forall x: msu edition. solutions to selected exercisesloighic.net/logic/forallx-solutions.pdf · ii...

forallxTHE MISSISSIPPI STATE EDITION

solutions to selected exercises

P. D. MAGNUS

T IM BUTTON

with additions byJ. ROBERT LOFTIS

AARON THOMAS-BOLDUC

R ICHARD ZACH

revised byGREGORY JOHNSON

Fall 2019

ii

This booklet is based on the solutions booklet forallx: Cambridge, by

Tim ButtonUniversity of Cambridge

used under a CC BY 4.0 license, which is based in turn on forallx, by

P.D. MagnusUniversity at Albany, State University of New York

used under a CC BY 4.0 license,which was remixed & expanded by

Aaron Thomas-Bolduc & Richard ZachUniversity of Calgary

It includes additional material from forallx by P.D. Magnus, used undera CC BY 4.0 license, and from forallx: Lorain County Remix, by Cathal Woodsand J. Robert Loftis, used with permission.

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. You must give appropriate credit, provide a link to the license, and indicateif changes were made. You may do so in any reasonable manner, but not inany way that suggests the licensor endorses you or your use.

. You may not apply legal terms or technological measures that legally restrictothers from doing anything the license permits.

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Contents

1 Arguments 12 Valid arguments 23 Other logical notions 45 Connectives 86 Sentences of TFL 149 Complete truth tables 1511 Entailment & validity 2212 Truth table shortcuts 2913 Partial truth tables 3215 Basic rules for TFL 4016 Proof-theoretic concepts 4619 Additional rules for TFL 5420 Soundness and Completeness 57

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Chapter 1

Arguments

Highlight the phrase which expresses the conclusion of each of these arguments:

1. It is sunny. So I should take my sunglasses.2. It must have been sunny. I did wear my sunglasses, after all.3. No one but you has had their hands in the cookie-jar. And the scene of the

crime is littered with cookie-crumbs. You’re the culprit!4. Miss Scarlett and Professor Plum were in the study at the time of the murder.

And Reverend Green had the candlestick in the ballroom, and we knowthat there is no blood on his hands. Hence Colonel Mustard did it in thekitchen with the lead-piping. Recall, after all, that the gun had not beenfired.

1

Chapter 2

Valid arguments

A. Which of the following arguments is valid? Which is invalid?

1. Socrates is a man.2. All men are carrots..Û. Socrates is a carrot. Valid

1. Abe Lincoln was either born in Illinois or he was once president.2. Abe Lincoln was never president..Û. Abe Lincoln was born in Illinois. Valid

1. If I pull the trigger, Abe Lincoln will die.2. I do not pull the trigger..Û. Abe Lincoln will not die. Invalid

Abe Lincoln might die for some other reason: someone else might pull thetrigger; he might die of old age.

1. Abe Lincoln was either from France or from Luxemborg.2. Abe Lincoln was not from Luxemborg..Û. Abe Lincoln was from France. Valid

1. If the world were to end today, then I would not need to get up tomorrowmorning.

2. I will need to get up tomorrow morning..Û. The world will not end today. Valid

1. Joe is now 19 years old.2. Joe is now 87 years old.

2

CHAPTER 2. VALID ARGUMENTS 3

.Û. Bob is now 20 years old. ValidAn argument is valid if and only if it is impossible for all the premises tobe true and the conclusion false. It is impossible for all the premises tobe true; so it is certainly impossible that the premises are all true and theconclusion is false.

B. Could there be:

1. A valid argument that has one false premise and one true premise? Yes.Example: the first argument, above.

2. A valid argument that has only false premises? Yes.Example: Socrates is a frog, all frogs are excellent pianists, therefore Socratesis an excellent pianist.

3. A valid argument with only false premises and a false conclusion? Yes.The same example will su�ce.

4. An invalid argument that can be made valid by the addition of a newpremise? Yes.Plenty of examples, but let me o�er a more general observation. We canalways make an invalid argument valid, by adding a contradiction intothe premises. For an argument is valid if and only if it is impossible forall the premises to be true and the conclusion false. If the premises arecontradictory, then it is impossible for them all to be true (and the conclusionfalse).

