More Proofs
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Transcript of More Proofs
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More Proofs
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REVIEW
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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.
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&-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.
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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.
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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.”
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(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.
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(P → (P → Q)), P Q├
1 1. (P → (P → Q)) A2 2. P A
A second application of assumption.
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(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.
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(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.
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(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.
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(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.
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(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.
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(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.
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MORE RULES
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&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.
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(P & (Q & R)) ((P & Q) & R)├1 1. (P & (Q & R)) A
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(P & (Q & R)) ((P & Q) & R)├1 1. (P & (Q & R)) A1 2. P 1 &E
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(P & (Q & R)) ((P & Q) & R)├1 1. (P & (Q & R)) A1 2. P 1 &E1 3. (Q & R) 1 &E
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(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
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(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
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(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
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(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
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(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
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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.
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P (P & P)├1 1. P A
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P (P & P)├1 1. P A1 2. (P & P) 1,1 &I
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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.
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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.
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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)
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(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
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Zero Dependencies
→I, because it allows you to remove dependencies, allows you to prove formulas with no dependencies. These are called “theorems.”
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├ ((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
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├ ((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
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├ (P → P)
1 1. P A (for →I)2. (P → P) 1,1 →I
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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.
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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.
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(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)
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(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
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(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
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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.
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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.
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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!
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(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
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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
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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.
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((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
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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.
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P ~~P├1 1. P A2 2. ~P A (for ~I)1,2 3. (P & ~P) 1,2 &I1 4. ~~P 2,3 ~I
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(~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
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(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
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(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
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(~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
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(~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
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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.
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(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
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