More Proofs

REVIEW

The Rule of Assumption: A

Assumption is the easiest rule to learn. It says at any stage in the derivation, we may write down any WFF we want to. That WFF only depends on itself.

&-Elimination: &E

&E is also a very easy-to-learn rule.

If we have proved (φ & ψ) on some line, then on any future line we may write down φ, and on any future line we may write down ψ. The result depends on everything (φ & ψ) depended on.

Arrow Elimination: →E

The →E rule says that if on one line we have a conditional (φ → ψ) and on another line we have the antecedent of the conditional φ then on any future line, we may write down the consequent of the conditional ψ depending on everything (φ → ψ) and φ depended on.

Let’s look at an example proof using these rules. First lets show this:

(P → (P → Q)), P Q├

This means “From the assumptions (P → (P → Q)) and P, there is a proof of Q.”

(P → (P → Q)), P Q├

1 1. (P → (P → Q)) A

The rule of assumption tells us we can write down anything we want. It seems reasonable to write down what we want to assume.

(P → (P → Q)), P Q├

1 1. (P → (P → Q)) A2 2. P A

A second application of assumption.

(P → (P → Q)), P Q├

1 1. (P → (P → Q)) A2 2. P A1,2 3. (P → Q) 1,2 →E

→E allows us to write down the consequent of a conditional, if we have the conditional + its antecedent.

(P → (P → Q)), P Q├

1 1. (P → (P → Q)) A2 2. P A1,2 3. (P → Q) 1,2 →E

We cite the lines used (1 and 2) as well as the rule used (→E) to the right.

(P → (P → Q)), P Q├

1 1. (P → (P → Q)) A2 2. P A1,2 3. (P → Q) 1,2 →E

Finally, we copy all the dependencies that are to the left of the lines used (1 and 2) to the right of 3. These are lines 1 and 2.

(P → (P → Q)), P Q├

1 1. (P → (P → Q)) A2 2. P A1,2 3. (P → Q) 1,2 →E1,2 4. Q 3,2 →E

Again, we have a conditional + its antecedent.

(P → (P → Q)), P Q├

1 1. (P → (P → Q)) A2 2. P A1,2 3. (P → Q) 1,2 →E1,2 4. Q 3,2 →E

We applied →E to lines 3 and 2.

(P → (P → Q)), P Q├

1 1. (P → (P → Q)) A2 2. P A1,2 3. (P → Q) 1,2 →E1,2 4. Q 3,2 →E

4 depends on what 3 and 2 depended on.

MORE RULES

&I: &-Introduction

Connectives have “introduction” rules that allow us to introduce them, and “elimination” rules that allow us to get rid of them. So in addition to &E, there is also a rule &I:If on some line you have proven φ and on some other line you have proven ψ, then on any future line you may write (φ & ψ),Depending on what φ and ψ depended on.

(P & (Q & R)) ((P & Q) & R)├1 1. (P & (Q & R)) A

(P & (Q & R)) ((P & Q) & R)├1 1. (P & (Q & R)) A1 2. P 1 &E

(P & (Q & R)) ((P & Q) & R)├1 1. (P & (Q & R)) A1 2. P 1 &E1 3. (Q & R) 1 &E

(P & (Q & R)) ((P & Q) & R)├1 1. (P & (Q & R)) A1 2. P 1 &E1 3. (Q & R) 1 &E1 4. Q 3 &E

(P & (Q & R)) ((P & Q) & R)├1 1. (P & (Q & R)) A1 2. P 1 &E1 3. (Q & R) 1 &E1 4. Q 3 &E1 5. R 3 &E

(P & (Q & R)) ((P & Q) & R)├1 1. (P & (Q & R)) A1 2. P 1 &E1 3. (Q & R) 1 &E1 4. Q 3 &E1 5. R 3 &E1 6. (P & Q) 2,4 &I

(P & (Q & R)) ((P & Q) & R)├1 1. (P & (Q & R)) A1 2. P 1 &E1 3. (Q & R) 1 &E1 4. Q 3 &E1 5. R 3 &E1 6. (P & Q) 2,4 &I1 7. ((P & Q) & R) 6,5 &I