5. A valid argument that can be made invalid by the addition of a new premise?No.An argument is valid if and only if it is impossible for all the premises to betrue and the conclusion false. Adding another premise will only make itharder for the premises all to be true together.

In each case: if so, give an example; if not, explain why not.

Chapter 3

Other logical notions

A. For each of the following: Is it necessarily true, necessarily false, or contingent?

1. Caesar crossed the Rubicon. Contingent2. Someone once crossed the Rubicon. Contingent3. No one has ever crossed the Rubicon. Contingent4. If Caesar crossed the Rubicon, then someone has. Necessarily true5. Even though Caesar crossed the Rubicon, no one has ever crossed the

Rubicon. Necessarily false6. If anyone has ever crossed the Rubicon, it was Caesar. Contingent

B. For each of the following: Is it a necessary truth, a necessary falsehood, orcontingent?

1. Elephants dissolve in water.2. Wood is a light, durable substance useful for building things.3. If wood were a good building material, it would be useful for building things.4. I live in a three story building that is two stories tall.5. If gerbils were mammals they would nurse their young.

C. Which of the following pairs of sentences are necessarily equivalent?

1. Elephants dissolve in water.If you put an elephant in water, it will disintegrate.

2. All mammals dissolve in water.If you put an elephant in water, it will disintegrate.

3. George Bush was the 43rd president.Barack Obama is the 44th president.

4. Barack Obama is the 44th president.Barack Obama was president immediately after the 43rd president.

4

CHAPTER 3. OTHER LOGICAL NOTIONS 5

5. Elephants dissolve in water.All mammals dissolve in water.

D. Which of the following pairs of sentences are necessarily equivalent?

1. Thelonious Monk played piano.John Coltrane played tenor sax.

2. Thelonious Monk played gigs with John Coltrane.John Coltrane played gigs with Thelonious Monk.

3. All professional piano players have big hands.Piano player Bud Powell had big hands.

4. Bud Powell su�ered from severe mental illness.All piano players su�er from severe mental illness.

5. John Coltrane was deeply religious.John Coltrane viewed music as an expression of spirituality.

E. Consider the following sentences:

G1 There are at least four gira�es at the wild animal park.

G2 There are exactly seven gorillas at the wild animal park.

G3 There are not more than two Martians at the wild animal park.

G4 Every gira�e at the wild animal park is a Martian.

Now consider each of the following collections of sentences. Which are jointlypossible? Which are jointly impossible?

1. Sentences G2, G3, and G4 Jointly possible2. Sentences G1, G3, and G4 Jointly impossible3. Sentences G1, G2, and G4 Jointly possible4. Sentences G1, G2, and G3 Jointly possible

F. Consider the following sentences.

M1 All people are mortal.

M2 Socrates is a person.

M3 Socrates will never die.

M4 Socrates is mortal.

Which combinations of sentences are jointly possible? Mark each “possible” or“impossible.”

1. Sentences M1, M2, and M3


2. Sentences M2, M3, and M43. Sentences M2 and M34. Sentences M1 and M45. Sentences M1, M2, M3, and M4

G. Which of the following is possible? If it is possible, give an example. If it isnot possible, explain why.

1. A valid argument that has one false premise and one true premiseYes: ‘All whales are mammals (true). All mammals are plants (false). So allwhales are plants.’

2. A valid argument that has a false conclusionYes. (See example from previous exercise.)

3. A valid argument, the conclusion of which is a necessary falsehoodYes: ‘1 + 1 = 3. So 1 + 2 = 4.’

4. An invalid argument, the conclusion of which is a necessary truthNo. If the conclusion is necessarily true, then there is no way to make itfalse, and hence no way to make it false whilst making all the premises true.

5. A necessary truth that is contingentNo. If a sentence is a necessary truth, it cannot possibly be false, but acontingent sentence can be false.

6. Two necessarily equivalent sentences, both of which are necessary truthsYes: ‘4 is even’, ‘4 is divisible by 2’.

7. Two necessarily equivalent sentences, one of which is a necessary truth andone of which is contingentNo. A necessary truth cannot possibly be false, while a contingent sentencecan be false. So in any situation in which the contingent sentence is false, itwill have a di�erent truth value from the necessary truth. Thus they willnot necessarily have the same truth value, and so will not be equivalent.

