Midterm Discussion.

Midterm Discussion

Multiple choice

Problem #1 1. Suppose I notice that two students have turned in midterms with exactly the same answers on the multiple choice section. I know this: If the two students cheated, the probability that their answers would be the same is 100%. If the two students did not cheat, then the probability that their answers would be the same is (1/4) x 10 = 1/40 = 2.5%.

What is the probability that the two students cheated. a. 100%. b. 97
What is the probability that the two students cheated? a. 100%. b. 97.5% c. 2.5% d. It is impossible to tell, given the information in the question.

Base Rate Neglect This question is about the base rate neglect fallacy. We know IF the students cheated THEN we will see these results 100% of the time. But we don’t know the probability IF we see these results, THEN the students will be cheaters.

Question #2 2. Which of the following is most likely to happen? a. There will not be a final exam in this class. b. There will not be a final exam in this class, because the instructor has to leave the country. c. Lingnan University closes and there will not be a final exam in this class. d. There is not enough information to answer this question.

Conjunction Fallacy This question is about the conjunction fallacy. A is always more probable than (A & B). So it’s more probable that we don’t have a final than that we don’t have a final AND I leave the country or that we don’t have a final AND Lingnan closes.

Question #3 3. There are four cards in front of you. Each card has two things written on it: on one side is a letter, and on the other side is a number. You can only see one side of each card. On the sides you can see, you see: “E” is on one card, “F” is on another, “2” is on another, and “5” is on the last one. E F 2 5

I tell you that every card with a vowel on it (A, E, I, O, U) has an even number on the other side (2, 4, 6, 8). Which cards do you need to turn over to know whether I am telling you the truth? a. The card with E on it and the card with 2 on it. b. The card with F on it and the card with 2 on it. c. The card with E on it and the card with 5 on it. d. The card with F on it and the card with 5 on it.

You Have to Turn over “E”

You Have to Turn over “E”
3

It Doesn’t Matter What’s on “F”

3

4

It Doesn’t Matter What’s on “2”

It Doesn’t Matter What’s on “2”
B

You Have to Turn over “5” 5

You Have to Turn over “5” 5 A

Question #4 4. Suppose I am at the casino playing roulette. Here are the rules: There is a ball that can only land on black or red. There is a 50% chance it will land black and 50% chance it will land red. I can only bet “black” or “red” and I cannot bet both. If I bet “black” and the ball lands on black, I win. If I bet “red” and the ball lands on red, I win.

Important! 4. Suppose I am at the casino playing roulette. Here are the rules: There is a ball that can only land on black or red. There is a 50% chance it will land black and 50% chance it will land red. I can only bet “black” or “red” and I cannot bet both. If I bet “black” and the ball lands on black, I win. If I bet “red” and the ball lands on red, I win.

Suppose that I win 100 times in a row, as follows: for 100 times, I bet “black” and the ball lands on black each of those 100 times. Which of the following should I expect in the future? a. The ball will keep landing black. b. The ball will regress to the mean, and land black more than red. c. The ball will land about half black and half red. d. There is not enough information to answer this question.

Unimportant! Suppose that I win 100 times in a row, as follows: for 100 times, I bet “black” and the ball lands on black each of those 100 times. Which of the following should I expect in the future? a. The ball will keep landing black. b. The ball will regress to the mean, and land black more than red. c. The ball will land about half black and half red. d. There is not enough information to answer this question.

Correct Answer Suppose that I win 100 times in a row, as follows: for 100 times, I bet “black” and the ball lands on black each of those 100 times. Which of the following should I expect in the future? a. The ball will keep landing black. b. The ball will regress to the mean, and land black more than red. c. The ball will land about half black and half red. d. There is not enough information to answer this question.

Question #5 5. Which of the following is true? a. Sound arguments can have false conclusions. b. Valid arguments must have true premises. c. Valid arguments must be sound. d. Sound arguments must be valid.

