Tag: reductio

Baby Logic Practice 2: Derive the Conclusions Using Reductio (Answers)

Difficulty: What the hell

Here are the answers to Baby Logic Practice 2. As promised, each proof uses reductio ad absurdum, even in cases where there are quicker ways to get the conclusion.

Exercise 1

1. P → Q

2. ¬Q

∴ ¬P

3. Assume P

4. Q modus ponens 1, 3

5. ⊥ contradiction 2, 4

6. ¬P reductio 3–5

Exercise 2

1. P ∨ Q

2. ¬P

∴ Q

3. Assume ¬Q

4. ¬P ∧ ¬Q conjunction 2, 3

5. ¬(P ∨ Q) De Morgan’s Law, 4

6. ⊥ contradiction 1, 5

7. Q reductio 3–6

Exercise 3

1. (P ∧ Q) → R

2. P

3. ¬R

∴ ¬Q

4. Assume Q

5. P ∧ Q conjunction 2, 4

6. R modus ponens 1, 5

7. ⊥ contradiction 3, 6

8. ¬Q reductio 4–7

Exercise 4

1. P → Q

2. Q → R

3. ¬R

∴ ¬P

4. Assume P

5. Q modus ponens 1, 4

6. R modus ponens 2, 5

7. ⊥ contradiction 3, 6

8. ¬P reductio 4–7

Exercise 5

1. P → (Q ∨ R)

2. ¬Q

3. ¬R

∴ ¬P

4. Assume P

5. Q ∨ R modus ponens 1, 4

6. ¬Q ∧ ¬R conjunction 2, 3

7. ¬(Q ∨ R) De Morgan’s Law, 6

8. ⊥ contradiction 5, 7

9. ¬P reductio 4–8

Exercise 6

1. P ↔ Q

2. ¬Q

∴ ¬P

3. P → Q biconditional elimination, 1

4. Assume P

5. Q modus ponens 3, 4

6. ⊥ contradiction 2, 5

7. ¬P reductio 4–6

Exercise 7

1. P ∨ Q

2. P → R

3. Q → R

4. ¬R

∴ ¬P ∧ ¬Q

5. Assume P

6. R modus ponens 2, 5

7. ⊥ contradiction 4, 6

8. ¬P reductio 5–7

9. Assume Q

10. R modus ponens 3, 9

11. ⊥ contradiction 4, 10

12. ¬Q reductio 9–11

13. ¬P ∧ ¬Q conjunction 8, 12

Exercise 8

1. (P ∨ Q) → R

2. ¬R

∴ ¬P

3. Assume P

4. P ∨ Q addition 3

5. R modus ponens 1, 4

6. ⊥ contradiction 2, 5

7. ¬P reductio 3–6

So there you have it—the answers to our second baby logic quiz. Reductio can feel a little roundabout, but that’s part of what makes it useful: it forces you to see exactly where an assumption blows up.

Baby Logic Practice 2: Derive the Conclusions Using Reductio

Difficulty: What the hell

Today’s baby logic practice is on reductio ad absurdum, also known as indirect proof. The basic idea is simple: if assuming the opposite of what you want to prove leads to a contradiction, then the thing you wanted to prove must be true.

So, for each of the following exercises, derive the indicated conclusion using reductio. No quantifiers. No modal logic. No funny business. Just contradiction, negation, and a little patience.

Exercise 1

1. P → Q

2. ¬Q

∴ ¬P

Exercise 2

1. P ∨ Q

2. ¬P

∴ Q

Exercise 3

1. (P ∧ Q) → R

2. P

3. ¬R

∴ ¬Q

Exercise 4

1. P → Q

2. Q → R

3. ¬R

∴ ¬P

Exercise 5

1. P → (Q ∨ R)

2. ¬Q

3. ¬R

∴ ¬P

Exercise 6

1. P ↔ Q

2. ¬Q

∴ ¬P

Exercise 7

1. P ∨ Q

2. P → R

3. Q → R

4. ¬R

∴ ¬P ∧ ¬Q

Exercise 8

1. (P ∨ Q) → R

2. ¬R

∴ ¬P

In some of these, reductio will feel a little artificial because there are quicker ways to get the conclusion. Too bad. The point today is to practice reductio, not to be efficient.

I’ll probably post the answers tomorrow. Have fun.