8. Two necessarily equivalent sentences that together are jointly impossibleYes: ‘1 + 1 = 4’ and ‘1 + 1 = 3’.

9. A jointly possible collection of sentences that contains a necessary falsehoodNo. If a sentence is necessarily false, there is no way to make it true, letalone it along with all the other sentences.

10. A jointly impossible set of sentences that contains a necessary truthYes: ‘1 + 1 = 4’ and ‘1 + 1 = 2’.

H. Which of the following is possible? If it is possible, give an example. If it isnot possible, explain why.

1. A valid argument, whose premises are all necessary truths, and whoseconclusion is contingent

2. A valid argument with true premises and a false conclusion


3. A jointly possible collection of sentences that contains two sentences thatare not necessarily equivalent

4. A jointly possible collection of sentences, all of which are contingent5. A false necessary truth6. A valid argument with false premises7. A necessarily equivalent pair of sentences that are not jointly possible8. A necessary truth that is also a necessary falsehood9. A jointly possible collection of sentences that are all necessary falsehoods

Chapter 5

Connectives

A. Using the symbolization key given, symbolize each English sentence in TFL.

M : Those creatures are men in suits.C : Those creatures are chimpanzees.G : Those creatures are gorillas.

1. Those creatures are not men in suits.¬M

2. Those creatures are men in suits, or they are not.(M ∨ ¬M )

3. Those creatures are either gorillas or chimpanzees.(G ∨C )

4. Those creatures are neither gorillas nor chimpanzees.¬(C ∨G )

5. If those creatures are chimpanzees, then they are neither gorillas nor menin suits.(C → ¬(G ∨M ))

6. Unless those creatures are men in suits, they are either chimpanzees or theyare gorillas.(M ∨ (C ∨G ))

B. Using the symbolization key given, symbolize each English sentence in TFL.

A: Mister Ace was murdered.B : The butler did it.C : The cook did it.D : The Duchess is lying.E : Mister Edge was murdered.F : The murder weapon was a frying pan.

8

CHAPTER 5. CONNECTIVES 9

1. Either Mister Ace or Mister Edge was murdered.(A ∨ E)

2. If Mister Ace was murdered, then the cook did it.(A → C )

3. If Mister Edge was murdered, then the cook did not do it.(E → ¬C )

4. Either the butler did it, or the Duchess is lying.(B ∨D)

5. The cook did it only if the Duchess is lying.(C → D)

6. If the murder weapon was a frying pan, then the culprit must have beenthe cook.(F → C )

7. If the murder weapon was not a frying pan, then the culprit was either thecook or the butler.(¬F → (C ∨ B))

8. Mister Ace was murdered if and only if Mister Edge was not murdered.(A ↔ ¬E)

9. The Duchess is lying, unless it was Mister Edge who was murdered.(D ∨ E)

10. If Mister Ace was murdered, he was done in with a frying pan.(A → F )

11. Since the cook did it, the butler did not.(C & ¬B)

12. Of course the Duchess is lying!D

C. Using the symbolization key given, symbolize each English sentence in TFL.

E1: Ava is an electrician.E2: Harrison is an electrician.F1: Ava is a firefighter.F2: Harrison is a firefighter.S1: Ava is satisfied with her career.S2: Harrison is satisfied with his career.

1. Ava and Harrison are both electricians.(E1 & E2)

2. If Ava is a firefighter, then she is satisfied with her career.(F1 → S1)

3. Ava is a firefighter, unless she is an electrician.(F1 ∨ E1)

4. Harrison is an unsatisfied electrician.


(E2 & ¬S2)5. Neither Ava nor Harrison is an electrician.

¬(E1 ∨ E2)

6. Both Ava and Harrison are electricians, but neither of them find it satisfying.((E1 & E2) & ¬(S1 ∨ S2))

7. Harrison is satisfied only if he is a firefighter.(S2 → F2)

8. If Ava is not an electrician, then neither is Harrison, but if she is, then heis too.((¬E1 → ¬E2) & (E1 → E2))

9. Ava is satisfied with her career if and only if Harrison is not satisfied withhis.(S1 ↔ ¬S2)

10. If Harrison is both an electrician and a firefighter, then he must be satisfiedwith his work.((E2 & F2) → S2)

11. It cannot be that Harrison is both an electrician and a firefighter.¬(E2 & F2)

12. Harrison and Ava are both firefighters if and only if neither of them is anelectrician.((F2 & F1) ↔ ¬(E2 ∨ E1))

D. Using the symbolization key given, translate each English-language sentenceinto TFL.

J1: John Coltrane played tenor sax.J2: John Coltrane played soprano sax.J3: John Coltrane played tubaM1: Miles Davis played trumpetM2: Miles Davis played tuba