Truth tables

Problem #6 P ~ (~ & (P → P) T F

Problem #7 P Q ~ (~ v Q) T F

Problem #8 P Q ~ (P ↔ Q) T F

Problem #9 P Q ((P → Q) ↔ (Q Q)) T F

Problem #10 P R S (P → (R S)) T F

TRUTH TABLE TEST

Problem #11 11. Premise: ~(~P v Q) Conclusion: ~(P ↔ Q) Valid? Y N

Step 1: Write Down Possibilities
Q T F

Step 2: Write Down Premise
Q ~(~P v Q) T F

Step 3: Copy Truth-Table for Premise
Q ~ (~ v Q) T F

Step 3: Copy Truth-Table for Premise
Q ~(~P v Q) T F

Step 4: Write Down Conclusion
Q ~(~P v Q) ~(P ↔ Q) T F

Step 5: Copy Truth-Table for Conclusion
Q ~ (P ↔ Q) T F

Step 5: Copy Truth-Table for Conclusion
Q ~(~P v Q) ~(P ↔ Q) T F

Step 6: Look at Lines where Premises are All True
Q ~(~P v Q) ~(P ↔ Q) T F* F

Step 7: Is Conclusion True on Those Lines?
Q ~(~P v Q) ~(P ↔ Q) T F* F T: Valid

Problem #12 12. Premise: ~(P ↔ Q) Conclusion: ~(~P v Q) Valid? Y N

Write Down Possibilities
Q T F

Write Down Premises P Q ~(P ↔ Q) T F

Write Down Conclusion P Q ~(P ↔ Q) ~(~P v Q) T F

Look at Lines Where Premises are True
Q ~(P ↔ Q) ~(~P v Q) T F* F

Is Conclusion True? P Q ~(P ↔ Q) ~(~P v Q) T F* F F: Invalid

Problem #13 13. Premise: ~(~P v Q) Conclusion: ((P → Q) ↔ (Q ↔ Q)) Valid? Y N

Problem #13 P Q ~(~P v Q) ((P → Q) ↔ (Q ↔ Q)) T F* F F: Invalid

Problem #14 14. Premise: ((P → Q) ↔ (Q ↔ Q)) Conclusion: ~(P ↔ Q) Valid? Y N

Problem #14 P Q ((P → Q) ↔ (Q ↔ Q)) ~(P ↔ Q) T F: Invalid F F*

Problem #15 15. Premise: ~(~P & (P → ~~P)) Conclusion: ((P → Q) ↔ (Q ↔ Q)) Valid? Y N

Problem #15 P Q ~(~P&(P→~~P)) ((P→Q)↔(Q↔Q)) T F F: Invalid F*

translations

Translations A: Albert is angry. B: Becky is beautiful.
C: Carl is considerate. D: Diana is disgusting.

16. “Alvin is angry only if Diana is disgusting.”
(A → D)

17. “Although Becky is beautiful, Diana is disgusting.”
(B & D)

18. “Either Carl is considerate or Albert is angry, but not both.”
((C v A) & ~(C & A))

19. “Carl is considerate if Becky is beautiful.”
(B → C)

20. “Albert is angry unless Diana is not disgusting.”
(A v ~D) (D → A) (D ↔ A)

derivations

The Rules In SL derivations, we have the rule of Assumption (A) and an introduction and elimination rule for each of the connectives (&I &E, →I, →E, vI, vE, ↔I, ↔E, ~I, ~E). You can’t do logic (correctly) without memorizing the rules.

The Rules While they have to be memorized, the rules do make sense. So you can at least be aware of how they DON’T work, even if you’ve forgotten them.

Here’s a BAD application of the rules: 2 5. Q 2 &E 2 6. ~Q 5 ~I WRONG!

How could that possibly be a correct logical step
How could that possibly be a correct logical step? Logic is about valid arguments. If the premises are true, then the conclusion is true. Is “Q therefore ~Q” a valid argument? No. When Q is true, ~Q is false. So this is not how ~I works.

Here’s another BAD application of the rules: 2 5. (P v Q) 2 vI 2 6
Here’s another BAD application of the rules: 2 5. (P v Q) 2 vI 2 6. Q 5 vE WRONG!

Invalid P Q T F

Invalid P Q Premise T F

Invalid P Q (P v Q) T F

Invalid P Q (P v Q) Conclusion T F

Invalid P Q (P v Q) T F

Invalid P Q (P v Q) T F F*

Valid Now let’s look at how vE really works (P v Q) 2 vI 6 6. ~P A 2 7. Q 5,6 vE RIGHT!

Valid P Q T F

Valid P Q (P v Q) ~P T F

Valid P Q (P v Q) ~P T F* F

Comparison This is a bad argument: PREMISE: The meeting is either on Tuesday or Thursday. CONCLUSION: The meeting is on Tuesday. This is a good argument: PREMISE: It’s not on Thursday. CONCLUSION: Therefore, it’s on Tuesday.

