Tag: propositional logic

Baby Logic Practice 3: Conditional Proofs (Answers)

Difficulty: What the hell

Here are the answers to Baby Logic Practice 3. As promised, each proof uses conditional proof, even in cases where there are quicker or more boring ways to get the conclusion.

Exercise 1

1. P → Q

∴ P → Q

2. Assume P

3. Q modus ponens 1, 2

4. P → Q conditional proof 2–3

Exercise 2

1. P → Q

2. Q → R

∴ P → R

3. Assume P

4. Q modus ponens 1, 3

5. R modus ponens 2, 4

6. P → R conditional proof 3–5

Exercise 3

1. P ∧ Q

∴ P → Q

2. Assume P

3. Q simplification 1

4. P → Q conditional proof 2–3

Exercise 4

1. P → (Q → R)

2. P

∴ Q → R

3. Assume Q

4. Q → R modus ponens 1, 2

5. R modus ponens 3, 4

6. Q → R conditional proof 3–5

Exercise 5

1. P → Q

2. R → S

3. P ∧ R

∴ P → S

4. Assume P

5. R simplification 3

6. S modus ponens 2, 5

7. P → S conditional proof 4–6

Exercise 6

1. P → Q

2. Q → R

3. R → S

∴ P → S

4. Assume P

5. Q modus ponens 1, 4

6. R modus ponens 2, 5

7. S modus ponens 3, 6

8. P → S conditional proof 4–7

Exercise 7

1. P → Q

2. P → R

∴ P → (Q ∧ R)

3. Assume P

4. Q modus ponens 1, 3

5. R modus ponens 2, 3

6. Q ∧ R conjunction 4, 5

7. P → (Q ∧ R) conditional proof 3–6

Exercise 8

1. P → Q

2. Q → (R ∨ S)

3. ¬R

∴ P → S

4. Assume P

5. Q modus ponens 1, 4

6. R ∨ S modus ponens 2, 5

7. S disjunctive syllogism 3, 6

8. P → S conditional proof 4–7

So there you have it—the answers to our third baby logic quiz. Conditional proof is simple in principle, but it gets nicer once assuming the antecedent starts to feel automatic.

Baby Logic Practice 3: Conditional Proofs

Difficulty: What the hell

Today’s baby logic practice is on conditional proof. The basic idea is simple: if assuming P lets you derive Q, then you are allowed to conclude that P → Q.

So, for each of the following exercises, derive the indicated conclusion using conditional proof. No quantifiers. No modal logic. No funny business. Just assumptions, a little patience, and the sweet smell of valid inference.

Exercise 1

1. P → Q

∴ P → Q

Exercise 2

1. P → Q

2. Q → R

∴ P → R

Exercise 3

1. P ∧ Q

∴ P → Q

Exercise 4

1. P → (Q → R)

2. P

∴ Q → R

Exercise 5

1. P → Q

2. R → S

3. P ∧ R

∴ P → S

Exercise 6

1. P → Q

2. Q → R

3. R → S

∴ P → S

Exercise 7

1. P → Q

2. P → R

∴ P → (Q ∧ R)

Exercise 8

1. P → Q

2. Q → (R ∨ S)

3. ¬R

∴ P → S

Some of these are pretty easy. Good. Conditional proof should start to feel almost annoyingly natural after enough practice. That is the point.

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