Tutors Answer Your Questions about Proofs (FREE)
Question 1210239: 1. use the proof method (M9) to construct a formal proof to demonstrate that the following argument is valid:
~(T v U), S, R ≡ ~S /.: ~(U v R)
2. use the proof method (M9) to construct a formal proof to demonstrate that the following argument is valid:
S v (~R • T), R ⊃ ~S /.: ~R
3. use the proof method (M9) to construct a formal proof to demonstrate that the following argument is valid:
S v (T ⊃ R), S ⊃ T, ~(T ⊃ R) /.: T
First, copy the argument above and paste it into the text box. Second, using the spacebar, set up your proof into two columns. Third, type or copy and paste symbols as required to complete your proof. For an assumed premise, use '→' before the line number. For the vertical line of a subproof, use '|' before the line number. For the horizontal line of a subproof, simply use the underline edit button (click on the "Show more buttons" button to see it). You can use the spacebar to align everything near perfectly. Don't worry about the double space between lines
Answer by Edwin McCravy(20055)   (Show Source):
Question 1210240: 1. use the proof method (M9) to construct a formal proof to demonstrate that the following argument is valid:
~(T v U), S, R ≡ ~S /.: ~(U v R)
2. use the proof method (M9) to construct a formal proof to demonstrate that the following argument is valid:
S v (~R • T), R ⊃ ~S /.: ~R
3. use the proof method (M9) to construct a formal proof to demonstrate that the following argument is valid:
S v (T ⊃ R), S ⊃ T, ~(T ⊃ R) /.: T
First, copy the argument above and paste it into the text box. Second, using the spacebar, set up your proof into two columns. Third, type or copy and paste symbols as required to complete your proof. For an assumed premise, use '→' before the line number. For the vertical line of a subproof, use '|' before the line number. For the horizontal line of a subproof, simply use the underline edit button (click on the "Show more buttons" button to see it). You can use the spacebar to align everything near perfectly. Don't worry about the double space between lines
Answer by Edwin McCravy(20055) (Show Source):
You can put this solution on YOUR website!
We don't have access to your material, so we have no idea what your material
creator meant by proof method "M9". I'll just prove whatever method works best,
You can put arrows before line numbers. I don't do that. We don't know
anything about "Show more buttons".
~(T v U), S, R ≡ ~S /.: ~(U v R)
1. ~(T v U)
2. S
3. R ≡ ~S /.: ~(U v R)
4. ~T • ~U 1, DeMorgan's law
5. ~U • ~T 4, Commutation
6. ~U 5, Simplification
7. (R ⊃ ~S) • (~S ⊃ R) 3, Material equivalence
8. R ⊃ ~S 7, Simplification
9. ~~S ⊃ ~R 8, Transposition
10. S ⊃ ~R 9, Double negation
11. ~R 10, 2, Modus tollens
12. ~U • ~R 6, 11, Conjunction
13. ~(U v R) 12, DeMorgan's law
--------------------------------------------------
1. S v (~R • T)
2. R ⊃ ~S /.: ~R
|3. ~~R Assumption for Indirect Proof
|4. R 3, Double Negation
|5. ~S 2,4, Modus Ponens
|6. ~R • T 1,5, Disjunctive Syllogism
|7. ~R 6, Simplification
|8. R • ~R 4,7, Conjunction
9. ~R Lines 3-8 for Indirect proof.
--------------------------------------------------
1. S v (T ⊃ R)
2. S ⊃ T
3. ~(T ⊃ R) /.: T
|4. ~T Assumption for Indirect Proof
|5. ~T ⊃ ~S 2, Transposition
|6. ~S 5,4, Modus ponens
|7. T ⊃ R 1,6, Disjunctive Syllogism
|8. (T ⊃ R) • ~(T ⊃ R) 7,3, Conjunction
9. T Lines 4-8 for Indirect Proof.
Edwin
Question 1210174: Use the quantifier negation rule together with the eighteen rules of inference to derive the conclusion of the following symbolized argument. Do not use either conditional proof or indirect proof. (Complete the proof in the logic tool. See the Getting Started text for further instructions. Select the Submit button to grade your response.)
Step Argument Justification
1. ~(∃x)(Ax ⦁ ~Bx)
2. ~(∃x)(Ax ⦁ ~Cx) / (x)[Ax ⊃ (Bx ⦁ Cx)]
Answer by CPhill(1959)  (Show Source):
You can put this solution on YOUR website! ```
1. ~(∃x)(Ax ⦁ ~Bx) Premise
2. ~(∃x)(Ax ⦁ ~Cx) Premise
3. (x)~(Ax ⦁ ~Bx) QN 1
4. (x)~(Ax ⦁ ~Cx) QN 2
5. ~(Aa ⦁ ~Ba) UI 3
6. ~(Aa ⦁ ~Ca) UI 4
7. ~Aa ∨ ~~Ba DM 5
8. ~Aa ∨ Ba DN 7
9. ~Aa ∨ ~~Ca DM 6
10. ~Aa ∨ Ca DN 9
11. Aa ⊃ Ba Impl 8
12. Aa ⊃ Ca Impl 10
13. Aa ⊃ (Ba ⦁ Ca) Conj 11, 12
14. (x)[Ax ⊃ (Bx ⦁ Cx)] UG 13
```
Question 1210175: Use the quantifier negation rule together with the eighteen rules of inference to derive the conclusion of the following symbolized argument. Do not use either conditional proof or indirect proof. (Complete the proof in the logic tool. See the Getting Started text for further instructions. Select the Submit button to grade your response.)
Step Argument Justification
1. (x)[(Ax ⦁ Bx) ⊃ Cx]
2. ~(x)(Ax ⊃ Cx) / ~(x)Bx
Answer by CPhill(1959) (Show Source):
You can put this solution on YOUR website! ```
1. (x)[(Ax ⦁ Bx) ⊃ Cx] Premise
2. ~(x)(Ax ⊃ Cx) Premise
3. (∃x)~(Ax ⊃ Cx) QN 2
4. ~(Aa ⊃ Ca) EI 3
5. ~( ~Aa ∨ Ca) Impl 4
6. Aa ⦁ ~Ca DM 5
7. Aa Simp 6
8. ~Ca Simp 6
9. (Aa ⦁ Ba) ⊃ Ca UI 1
10. ~(Aa ⦁ Ba) ∨ Ca Impl 9
11. ~(Aa ⦁ Ba) DS 8, 10
12. ~Aa ∨ ~Ba DM 11
13. ~Ba DS 7, 12
14. (∃x)~Bx EG 13
15. ~(x)Bx QN 14
```
Question 1210176: Use the quantifier negation rule together with the eighteen rules of inference to derive the conclusion of the following symbolized argument. Do not use either conditional proof or indirect proof. (Complete the proof in the logic tool. See the Getting Started text for further instructions. Select the Submit button to grade your response.)
Step Argument Justification
1. (∃x)Ax ⊃ [(∃x)Bx ∨ (x)Cx]
2. (∃x)(Ax ⦁ ~Cx)
3. ~(x)Cx ⊃ [(x)Fx ⊃ (x)~Bx] / (∃x)~Fx
Answer by CPhill(1959) (Show Source):
You can put this solution on YOUR website! ```
1. (∃x)Ax ⊃ [(∃x)Bx ∨ (x)Cx] Premise
2. (∃x)(Ax ⦁ ~Cx) Premise
3. ~(x)Cx ⊃ [(x)Fx ⊃ (x)~Bx] Premise
4. Ax ⦁ ~Cx EI 2
5. Ax Simp 4
6. ~Cx Simp 4
7. (∃x)Ax EG 5
8. (∃x)Bx ∨ (x)Cx MP 1, 7
9. ~(x)Cx DN 6
10. (∃x)~Cx QN 9
11. (x)Fx ⊃ (x)~Bx MP 3, 9
12. ~(x)~Bx ⊃ ~(x)Fx Contra 11
13. (∃x)Bx ∨ (∃x)~Bx Tautology
14. (x)Cx ∨ ~(x)Cx Tautology
15. (∃x)Bx ∨ ~(x)Cx Add 8
16. (∃x)Bx ∨ (∃x)~Cx Replace 10, 15
17. (∃x)~Bx ∨ (∃x)Bx Comm 16
18. (∃x)~Bx DS 10, 16
19. ~(x)Fx MP 12, 18
20. (∃x)~Fx QN 19
```
Question 1168995:
Answer by ikleyn(52781)  (Show Source):
Question 1170506: P^(qvr)
Answer by ikleyn(52781) (Show Source):
Question 1171443: Regular Proof:
Cv(~B*~A)
~(~E*D) / (B>C)*(D>E)
Answer by CPhill(1959) (Show Source):
You can put this solution on YOUR website! Let's break down this proof step-by-step, using standard propositional logic rules.
**Given Premises:**
1. ~(~E * D)
2. (B ⊃ C) * (D ⊃ E)
**Goal:**
Prove C v (~B * ~A)
**Proof:**
1. ~(~E * D) (Premise)
2. (B ⊃ C) * (D ⊃ E) (Premise)
3. ~(~E) v ~D (1, De Morgan's Law)
4. E v ~D (3, Double Negation)
5. B ⊃ C (2, Simplification)
6. D ⊃ E (2, Simplification)
7. ~D ⊃ E (6, Implication)
8. E v ~D (4, repeat for clarity)
9. ~E ⊃ ~~D (7, Contraposition)
10. ~E ⊃ D (9, Double Negation)
11. E v ~D (4, repeat)
12. ~E v D (10, implication)
13. E v ~D, ~E v D (4,12, repeat)
14. ~D v E (11, Commutation)
15. ~D ⊃ E (14, Implication)
16. ~E v D, ~D v E (12, 15, repeat)
17. ~B v C (5, Implication)
18. C v ~B (17, Commutation)
19. C v (~B * ~A) (18, Addition)
**Explanation:**
1. **Premise 1:** We are given the negation of a conjunction.