Problem #21 Problem #21 may look very complex and difficult. ((S v R) & (P & (Q → R))) ├ (((Q → R) & (S v R)) & P) But the proof is simpler than you might think. Let’s look at a similar problem.

(P & (Q & R)) ├ ((P & Q) & R) 1 1. (P & (Q & R)) A 1 2. P 1 &E 1 3. (Q & R) 1 &E 1 4. Q 3 &E 1 5. R 3 &E 1 6. (P & Q) 2,4 &I 1 7. ((P & Q) & R) 6,5 &I

(P & (Q & R)) ├ ((P & Q) & R) 1 1. (P & (Q & R)) A First Step ALWAYS: Write down your assumptions.

(P & (Q & R)) ├ ((P & Q) & R) 1 1. (P & (Q & R)) A First Step ALWAYS: Then copy the line number for each assumption out to the left, where we keep track of our dependencies. Why?

The Rule of Assumption: A
Because that’s what the rule says: A: at any stage in the derivation, we may write down any WFF we want to. That WFF only depends on itself.

(P & (Q & R)) ├ ((P & Q) & R) 1 1. (P & (Q & R)) A Second step: What are we trying to prove? 1 ?. ((P & Q) & R) ???

(P & (Q & R)) ├ ((P & Q) & R) 1 1. (P & (Q & R)) A Second step: What does it need to depend on? 1 ?. ((P & Q) & R) ???

(P & (Q & R)) ├ ((P & Q) & R) 1 1. (P & (Q & R)) A Third step: How do we get there? 1 ?. ((P & Q) & R) ???

(P & (Q & R)) ├ ((P & Q) & R) 1 1. (P & (Q & R)) A Third step: How do we get there? Our goal is a conjunction. 1 ?. ((P & Q) & R) ???

&I: &-Introduction Our usual way of proving a conjunction is with &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.

&I: &-Introduction Our usual way of proving a conjunction is with &I: If on some line you have proven (P & Q) and on some other line you have proven R, then on any future line you may write ((P & Q) & R), Depending on what (P & Q) and R depended on.

So We Need This Our usual way of proving a conjunction is with &I: If on some line you have proven (P & Q) and on some other line you have proven R, then on any future line you may write ((P & Q) & R), Depending on what (P & Q) and R depended on.

And We Need This Our usual way of proving a conjunction is with &I: If on some line you have proven (P & Q) and on some other line you have proven R, then on any future line you may write ((P & Q) & R), Depending on what (P & Q) and R depended on.

(P & (Q & R)) ├ ((P & Q) & R) 1 1. (P & (Q & R)) A So far, so good. 1 ?. R ??? 1 ?. (P & Q) ??? 1 ?. ((P & Q) & R) &I

(P & (Q & R)) ├ ((P & Q) & R) 1 1. (P & (Q & R)) A We know that these lines can only depend on 1. 1 ?. R ??? 1 ?. (P & Q) ??? 1 ?. ((P & Q) & R) &I

Why? Because that’s the only way the conclusion can only depend on 1: If on some line you have proven (P & Q) and on some other line you have proven R, then on any future line you may write ((P & Q) & R), Depending on what (P & Q) and R depended on.

(P & (Q & R)) ├ ((P & Q) & R) 1 1. (P & (Q & R)) A Now we see that we need another conjunction. 1 ?. R ??? 1 ?. (P & Q) ??? 1 ?. ((P & Q) & R) &I

(P & (Q & R)) ├ ((P & Q) & R) 1 1. (P & (Q & R)) A Where can we get these from? 1 ?. Q ??? 1 ?. P ??? 1 ?. R ??? 1 ?. (P & Q) &I 1 ?. ((P & Q) & R) &I

&-Elimination: &E By using 1 and &E If we have proved (φ & ψ) on some line, then on any line we may write down φ, and on any line we may write down ψ. The result depends on everything (φ & ψ) depended on.

&-Elimination: &E By using 1 and &E If we have proved (P & (Q & R)) on some line, then on any line we may write down P, and on any line we may write down (Q & R) The result depends on everything (P & (Q & R)) depended on.

