Question

Помогите, пожалуйста.... Проверьте, является ли следующее высказывание тавтологией(по возможности не через таблицы истин. делать): (A → (C ∧ D)) → (((A → B) ∧ (E → ¬D)) → ((C → B) ∨ (D ∧ B ∧ ¬E))).

Answer 1

Ответ:

нет

Объяснение:

(A → (C ∧ D)) → (((A → B) ∧ (E → ¬D)) → ((C → B) ∨ (D ∧ B ∧ ¬E))) =

= (A → (C ∧ D)) → (((!A ∨ B) ∧ (!E ∨ !D)) → ((!C ∨ B) ∨ (D ∧ B ∧ !E))) =

= !(!A ∨ (C ∧ D)) ∨ (!((!A ∨ B) ∧ (!E ∨ !D)) ∨ ((!C ∨ B) ∨ (D ∧ B ∧ !E))) =

= (A ∧ !(C ∧ D)) ∨ ((!(!A ∨ B) ∨ !(!E ∨ !D)) ∨ ((!C ∨ B) ∨ (D ∧ B ∧ !E))) =

= (A ∧ (!C ∨ !D)) ∨ (((A ∧ !B) ∨ (E ∧ D)) ∨ ((!C ∨ B) ∨ (D ∧ B ∧ !E))) =

= (A ∧ (!C ∨ !D)) ∨ (A ∧ !B) ∨ (E ∧ D) ∨ !C ∨ B ∨ (D ∧ B ∧ !E) =

{

   B ∨ (D ∧ B ∧ !E) = (B ∨ D) ∧ (B ∨ B) ∧ (B ∨ !E) = (B ∨ D) ∧ B ∧ (B ∨ !E) = B ∨ (D ∧ 1 ∧ !E) = B ∨ (D ∧ !E)  

   (A ∧ (!C ∨ !D)) ∨ (A ∧ !B) = (A ∧ (!C ∨ !D ∨ !B))

}

= (A ∧ (!C ∨ !D ∨ !B)) ∨ (E ∧ D) ∨ !C ∨ B ∨ (D ∧ !E) =

{

   (E ∧ D) ∨ (D ∧ !E) = D ∧ (E ∨ !E) = D ∧ (1) = D

}

= (A ∧ (!C ∨ !D ∨ !B)) ∨ D ∨ !C ∨ B =

= (A ∧ !C) ∨ (A ∧ !D) ∨ (A ∧ !B) ∨ D ∨ !C ∨ B =

{

   (A ∧ !C) ∨ !C = (A ∧ !C) ∨ (!C ∧ 1) = !C ∧ (A ∨ 1) = !C ∧ (1) = !C

}

= !C ∨ (A ∧ !D) ∨ (A ∧ !B) ∨ D ∨ B =

{

   (A ∧ !B) ∨ B = (A ∨ B) ∧ (!B ∨ B) = (A ∨ B) ∧ (1) = (A ∨ B)

   (A ∧ !D) ∨ D = (A ∨ D) ∧ (!D ∨ D) = (A ∨ D) ∧ (1) = (A ∨ D)

}

= !C ∨ (A ∨ B) ∨ (A ∨ D) =

= !C ∨ A ∨ B ∨ A ∨ D =

= A ∨ B ∨ !C ∨ D

Answer 2

Checking for Tautology

To check if the given statement is a tautology, we can use logical equivalences and truth tables. The given statement is:

(A → (C ∧ D)) → (((A → B) ∧ (E → ¬D)) → ((C → B) ∨ (D ∧ B ∧ ¬E)))

Let's break it down step by step.

1. Step 1: We can start by simplifying the innermost expressions using logical equivalences.

- Innermost Expression 1: C → B This can be rewritten as ¬C ∨ B using the logical equivalence for implication.

- Innermost Expression 2: E → ¬D This can be rewritten as ¬E ∨ ¬D using the logical equivalence for implication.

2. Step 2: Substituting the simplified innermost expressions back into the original statement, we get:

(A → (C ∧ D)) → (((A → B) ∧ (¬E ∨ ¬D)) → ((¬C ∨ B) ∨ (D ∧ B ∧ ¬E)))

3. Step 3: We can continue simplifying the statement using logical equivalences and truth tables to check for tautology.

- Logical Equivalences: - Implication: A → B is equivalent to ¬A ∨ B.

- Truth Tables: We can construct a truth table to check if the given statement is a tautology by evaluating all possible combinations of truth values for A, B, C, D, and E.

Based on the provided search result, it seems that the specific logical equivalences and truth table evaluations are not explicitly mentioned. Therefore, I will proceed with the logical simplifications and truth table evaluation to determine if the given statement is a tautology.

Let's continue with the logical simplifications and truth table evaluation to determine if the given statement is a tautology.