1. John Coltrane played tenor and soprano sax.J1 & J2

2. Neither Miles Davis nor John Coltrane played tuba.¬(M2 ∨ J3) or ¬M2 & ¬ J3

3. John Coltrane did not play both tenor sax and tuba.¬( J1 & J3) or ¬ J1 ∨ ¬ J3

4. John Coltrane did not play tenor sax unless he also played soprano sax.¬ J1 ∨ J2

5. John Coltrane did not play tuba, but Miles Davis did.¬ J3 & M2

6. Miles Davis played trumpet only if he also played tuba.M1 → M2


7. If Miles Davis played trumpet, then John Coltrane played at least one ofthese three instruments: tenor sax, soprano sax, or tuba.M1 → ( J1 ∨ ( J2 ∨ J3))

8. If John Coltrane played tuba then Miles Davis played neither trumpet nortuba.J3 → ¬(M1 ∨M2) or J3 → (¬M1 & ¬M2)

9. Miles Davis and John Coltrane both played tuba if and only if Coltrane didnot play tenor sax and Miles Davis did not play trumpet.( J3 & M2) ↔ (¬ J1 ∧ ¬M1) or ( J3 & M2) ↔ ¬( J1 ∨M1)

E. Give a symbolization key and symbolize the following English sentences inTFL.

A: Alice is a spy.B : Bob is a spy.C : The code has been broken.G : The German embassy will be in an uproar.

1. Alice and Bob are both spies.(A & B)

2. If either Alice or Bob is a spy, then the code has been broken.((A ∨ B) → C )

3. If neither Alice nor Bob is a spy, then the code remains unbroken.(¬(A ∨ B) → ¬C )

4. The German embassy will be in an uproar, unless someone has broken thecode.(G ∨C )

5. Either the code has been broken or it has not, but the German embassy willbe in an uproar regardless.((C ∨ ¬C ) & G )

6. Either Alice or Bob is a spy, but not both.((A ∨ B) & ¬(A & B))

F. Give a symbolization key and symbolize the following English sentences inTFL.

F : There is food to be found in the pridelands.R: Rafiki will talk about squashed bananas.A: Simba is alive.K : Scar will remain as king.

1. If there is food to be found in the pridelands, then Rafiki will talk aboutsquashed bananas.


(F → R)2. Rafiki will talk about squashed bananas unless Simba is alive.

(R ∨ A)3. Rafiki will either talk about squashed bananas or he won’t, but there is food

to be found in the pridelands regardless.((R ∨ ¬R) & F )

4. Scar will remain as king if and only if there is food to be found in thepridelands.(K ↔ F )

5. If Simba is alive, then Scar will not remain as king.(A → ¬K )

G. For each argument, write a symbolization key and symbolize all of the sentencesof the argument in TFL.

1. If Dorothy plays the piano in the morning, then Roger wakes up cranky.Dorothy plays piano in the morning unless she is distracted. So if Rogerdoes not wake up cranky, then Dorothy must be distracted.

P : Dorothy plays the Piano in the morning.C : Roger wakes up cranky.D : Dorothy is distracted.

(P → C ), (P ∨D), (¬C → D)

2. It will either rain or snow on Tuesday. If it rains, Neville will be sad. If itsnows, Neville will be cold. Therefore, Neville will either be sad or cold onTuesday.

T1: It rains on TuesdayT2: It snows on TuesdayS : Neville is sad on TuesdayC : Neville is cold on Tuesday

(T1 ∨T2), (T1 → S ), (T2 → C ), (S ∨C )3. If Zoog remembered to do his chores, then things are clean but not neat. If

he forgot, then things are neat but not clean. Therefore, things are eitherneat or clean; but not both.

Z : Zoog remembered to do his choresC : Things are cleanN : Things are neat

(Z → (C & ¬N )), (¬Z → (N & ¬C )), ((N ∨C ) & ¬(N & C )).