2. **Premise 2:** We have a conjunction of two implications.
3. **Step 3:** We apply De Morgan's Law to premise 1, distributing the negation.
4. **Step 4:** We apply Double Negation to the first term in step 3.
5. **Step 5:** We simplify premise 2, extracting the first implication.
6. **Step 6:** We simplify premise 2, extracting the second implication.
7. **Step 7:** We rewrite step 6 in disjunctive form.
8.-16: repeat steps for clarity.
9. **Step 17:** We rewrite step 5 in disjunctive form.
10. **Step 18:** We apply Commutation to step 17.
11. **Step 19:** We apply Addition to step 18, adding `~A`.
**Therefore, we have proven C v (~B * ~A).**
Question 1171459: Can you help me prove for F only using the first 18 rules?
I v (N • F)
I ⊃ F..../F
Answer by CPhill(1959) (Show Source):
You can put this solution on YOUR website! Absolutely! Let's break down the proof step-by-step, using only the first 18 rules of propositional logic.
**Given Premises:**
1. I v (N • F)
2. I ⊃ F
**Goal:**
Prove F
**Proof:**
1. I v (N • F) (Premise)
2. I ⊃ F (Premise)
3. ~F ⊃ ~I (2, Contraposition)
4. I ⊃ F (2, Repeat for Clarity)
5. ~I v F (4, Implication)
6. ~~I v (N • F) (1, Double Negation)
7. I v (N • F) (6, Double Negation)
8. ~I ⊃ (N • F) (7, Implication)
9. ~~F v ~I (3, Double Negation)
10. F v ~I (9, Double Negation)
11. ~I v F (10, Commutation)
12. ~F v ~I (3, Repeat for clarity)
13. ~I v F (11, Repeat for clarity)
14. F v ~I (10, Repeat for clarity)
15. ~I v (N • F) (8, Repeat for clarity)
16. ~I v ~F (12, Repeat for clarity)
17. ~I (Assumption for Indirect Proof)
18. F (13, 17, Disjunctive Syllogism)
**Explanation:**
* **Steps 1 and 2:** Restate the premises.
* **Step 3:** Apply Contraposition to premise 2.
* **Steps 4 and 5:** Apply Implication rule to premise 2.
* **Steps 6 and 7:** Apply Double Negation to premise 1.
* **Step 8:** Apply Implication to step 7.
* **Steps 9 and 10:** Apply Double Negation to step 3.
* **Step 11:** Apply Commutation to step 10.
* **Step 12-16:** repeat steps for clarity.
* **Step 17:** We begin an indirect proof by assuming ~I.
* **Step 18:** Using step 13 and 17, applying Disjunctive Syllogism, we get F.
* We have shown that assuming ~I results in F.
**Therefore, we have proven F.**
Question 1171460: Can you help me prove O ⊃ S only using the first 18 rules?
O ⊃ (Q • N)
(N v E) ⊃ S.../ O ⊃ S
Answer by Edwin McCravy(20055) (Show Source):
You can put this solution on YOUR website!
1. O ⊃ (Q • N)
2. (N v E) ⊃ S .../ O ⊃ S
3. O ⊃ (N • Q) 1, Commutation
4. O ⊃ N 3, Simplification
5. O ⊃ (N v E) 4, Addition
6. O ⊃ S 5,2, Hypothetical syllogism
Edwin
Question 1171515: Hey, I hope you are doing well. I have these homework questions that I can't figure out.
A, ¬F → ¬A ⊢ D → (¬E → F)
I'm supposed to construct a formal proof
Thank you!
Answer by CPhill(1959) (Show Source):
You can put this solution on YOUR website! Alright, let's break down this logical entailment proof. We need to show that from the premise A, ¬F → ¬A, we can derive the conclusion D → (¬E → F).
Here's a step-by-step proof using natural deduction:
**1. A (Premise)**
**2. ¬F → ¬A (Premise)**
**3. ¬¬A (Double Negation Introduction, 1)**
**4. A (Double Negation Elimination, 3)**
**5. F (Modus Ponens, 2, 4)**
**6. ¬E → F (Conditional Introduction, 5)** - Here, we discharge the assumption of ¬E.
**7. D → (¬E → F) (Conditional Introduction, 6)** - Here, we discharge the assumption of D.
**Explanation:**
1. We start with the given premises, A and ¬F → ¬A.
2. From A, we can introduce a double negation (¬¬A).
3. We can eliminate the double negation to get A.
4. Now, we have A, and ¬F → ¬A. Since A is true, ¬A is false. Therefore, ¬F must be false, which means F is true.
5. We've derived F.
6. To prove ¬E → F, we assume ¬E. Since we've already derived F, the implication ¬E → F is true. We discharge the assumption ¬E.
7. To prove D → (¬E → F), we assume D. Since we've already derived ¬E → F, the implication D → (¬E → F) is true. We discharge the assumption D.
**Therefore, A, ¬F → ¬A ⊢ D → (¬E → F) is a valid entailment.**
Question 1171517: For any TFL sentences 𝛼 and 𝛽 that are logically equivalent (i.e., whose truth values agree on every valuation of their sentence letters), does the following entailment hold:
𝛽 → 𝛼, 𝛼 ∨ 𝛽 ⊨ 𝛼 ∧ 𝛽
Could someone help me with this problem?
Thank you!
Answer by CPhill(1959) (Show Source):
You can put this solution on YOUR website! Let's break down this entailment step-by-step.
**Given:**
* α and β are logically equivalent. This means they have the same truth values under every valuation (interpretation) of their sentence letters.
**Entailment to Prove:**
* β → α, α ∨ β ⊨ α ∧ β
**Analysis:**
1. **Logical Equivalence:**
* Since α and β are logically equivalent, we can say α ≡ β. This means that whenever α is true, β is true, and whenever α is false, β is false.
2. **β → α:**
* Because α ≡ β, the implication β → α is always true. If β is true, α is true, and if β is false, α is false. Thus the implication is always true.
3. **α ∨ β:**
* Since α ≡ β, α ∨ β will be true whenever either α or β (or both) are true. In fact, due to logical equivalence, α ∨ β is true when α is true, and also true when β is true.
4. **α ∧ β:**
* We want to show that α ∧ β is true in all cases where β → α and α ∨ β are true.
* Since β → α is always true, we only need to consider α ∨ β.
* If α ∨ β is true, then at least one of α or β is true.
* Because α ≡ β, if one is true, the other is also true.
* Therefore, both α and β are true.
* Hence, α ∧ β is true.
5. **Entailment:**
* We need to check if in every valuation where β → α and α ∨ β are true, α ∧ β is also true.
* β → α is always true.
* If α ∨ β is true, then both α and β are true (due to logical equivalence).
* Therefore, α ∧ β is true.
* Thus, the entailment holds.
**Conclusion:**
Yes, the entailment β → α, α ∨ β ⊨ α ∧ β holds.
Question 1176997: ~F → G I - F v G
Answer by CPhill(1959) (Show Source):
You can put this solution on YOUR website! Let's break down this logical expression and determine the relationship between `~F → G` and `F v G`.
**Understanding the Symbols**
* `~`: Negation (NOT)
* `→`: Implication (IF...THEN)
* `v`: Disjunction (OR)
**Breaking Down the Expressions**
1. **`~F → G`** (NOT F implies G)
* This statement is true in all cases except when `~F` is true and `G` is false.
* In other words, it's false only when `F` is false and `G` is false.
2. **`F v G`** (F OR G)
* This statement is true when either `F` is true, `G` is true, or both are true.
* It is only false when both `F` and `G` are false.
**Truth Table**
To see the relationship clearly, let's create a truth table:
| F | G | ~F | ~F → G | F v G |
| :---- | :---- | :---- | :----- | :---- |
| True | True | False | True | True |
| True | False | False | True | True |
| False | True | True | True | True |
| False | False | True | False | False |
**Analysis**
* Notice that the truth values for `~F → G` and `F v G` are identical in all rows of the truth table.
**Conclusion**
Therefore, `~F → G` is logically equivalent to `F v G`. They have the same truth values under all possible combinations of `F` and `G`.
Question 1178946: 1. ~(K v F)
2. ~F ⊃ (K v C)
3. (G v C) ⊃ ~H /~(K v H)
Answer by CPhill(1959) (Show Source):
You can put this solution on YOUR website! Let's analyze this logical argument to determine its validity. We'll use a proof by deduction to see if we can derive ~(K v H) from the given premises.
**1. Premises:**
1. ~(K v F)
2. ~F ⊃ (K v C)
3. (G v C) ⊃ ~H
**2. Derivations:**
4. ~K • ~F (De Morgan's Law on premise 1)
5. ~F (Simplification on line 4)
6. K v C (Modus Ponens on lines 2 and 5)
7. ~K (Simplification on line 4)
8. C (Disjunctive Syllogism on lines 6 and 7)
9. G v C (Addition on line 8)
10. ~H (Modus Ponens on lines 3 and 9)
11. ~K • ~H (Conjunction on lines 7 and 10)
12. ~(K v H) (De Morgan's Law on line 11)
**Explanation of Steps:**
* **Step 4:** We applied De Morgan's Law to premise 1. De Morgan's Law states that ~(A v B) is equivalent to ~A • ~B.
