Промежуточные таблицы истинности:
¬X2:

X3∧(¬X2):

(X3∧(¬X2))∨X1:

X3→((X3∧(¬X2))∨X1):

X3∧X1:

(¬X2)∨X0:

(X3∧X1)≡((¬X2)∨X0):

¬((X3∧X1)≡((¬X2)∨X0)):

X1∧X0:

(X1∧X0)∨(¬((X3∧X1)≡((¬X2)∨X0))):

(X3→((X3∧(¬X2))∨X1))→((X1∧X0)∨(¬((X3∧X1)≡((¬X2)∨X0)))):

Общая таблица истинности:

Логическая схема:

---

Совершенная дизъюнктивная нормальная форма (СДНФ):

По таблице истинности:
F_сднф = ¬X3∧¬X2∧¬X1∧¬X0 ∨ ¬X3∧¬X2∧¬X1∧X0 ∨ ¬X3∧¬X2∧X1∧¬X0 ∨ ¬X3∧¬X2∧X1∧X0 ∨ ¬X3∧X2∧¬X1∧X0 ∨ ¬X3∧X2∧X1∧X0 ∨ X3∧¬X2∧¬X1∧¬X0 ∨ X3∧¬X2∧¬X1∧X0 ∨ X3∧¬X2∧X1∧X0 ∨ X3∧X2∧¬X1∧¬X0 ∨ X3∧X2∧¬X1∧X0 ∨ X3∧X2∧X1∧¬X0 ∨ X3∧X2∧X1∧X0
Логическая cхема:

---

Совершенная конъюнктивная нормальная форма (СКНФ):

По таблице истинности:
F_скнф = (X3∨¬X2∨X1∨X0) ∧ (X3∨¬X2∨¬X1∨X0) ∧ (¬X3∨X2∨¬X1∨X0)
Логическая cхема:

---

Построение полинома Жегалкина:

По таблице истинности функции
Построим полином Жегалкина:
F_ж = C₀₀₀₀ ⊕ C₁₀₀₀∧X3 ⊕ C₀₁₀₀∧X2 ⊕ C₀₀₁₀∧X1 ⊕ C₀₀₀₁∧X0 ⊕ C₁₁₀₀∧X3∧X2 ⊕ C₁₀₁₀∧X3∧X1 ⊕ C₁₀₀₁∧X3∧X0 ⊕ C₀₁₁₀∧X2∧X1 ⊕ C₀₁₀₁∧X2∧X0 ⊕ C₀₀₁₁∧X1∧X0 ⊕ C₁₁₁₀∧X3∧X2∧X1 ⊕ C₁₁₀₁∧X3∧X2∧X0 ⊕ C₁₀₁₁∧X3∧X1∧X0 ⊕ C₀₁₁₁∧X2∧X1∧X0 ⊕ C₁₁₁₁∧X3∧X2∧X1∧X0

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 = 1 => С₁₀₀₀ = 1 ⊕ 1 = 0
F_ж(0100) = С₀₀₀₀ ⊕ С₀₁₀₀ = 0 => С₀₁₀₀ = 1 ⊕ 0 = 1
F_ж(0010) = С₀₀₀₀ ⊕ С₀₀₁₀ = 1 => С₀₀₁₀ = 1 ⊕ 1 = 0
F_ж(0001) = С₀₀₀₀ ⊕ С₀₀₀₁ = 1 => С₀₀₀₁ = 1 ⊕ 1 = 0
F_ж(1100) = С₀₀₀₀ ⊕ С₁₀₀₀ ⊕ С₀₁₀₀ ⊕ С₁₁₀₀ = 1 => С₁₁₀₀ = 1 ⊕ 0 ⊕ 1 ⊕ 1 = 1
F_ж(1010) = С₀₀₀₀ ⊕ С₁₀₀₀ ⊕ С₀₀₁₀ ⊕ С₁₀₁₀ = 0 => С₁₀₁₀ = 1 ⊕ 0 ⊕ 0 ⊕ 0 = 1
F_ж(1001) = С₀₀₀₀ ⊕ С₁₀₀₀ ⊕ С₀₀₀₁ ⊕ С₁₀₀₁ = 1 => С₁₀₀₁ = 1 ⊕ 0 ⊕ 0 ⊕ 1 = 0
F_ж(0110) = С₀₀₀₀ ⊕ С₀₁₀₀ ⊕ С₀₀₁₀ ⊕ С₀₁₁₀ = 0 => С₀₁₁₀ = 1 ⊕ 1 ⊕ 0 ⊕ 0 = 0
F_ж(0101) = С₀₀₀₀ ⊕ С₀₁₀₀ ⊕ С₀₀₀₁ ⊕ С₀₁₀₁ = 1 => С₀₁₀₁ = 1 ⊕ 1 ⊕ 0 ⊕ 1 = 1
F_ж(0011) = С₀₀₀₀ ⊕ С₀₀₁₀ ⊕ С₀₀₀₁ ⊕ С₀₀₁₁ = 1 => С₀₀₁₁ = 1 ⊕ 0 ⊕ 0 ⊕ 1 = 0
F_ж(1110) = С₀₀₀₀ ⊕ С₁₀₀₀ ⊕ С₀₁₀₀ ⊕ С₀₀₁₀ ⊕ С₁₁₀₀ ⊕ С₁₀₁₀ ⊕ С₀₁₁₀ ⊕ С₁₁₁₀ = 1 => С₁₁₁₀ = 1 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 1 = 1
F_ж(1101) = С₀₀₀₀ ⊕ С₁₀₀₀ ⊕ С₀₁₀₀ ⊕ С₀₀₀₁ ⊕ С₁₁₀₀ ⊕ С₁₀₀₁ ⊕ С₀₁₀₁ ⊕ С₁₁₀₁ = 1 => С₁₁₀₁ = 1 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 1 ⊕ 1 = 1
F_ж(1011) = С₀₀₀₀ ⊕ С₁₀₀₀ ⊕ С₀₀₁₀ ⊕ С₀₀₀₁ ⊕ С₁₀₁₀ ⊕ С₁₀₀₁ ⊕ С₀₀₁₁ ⊕ С₁₀₁₁ = 1 => С₁₀₁₁ = 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 = 1
F_ж(0111) = С₀₀₀₀ ⊕ С₀₁₀₀ ⊕ С₀₀₁₀ ⊕ С₀₀₀₁ ⊕ С₀₁₁₀ ⊕ С₀₁₀₁ ⊕ С₀₀₁₁ ⊕ С₀₁₁₁ = 1 => С₀₁₁₁ = 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 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 = 1 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 1 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 1 = 1

Таким образом, полином Жегалкина будет равен:
F_ж = 1 ⊕ X2 ⊕ X3∧X2 ⊕ X3∧X1 ⊕ X2∧X0 ⊕ X3∧X2∧X1 ⊕ X3∧X2∧X0 ⊕ X3∧X1∧X0 ⊕ X3∧X2∧X1∧X0
Логическая схема, соответствующая полиному Жегалкина: