕ C₀₁₁₀₀∧X1∧X3 ⊕ C₀₁₀₁₀∧X1∧X2 ⊕ C₀₁₀₀₁∧X1∧X4 ⊕ C₀₀₁₁₀∧X3∧X2 ⊕ C₀₀₁₀₁∧X3∧X4 ⊕ C₀₀₀₁₁∧X2∧X4 ⊕ C₁₁₁₀₀∧Y∧X1∧X3 ⊕ C₁₁₀₁₀∧Y∧X1∧X2 ⊕ C₁₁₀₀₁∧Y∧X1∧X4 ⊕ C₁₀₁₁₀∧Y∧X3∧X2 ⊕ C₁₀₁₀₁∧Y∧X3∧X4 ⊕ C₁₀₀₁₁∧Y∧X2∧X4 ⊕ C₀₁₁₁₀∧X1∧X3∧X2 ⊕ C₀₁₁₀₁∧X1∧X3∧X4 ⊕ C₀₁₀₁₁∧X1∧X2∧X4 ⊕ C₀₀₁₁₁∧X3∧X2∧X4 ⊕ C₁₁₁₁₀∧Y∧X1∧X3∧X2 ⊕ C₁₁₁₀₁∧Y∧X1∧X3∧X4 ⊕ C₁₁₀₁₁∧Y∧X1∧X2∧X4 ⊕ C₁₀₁₁₁∧Y∧X3∧X2∧X4 ⊕ C₀₁₁₁₁∧X1∧X3∧X2∧X4 ⊕ C₁₁₁₁₁∧Y∧X1∧X3∧X2∧X4

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 = 0 => С₁₀₀₀₀ = 1 ⊕ 0 = 1
F_ж(01000) = С₀₀₀₀₀ ⊕ С₀₁₀₀₀ = 1 => С₀₁₀₀₀ = 1 ⊕ 1 = 0
F_ж(00100) = С₀₀₀₀₀ ⊕ С₀₀₁₀₀ = 0 => С₀₀₁₀₀ = 1 ⊕ 0 = 1
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 ⊕ 1 ⊕ 0 ⊕ 0 = 0
F_ж(10100) = С₀₀₀₀₀ ⊕ С₁₀₀₀₀ ⊕ С₀₀₁₀₀ ⊕ С₁₀₁₀₀ = 1 => С₁₀₁₀₀ = 1 ⊕ 1 ⊕ 1 ⊕ 1 = 0
F_ж(10010) = С₀₀₀₀₀ ⊕ С₁₀₀₀₀ ⊕ С₀₀₀₁₀ ⊕ С₁₀₀₁₀ = 0 => С₁₀₀₁₀ = 1 ⊕ 1 ⊕ 0 ⊕ 0 = 0
F_ж(10001) = С₀₀₀₀₀ ⊕ С₁₀₀₀₀ ⊕ С₀₀₀₀₁ ⊕ С₁₀₀₀₁ = 0 => С₁₀₀₀₁ = 1 ⊕ 1 ⊕ 0 ⊕ 0 = 0
F_ж(01100) = С₀₀₀₀₀ ⊕ С₀₁₀₀₀ ⊕ С₀₀₁₀₀ ⊕ С₀₁₁₀₀ = 1 => С₀₁₁₀₀ = 1 ⊕ 0 ⊕ 1 ⊕ 1 = 1
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 = 0 => С₀₀₁₁₀ = 1 ⊕ 1 ⊕ 0 ⊕ 0 = 0
F_ж(00101) = С₀₀₀₀₀ ⊕ С₀₀₁₀₀ ⊕ С₀₀₀₀₁ ⊕ С₀₀₁₀₁ = 0 => С₀₀₁₀₁ = 1 ⊕ 1 ⊕ 0 ⊕ 0 = 0
F_ж(00011) = С₀₀₀₀₀ ⊕ С₀₀₀₁₀ ⊕ С₀₀₀₀₁ ⊕ С₀₀₀₁₁ = 0 => С₀₀₀₁₁ = 1 ⊕ 0 ⊕ 0 ⊕ 0 = 1
F_ж(11100) = С₀₀₀₀₀ ⊕ С₁₀₀₀₀ ⊕ С₀₁₀₀₀ ⊕ С₀₀₁₀₀ ⊕ С₁₁₀₀₀ ⊕ С₁₀₁₀₀ ⊕ С₀₁₁₀₀ ⊕ С₁₁₁₀₀ = 0 => С₁₁₁₀₀ = 1 ⊕ 1 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 = 0
F_ж(11010) = С₀₀₀₀₀ ⊕ С₁₀₀₀₀ ⊕ С₀₁₀₀₀ ⊕ С₀₀₀₁₀ ⊕ С₁₁₀₀₀ ⊕ С₁₀₀₁₀ ⊕ С₀₁₀₁₀ ⊕ С₁₁₀₁₀ = 0 => С₁₁₀₁₀ = 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
F_ж(11001) = С₀₀₀₀₀ ⊕ С₁₀₀₀₀ ⊕ С₀₁₀₀₀ ⊕ С₀₀₀₀₁ ⊕ С₁₁₀₀₀ ⊕ С₁₀₀₀₁ ⊕ С₀₁₀₀₁ ⊕ С₁₁₀₀₁ = 0 => С₁₁₀₀₁ = 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
F_ж(10110) = С₀₀₀₀₀ ⊕ С₁₀₀₀₀ ⊕ С₀₀₁₀₀ ⊕ С₀₀₀₁₀ ⊕ С₁₀₁₀₀ ⊕ С₁₀₀₁₀ ⊕ С₀₀₁₁₀ ⊕ С₁₀₁₁₀ = 1 => С₁₀₁₁₀ = 1 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 = 0
F_ж(10101) = С₀₀₀₀₀ ⊕ С₁₀₀₀₀ ⊕ С₀₀₁₀₀ ⊕ С₀₀₀₀₁ ⊕ С₁₀₁₀₀ ⊕ С₁₀₀₀₁ ⊕ С₀₀₁₀₁ ⊕ С₁₀₁₀₁ = 1 => С₁₀₁₀₁ = 1 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 = 0