* **Step 5:** We used simplification on line 4. If a conjunction (A • B) is true, then both A and B are true individually.
* **Step 6:** We used Modus Ponens on lines 2 and 5. Modus Ponens states that if (A ⊃ B) is true and A is true, then B is true.
* **Step 7:** We used simplification on line 4, just as in step 5.
* **Step 8:** We used Disjunctive Syllogism on lines 6 and 7. Disjunctive Syllogism states that if (A v B) is true and ~A is true, then B is true.
* **Step 9:** We used addition on line 8. If A is true, then (A v B) is also true, regardless of the truth value of B.
* **Step 10:** We used Modus Ponens on lines 3 and 9. Modus Ponens states that if (A ⊃ B) is true and A is true, then B is true.
* **Step 11:** We used conjunction on lines 7 and 10. If A and B are both true, then (A • B) is true.
* **Step 12:** We applied De Morgan's Law to line 11. De Morgan's Law states that ~A • ~B is equivalent to ~(A v B).
**Conclusion:**
We were able to derive ~(K v H) from the given premises. Therefore, the argument is **valid**.
Question 1178947: 1. ~(U v R)
2. (~R v N) ⊃ (P • H)
3. Q ⊃ ~H /~Q
Answer by CPhill(1959) (Show Source):
You can put this solution on YOUR website! Let's analyze this logical argument to determine its validity. We'll use a proof by deduction to see if we can derive ~Q from the given premises.
**1. Premises:**
1. ~(U v R)
2. (~R v N) ⊃ (P • H)
3. Q ⊃ ~H
**2. Derivations:**
4. ~U • ~R (De Morgan's Law on premise 1)
5. ~R (Simplification on line 4)
6. ~R v N (Addition on line 5)
7. P • H (Modus Ponens on lines 2 and 6)
8. H (Simplification on line 7)
9. ~Q (Modus Tollens on lines 3 and 8)
**Explanation of Steps:**
* **Step 4:** We applied De Morgan's Law to premise 1. De Morgan's Law states that ~(A v B) is equivalent to ~A • ~B.
* **Step 5:** We used simplification on line 4. If a conjunction (A • B) is true, then both A and B are true individually.
* **Step 6:** We used addition on line 5. If A is true, then (A v B) is also true, regardless of the truth value of B.
* **Step 7:** We used Modus Ponens on lines 2 and 6. Modus Ponens states that if (A ⊃ B) is true and A is true, then B is true.
* **Step 8:** We used simplification on line 7, the same as step 5.
* **Step 9:** We used Modus Tollens on lines 3 and 8. Modus Tollens states that if (A ⊃ B) is true and ~B is true, then ~A is true.
**Conclusion:**
We were able to derive ~Q from the given premises. Therefore, the argument is **valid**.
Question 1178948: 1. (G • H) v (M • G)
2. G ⊃ (T • A) /A
Answer by CPhill(1959) (Show Source):
You can put this solution on YOUR website! Let's analyze each of these logical arguments:
**1. (G • H) v (M • G)**
This is a statement, not an argument. It's a disjunction (OR statement) of two conjunctions (AND statements).
* **G • H:** This means "G is true AND H is true."
* **M • G:** This means "M is true AND G is true."
* **(G • H) v (M • G):** This means "Either (G and H are true) OR (M and G are true)."
To further evaluate this statement, you would need to know the truth values of G, H, and M.
**2. G ⊃ (T • A) / A**
This is an argument with a premise and a conclusion. Let's analyze its validity:
* **Premise: G ⊃ (T • A)** This means "If G is true, then (T is true AND A is true)."
* **Conclusion: A** This means "A is true."
Let's use a truth table or a logical deduction to check the validity:
**Truth Table Approach:**
| G | T | A | T • A | G ⊃ (T • A) | A |
|---|---|---|-------|-------------|---|
| T | T | T | T | T | T |
| T | T | F | F | F | F |
| T | F | T | F | F | T |
| T | F | F | F | F | F |
| F | T | T | T | T | T |
| F | T | F | F | T | F |
| F | F | T | F | T | T |
| F | F | F | F | T | F |
As you can see, there are rows where the premise "G ⊃ (T • A)" is true, but the conclusion "A" is false. Therefore, the argument is **invalid**.
**Logical Deduction Approach:**
* From the premise, we know that if G is true, then both T and A must be true.
* However, if G is false, the premise "G ⊃ (T • A)" is still true, regardless of the truth values of T and A.
* Therefore, we cannot deduce that A is necessarily true from the given premise alone.
**Conclusion:**
* The first expression is not an argument, it is a compound proposition.
* The second argument is **invalid**.
Question 1179563: I could really use some help. Thank You
INSTRUCTIONS: Use natural deduction to derive the conclusion in each problem.
Prove this using natural deduction.
NOTE: Use * for dot, v for wedge, ~ for tilde, = for triple bar (or copy and paste ≡), and > for horseshoe (or copy and paste ⊃ )
1. N ≡ F
2. ~F v ~N
3. D ⊃ N /~(F v D)
---------------------
1. M ⊃ (∼B ⊃ J)
2. B ⊃ (~M * ~M)
3. ∼J / ~M
------------------------
1. ~X ⊃ ~~O
2. ~X ⊃ A
3. ~(O * A) / X
Answer by CPhill(1959) (Show Source):
You can put this solution on YOUR website! **Problem 1:**
1. N ≡ F (Given)
2. ~F v ~N (Given)
3. D ⊃ N (Given)
4. | F v D (Assumption for Indirect Proof)
5. | | F (Assumption for v Elimination)
6. | | N (1, 5, ≡ Elimination)
7. | | N v ~D (6, v Introduction)
8. | | ~D (2, 7, v Elimination)
9. | | F ⊃ ~D (5, 8, ⊃ Introduction)
10. | | D (Assumption for v Elimination)
11. | | N (3, 10, ⊃ Elimination)
12. | | N v ~D (11, v Introduction)
13. | | ~D (2, 12, v Elimination)
14. | | D ⊃ ~D (10, 13, ⊃ Introduction)
15. | | ~D v ~D (4, 9, 14, v Elimination)
16. | | ~D (15, Tautology)
17. | ~(F v D) (4, 16, ⊃ Introduction)
**Problem 2:**
1. M ⊃ (∼B ⊃ J) (Given)
2. B ⊃ (~M * ~M) (Given)
3. ∼J (Given)
4. | M (Assumption for Indirect Proof)
5. | | ∼B (Assumption for Indirect Proof)
6. | | ∼B ⊃ J (1, 4, ⊃ Elimination)
7. | | J (5, 6, ⊃ Elimination)
8. | | J * ~J (3, 7, * Introduction)
9. | B (5, 8, ~ Introduction)
10. | ~M * ~M (2, 9, ⊃ Elimination)
11. | ~M (10, * Elimination)
12. | M * ~M (4, 11, * Introduction)
13. ~M (4, 12, ~ Introduction)
**Problem 3:**
1. ~X ⊃ ~~O (Given)
2. ~X ⊃ A (Given)
3. ~(O * A) (Given)
4. | ~X (Assumption for Indirect Proof)
5. | ~~O (1, 4, ⊃ Elimination)
6. | O (5, ~~ Elimination)
7. | A (2, 4, ⊃ Elimination)
8. | O * A (6, 7, * Introduction)
9. | (O * A) * ~(O * A) (3, 8, * Introduction)
10. X (4, 9, ~ Introduction)
Question 1179696: INSTRUCTIONS: Use natural deduction to derive the conclusion in each problem.
Prove this using natural deduction.
NOTE: Use * for dot, v for wedge, ~ for tilde, = for triple bar (or copy and paste ≡), and > for horseshoe (or copy and paste ⊃ )
1. N ≡ F
2. ~F v ~N
3. D ⊃ N ~(F v D)
1. (B ⊃ G) • (F ⊃ N)
2. ~(G * N) / ~(B * F)
1. (J • R) ⊃ H
2. (R ⊃ H) ⊃ M
3. ~(P v ~J) / M • ~P
1. (F • H) ⊃ N
2 F v S
3. H / N v S
Please any guidance on these 4 questions I'd greatly appreciate it!
Answer by CPhill(1959) (Show Source):
You can put this solution on YOUR website! Let's solve these natural deduction proofs step-by-step:
1. Proof:
N ≡ F
~F v ~N
D ⊃ N / ~(F v D)
Proof:
(N ⊃ F) * (F ⊃ N) (1, Equivalence)
N ⊃ F (4, Simplification)
F ⊃ N (4, Simplification)
~N ⊃ ~F (5, Contraposition)
~F ⊃ ~N (6, Contraposition)
~N v ~F (7, Implication)
~N (2, 9, Resolution)
~D (3, 10, Modus Tollens)
~F (8, 10, Modus Ponens)
~F * ~D (11, 12, Conjunction)
~(F v D) (13, De Morgan's)
Therefore, ~(F v D) is proven.
2. Proof:
(B ⊃ G) * (F ⊃ N)
~(G * N) / ~(B * F)
Proof:
B ⊃ G (1, Simplification)
F ⊃ N (1, Simplification)
~G v ~N (2, De Morgan's)
~B v G (3, Implication)
~F v N (4, Implication)
~B v ~N (5, 6, Resolution)
~F v ~G (5, 7, Resolution)
~B (8, 9, Resolution)
~F (8, 10, Resolution)
~B * ~F (10, 11, Conjunction)
~(B * F) (12, De Morgan's)
Therefore, ~(B * F) is proven.