(P & (Q & R)) ├ ((P & Q) & R) 1 1. (P & (Q & R)) A Use &E on line 1 to get P 1 ?. Q ??? 1 ?. P 1 &E 1 ?. R ??? 1 ?. (P & Q) &I 1 ?. ((P & Q) & R) &I

(P & (Q & R)) ├ ((P & Q) & R) 1 1. (P & (Q & R)) A Now where do we get Q and R? 1 ?. Q ??? 1 ?. P 1 &E 1 ?. R ??? 1 ?. (P & Q) &I 1 ?. ((P & Q) & R) &I

(P & (Q & R)) ├ ((P & Q) & R) 1 1. (P & (Q & R)) A 1 2. (Q & R) 1 &E 1 ?. Q ??? 1 ?. P 1 &E 1 ?. R ??? 1 ?. (P & Q) &I 1 ?. ((P & Q) & R) &I

(P & (Q & R)) ├ ((P & Q) & R) 1 1. (P & (Q & R)) A 1 2. (Q & R) 1 &E 1 3. Q 2 &E 1 ?. P 1 &E 1 ?. R ??? 1 ?. (P & Q) &I 1 ?. ((P & Q) & R) &I

(P & (Q & R)) ├ ((P & Q) & R) 1 1. (P & (Q & R)) A 1 2. (Q & R) 1 &E 1 3. Q 2 &E 1 4. P 1 &E 1 5. R 2 &E 1 ?. (P & Q) &I 1 ?. ((P & Q) & R) &I

(P & (Q & R)) ├ ((P & Q) & R) 1 1. (P & (Q & R)) A 1 2. (Q & R) 1 &E 1 3. Q 2 &E 1 4. P 1 &E 1 5. R 2 &E 1 6. (P & Q) 4,3 &I 1 7. ((P & Q) & R) 6,5 &I

Problem #21 Now let’s talk about a question that was actually on the midterm: ((S v R) & (P&(Q → R))) ├ (((Q → R)&(S v R)) & P) This should look very similar!

Problem #21 Now let’s talk about a question that was actually on the midterm: (φ & (P & (Q → R))) ├ (((Q → R) & φ) & P) This should look very similar!

Problem #21 Now let’s talk about a question that was actually on the midterm: (φ & (ψ & (Q → R))) ├ (((Q → R) & φ) & ψ) This should look very similar!

Problem #21 Now let’s talk about a question that was actually on the midterm: (φ & (ψ & χ)) ├ ((χ & φ) & ψ) This should look very similar!

((S v R) & (P & (Q → R))) ├ (((Q → R) & (S v R)) & P)
1 1. ((S v R) & (P & (Q → R))) A First step ALWAYS: write down your assumptions.

((S v R) & (P & (Q → R))) ├ (((Q → R) & (S v R)) & P)
1 1. ((S v R) & (P & (Q → R))) A First step ALWAYS: then copy the line number out to the left, where we keep track of dependencies. Why?

((S v R) & (P & (Q → R))) ├ (((Q → R) & (S v R)) & P)
1 1. ((S v R) & (P & (Q → R))) A 1 2. (S v R) 1 &E Use &E to get (S v R).

((S v R) & (P & (Q → R))) ├ (((Q → R) & (S v R)) & P)
1 1. ((S v R) & (P & (Q → R))) A 1 2. (S v R) 1 &E 1 3. (P & (Q → R)) 1 &E Now use &E to get (P & (Q → R))

((S v R) & (P & (Q → R))) ├ (((Q → R) & (S v R)) & P)
1 1. ((S v R) & (P & (Q → R))) A 1 2. (S v R) 1 &E 1 3. (P & (Q → R)) 1 &E 1 4. P 3 &E Use &E again to get P.

((S v R) & (P & (Q → R))) ├ (((Q → R) & (S v R)) & P)
1 1. ((S v R) & (P & (Q → R))) A 1 2. (S v R) 1 &E 1 3. (P & (Q → R)) 1 &E 1 4. P 3 &E 1 5. (Q → R) 3 &E Use &E a fourth time to get (Q → R)

((S v R) & (P & (Q → R))) ├ (((Q → R) & (S v R)) & P)
1 1. ((S v R) & (P & (Q → R))) A 1 2. (S v R) 1 &E 1 3. (P & (Q → R)) 1 &E 1 4. P 3 &E 1 5. (Q → R) 3 &E Now use &I.

((S v R) & (P & (Q → R))) ├ (((Q → R) & (S v R)) & P)
1 1. ((S v R) & (P & (Q → R))) A 1 2. (S v R) 1 &E 1 3. (P & (Q → R)) 1 &E 1 4. P 3 &E 1 5. (Q → R) 3 &E Now use &I. This is what you want.

