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.