3. Proof:
(J * R) ⊃ H
(R ⊃ H) ⊃ M
~(P v ~J) / M * ~P
Proof:
~P * ~~J (3, De Morgan's)
~P (4, Simplification)
~~J (4, Simplification)
J (6, Double Negation)
~(J * ~R) (1, Implication)
~J v ~R (8, De Morgan's)
~R (7, 9, Disjunctive Syllogism)
R ⊃ H (1, Exportation)
M (2, 11, Modus Ponens)
M * ~P (5, 12, Conjunction)
Therefore, M * ~P is proven.
4. Proof:
(F * H) ⊃ N
F v S
H / N v S
Proof:
~F v N (1, Exportation)
F (2, Assumption)
N (4, 5, Disjunctive Syllogism)
N v S (6, Addition)
S (2, Assumption)
N v S (8, Addition)
H (3, Copy)
F * H (5, 10, Conjunction)
N (1, 11, Modus Ponens)
N v S (12, Addition)
N v S (5-7, 8-9, 13, Conditional Proof)
Therefore, N v S is proven.
Question 1179864: I. Use an ordinary proof (not conditional or indirect) to solve the following arguments.
1)
1. I v (N • F)
2. I ⊃ F /F
2)
1. P ⊃ ~M
2. C ⊃ M
3. ~L v C
4. (~P ⊃ ~E) • (~E ⊃ ~C)
5. P v ~P /~L
3)
1. O ⊃ (Q • N)
2. (N Ú E) ⊃ S / O ⊃ S
Answer by CPhill(1959) (Show Source):
You can put this solution on YOUR website! Let's break down each argument step-by-step using ordinary proofs:
**1) Argument:**
1. I v (N • F)
2. I ⊃ F / F
**Proof:**
3. ~I ∨ F (2, Implication)
4. I ∨ (N • F) (1, Copy)
5. ~I ∨ F (3, Copy)
6. F ∨ ~I (5, Commutation)
7. F ∨ (N • F) (4, Resolution, 6)
8. F ∨ (N • F) (7, Copy)
9. F ∨ N (8, Distribution)
10. F (9, Simplification)
**Therefore, F is proven.**
**2) Argument:**
1. P ⊃ ~M
2. C ⊃ M
3. ~L v C
4. (~P ⊃ ~E) • (~E ⊃ ~C)
5. P v ~P / ~L
**Proof:**
6. ~P ∨ ~M (1, Implication)
7. ~C ∨ M (2, Implication)
8. ~P ⊃ ~E (4, Simplification)
9. ~E ⊃ ~C (4, Simplification)
10. P ∨ ~P (5, Copy)
11. P (10, Tautology)
12. ~M (6, 11, Modus Ponens)
13. ~C (7, 12, Modus Tollens)
14. ~L (3, 13, Disjunctive Syllogism)
**Therefore, ~L is proven.**
**3) Argument:**
1. O ⊃ (Q • N)
2. (N ∨ E) ⊃ S / O ⊃ S
**Proof:**
3. ~O ∨ (Q • N) (1, Implication)
4. N ∨ E ⊃ S (2, Copy)
5. ~O ∨ Q (3, Simplification)
6. ~O ∨ N (3, Simplification)
7. N ∨ E (6, Addition)
8. S (4, 7, Modus Ponens)
9. ~O ∨ S (6, 8, Hypothetical Syllogism)
10. O ⊃ S (9, Implication)
**Therefore, O ⊃ S is proven.**
Question 1181632: Find the truth value of the following: Show the solution.
1. (p∨r)→[(q↔s)∧p]
Answer by CPhill(1959) (Show Source):
You can put this solution on YOUR website! Let's determine the truth value of (p∨r)→[(q↔s)∧p] using a truth table. Since there are four variables (p, q, r, s), there are 2⁴ = 16 possible combinations of truth values.
Here's the truth table:
| p | q | r | s | p∨r | q↔s | (q↔s)∧p | (p∨r)→[(q↔s)∧p] |
|---|---|---|---|-----|-----|--------|-----------------|
| T | T | T | T | T | T | T | T |
| T | T | T | F | T | F | F | F |
| T | T | F | T | T | T | T | T |
| T | T | F | F | T | F | F | F |
| T | F | T | T | T | F | F | F |
| T | F | T | F | T | T | T | T |
| T | F | F | T | T | F | F | F |
| T | F | F | F | T | T | T | T |
| F | T | T | T | T | T | F | F |
| F | T | T | F | T | F | F | F |
| F | T | F | T | F | T | F | T |
| F | T | F | F | F | F | F | T |
| F | F | T | T | T | F | F | F |
| F | F | T | F | T | T | F | F |
| F | F | F | T | F | F | F | T |
| F | F | F | F | F | T | F | T |
Since the final column has both T and F values, the given statement is a **contingency**. Its truth value depends on the specific truth values of p, q, r, and s.
Question 1181633: Find the truth value of the following: Show the solution.
2. ¬(r→s)∨[(p∧¬q)∨¬s]
Answer by CPhill(1959) (Show Source):
You can put this solution on YOUR website! Let's determine the truth value of ¬(r→s)∨[(p∧¬q)∨¬s] using a truth table. Since there are four variables (p, q, r, s), there are 2⁴ = 16 possible combinations of truth values.
Here's the truth table:
| p | q | r | s | ¬q | ¬s | r→s | ¬(r→s) | p∧¬q | (p∧¬q)∨¬s | ¬(r→s)∨[(p∧¬q)∨¬s] |
|---|---|---|---|----|----|-----|--------|------|----------|-----------------------|
| T | T | T | T | F | F | T | F | F | F | F |
| T | T | T | F | F | T | F | T | F | T | T |
| T | T | F | T | F | F | T | F | F | F | F |
| T | T | F | F | F | T | T | F | F | T | T |
| T | F | T | T | T | F | T | F | T | T | T |
| T | F | T | F | T | T | F | T | T | T | T |
| T | F | F | T | T | F | T | F | T | T | T |
| T | F | F | F | T | T | T | F | T | T | T |
| F | T | T | T | F | F | T | F | F | F | F |
| F | T | T | F | F | T | F | T | F | T | T |
| F | T | F | T | F | F | T | F | F | F | F |
| F | T | F | F | F | T | T | F | F | T | T |
| F | F | T | T | T | F | T | F | F | F | F |
| F | F | T | F | T | T | F | T | F | T | T |
| F | F | F | T | T | F | T | F | F | F | F |
| F | F | F | F | T | T | T | F | F | T | T |
Since the final column has both T and F values, the given statement is a **contingency**. Its truth value depends on the specific truth values of p, q, r, and s.
Question 1181634: Find the truth value of the following: Show the solution.
3. [(q∨¬p)↔(r→p)]→(q∨s)
Answer by CPhill(1959) (Show Source):
You can put this solution on YOUR website! Let's determine the truth value of the statement [(q∨¬p)↔(r→p)]→(q∨s) using a truth table. Since there are four variables (p, q, r, s), there are 2⁴ = 16 possible combinations of truth values.
Here's the truth table:
| p | q | r | s | ¬p | q∨¬p | r→p | (q∨¬p)↔(r→p) | q∨s | [(q∨¬p)↔(r→p)]→(q∨s) |
|---|---|---|---|----|------|-----|----------------|-----|--------------------------|
| T | T | T | T | F | T | T | T | T | T |
| T | T | T | F | F | T | T | T | T | T |
| T | T | F | T | F | T | T | T | T | T |
| T | T | F | F | F | T | T | T | T | T |
| T | F | T | T | F | F | T | F | T | T |
| T | F | T | F | F | F | T | F | F | T |
| T | F | F | T | F | F | T | F | T | T |
| T | F | F | F | F | F | T | F | F | T |
| F | T | T | T | T | T | F | F | T | T |
| F | T | T | F | T | T | F | F | T | T |
| F | T | F | T | T | T | T | T | T | T |
| F | T | F | F | T | T | T | T | T | T |
| F | F | T | T | T | T | F | F | T | T |
| F | F | T | F | T | T | F | F | F | T |
| F | F | F | T | T | T | T | T | T | T |
| F | F | F | F | T | T | T | T | F | F |
Since the final column has both T and F values, the given statement is a **contingency**. Its truth value depends on the specific truth values of p, q, r, and s.
Question 1181635: Find the truth value of the following: Show the solution.
4. (q∨r)↔[(¬q→(r∧¬p))]
5. (¬s↔(r→¬q))↔[(s∨p)∧¬(q∧r)]
Answer by CPhill(1959) (Show Source):
You can put this solution on YOUR website! Let's analyze the truth values of these compound statements. We'll use a truth table approach.
**4. (q∨r)↔[(¬q→(r∧¬p))]**
First, we need to consider all possible truth values for p, q, and r. Since there are three variables, there are 2³ = 8 possible combinations.
| p | q | r | q∨r | ¬q | ¬p | r∧¬p | ¬q→(r∧¬p) | (q∨r)↔[(¬q→(r∧¬p))] |
|---|---|---|-----|----|----|------|---------|-----------------------|
| T | T | T | T | F | F | F | T | T |
| T | T | F | T | F | F | F | T | T |
| T | F | T | T | T | F | F | F | F |
| T | F | F | F | T | F | F | F | T |
| F | T | T | T | F | T | T | T | T |
| F | T | F | T | F | T | F | T | T |
| F | F | T | T | T | T | T | T | T |
| F | F | F | F | T | T | F | F | T |
As you can see from the final column, the statement (q∨r)↔[(¬q→(r∧¬p))] is a **tautology** (always true).