(P & (Q & R)) ((P & Q) & R)├1 1. (P & (Q & R)) A1 2. P 1 &E1 3. (Q & R) 1 &E1 4. Q 3 &E1 5. R 3 &E1 6. (P & Q) 2,4 &I1 7. ((P & Q) & R) 6,5 &I

Important

Importantly, the last line of the proof (line 7) only depends on line 1 (P & (Q & R)). Since we were trying to show that ((P & Q) & R) followed from this assumption, that’s OK. Our proof would not be done, however, if line 7 depended on something other than line 1.

P (P & P)├1 1. P A

P (P & P)├1 1. P A1 2. (P & P) 1,1 &I

As you should have expected, there’s a rule →I that allows us to introduce conditionals. This rule is noticeably more difficult to use than any of the previous rules we learned.

If on some line you have assumed φ,And on some other line you have proved ψ,And ψ depends on your assumption φ,Then on any future line you may write (φ→ψ),Depending on everything ψ depended onExcept φ.

Strategies for Proving Conditionals

If you want to prove a conditional statement (φ→ψ), the best way to go about it is (usually, but not always):

1. Assume the antecedent φ2. Prove the consequent ψ3. Use →I to conclude (φ→ψ)

(P → Q), (Q → R) (P → R)├1 1. (P → Q) A2 2. (Q → R) A3 3. P A (for →I)1,3 4. Q 1,3 →E1,2,3 5. R 2,4 →E1,2 6. (P → R) 3,5 →I

Zero Dependencies

→I, because it allows you to remove dependencies, allows you to prove formulas with no dependencies. These are called “theorems.”

├ ((P→(Q→R))→((P →Q)→(P →R)))

1 1. (P → (Q → R)) A (for →I)2 2. (P → Q) A (for →I)3 3. P A (for →I)1,3 4. (Q → R) 1,3 →E1,2 5. Q 2,3 →E1,2,3 6. R 4,5 →E1,2 7. (P → R) 3,6 →I

├ ((P→(Q→R))→((P→Q)→(P→R)))

1 8. ((P → Q) → (P → R)) 2,7 →I9. ((P → (Q → R)) → ((P → Q) → (P → R)))

1,8 →I

├ (P → P)

1 1. P A (for →I)2. (P → P) 1,1 →I

The Biconditional

↔ has an easy set of rules to learn:

↔I: If on some line you have proved ((φ→ψ) & (ψ→φ))

Then on any future line you may write(φ↔ψ)

Depending on what ((φ→ψ) & (ψ→φ)) did.

The Biconditional

↔ has an easy set of rules to learn:

↔E: If on some line you have proved (φ↔ψ)

Then on any future line you may write((φ→ψ) & (ψ→φ))

Depending on what (φ↔ψ) depended on.

(P ↔ Q), (Q ↔ R) (P ↔ R)├1 1. (P ↔ Q) A2 2. (Q ↔ R) A1 3. ((P → Q) & (Q → P)) 1 ↔E2 4. ((Q → R) & (R → Q)) 2 ↔E1 5. (P → Q) 3 &E2 6. (Q → R) 4 &E7 7. P A (for →I)

(P ↔ Q), (Q ↔ R) (P ↔ R)├1,7 8. Q 5,7 →E1,2,7 9. R 6,8 →E1,2 10. (P → R) 7,9 →I2 11. (R → Q) 4 &E1 12. (Q → P) 3 &E13 13. R A (for →I)2,13 14. Q 11,13 →E

(P ↔ Q), (Q ↔ R) (P ↔ R)├1,2,1315. P 12,14 →E1,2 16. (R → P) 13,15 →I1,2 17. ((P → R) & (R → P)) 10,16 &I1,2 18. (P ↔ R) 17 ↔I

Rules for Wedge: vI

vI: If on some line you have proved φThen on any future line you may write (φ v ψ)And on any future line you may write (ψ v φ)The result depends on what φ depends on.