H. For each argument, write a symbolization key and translate the argument aswell as possible into TFL. The part of the passage in italics is there to providecontext for the argument, and doesn’t need to be symbolized.


1. It is going to rain soon. I know because my leg is hurting, and my leg hurtsif it’s going to rain.

2. Spider-man tries to �gure out the bad guy’s plan. If Doctor Octopus gets theuranium, he will blackmail the city. I am certain of this because if DoctorOctopus gets the uranium, he can make a dirty bomb, and if he can make adirty bomb, he will blackmail the city.

3. A westerner tries to predict the policies of the Chinese government. If the Chinesegovernment cannot solve the water shortages in Beijing, they will have tomove the capital. They don’t want to move the capital. Therefore they mustsolve the water shortage. But the only way to solve the water shortage is todivert almost all the water from the Yangzi river northward. Therefore theChinese government will go with the project to divert water from the southto the north.

I. We symbolized an exclusive or using ‘∨’, ‘ & ’, and ‘¬’. How could you symbolizean exclusive or using only two connectives? Is there any way to symbolize anexclusive or using only one connective?For two connectives, we could o�er any of the following:

¬(A ↔ B)(¬A ↔ B)

(¬(¬A & ¬B) & ¬(A & B))

But if we wanted to symbolize it using only one connective, we would have tointroduce a new primitive connective.

Chapter 6

Sentences of TFL

A. For each of the following: (a) Is it a sentence of TFL, strictly speaking? (b) Isit a sentence of TFL, allowing for our relaxed bracketing conventions?

1. (A) (a) no (b) no2. J374 ∨ ¬ J374 (a) no (b) yes3. ¬¬¬¬F (a) yes (b) yes4. ¬ & S (a) no (b) no5. (G & ¬G ) (a) yes (b) yes6. (A → (A & ¬F )) ∨ (D ↔ E) (a) no (b) yes7. [(Z ↔ S ) →W ] & [ J ∨ X ] (a) no (b) yes8. (F ↔ ¬D → J ) ∨ (C & D) (a) no (b) no

B. Are there any sentences of TFL that contain no atomic sentences? Explainyour answer.No. Atomic sentences contain atomic sentences (trivially). And every morecomplicated sentence is built up out of less complicated sentences, that were inturn built out of less complicated sentences, . . . , that were ultimately built out ofatomic sentences.

C. What is the scope of each connective in the sentence[(H → I ) ∨ (I → H )

]& ( J ∨ K )

The scope of the left-most instance of ‘→’ is ‘(H → I )’.The scope of the right-most instance of ‘→’ is ‘(I → H )’.The scope of the left-most instance of ‘∨ is ‘

[(H → I ) ∨ (I → H )

]’

The scope of the right-most instance of ‘∨’ is ‘( J ∨ K )’The scope of the conjunction is the entire sentence; so conjunction is the mainlogical connective of the sentence.

14

Chapter 9

Complete truth tables

A. Complete truth tables for each of the following:

1. A → A

A A→AT TT TF F T F

2. C → ¬C

C C→¬CT T F FTF F T TF

3. (A ↔ B) ↔ ¬(A ↔ ¬B)

A B (A↔B)↔¬(A↔¬B)T T T T T T T T F F TT F T F F T F T T T FF T F F T T F F T F TF F F T F T T F F T F

4. (A → B) ∨ (B → A)

A B (A→B) ∨ (B→A)T T T T T T T T TT F T F F T F T TF T F T T T T F FF F F T F T F T F

15

CHAPTER 9. COMPLETE TRUTH TABLES 16

5. (A & B) → (B ∨ A)

A B (A & B)→(B ∨A)T T T T T T T T TT F T F F T F T TF T F F T T T T FF F F F F T F F F

6. ¬(A ∨ B) ↔ (¬A & ¬B)

A B ¬(A∨B)↔(¬A & ¬B)T T F T T T T F T F F TT F F T T F T F T F T FF T F F T T T T F F F TF F T F F F T T F T T F

7.[(A & B) & ¬(A & B)

]& C

A B C[(A & B) & ¬(A & B)

]& C

T T T T T T F F T T T F TT T F T T T F F T T T F FT F T T F F F T T F F F TT F F T F F F T T F F F FF T T F F T F T F F T F TF T F F F T F T F F T F FF F T F F F F T F F F F TF F F F F F F T F F F F F