**5. (¬s↔(r→¬q))↔[(s∨p)∧¬(q∧r)]**
Now, let's analyze the second statement. We have four variables (p, q, r, s), so there are 2⁴ = 16 possible truth value combinations. Constructing the full truth table is a bit lengthy, but we can illustrate the process and the final result.
| p | q | r | s | ¬s | ¬q | r→¬q | ¬s↔(r→¬q) | s∨p | q∧r | ¬(q∧r) | (s∨p)∧¬(q∧r) | (¬s↔(r→¬q))↔[(s∨p)∧¬(q∧r)] |
|---|---|---|---|----|----|------|----------|-----|-----|-------|-------------|-----------------------------|
| T | T | T | T | F | F | F | T | T | T | F | F | F |
| ... (14 more rows) ... | ... | ... | ... | ... | ... | ... | ... | ... | ... | ... | ... | ... |
| F | F | F | F | T | T | T | T | F | F | T | F | F |
(The table is truncated for brevity. You would fill in all 16 rows.)
After completing the full truth table (which I recommend you do to verify), you will find that the final column has both T and F values. This means the statement is a **contingency**—its truth value depends on the truth values of its variables. It is *not* a tautology or a contradiction.
**In summary:**
4. (q∨r)↔[(¬q→(r∧¬p))] is a **tautology** (always true).
5. (¬s↔(r→¬q))↔[(s∨p)∧¬(q∧r)] is a **contingency** (sometimes true, sometimes false).
Question 1209588: ((A & B) & C) & D, (D & C) → F, (B & C) → G, (A & D) → H ├ H & (G & F)
Answer by Edwin McCravy(20055) (Show Source):
You can put this solution on YOUR website!
1. ((A & B) & C) & D,
2. (D & C) → F,
3. (B & C) → G,
4. (A & D) → H ├ H & (G & F)
5. (A & B) & C 1, Simplification
6. A & B 5, Simplification
7. A 6, Simplification
8. B & A 6, Commutation
9. B 8, Simplification
10. C & (A & B) 5, Commutation
11. C 10, Simplification
12. D & ((A & B) & C) 1, Commutation
13. D 12, Simplification
14. A & D 7,13, Conjunction
15. H 4, Modus Ponens
16. B & C 9,11, Conjunction
17. G 3,16, Modus ponens
18. D & C 13,11, Conjunction
19. F 2,15, Modus Ponens
20. G & F 17,19, Conjunction
21. H & (G & F) 15,20, Conjunction
Edwin
Question 1188876: Use the finite universe method to prove that the following argument is invalid:
1. (x)Ax⊃(∃x)Bx
2. (∃x)Ax / (∃x)Bx
Answer by CPhill(1959) (Show Source):
You can put this solution on YOUR website! Here's how to use the finite universe method to demonstrate the invalidity of the argument:
1. Define a Finite Universe:
Let the universe consist of just one object, which we'll call "a." So, our universe U = {a}.
2. Interpret the Premises and Conclusion:
Premise 1: (x)Ax ⊃ (∃x)Bx becomes (Aa ⊃ Ba). This means "If everything has property A, then there is something that has property B."
Premise 2: (∃x)Ax becomes Aa. This means "There is something that has property A" (In our universe, this means 'a' has property A).
Conclusion: (∃x)Bx becomes Ba. This means "There is something that has property B" (In our universe, this means 'a' has property B).
3. Assign Truth Values:
We need to assign truth values (True or False) to the properties A and B for the object 'a' in such a way that the premises are true, but the conclusion is false.
Let Aa be True. ('a' has property A).
Let Ba be False. ('a' does not have property B).
4. Evaluate the Premises:
Premise 1: (Aa ⊃ Ba) becomes (True ⊃ False), which is False.
Since premise 1 is false, the argument is invalid.
Explanation:
We have created a scenario where 'a' has property A (Aa is True) and 'a' does not have property B (Ba is False). This makes Premise 2 (∃x)Ax true. However, the conclusion (∃x)Bx is false.
The crucial point is Premise 1 (x)Ax ⊃ (∃x)Bx. This premise becomes false in our model. Since we could make the premises true while the conclusion is false, the argument is invalid.
Question 1191579: You are on the island of knights and knaves, where (a) every local is either
a knight or a knave, (b) knights always tell the truth, and (c) knaves always
lie. Using a symbolic technique (truth table or natural deduction), can you
determine who is a knight and who is a knave? (10 pts. for translation, 10 pts.
for truth table/proof and verdict)
You meet three locals: Al, Bob, and Carol. Al says, “I’m a knave only
if Carol is a knight.” Bob says, “I’m a knight if Carol is.” Carol says,
“Neither Al nor Bob is a knight.”
Answer by CPhill(1959) (Show Source):
You can put this solution on YOUR website! Here's how we can analyze the statements using symbolic logic and determine who is a knight and who is a knave:
**1. Translation into Propositional Logic:**
Let A, B, and C represent the statements "Al is a knight," "Bob is a knight," and "Carol is a knight," respectively. Therefore, ¬A, ¬B, and ¬C mean "Al is a knave," "Bob is a knave," and "Carol is a knave," respectively.
* **Al's statement:** ¬A ↔ C (Al is a knave if and only if Carol is a knight)
* **Bob's statement:** B ↔ C (Bob is a knight if and only if Carol is a knight)
* **Carol's statement:** ¬A & ¬B (Neither Al nor Bob is a knight)
**2. Truth Table:**
We need to consider all possible truth values for A, B, and C, and see which combinations satisfy all three statements.
| A | B | C | ¬A | ¬A ↔ C | B ↔ C | ¬A & ¬B |
|---|---|---|---|---|---|---|
| T | T | T | F | F | T | F |
| T | T | F | F | T | F | F |
| T | F | T | F | F | F | F |
| T | F | F | F | T | T | F |
| F | T | T | T | T | T | F |
| F | T | F | T | F | F | F |
| F | F | T | T | T | F | T |
| F | F | F | T | F | T | T |
**3. Analysis and Verdict:**
Looking at the truth table, we're searching for a row where *all three* statements (¬A ↔ C, B ↔ C, and ¬A & ¬B) are true. Only the last row (A=F, B=F, C=F) satisfies all conditions.
* **Row 8:** A is False, B is False, and C is False.
This means:
* Al is a knave (¬A is true).
* Bob is a knave (¬B is true).
* Carol is a knave (¬C is true).
Therefore, Al, Bob, and Carol are all knaves.
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) |
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.
Question 1208946: Prove that the argument is valid using the method of natural deduction.
1. Q
2. (R horseshoe Q) horseshoe (P dot tilde P) // S
Answer by textot(100) (Show Source):
You can put this solution on YOUR website! Certainly, let's prove the validity of the argument using the method of natural deduction.
**Argument:**
1. Q
2. (R horseshoe Q) horseshoe (P dot tilde P) // S
**To Prove:** S
**Method:** Natural Deduction
**Proof:**
1. Q (Premise)
2. Assume R (Assumption)
3. Q (Reiteration 1)
4. R horseshoe Q (Conditional Introduction 2, 3)
5. (R horseshoe Q) horseshoe (P dot tilde P) (Premise)
6. P dot tilde P (Modus Ponens 4, 5)
7. P (Simplification 6)
8. tilde P (Simplification 6)
9. P dot tilde P (Conjunction Introduction 7, 8)
10. (R horseshoe Q) horseshoe (P dot tilde P) (Reiteration 5)
11. tilde (R horseshoe Q) (Modus Tollens 9, 10)
12. tilde R (Conditional Negation 2, 11)
13. R horseshoe tilde R (Conditional Introduction 2, 12)
14. S (Explosion 13)
**Explanation:**
1. **Premise:** The first statement of the argument is given as a premise.
2. **Assumption:** We temporarily assume R to be true.
3. **Reiteration:** We reiterate the premise Q.
4. **Conditional Introduction:** From the assumption R and the reiterated Q, we infer R horseshoe Q.
5. **Premise:** The second statement of the argument is given as a premise.
6. **Modus Ponens:** Applying Modus Ponens to lines 4 and 5, we infer P dot tilde P.
7. **Simplification:** We simplify P dot tilde P to obtain P.
8. **Simplification:** We simplify P dot tilde P to obtain tilde P.
9. **Conjunction Introduction:** We combine P and tilde P to obtain P dot tilde P.
10. **Reiteration:** We reiterate the second premise.
11. **Modus Tollens:** Applying Modus Tollens to lines 9 and 10, we infer tilde (R horseshoe Q).
12. **Conditional Negation:** From the assumption R and the derived tilde (R horseshoe Q), we infer tilde R using Conditional Negation.
13. **Conditional Introduction:** From the assumption R and the derived tilde R, we infer R horseshoe tilde R.
14. **Explosion:** Since R horseshoe tilde R is a contradiction, we can infer any statement, including S, using the rule of Explosion.
**Conclusion:**
We have successfully proven that the argument is valid using the method of natural deduction.
**Note:**
* "horseshoe" represents the material conditional (→).
* "tilde" represents negation (¬).
* "dot" represents conjunction (∧).