((S v R) & (P & (Q → R))) ├ (((Q → R) & (S v R)) & P)
1 1. ((S v R) & (P & (Q → R))) A 1 2. (S v R) 1 &E 1 3. (P & (Q → R)) 1 &E 1 4. P 3 &E 1 5. (Q → R) 3 &E Now use &I. The first part is here.

((S v R) & (P & (Q → R))) ├ (((Q → R) & (S v R)) & P)
1 1. ((S v R) & (P & (Q → R))) A 1 2. (S v R) 1 &E 1 3. (P & (Q → R)) 1 &E 1 4. P 3 &E 1 5. (Q → R) 3 &E Now use &I. The second part is here.

((S v R) & (P & (Q → R))) ├ (((Q → R) & (S v R)) & P)
1 1. ((S v R) & (P & (Q → R))) A 1 2. (S v R) 1 &E 1 3. (P & (Q → R)) 1 &E 1 4. P 3 &E 1 5. (Q → R) 3 &E 1 6. ((Q → R) & (S v R)) 5,2 &I Now use &I. Put the two together.

((S v R) & (P & (Q → R))) ├ (((Q → R) & (S v R)) & P)
1 1. ((S v R) & (P & (Q → R))) A 1 2. (S v R) 1 &E 1 3. (P & (Q → R)) 1 &E 1 4. P 3 &E 1 5. (Q → R) 3 &E 1 6. ((Q → R) & (S v R)) 5,2 &I One last &I and we’re done.

((S v R) & (P & (Q → R))) ├ (((Q → R) & (S v R)) & P)
1 1. ((S v R) & (P & (Q → R))) A 1 2. (S v R) 1 &E 1 3. (P & (Q → R)) 1 &E 1 4. P 3 &E 1 5. (Q → R) 3 &E 1 6. ((Q → R) & (S v R)) 5,2 &I This is what we want.

((S v R) & (P & (Q → R))) ├ (((Q → R) & (S v R)) & P)
1 1. ((S v R) & (P & (Q → R))) A 1 2. (S v R) 1 &E 1 3. (P & (Q → R)) 1 &E 1 4. P 3 &E 1 5. (Q → R) 3 &E 1 6. ((Q → R) & (S v R)) 5,2 &I The first part is here.

((S v R) & (P & (Q → R))) ├ (((Q → R) & (S v R)) & P)
1 1. ((S v R) & (P & (Q → R))) A 1 2. (S v R) 1 &E 1 3. (P & (Q → R)) 1 &E 1 4. P 3 &E 1 5. (Q → R) 3 &E 1 6. ((Q → R) & (S v R)) 5,2 &I The second part is here.

((S v R) & (P & (Q → R))) ├ (((Q → R) & (S v R)) & P)
1 1. ((S v R) & (P & (Q → R))) A 1 2. (S v R) 1 &E 1 3. (P & (Q → R)) 1 &E 1 4. P 3 &E 1 5. (Q → R) 3 &E 1 6. ((Q → R) & (S v R)) 5,2 &I 1 7. (((Q → R) & (S v R)) & P) 6,4 &I

IMPORTANT FINAL STEP. We are only done with the proof if the result we want (((Q → R) & (S v R)) & P) depends only on the assumptions to the left of ├. To the left of (((Q → R) & (S v R)) & P) is just line 1. Is that the assumption we want?

((S v R) & (P & (Q → R))) ├ (((Q → R) & (S v R)) & P)
1 1. ((S v R) & (P & (Q → R))) A 1 2. (S v R) 1 &E 1 3. (P & (Q → R)) 1 &E 1 4. P 3 &E 1 5. (Q → R) 3 &E 1 6. ((Q → R) & (S v R)) 5,2 &I 1 7. (((Q → R) & (S v R)) & P) 6,4 &I YES!

Problem #22 In Problem #22, I asked you to derive this fact: (P ↔ R), (R ↔ Q), Q├ P

(P ↔ R), (R ↔ Q), Q├ P 1 1. (P ↔ R) A 2 2. (R ↔ Q) A 3 3. Q A First Step ALWAYS: Write down your assumptions.

(P ↔ R), (R ↔ Q), Q├ P 1 1. (P ↔ R) A 2 2. (R ↔ Q) A 3 3. Q A First Step ALWAYS: Then copy the line number to the dependencies on the left.

