Question 1208945: Prove that the argument is valid using the method of natural deduction.
1. tilde(S dot tilde R)
2. tilde (P tribar S)
3. R wedge V / therefore tilde R horseshoe (P dot Q)
Answer by textot(100)   (Show Source):
You can put this solution on YOUR website! **1. Premises:**
* 1. ~ (S ∧ ~R)
* 2. ~ (P ≡ S)
* 3. R ∧ V
**2. Goal:**
* Prove: ~R → (P ∧ Q)
**3. Natural Deduction Proof:**
1. **~ (S ∧ ~R)** (Premise 1)
2. **~ (P ≡ S)** (Premise 2)
3. **R ∧ V** (Premise 3)
4. **R** (Simplification from line 3)
5. **Assume ~R** (Assumption for Conditional Proof)
6. **R ∧ ~R** (Conjunction Introduction from lines 4 and 5)
7. **⊥** (Contradiction from line 6)
8. **P ∧ Q** (Ex Falso Quodlibet from line 7)
9. **~R → (P ∧ Q)** (Conditional Introduction from lines 5 and 8)
**Explanation:**
* **Lines 1-3:** These are the given premises.
* **Line 4:** We use Simplification on line 3 to extract R.
* **Line 5:** We assume ~R for the purpose of Conditional Proof.
* **Line 6:** We combine R (from line 4) and the assumed ~R to create a contradiction.
* **Line 7:** Ex Falso Quodlibet states that any proposition can be derived from a contradiction. Here, we derive P ∧ Q.
* **Line 8:** Conditional Introduction allows us to conclude that ~R → (P ∧ Q), as we have shown that if we assume ~R, we can derive P ∧ Q.
**Therefore, the argument is valid.**
**Note:**
* "∧" represents conjunction (AND)
* "→" represents conditional (IF...THEN)
* "≡" represents biconditional (IF AND ONLY IF)
* "~" represents negation (NOT)
* "⊥" represents contradiction
This proof demonstrates that the conclusion (~R → (P ∧ Q)) logically follows from the given premises.
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