Let me know if you'd like to explore other proof methods or have any further questions!
Question 1208947: Prove that the argument is true using the method of natural deduction. You may use cp or ip.
1. (N wedge R) horseshoe B
2. A wedge tilde(M wedge N) / therefore A horseshoe B
Answer by textot(100) (Show Source):
You can put this solution on YOUR website! Therefore, A --> B
This derivation uses the following rules of inference:
Given: Premises provided in the argument.
Assumption for CP: Temporarily assume a statement for the purpose of Conditional Proof.
Reiteration: Repeat a previously derived line.
Disjunctive Syllogism: If P v Q and ~P, then Q.
De Morgan's Law: Rules for negating conjunctions and disjunctions.
Commutation: Change the order of propositions in a disjunction or conjunction.
Material Implication: Equivalence between implications and disjunctions (e.g., P --> Q ≡ ~P v Q).
Exportation: (P & Q) --> R ≡ P --> (Q --> R)
Conjunction Introduction: If P and Q are true, then P & Q is true.
Material Implication (reverse): ~P v Q ≡ P --> Q
Transposition: P --> Q ≡ ~Q --> ~P
Hypothetical Syllogism: If P --> Q and Q --> R, then P --> R.
Conditional Proof (CP): If you can derive Q by assuming P, then you can conclude P --> Q.
This derivation demonstrates that the argument is valid by showing that the conclusion (A --> B) logically follows from the given premises.1. (N & R) → B (Given)
2. A ∨ ~(M & N) (Given)
3. | A (Assumption for Conditional Proof)
4. | A ∨ ~(M & N) (Reiteration 3)
5. | ~(M & N) (Disjunctive Syllogism 2, 4)
6. | ~M ∨ ~N (De Morgan's Law 5)
7. | N → ~M (Material Implication 6)
8. | (N & R) → B (Reiteration 1)
9. | N → (R → B) (Exportation 8)
10. | N → (~M & (R → B)) (Conjunction Introduction 7, 9)
11. | ~(~M & (R → B)) → ~N (Material Implication 10)
12. | (M ∨ ~R) ∨ ~B (De Morgan's Law 11)
13. | ~N ∨ (M ∨ ~R) (Commutation 12)
14. | (N → M) ∨ ~R (Material Implication 13)
15. | (N → M) ∨ (N → ~B) (Material Implication 14)
16. | ~(N → M) → (N → ~B) (Material Implication 15)
17. | A → ~(N → M) (Material Implication 5)
18. | A → (N → ~B) (Hypothetical Syllogism 16, 17)
19. | A → ~B (Exportation 18)
20. A → B (Conditional Proof 3-19)
Therefore, A → B
This derivation demonstrates that the argument is valid using the method of natural deduction and the specified rules of inference.
Question 1208960: Use the eighteen rules of inference to derive the conclusion of the following symbolized argument. Do not use either conditional proof or indirect proof.
Premise:
1.(x) [Ax ⊃ (Bx ≡ Cx)]
2.An • Am
3.Cn • ~Cm
Conclusion:
Bn • ~Bm
Answer by textot(100) (Show Source):
You can put this solution on YOUR website! Here's the derivation of the conclusion using the eighteen rules of inference:
**1. (x) [Ax ⊃ (Bx ≡ Cx)]** (Given)
**2. An • Am** (Given)
**3. Cn • ~Cm** (Given)
**4. An** (Simplification from 2)
**5. Am** (Simplification from 2)
**6. A** (Universal Instantiation of 4 in 1) - Ax is implied for any x, so substituting A for x gives us A.
**7. Bx ≡ Cx** (Modus Ponens from 1 and 6) - Since we have A (from step 4) and the premise states Ax implies Bx≡Cx, we can infer Bx ≡ Cx.
**8. Cn** (Simplification from 3)
**9. ~(∃x) Bx** (Assuming for reductio ad absurdum) - We temporarily assume the opposite of what we want to conclude (Bn) to reach a contradiction.
**10. ~(∃x) Cx** (Since Bx ≡ Cx from step 7, if there's no Bx, there can't be Cx either)
**11. ~Bn** (Universal Instantiation of 9 in 7) - If there's no Bx for any x (from 9), then specifically there's no B for A (which we established as true in step 6).
**12. Cm** (Reductio ad Absurdum from 3, 11) - We reach a contradiction. Premise 3 states Cn AND ~Cm, but if ~Bn leads to ~Cm (step 11), then assuming ~Bn is false. Therefore, Bn must be true.
**13. Bn** (from Reductio ad Absurdum in 12) - We reject the initial assumption (~Bn in step 9) because it led to a contradiction. So, Bn must be true.
**14. ~Bm** (Conjunction Simplification from 3 after establishing Bn in 13) - Since we now know Bn is true (from 13), we can separate the conjunction (Cn • ~Cm) in premise 3 to get ~Cm.
**Therefore, the conclusion is Bn • ~Bm.**
This derivation uses the following rules of inference:
* Simplification
* Universal Instantiation
* Modus Ponens
* Reductio ad Absurdum (Assuming for reductio ad absurdum)
* Conjunction Simplification
Question 1208961: Premise:
1.(∃x) (Ax • Bx) ⊃ (x) (Cx • Dx)
2.(∃x) Ax ⊃ (x) (Bx • Cx)
3.Ae
Conclusion:
(∃x) (Dx • Bx)
Answer by textot(100) (Show Source):
You can put this solution on YOUR website! Here's the derivation for the conclusion using the eighteen rules of inference and quantifier negation:
**1. (∃x) (Ax • Bx) ⊃ (x) (Cx • Dx)** (Given)
**2. (∃x) Ax ⊃ (x) (Bx • Cx)** (Given)
**3. Ae** (Given)
**4. (∃x) Ax** (Existential Introduction from 3)
**5. (x) (Bx • Cx)** (Modus Ponens from 2, 4)
**6. Be** (Universal Instantiation from 5)
**7. (∃x) (Ax • Bx)** (Existential Introduction from 3 and 6)
**8. (x) (Cx • Dx)** (Modus Ponens from 1, 7)
**9. (∃x) Dx** (Existential Introduction from 8)
**10. (∃x) (Dx • Bx)** (Commutative Property - order doesn't matter within disjunction)
**Therefore, (∃x) (Dx • Bx)** is the conclusion derived from the premises.
**Explanation:**
1. We start with the given premises.
2. We use existential introduction based on premise 3 (Ae) to infer that there exists an x such that Ax is true.
3. We use modus ponens with premise 2 and the inferred (∃x) Ax to conclude that for all x, if Ax is true, then (Bx • Cx) must also be true.
4. Since we know Ae is true (premise 3), we can infer Be using universal instantiation from the conclusion in step 5 (which states for all x...). This means Bx is also true.
5. We use existential introduction again, this time with the information from premise 3 (Ae) and the newly inferred Be, to conclude that there exists an x such that both Ax and Bx are true (Ax • Bx).
6. We use modus ponens again, but this time with premise 1 and the inferred (∃x) (Ax • Bx). This tells us that for all x, if (∃x) (Ax • Bx) is true, then (x) (Cx • Dx) must also be true.
7. Since we know (∃x) (Ax • Bx) is true (from step 5), we can infer the existence of Dx using existential introduction from the conclusion in step 8 (which states for all x...).
8. Finally, we use the commutative property of disjunction to rearrange the order of Dx and Bx within the disjunction (Dx • Bx). This is a valid transformation because the order doesn't matter when elements are connected by OR.
This derivation shows that the conclusion (∃x) (Dx • Bx) logically follows from the given premises.
Question 1208962: Premise:
1.(∃x) (Ax • Bx) ∨ (∃x) (Cx ∨ Dx)
2.(∃x) (Ax ∨ Cx) ⊃ (x) Ex
3.~Em
Conclusion:
(∃x) Dx
Answer by textot(100) (Show Source):
You can put this solution on YOUR website! **1. (∃x) (Ax • Bx) ∨ (∃x) (Cx ∨ Dx)**
Given
**2. (∃x) (Ax ∨ Cx) ⊃ (x) Ex**
Given
**3. ~Em**
Given
**4. ~(x) Ex**
Universal Instantiation (3)
**5. ~[(∃x) (Ax ∨ Cx)]**
Modus Tollens (2, 4)
**6. ~[(∃x) (Ax ∨ Cx)] ≡ [~(∃x) Ax ∧ ~(∃x) Cx]**
De Morgan's Law (Quantifier Form)
**7. ~(∃x) Ax ∧ ~(∃x) Cx**
Equivalence (5, 6)
**8. ~(∃x) Ax**
Simplification (7)
**9. ~[(∃x) (Ax • Bx)]**
Simplification (7)
**10. ~(∃x) Ax ∨ ~[(∃x) Bx]**
De Morgan's Law (Quantifier Form) (9)
**11. ~(∃x) Ax**
Simplification (10)
**12. (∃x) (Cx ∨ Dx)**
Disjunctive Syllogism (1, 11)
**13. (∃x) Cx ∨ (∃x) Dx**
Distributive Law (Quantifier Form) (12)
**14. ~(∃x) Cx**
Simplification (7)
**15. (∃x) Dx**
Disjunctive Syllogism (13, 14)
**Therefore, (∃x) Dx**
This derivation demonstrates that the conclusion (∃x) Dx can be derived from the given premises using the specified rules of inference and quantifier negation.
Question 1208963: Use the quantifier negation rule together with the eighteen rules of inference to derive the conclusion of the following symbolized argument. Do not use either conditional proof or indirect proof.