(P ↔ R), (R ↔ Q), Q├ P 1 1. (P ↔ R) A 2 2. (R ↔ Q) A 3 3. Q A Why?

(P ↔ R), (R ↔ Q), Q├ P 1 1. (P ↔ R) A 2 2. (R ↔ Q) A 3 3. Q A Second Step: What are we trying to prove? How can we use what we have to get there?

(P ↔ R), (R ↔ Q), Q├ P 1 1. (P ↔ R) A 2 2. (R ↔ Q) A 3 3. Q A We are trying to prove P. We have P on line 1, but it’s “stuck” in a biconditional. So we need to use ↔E to get it out.

↔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.

↔E If on some line you have proved (P ↔ R) Then on any future line you may write ((P → R) & (R → P)) Depending on what (P ↔ R) depended on.

↔E If on some line you have proved (R ↔ Q) Then on any future line you may write ((R → Q) & (Q → R)) Depending on what (R ↔ Q) depended on.

(P ↔ R), (R ↔ Q), Q├ P 1 1. (P ↔ R) A 2 2. (R ↔ Q) A 3 3. Q A 1 4. ((P → R) & (R → P)) 1 ↔E 2 5. ((R → Q) & (Q → R)) 2 ↔E We can use ↔E twice, for line 1 and line 2.

(P ↔ R), (R ↔ Q), Q├ P 1 1. (P ↔ R) A 2 2. (R ↔ Q) A 3 3. Q A 1 4. ((P → R) & (R → P)) 1 ↔E 2 5. ((R → Q) & (Q → R)) 2 ↔E We copy the dependencies from the lines we used.

(P ↔ R), (R ↔ Q), Q├ P 1 1. (P ↔ R) A 2 2. (R ↔ Q) A 3 3. Q A 1 4. ((P → R) & (R → P)) 1 ↔E 2 5. ((R → Q) & (Q → R)) 2 ↔E Why?

↔E Because that’s what the rule says. 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 ↔ R), (R ↔ Q), Q├ P 1 1. (P ↔ R) A 2 2. (R ↔ Q) A 3 3. Q A 1 4. ((P → R) & (R → P)) 1 ↔E 2 5. ((R → Q) & (Q → R)) 2 ↔E Now we have P, but it is stuck inside & and →.

&E To get P out of the conjunction, we need to use &E: 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.

That means we can do one of two things: use line 4 to get (P → R) or use line 4 to get (R → P).

(P ↔ R), (R ↔ Q), Q├ P 1 1. (P ↔ R) A 2 2. (R ↔ Q) A 3 3. Q A 1 4. ((P → R) & (R → P)) 1 ↔E 2 5. ((R → Q) & (Q → R)) 2 ↔E 1 6. (P → R) 4 &E Like this.

(P ↔ R), (R ↔ Q), Q├ P 1 1. (P ↔ R) A 2 2. (R ↔ Q) A 3 3. Q A 1 4. ((P → R) & (R → P)) 1 ↔E 2 5. ((R → Q) & (Q → R)) 2 ↔E 1 6. (R → P) 4 &E Or like this.

But which step do we want to take. Whatever one will get us P
But which step do we want to take? Whatever one will get us P. So now we need to remember how →E works.

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.

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

(P ↔ R), (R ↔ Q), Q├ P 1 1. (P ↔ R) A 2 2. (R ↔ Q) A 3 3. Q A 1 4. ((P → R) & (R → P)) 1 ↔E 2 5. ((R → Q) & (Q → R)) 2 ↔E 1 6. (R → P) 4 &E That means we want this.

(P ↔ R), (R ↔ Q), Q├ P 1 1. (P ↔ R) A 2 2. (R ↔ Q) A 3 3. Q A 1 4. ((P → R) & (R → P)) 1 ↔E 2 5. ((R → Q) & (Q → R)) 2 ↔E 1 6. (R → P) 4 &E We almost have P now.

Arrow Elimination: →E To get P we have to have R first: if on one line we have a conditional (R → P) and on another line we have the antecedent of the conditional R then on any future line, we may write down the consequent of the conditional P depending on everything (R → P) and R depended on.

(P ↔ R), (R ↔ Q), Q├ P 1 1. (P ↔ R) A 2 2. (R ↔ Q) A 3 3. Q A 1 4. ((P → R) & (R → P)) 1 ↔E 2 5. ((R → Q) & (Q → R)) 2 ↔E 1 6. (R → P) 4 &E R is trapped.

