Adapte os exercícios resolvidos à sua disciplina. Registe-se gratuitamente
Demonstre atraves de tabelas de verdade se as seguintes formulas
sao equivalentes ou nao:
21. [Exercicio 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 |
NAO HA EQUIVALENCIA LOGICA. CONTRAEXEMPLO: I(p)= 1, I(q)=1 e I(r)= 0.
22. [Exercicio 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 |
HA EQUIVALENCIA LOGICA. As formulas sao logicamente equivalentes (Lei distributiva).
23. [Exercicio 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 |
NAO HA EQUIVALENCIA LOGICA. CONTRAEXEMPLO: I(p)= 1, I (q)=1, I(r)=1
24. [Exercicio 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 |
NAO HA EQUIVALENCIA LOGICA. CONTRAEXEMPLO: I(p)=1, I(q)= 1, I(r) = 0
25. [Exercicio 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 |
NAO HA EQUIVALENCIA LOGICA. CONTRAEXEMPLO: I(p)=1, I(q)= 1, I(r) = 1
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