Adapta los ejercicios resueltos a tu asignatura. Regístrate gratis
Demuestre mediante tablas de verdad si las siguientes fórmulas
son equivalentes o no:
21. [Ejercicio 21] p ∧ q → p , p ∨ p → r
| p | q | r | p ∧ q | p ∧ q → p | p ∨ p | p ∨ p → r | (p ∧ q → p) ∧ (p ∨ p → r) |
|---|---|---|---|---|---|---|---|
| 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |
| 1 | 1 | 0 | 1 | 1 | 1 | 0 | 0 |
| 1 | 0 | 1 | 0 | 1 | 1 | 1 | 1 |
| 1 | 0 | 0 | 0 | 1 | 1 | 0 | 0 |
| 0 | 1 | 1 | 0 | 1 | 0 | 1 | 1 |
| 0 | 1 | 0 | 0 | 1 | 0 | 1 | 1 |
| 0 | 0 | 1 | 0 | 1 | 0 | 1 | 1 |
| 0 | 0 | 0 | 0 | 1 | 0 | 1 | 1 |
NO HAY EQUIVALENCIA LÓGICA. CONTRAEJEMPLO: I(p)= 1, I(q)=1 y I(r)= 0.
22. [Ejercicio 22] p ∧ (q ∨ r) , (p ∧ q) ∨ (p ∧ r)
| p | q | r | q ∨ r | p ∧ (q ∨ r) | p ∧ q | p ∧ r | (p ∧ q) ∨ (p ∧ r) | p ∧ (q ∨ r) ↔ (p ∧ q) ∨ (p ∧ r) |
|---|---|---|---|---|---|---|---|---|
| 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |
| 1 | 1 | 0 | 1 | 1 | 1 | 0 | 1 | 1 |
| 1 | 0 | 1 | 1 | 1 | 0 | 1 | 1 | 1 |
| 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 |
| 0 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 1 |
| 0 | 1 | 0 | 1 | 0 | 0 | 0 | 0 | 1 |
| 0 | 0 | 1 | 1 | 0 | 0 | 0 | 0 | 1 |
| 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 |
HAY EQUIVALENCIA LÓGICA. Las fórmulas son lógicamente equivalentes (Ley distributiva).
23. [Ejercicio 23] p ∨ (q → r) , p → ¬¬(q → ¬r)
| p | q | r | ¬r | q → r | p ∨ (q → r) | q → ¬r | ¬(q → ¬r) | ¬¬(q → ¬r) | p →¬¬(q → ¬r) | p ∨ (q → r) ∧ p →¬¬(q → ¬r) |
|---|---|---|---|---|---|---|---|---|---|---|
| 1 | 1 | 1 | 0 | 1 | 1 | 0 | 1 | 0 | 0 | 0 |
| 1 | 1 | 0 | 1 | 0 | 1 | 1 | 0 | 1 | 1 | 1 |
| 1 | 0 | 1 | 0 | 1 | 1 | 1 | 0 | 1 | 1 | 1 |
| 1 | 0 | 0 | 1 | 1 | 1 | 1 | 0 | 1 | 1 | 1 |
| 0 | 1 | 1 | 0 | 0 | 0 | 0 | 1 | 0 | 1 | 0 |
| 0 | 1 | 0 | 1 | 1 | 1 | 1 | 0 | 1 | 1 | 1 |
| 0 | 0 | 1 | 0 | 1 | 1 | 1 | 0 | 1 | 1 | 1 |
| 0 | 0 | 0 | 1 | 1 | 1 | 1 | 0 | 1 | 1 | 1 |
NO HAY EQUIVALENCIA LOGICA. CONTRAEJEMPLO: I(p)= 1, I (q)=1, I(r)=1
24. [Ejercicio 24] (p → q) ∧ (q → r) → (p → r) , p → (q → r)
| p | q | r | p → q | q → r | (p → q) ∧ (p → r) | p → r | (p → q) ∧ (p → r) → (p → r) | (p → (q → r) | (p → q) ∧ (p → r) ↔ (p → (q → r) |
|---|---|---|---|---|---|---|---|---|---|
| 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |
| 1 | 1 | 0 | 1 | 0 | 0 | 0 | 1 | 0 | 0 |
| 1 | 0 | 1 | 0 | 1 | 0 | 1 | 1 | 1 | 1 |
| 1 | 0 | 0 | 0 | 1 | 0 | 0 | 1 | 1 | 1 |
| 0 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |
| 0 | 1 | 0 | 1 | 0 | 0 | 1 | 1 | 1 | 1 |
| 0 | 0 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |
| 0 | 0 | 0 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |
NO HAY EQUIVALENCIA LÓGICA. CONTRAEJEMPLO: I(p)=1, I(q)= 1, I(r) = 0
25. [Ejercicio 25] ¬(p ∨ q) ↔ ¬p ∨ ¬r , ¬(p → q) ↔ p ∧ r
| p | q | r | p ∨ q | ¬(p ∨ q) | ¬p ∨ ¬r | ¬(p ∨ q) ↔ ¬p ∨ ¬r | p → q | ¬(p → q) | p ∧ r | ¬(p → q) ↔ p ∧ r | Bicondicional final |
|---|---|---|---|---|---|---|---|---|---|---|---|
| 1 | 1 | 1 | 1 | 0 | 0 | 1 | 1 | 0 | 1 | 0 | 0 |
| 1 | 1 | 0 | 1 | 0 | 1 | 0 | 1 | 0 | 0 | 1 | 0 |
| 1 | 0 | 1 | 1 | 0 | 0 | 1 | 0 | 1 | 1 | 1 | 1 |
| 1 | 0 | 0 | 1 | 0 | 1 | 0 | 0 | 1 | 0 | 0 | 1 |
| 0 | 1 | 1 | 1 | 0 | 1 | 0 | 1 | 0 | 0 | 1 | 0 |
| 0 | 1 | 0 | 1 | 0 | 1 | 0 | 1 | 0 | 0 | 1 | 0 |
| 0 | 0 | 1 | 0 | 1 | 1 | 1 | 1 | 0 | 0 | 1 | 1 |
| 0 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 0 | 0 | 1 | 1 |
NO HAY EQUIVALENCIA LÓGICA. CONTRAEJEMPLO: I(p)=1, I(q)= 1, I(r) = 1
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