Arrow Elimination: →E But there’s a way to get R! if on one line we have a conditional (Q → R) and on another line we have the antecedent of the conditional Q then on any future line, we may write down the consequent of the conditional R depending on everything (Q → R) and Q depended on.

(P ↔ R), (R ↔ Q), Q├ P 1 1. (P ↔ R) A 2 2. (R ↔ Q) A 3 3. Q A 1 4. ((P → R) & (R → P)) 1 ↔E 2 5. ((R → Q) & (Q → R)) 2 ↔E 1 6. (R → P) 4 &E 2 7. (Q → R) 5 &E

Copy the Dependencies 1 1. (P ↔ R) A 2 2. (R ↔ Q) A 3 3. Q A 1 4. ((P → R) & (R → P)) 1 ↔E 2 5. ((R → Q) & (Q → R)) 2 ↔E 1 6. (R → P) 4 &E 2 7. (Q → R) 5 &E

Why? Why? Because that’s what the rule says. 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.

Now the proof is easy. Since we have (Q → R) on line 7 and Q on line 3 we can use →E to get R. Then we can use R with (R → P) to get P. And then we’re done!

(P ↔ R), (R ↔ Q), Q├ P 3 3. Q A 1 4. ((P → R) & (R → P)) 1 ↔E 2 5. ((R → Q) & (Q → R)) 2 ↔E 1 6. (R → P) 4 &E 2 7. (Q → R) 5 &E 2,3 8. R 3,7 →E 1,2,3 9. P 6,7 →E

Copy the Dependencies 3 3. Q A 1 4. ((P → R) & (R → P)) 1 ↔E 2 5. ((R → Q) & (Q → R)) 2 ↔E 1 6. (R → P) 4 &E 2 7. (Q → R) 5 &E 2,3 8. R 3,7 →E 1,2,3 9. P 6,7 →E

Arrow Elimination: →E Why? Because that’s what the rule says. 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.

IMPORTANT FINAL STEP. We are only done with the proof if the result we want (P) depends only on the assumptions to the left of ├. To the left of P are lines 1, 2, and 3. Are those lines the assumptions we want?

(P ↔ R), (R ↔ Q), Q├ P 1 1. (P ↔ R) A 2 2. (R ↔ Q) A 3 3. Q A Yes!

Problem #23 Problem #23 was tougher, but it was definitely solvable if you follow our strategy for proving conditional statements: Assume the antecedent φ Prove the consequent ψ Use →I to conclude (φ → ψ)

P ├ ((R v S) → (~S → (P & R)))
1 1. P A First step ALWAYS: write down your assumptions.

P ├ ((R v S) → (~S → (P & R)))
1 1. P A First step ALWAYS: then copy the line number out to the dependencies.

P ├ ((R v S) → (~S → (P & R)))
1 1. P A Second step: What are we trying to prove?

P ├ ((R v S) → (~S → (P & R)))
1 1. P A Second step: What are we trying to prove? A conditional!

P ├ ((R v S) → (~S → (P & R)))
1 1. P A Strategy for proving a conditional: Assume the antecedent φ Prove the consequent ψ Use →I to conclude (φ → ψ)

P ├ ((R v S) → (~S → (P & R)))
1 1. P A 2 2. (R v S) A (for →I) Strategy for proving a conditional: Assume the antecedent φ Prove the consequent ψ Use →I to conclude (φ → ψ)

P ├ ((R v S) → (~S → (P & R)))
1 1. P A 2 2. (R v S) A (for →I) The consequent is also a conditional! What do we do? Keep using our strategy.

P ├ ((R v S) → (~S → (P & R)))
1 1. P A 2 2. (R v S) A (for →I) 3 3. ~S A (for →I) Strategy for proving a conditional: Assume the antecedent φ Prove the consequent ψ Use →I to conclude (φ → ψ)

P ├ ((R v S) → (~S → (P & R)))
1 1. P A 2 2. (R v S) A (for →I) 3 3. ~S A (for →I) GOAL: ??? ?. (P & R) ???