F_ж(10011) = С₀₀₀₀₀ ⊕ С₁₀₀₀₀ ⊕ С₀₀₀₁₀ ⊕ С₀₀₀₀₁ ⊕ С₁₀₀₁₀ ⊕ С₁₀₀₀₁ ⊕ С₀₀₀₁₁ ⊕ С₁₀₀₁₁ = 1 => С₁₀₀₁₁ = 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 1 = 0
F_ж(01110) = С₀₀₀₀₀ ⊕ С₀₁₀₀₀ ⊕ С₀₀₁₀₀ ⊕ С₀₀₀₁₀ ⊕ С₀₁₁₀₀ ⊕ С₀₁₀₁₀ ⊕ С₀₀₁₁₀ ⊕ С₀₁₁₁₀ = 1 => С₀₁₁₁₀ = 1 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 = 0
F_ж(01101) = С₀₀₀₀₀ ⊕ С₀₁₀₀₀ ⊕ С₀₀₁₀₀ ⊕ С₀₀₀₀₁ ⊕ С₀₁₁₀₀ ⊕ С₀₁₀₀₁ ⊕ С₀₀₁₀₁ ⊕ С₀₁₁₀₁ = 1 => С₀₁₁₀₁ = 1 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 = 0
F_ж(01011) = С₀₀₀₀₀ ⊕ С₀₁₀₀₀ ⊕ С₀₀₀₁₀ ⊕ С₀₀₀₀₁ ⊕ С₀₁₀₁₀ ⊕ С₀₁₀₀₁ ⊕ С₀₀₀₁₁ ⊕ С₀₁₀₁₁ = 0 => С₀₁₀₁₁ = 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 = 0
F_ж(00111) = С₀₀₀₀₀ ⊕ С₀₀₁₀₀ ⊕ С₀₀₀₁₀ ⊕ С₀₀₀₀₁ ⊕ С₀₀₁₁₀ ⊕ С₀₀₁₀₁ ⊕ С₀₀₀₁₁ ⊕ С₀₀₁₁₁ = 0 => С₀₀₁₁₁ = 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 = 1
F_ж(11110) = С₀₀₀₀₀ ⊕ С₁₀₀₀₀ ⊕ С₀₁₀₀₀ ⊕ С₀₀₁₀₀ ⊕ С₀₀₀₁₀ ⊕ С₁₁₀₀₀ ⊕ С₁₀₁₀₀ ⊕ С₁₀₀₁₀ ⊕ С₀₁₁₀₀ ⊕ С₀₁₀₁₀ ⊕ С₀₀₁₁₀ ⊕ С₁₁₁₀₀ ⊕ С₁₁₀₁₀ ⊕ С₁₀₁₁₀ ⊕ С₀₁₁₁₀ ⊕ С₁₁₁₁₀ = 0 => С₁₁₁₁₀ = 1 ⊕ 1 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 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 ⊕ 1 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 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 = 1 => С₁₁₀₁₁ = 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 = 0
F_ж(10111) = С₀₀₀₀₀ ⊕ С₁₀₀₀₀ ⊕ С₀₀₁₀₀ ⊕ С₀₀₀₁₀ ⊕ С₀₀₀₀₁ ⊕ С₁₀₁₀₀ ⊕ С₁₀₀₁₀ ⊕ С₁₀₀₀₁ ⊕ С₀₀₁₁₀ ⊕ С₀₀₁₀₁ ⊕ С₀₀₀₁₁ ⊕ С₁₀₁₁₀ ⊕ С₁₀₁₀₁ ⊕ С₁₀₀₁₁ ⊕ С₀₀₁₁₁ ⊕ С₁₀₁₁₁ = 1 => С₁₀₁₁₁ = 1 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 1 = 0
F_ж(01111) = С₀₀₀₀₀ ⊕ С₀₁₀₀₀ ⊕ С₀₀₁₀₀ ⊕ С₀₀₀₁₀ ⊕ С₀₀₀₀₁ ⊕ С₀₁₁₀₀ ⊕ С₀₁₀₁₀ ⊕ С₀₁₀₀₁ ⊕ С₀₀₁₁₀ ⊕ С₀₀₁₀₁ ⊕ С₀₀₀₁₁ ⊕ С₀₁₁₁₀ ⊕ С₀₁₁₀₁ ⊕ С₀₁₀₁₁ ⊕ С₀₀₁₁₁ ⊕ С₀₁₁₁₁ = 0 => С₀₁₁₁₁ = 1 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 = 1
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 ⊕ 1 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 1 = 0

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