Rules for Wedge: vE

vE: If you have proved (φ v ψ)And you have proved ~φThen you may write ψDepending on whatever (φ v ψ) depended on.If you have proved (φ v ψ)And you have proved ~ψThen you may write φDepending on whatever (φ v ψ) depended on.

Anything Provable from Contradiction

In logic, anything is provable from a contradiction. So, for example:

(P & ~P) Q├This might mean “Michael is happy AND Michael is not happy. Therefore, 2 + 2 =17.” That’s a valid argument!One way to see why it’s valid is to look at the truth-tables. Q is never F when (P & ~P) is true, because (P & ~P) is never true!

(P & ~P) Q├1 1. (P & ~P) A1 2. P 1 &E1 3. (P v Q) 2 vI1 4. ~P 1 &E1 5. Q 3 vE

Proof by Cases

PC: First, you must have three formulas:(φ v ψ)(φ → α)(ψ → β)

Then on any future line you may write(α v β)

Depending on what the 3 formulas depended on

Proof by Cases: Example

P1: Either I’ll have free time this weekend, or I’ll still have work to do.P2: If I have free time, I’ll go to the museum.P3: If I have work to do, I’ll grade papers.Therefore,C: Either I’ll go to the museum this weekend, or I’ll grade papers.

((P & P) v (Q & Q)) (P v Q)├1 1. ((P & P) v (Q & Q)) A2 2. (P & P) A (for →I)2 3. P 2 &E

4. ((P & P) → P) 2,3 →I5 5. (Q & Q) A (for →I)5 6. Q 5 &E

7. ((Q & Q) → Q)5,6 →I1 8. (P v Q) 1,4,7 PC

Negation Introduction

Our two other rules have to do with ~, and they are (predictably) ~I and ~E. ~I (Negation Introduction) If you have assumed ψ, and you have derived (φ&~φ), then you can write down ~ψ, depending on everything (φ&~φ) depends on except ψ.

P ~~P├1 1. P A2 2. ~P A (for ~I)1,2 3. (P & ~P) 1,2 &I1 4. ~~P 2,3 ~I

(~P v Q) (P → Q)├1 1. (~P v Q) A2 2. P A3 3. ~P A (for ~I)2,3 4. (P & ~P) 2,3 &I2 5. ~~P 3,4 ~I1,2 6. Q 1,5 vE1 7. (P → Q) 2,6 →I

(P → Q) ~(P & ~Q)├1 1. (P → Q) A2 2. (P & ~Q) A (for ~I)2 3. P 2 &E2 4. ~Q 2 &E1,2 5. Q 1,3 →E1,2 6. (Q & ~Q) 5,4 &I1 7. ~(P & ~Q) 2,6 ~I

(P & ~Q) ~(P → Q)├1 1. (P & ~Q) A2 2. (P → Q) A (for ~I)1 3. P 1 &E1 4. ~Q 1 &E1,2 5. Q 2,3 →E1,2 6. (Q & ~Q) 5,4 &I1 7. ~(P → Q) 2,6 ~I

(~P v ~Q) ~(P & Q)├1 1. (~P v ~Q) A2 2. (P & Q) A (for ~I)2 3. P 2 &E4 4. ~P A (for ~I)2,4 5. (P & ~P) 3,4 &I2 6. ~~P 4,5 ~I1,2 7. ~Q 1,6 vE

(~P v ~Q) ~(P & Q)├2 8. Q 2 &E1,2 9. (Q & ~Q) 7,8 &I1 3. ~(P & Q) 2,9 ~I4 4. ~P A (for ~I)2,4 5. (P & ~P) 3,4 &I2 6. ~~P 4,5 ~I1,2 7. ~Q 1,6 vE

Negation Elimination

~E (Negation Elimination) If you have assumed ~ψ, and you have derived (φ&~φ), then you can write down ψ, depending on everything (φ&~φ) depends on except ~ψ.

(P v P) P├1 1. (P v P) A2 2. ~P A (for ~E)1,2 3. P 1,2 vE1,2 4. (P & ~P) 2,3 &I1 5. P 2,4 ~E