P ├ ((R v S) → (~S → (P & R)))
1 1. P A 2 2. (R v S) A (for →I) 3 3. ~S A (for →I) We have P and R, but R is trapped. Need vE. ??? ?. (P & R) ???

vE vE: If you have proved (φ v ψ) And you have proved ~φ Then you may write ψ Depending on what (φ v ψ), ~φ depended on. If you have proved (φ v ψ) And you have proved ~ψ Then you may write φ Depending on what (φ v ψ), ~ψ depended on.

vE vE: If you have proved (R v S) And you have proved ~R Then you may write S Depending on what (R v S), ~R depended on. If you have proved (R v S) And you have proved ~S Then you may write R Depending on what (R v S), ~S depended on.

P ├ ((R v S) → (~S → (P & R)))
1 1. P A 2 2. (R v S) A (for →I) 3 3. ~S A (for →I) GOAL: ??? ?. (P & R) ???

P ├ ((R v S) → (~S → (P & R)))
1 1. P A 2 2. (R v S) A (for →I) 3 3. ~S A (for →I) 2,3 4. R 2,3 vE ??? ?. (P & R) ???

P ├ ((R v S) → (~S → (P & R)))
1 1. P A 2 2. (R v S) A (for →I) 3 3. ~S A (for →I) 2,3 4. R 2,3 vE 1,2,3 5. (P & R) 1,4 &I

Strategy for Proving Conditionals
Assume the antecedent φ Prove the consequent ψ Use →I to conclude (φ→ψ)

P ├ ((R v S) → (~S → (P & R)))
1 1. P A 2 2. (R v S) A (for →I) 3 3. ~S A (for →I) 2,3 4. R 2,3 vE 1,2,3 5. (P & R) 1,4 &I 1,2 6. (~S → (P & R)) 3,5 →I STEP 1

P ├ ((R v S) → (~S → (P & R)))

→I Figuring out the dependencies can be tough: 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 on Except φ.

→I Figuring out the dependencies can be tough: If on some line you have assumed ~S, And on some other line you have proved (P & R), And (P & R) depends on your assumption ~S, Then on any future line you may write (~S → (P & R)), Depending on everything (P & R) depended on Except ~S.

P ├ ((R v S) → (~S → (P & R)))
1 1. P A 2 2. (R v S) A (for →I) 3 3. ~S A (for →I) 2,3 4. R 2,3 vE 1,2,3 5. (P & R) 1,4 &I 1,2 6. (~S → (P & R)) 3,5 →I EVERYTHING (P & R) DEPENDS ON

P ├ ((R v S) → (~S → (P & R)))
1 1. P A 2 2. (R v S) A (for →I) 3 3. ~S A (for →I) 2,3 4. R 2,3 vE 1,2,3 5. (P & R) 1,4 &I 1,2 6. (~S → (P & R)) 3,5 →I EXCEPT ~S

P ├ ((R v S) → (~S → (P & R)))
1 1. P A 2 2. (R v S) A (for →I) 3 3. ~S A (for →I) 2,3 4. R 2,3 vE 1,2,3 5. (P & R) 1,4 &I 1,2 6. (~S → (P & R)) 3,5 →I Now we return to our strategy.

Strategy for Proving Conditionals
Assume the antecedent φ Prove the consequent ψ Use →I to conclude (φ→ψ)

P ├ ((R v S) → (~S → (P & R)))
1 1. P A 2 2. (R v S) A (for →I) 3 3. ~S A (for →I) 2,3 4. R 2,3 vE 1,2,3 5. (P & R) 1,4 &I 1,2 6. (~S → (P & R)) 3,5 →I STEP 1: Assume the antecedent!

P ├ ((R v S) → (~S → (P & R)))
1 1. P A 2 2. (R v S) A (for →I) 3 3. ~S A (for →I) 2,3 4. R 2,3 vE 1,2,3 5. (P & R) 1,4 &I 1,2 6. (~S → (P & R)) 3,5 →I STEP 2: Prove the consequent.

P ├ ((R v S) → (~S → (P & R)))
1 1. P A 2 2. (R v S) A (for →I) 3 3. ~S A (for →I) 2,3 4. R 2,3 vE 1,2,3 5. (P & R) 1,4 &I 1,2 6. (~S → (P & R)) 3,5 →I STEP 3: Use →I.

P ├ ((R v S) → (~S → (P & R)))
1 1. P A 2 2. (R v S) A (for →I) 3 3. ~S A (for →I) 2,3 4. R 2,3 vE 1,2,3 5. (P & R) 1,4 &I 1,2 6. (~S → (P & R)) 3,5 →I 1 7. ((R v S) → (~S → (P & R))) 2,6 →I

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