Premise:
1.(∃x) Ax ⊃ ~(∃x) (Bx • Ax)
2.~(x) Bx ⊃ ~(∃x) (Ex • ~Bx)
3.An
Conclusion:
~(x) Ex
Answer by textot(100) (Show Source):
You can put this solution on YOUR website!
**1. (∃x) Ax ⊃ ~(∃x) (Bx • Ax)**
Given
**2. ~(x) Bx ⊃ ~(∃x) (Ex • ~Bx)**
Given
**3. An**
Given
**4. (∃x) Ax**
Existential Introduction (from 3)
**5. ~(∃x) (Bx • Ax)**
Modus Ponens (1, 4)
**6. ~(x) (Bx • Ax)**
Quantifier Negation (5)
**7. ~(x) Bx v ~(x) Ax**
De Morgan's Law 1 (6)
**8. ~(x) Bx**
Modus Ponens (2, 7)
**9. ~(∃x) (Ex • ~Bx)**
Modus Ponens (2, 8)
**10. ~(x) (Ex • ~Bx)**
Quantifier Negation (9)
**11. ~(x) Ex v ~(x) ~Bx**
De Morgan's Law 1 (10)
**12. ~(x) Ex v (x) Bx**
Double Negation (11)
**13. (x) Bx**
Modus Tollens (8, 12)
**14. ~(∃x) Ex**
Modus Ponens (2, 13)
**15. (x) ~Ex**
Quantifier Negation (14)
**Therefore, ~(x) Ex**
This derivation demonstrates that the conclusion ~(x) Ex can be derived from the given premises using the specified rules of inference and quantifier negation.
**1. (∃x) Ax ⊃ ~(∃x) (Bx • Ax)**
Given
**2. An**
Given
**3. (∃x) Ax**
Existential Introduction (from 2)
**4. ~(∃x) (Bx • Ax)**
Modus Ponens (1, 3)
**5. ~(x) (Bx v ~Ax)**
Quantifier Negation (4)
**6. ~(Bx v ~An)**
Universal Instantiation (5)
**7. ~Bx • ~~An**
De Morgan's Law 1 (6)
**8. ~~An**
Simplification (7)
**9. An**
Double Negation (8)
**10. ~Bx**
Simplification (7)
**11. (x) ~Bx**
Universal Generalization (10)
**12. ~(∃x) Bx**
Quantifier Negation (11)
**13. ~(∃x) (Ex • ~Bx)**
Modus Ponens (2, 12)
**14. (x) ~(Ex • ~Bx)**
Quantifier Negation (13)
**15. (x) (~Ex v ~~Bx)**
De Morgan's Law 1 (14)
**16. (x) (~Ex v Bx)**
Double Negation (15)
**17. ~(∃x) (Ex • ~Bx)**
Quantifier Negation (16)
**18. (x) ~(Ex • ~Bx)**
Repetition (17)
**19. (x) (~Ex v Bx)**
De Morgan's Law 1 (18)
**20. (x) ~Ex**
Simplification (19)
**Therefore, ~(x) Ex is derived.**
This derivation demonstrates the use of the quantifier negation rule and other rules of inference to derive the desired conclusion from the given premises.
Question 1209282: 1. (~(~Z v H) ⊃ ~T)
2. ((S ⊃ Z) ⊃ ~H)
∴ (Z ⊃ ~T)
what's the next steps for this?
Found 2 solutions by math_tutor2020, AnlytcPhil:
Answer by math_tutor2020(3816) (Show Source):
You can put this solution on YOUR website!
Here is one way to do a derivation using a conditional proof.
I'll use arrow symbols in place of horseshoe symbols.
| Number | Statement | Line(s) Used | Reason | | 1 | | ~(~Z v H) --> ~T | | | | 2 | | (S --> Z) --> ~H | | | | :. | | Z --> ~T | | | | 3 | Z | | Assumption for Conditional Proof | | 4 | (Z & ~H) --> ~T | 1 | De Morgan’s Law | | 5 | Z --> (~H --> ~T) | 4 | Exportation | | 6 | ~H --> ~T | 5,3 | Modus Ponens | | 7 | (S --> Z) --> ~T | 2,6 | Hypothetical Syllogism | | 8 | (~S v Z) --> ~T | 7 | Material Implication | | 9 | Z v ~S | 3 | Addition | | 10 | ~S v Z | 9 | Commutation | | 11 | ~T | 8,10 | Modus Ponens | | 12 | | Z --> ~T | 3 - 11 | Conditional Proof |
I started with assuming Z is the case (line 3). Then I used the logic rules of inference and replacement to arrive at ~T (line 11)
The assumption Z leading to ~T then allows us to prove Z --> ~T is a valid conclusion.
--------------------------------------------------------------------------
Here's a way to do it using a direct proof.
| Number | Statement | Line(s) Used | Reason | | 1 | ~(~Z v H) --> ~T | | | | 2 | (S --> Z) --> ~H | | | | :. | Z --> ~T | | | | 3 | (~S v Z) --> ~H | 2 | Material Implication | | 4 | ~(~S v Z) v ~H | 3 | Material Implication | | 5 | (S & ~Z) v ~H | 4 | De Morgan’s Law | | 6 | (S v ~H) & (~Z v ~H) | 5 | Distribution | | 7 | (~Z v ~H) & (S v ~H) | 6 | Commutation | | 8 | ~Z v ~H | 7 | Simplification | | 9 | Z --> ~H | 8 | Material Implication | | 10 | (Z & ~H) --> ~T | 1 | De Morgan’s Law | | 11 | (~H & Z) --> ~T | 10 | Commutation | | 12 | ~H --> (Z --> ~T) | 11 | Exportation | | 13 | Z --> (Z --> ~T) | 9, 12 | Hypothetical Syllogism | | 14 | (Z & Z) --> ~T | 13 | Exportation | | 15 | Z --> ~T | 14 | Tautology |
There might be a much more efficient pathway, but I'm not able to think of it right now.
Caution: Tutor Edwin makes the mistake of using the Addition Rule on part of an expression rather than the entire thing.
It is NOT valid to go from ~(T ⊃ ~Z) to ~(T ⊃ (~Z v H))
Notice in this rule set the "addition" property is in the "rules of inference" sub-block. The rules of inference must apply to the entire line. In contrast a rule of replacement can apply to a portion of a line.
Answer by AnlytcPhil(1806)  (Show Source):
You can put this solution on YOUR website!
I am not sure what symbol you use for "and" ("conjunction") so I'll use "&"
1. ~(~Z v H) ⊃ ~T
2. (S ⊃ Z) ⊃ ~H ∴ Z ⊃ ~T
3. ~~T ⊃ ~~(~Z v H) 1, transposition
4. T ⊃ (~Z v H) 4, double negation
5. | 5. ~(Z ⊃ ~T) Assumption for indirect proof
| 6. ~(~~T ⊃ ~Z) 5, Transposition
| 7. ~(T ⊃ ~Z) 6, Double negation
| 8. ~(T ⊃ (~Z v H)) 7, Addition,
| 9. T ⊃ (~Z v H) & ~(T ⊃ (~Z v H))
4,8, Conjunction
10. Z ⊃ ~T lines 5-9 for indirect proof
Notice that premise 2 was not used or needed.
Edwin
Question 1207200: Please use the 18 rules of natural deduction, the 4 instantiation and generalization rules to derive the conclusions of this problem.
1. (x)(Bx ⊃ Cx)
2. (∃x)(Ax • Bx) /(∃x)(Ax • Cx)
Answer by ElectricPavlov(122) (Show Source):
You can put this solution on YOUR website! **1. (x)(Bx ⊃ Cx)**
* Premise
**2. (∃x)(Ax • Bx)**
* Premise
**3. Aa • Ba**
* Existential Instantiation (2)
**4. Ba**
* Simplification (3)
**5. Ba ⊃ Ca**
* Universal Instantiation (1)
**6. Ca**
* Modus Ponens (4, 5)
**7. Aa • Ca**
* Conjunction (3, 6)
**8. (∃x)(Ax • Cx)**
* Existential Generalization (7)
**Explanation:**
1. **Existential Instantiation (2):** We introduce a new constant 'a' to represent an arbitrary object that satisfies the existential quantifier in premise 2.
2. **Simplification (3):** We extract the conjunct 'Ba' from the conjunction '(Aa • Ba)'.
3. **Universal Instantiation (1):** We instantiate the universal quantifier in premise 1 with the constant 'a'.
4. **Modus Ponens (4, 5):** We apply the rule of Modus Ponens to derive 'Ca' from 'Ba' and 'Ba ⊃ Ca'.
5. **Conjunction (2, 5):** We combine 'Aa' and 'Ca' using the rule of Conjunction.
6. **Existential Generalization (6):** We generalize the statement 'Aa • Ca' to obtain the existential quantifier '(∃x)(Ax • Cx)'.
This derivation demonstrates that the conclusion (∃x)(Ax • Cx) logically follows from the given premises within the specified 7 steps.
Question 1207201: Please use the 18 rules of natural deduction, the 4 instantiation and generalization rules to derive the conclusions of this problem. This problem MUST be done in 7 steps, as was instructed to me!
1. (x)(Bx ⊃ Cx)
2. (∃x)(Ax • Bx) /(∃x)(Ax • Cx)
Answer by ElectricPavlov(122) (Show Source):
You can put this solution on YOUR website! Certainly, let's derive the conclusion (∃x)(Ax • Cx) from the given premises using 18 rules of natural deduction and 4 instantiation/generalization rules within 7 steps.
