Question 1191241: Please prove the following arguments (10 questions, 1 point each):
(1) P ∨ P ⊢ P
(2) P ⊢ (P → Q) → Q
(3) ∼(P & Q), P ⊢ ∼ Q
(4) P ⊢ (∼(Q → R) → ∼ P) → (∼ R → ∼ Q))(4)
(5) (P ∨ Q) → R ⊢ (P → R) & (Q → R)(5)
(6) ⊢ (P ∨ Q) → (Q ∨ P)(6)
(7) ∼(P & ∼ Q) ⊢ P → Q
(8) (P ∨ Q) ↔ P ⊢ Q → P
(9) P ↔ Q, Q ↔ R ⊢ P ↔ R
(10) ⊢ (P → Q) → (∼ Q → ∼ P)
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Here are the proofs for each argument:
**(1) P ∨ P ⊢ P**
| Statement | Reason |
|---|---|
| 1. P ∨ P | Premise |
| 2. P | Idempotent Law |
**(2) P ⊢ (P → Q) → Q**
| Statement | Reason |
|---|---|
| 1. P | Premise |
| 2. P → Q | Assumption |
| 3. Q | Modus Ponens (1, 2) |
| 4. (P → Q) → Q | Conditional Proof (2-3) |
**(3) ∼(P & Q), P ⊢ ∼Q**
| Statement | Reason |
|---|---|
| 1. ∼(P & Q) | Premise |
| 2. P | Premise |
| 3. ∼P ∨ ∼Q | De Morgan's Law (1) |
| 4. ∼Q | Disjunctive Syllogism (2, 3) |
**(4) P ⊢ (∼(Q → R) → ∼P) → (∼R → ∼Q)**
| Statement | Reason |
|---|---|
| 1. P | Premise |
| 2. ∼(Q → R) → ∼P | Assumption |
| 3. ∼P | Modus Ponens (1,2) |
| 4. ∼(Q → R) | Implication Elimination (2,3) |
| 5. ∼(∼Q ∨ R) | Implication Equivalence (4)|
| 6. Q & ∼R | DeMorgan's Law (5) |
| 7. ∼R | Simplification (6) |
| 8. ∼R → ∼Q | Conditional Proof (7) |
| 9. (∼(Q → R) → ∼P) → (∼R → ∼Q) | Conditional Proof (2-8) |
**(5) (P ∨ Q) → R ⊢ (P → R) & (Q → R)**
| Statement | Reason |
|---|---|
| 1. (P ∨ Q) → R | Premise |
| 2. P | Assumption |
| 3. P ∨ Q | Introduction (2) |
| 4. R | Modus Ponens (1, 3) |
| 5. P → R | Conditional Proof (2-4) |
| 6. Q | Assumption |
| 7. P ∨ Q | Introduction (6) |
| 8. R | Modus Ponens (1, 7) |
| 9. Q → R | Conditional Proof (6-8) |
| 10. (P → R) & (Q → R) | Conjunction (5, 9) |
**(6) ⊢ (P ∨ Q) → (Q ∨ P)**
| Statement | Reason |
|---|---|
| 1. P ∨ Q | Assumption |
| 2. P | Sub-assumption |
| 3. Q ∨ P | Introduction (2) |
| 4. Q | Sub-assumption |
| 5. Q ∨ P | Introduction (4) |
| 6. (P ∨ Q) → (Q ∨ P) | Conditional Proof (1-5) |
**(7) ∼(P & ∼Q) ⊢ P → Q**
| Statement | Reason |
|---|---|
| 1. ∼(P & ∼Q) | Premise |
| 2. ∼P ∨ ∼∼Q | De Morgan's Law (1) |
| 3. ∼P ∨ Q | Double Negation (2) |
| 4. P → Q | Implication Equivalence (3) |
**(8) (P ∨ Q) ↔ P ⊢ Q → P**
| Statement | Reason |
|---|---|
| 1. (P ∨ Q) ↔ P | Premise |
| 2. (P ∨ Q) → P | Biconditional Elimination (1) |
| 3. Q | Assumption |
| 4. P ∨ Q | Introduction (3) |
| 5. P | Modus Ponens (2, 4) |
| 6. Q → P | Conditional Proof (3-5) |
**(9) P ↔ Q, Q ↔ R ⊢ P ↔ R**
| Statement | Reason |
|---|---|
| 1. P ↔ Q | Premise |
| 2. Q ↔ R | Premise |
| 3. P → Q | Biconditional Elimination (1) |
| 4. Q → R | Biconditional Elimination (2) |
| 5. P → R | Hypothetical Syllogism (3, 4) |
| 6. R → Q | Biconditional Elimination (2) |
| 7. Q → P | Biconditional Elimination (1) |
| 8. R → P | Hypothetical Syllogism (6, 7) |
| 9. P ↔ R | Biconditional Introduction (5, 8) |
**(10) ⊢ (P → Q) → (∼Q → ∼P)**
| Statement | Reason |
|---|---|
| 1. P → Q | Assumption |
| 2. ∼Q | Assumption |
| 3. ∼P | Modus Tollens (1, 2) |
| 4. ∼Q → ∼P | Conditional Proof (2-3) |
| 5. (P → Q) → (∼Q → ∼P) | Conditional Proof (1-4) |
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