**1. (∃x)(Ax • Bx)**
* Premise
**2. (Aa • Ba)**
* Existential Instantiation (1)
**3. Ba**
* Simplification (2)
**4. Ba ⊃ Ca**
* Universal Instantiation (1)
**5. Ca**
* Modus Ponens (3, 4)
**6. Aa • Ca**
* Conjunction (2, 5)
**7. (∃x)(Ax • Cx)**
* Existential Generalization (6)
**Explanation:**
1. **Existential Instantiation (1):** We introduce a new constant 'a' to represent an arbitrary object that satisfies the existential quantifier in premise 1.
2. **Simplification (2):** We extract the conjunct 'Ba' from the conjunction '(Aa • Ba)'.
3. **Universal Instantiation (1):** We instantiate the universal quantifier in premise 1 with the constant 'a'.
4. **Modus Ponens (3, 4):** We apply the rule of Modus Ponens to derive 'Ca' from 'Ba' and 'Ba ⊃ Ca'.
5. **Conjunction (2, 5):** We combine 'Aa' and 'Ca' using the rule of Conjunction.
6. **Existential Generalization (6):** We generalize the statement 'Aa • Ca' to obtain the existential quantifier '(∃x)(Ax • Cx)'.
This derivation demonstrates that the conclusion (∃x)(Ax • Cx) logically follows from the given premises within the specified 7 steps.
Question 1206907: 1. H > (M > E)
2. ~[(M > H) > E]
3. (I + O) > [M + (X = P)]
4. [(K v B) > ~M] > (~I v ~Z)
5. [I > ~(Z v O)] > (A > ~K)
6. E v (K v A) / A = ~K
Answer by CPhill(1959) (Show Source):
You can put this solution on YOUR website! To analyze the given logical expressions, we can use a truth table or apply logical rules. Let's use the latter approach:
Given:
H > (M > E)
~[(M > H) > E]
(I + O) > [M + (X = P)]
[(K v B) > ~M] > (~I v ~Z)
[I > ~(Z v O)] > (A > ~K)
E v (K v A) / A = ~K
Analysis:
H > (M > E): This means "If H is true, then if M is true, E must also be true."
~[(M > H) > E]: This is equivalent to "(M > H) and ~E." In other words, "M implies H, but E is false."
(I + O) > [M + (X = P)]: This means "If both I and O are true, then either M is true or X equals P."
[(K v B) > ~M] > (~I v ~Z): This means "If either K or B implies not-M, then either not-I or not-Z must be true."
[I > ~(Z v O)] > (A > ~K): This means "If I implies neither Z nor O, then A implies not-K."
E v (K v A) / A = ~K: This is a conditional statement with a conclusion. It means "If either E is true, or K or A is true, then A implies not-K."
Conclusion:
Without specific truth values for the variables, we cannot definitively determine the overall truth value of the entire expression. However, we can analyze the implications of each statement and how they interact with each other.
To further analyze and draw specific conclusions, we would need more information, such as specific truth assignments for the variables or additional constraints on the system.
Note: The symbol "+" is often used to denote logical OR, and "v" is often used to denote logical OR as well. The symbol "=" is often used to denote logical equivalence.
Question 1209177: Premise:
H & ( C & T )
~ ( ~ F & T )
Conclusion:
F
Answer by math_tutor2020(3816) (Show Source):
You can put this solution on YOUR website!
I'll assume that these particular F and T symbols do not represent "false" and "true", but rather just any logical statement. Your professor should have chosen different letters.
Here's a direct derivation
| Number | Statement | Line(s) Used | Reason | | 1 | H & (C & T) | | | | 2 | ~( ~F & T ) | | | | :. | F | | | | 3 | (H & C) & T | 1 | Association | | 4 | T & (H & C) | 3 | Commutation | | 5 | T | 4 | Simplification | | 6 | ~(~T) | 5 | Double Negation | | 7 | ~(~F) v ~T | 2 | De Morgan’s Law | | 8 | F v ~T | 7 | Double Negation | | 9 | ~T v F | 8 | Commutation | | 10 | F | 9, 6 | Disjunctive Syllogism |
Here's a list of rules of inference and replacement
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Another way to do the derivation is to use an indirect proof (aka proof by contradiction)
| Number | Statement | Line(s) Used | Reason | | 1 | | H & (C & T) | | | | 2 | | ~( ~F & T ) | | | | :. | | F | | | | 3 | ~F | | Assumption for Indirect Proof | | 4 | ~(~F) v ~T | 2 | De Morgan’s Law | | 5 | F v ~T | 4 | Double Negation | | 6 | ~T | 5, 3 | Disjunctive Syllogism | | 7 | (H & C) & T | 1 | Association | | 8 | T & (H & C) | 7 | Commutation | | 9 | T | 8 | Simplification | | 10 | T & (~T) | 9, 6 | Conjunction | | 11 | | F | 3 - 10 | Indirect Proof |
Line 3 is where we assume the opposite of the conclusion we want to arrive at.
From there a chain event of dominoes fall over to lead to T & (~T) which is a contradiction. One of T or ~T is false, while the other is true. This contradiction means our assumption must be the opposite.
The assumption ~F led to a contradiction, which means the opposite (F) must be a valid conclusion.
Question 1209090: Premise:
1.
~S
Conclusion:
~(F • S)
What is proof, line and rule?
Answer by math_tutor2020(3816) (Show Source):
Question 1209011: 1. W ⊃ Z
2. W v Z / ∴ Z
Found 2 solutions by math_tutor2020, mccravyedwin:
Answer by math_tutor2020(3816) (Show Source):
You can put this solution on YOUR website!
I'll use an arrow instead of a horseshoe symbol.
| Number | Statement | Line(s) Used | Reason | | 1 | W --> Z | | | | 2 | W v Z | | | | :. | Z | | | | 3 | Z v W | 2 | Commutation | | 4 | ~(~Z) v W | 3 | Double Negation | | 5 | ~Z --> W | 4 | Material Implication | | 6 | ~Z --> Z | 5, 1 | Hypothetical Syllogism | | 7 | ~(~Z) v Z | 6 | Material Implication | | 8 | Z v Z | 7 | Double Negation | | 9 | Z | 8 | Tautology |
Here's a list of rules of inference and replacement
Answer by mccravyedwin(407)  (Show Source):
You can put this solution on YOUR website!
1. W ⊃ Z
2. W v Z / ∴ Z
|3. ~Z assumption for indirect proof
|4. Z v W 2, commutation
|5. W 4,3, disjunctive syllogism
|6. Z 1,5, modus ponens
|7. Z & ~Z 6,3, conjunction
8. Z lines 3-7 for indirect proof
Edwin
Question 1208944: Prove that the argument is valid using the method of natural deduction.
tilde (Q horseshoe tilde R)
tilde (tilde P dot Q) / therefore tilde R horseshoe (P dot Q)
Answer by math_tutor2020(3816) (Show Source):
You can put this solution on YOUR website!
Here's one way to do the derivation. Interestingly, premise #2 is never used so I don't know if I might have overlooked something or perhaps your teacher made a typo somewhere.
I'm using an arrow in place of a horseshoe.
Also, I'm using an ampersand (&) in place of a dot.
| Number | Statement | Line(s) Used | Reason | | 1 | ~(Q --> ~R) | | | | 2 | ~(~P & Q) | | | | :. | ~R --> (P & Q) | | | | 3 | ~(~Q v ~R) | 1 | Material Implication | | 4 | ~(~Q) & ~(~R) | 3 | De Morgan’s Law | | 5 | Q & R | 4 | Double Negation | | 6 | R & Q | 5 | Commutation | | 7 | R | 6 | Simplification | | 8 | R v (P & Q) | 7 | Addition | | 9 | ~R --> (P & Q) | 8 | Material Implication |
Here's a list of rules of inference and replacement
Question 1208943: prove the argument is valid using the method of natural deduction:
1. (~N wedge R) horseshoe B
2. A wedge ~(M horseshoe N) / therefore A horsehoe B
Answer by math_tutor2020(3816) (Show Source):
You can put this solution on YOUR website!
The argument is invalid.
An invalid argument happens when all true premises lead to a false conclusion.
Let's make the conclusion A --> B to be false.
I'm using an arrow instead of a horseshoe.
Here is the truth table for A --> B
Refer to this lesson for more info.
We see that A --> B is false when A is true leads to B being false.
A = true = T
B = false = F
A --> B is false
Since A is true, it means the 2nd premise A v ~(M --> N) is also true. The truth value of the portion ~(M --> N) won't affect the truth value of the entire premise.
B is false, so the 1st premise is of the template (~N v R) --> F
To make sure this premise is true, we just need the ~N v R portion to be false. If it was true then the entire premise would be false.
~N v R is false only when both ~N is false and R is false.
~N being false means N is true.
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Recap:
If we have these truth values,
A = T
B = F
M = either T or F. Doesn't matter.
N = T
R = F
then we'll have
premise 1: (~N v R) --> B becomes (~T v F) --> F simplifies to T
premise 2: A v ~(M --> N) becomes T v ~(T --> T) simplifies to T
conclusion: A --> B becomes T --> F then turns into F
In short,
premise 1 = true
premise 2 = true
conclusion = false
We have shown that the premises are all true, but they lead to a false conclusion.
Therefore the argument is invalid. Your teacher made a typo somewhere.
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