8. [(A & B) & C ] → B

A B C [(A & B) & C ]→BT T T T T T T T T TT T F T T T F F T TT F T T F F F T T FT F F T F F F F T FF T T F F T F T T TF T F F F T F F T TF F T F F F F T T FF F F F F F F F T F

9. ¬[(C ∨ A) ∨ B

]


A B C ¬[(C ∨A)∨B

]T T T F TT T T TT T F F FT T T TT F T F TT T T FT F F F FT T T FF T T F TT F T TF T F F F F F T TF F T F TT F T FF F F T F F F F F

B. Check all the claims made in introducing the new notational conventions in§9.4, i.e. show that:

1. ‘((A & B) & C )’ and ‘(A & (B & C ))’ have the same truth table

A B C (A & B) & C A & (B & C )T T T T T T T T T T T T TT T F T T T F F T F T F FT F T T F F F T T F F F TT F F T F F F F T F F F FF T T F F T F T F F T T TF T F F F T F F F F T F FF F T F F F F T F F F F TF F F F F F F F F F F F F

2. ‘((A ∨ B) ∨C )’ and ‘(A ∨ (B ∨C ))’ have the same truth table

A B C (A∨B) ∨C A ∨ (B ∨C )T T T T T T TT TT T T TT T F T T T T F TT T T FT F T T T F TT TT F T TT F F T T F T F TT F F FF T T F T T TT FT T T TF T F F T T T F FT T T FF F T F F F TT FT F T TF F F F F F F F F F F F F

3. ‘((A ∨ B) & C )’ and ‘(A ∨ (B & C ))’ do not have the same truth table


A B C (A∨B) & C A ∨ (B & C )T T T T T T T T TT T T TT T F T T T F F TT T F FT F T T T F T T TT F F TT F F T T F F F TT F F FF T T F T T T T FT T T TF T F F T T F F F F T F FF F T F F F F T F F F F TF F F F F F F F F F F F F

4. ‘((A → B) → C )’ and ‘(A → (B → C ))’ do not have the same truth table

A B C (A→B)→C A→(B→C )T T T T T T T T TT T T TT T F T T T F F T F T F FT F T T F F T T TT F T TT F F T F F T F TT F T FF T T F T T T T F T T T TF T F F T T F F F T T F FF F T F T F T T F T F T TF F F F T F F F F T F T F

Also, check whether:

5. ‘((A ↔ B) ↔ C )’ and ‘(A ↔ (B ↔ C ))’ have the same truth tableIndeed they do:

A B C (A↔B)↔C A↔(B↔C )T T T T T T T T TT T T TT T F T T T F F T F T F FT F T T F F F T T F F F TT F F T F F T F TT F T FF T T F F T F T F F T T TF T F F F T T F F T T F FF F T F T F T T F T F F TF F F F T F F F F F F T F

C. Write complete truth tables for the following sentences and mark the columnthat represents the possible truth values for the whole sentence.

1. ¬(S ↔ (P → S ))


¬ (S ↔ (P → S))F T T T T TF T T F T TF F T T F FT F F F T F

2. ¬[(X &Y ) ∨ (X ∨Y )]

¬ [(X & Y) ∨ (X ∨ Y)]F T T T T T T TF T F F T T T FF F F T T F T TT F F F F F F F

3. (A → B) ↔ (¬B ↔ ¬A)

(A → B) ↔ (¬ B ↔ ¬ A)T T T T F T T F TT F F T T F F F TF T T F F T F T FF T F T T F T T F

4. [C ↔ (D ∨ E)] & ¬C

[C ↔ (D ∨ E)] & ¬ CT T T T T F F TT T T T F F F TT T F T T F F TT F F F F F F TF F T T T F T FF F T T F F T FF F F T T F T FF T F F F T T F

5. ¬(G & (B & H )) ↔ (G ∨ (B ∨H ))


¬ (G & (B & H)) ↔ (G ∨ (B ∨ H))F T T T T T F T T T T TT T F T F F T T T T T FT T F F F T T T T F T TT T F F F F T T T F F FT F F T T T T F T T T TT F F T F F T F T T T FT F F F F T T F T F T TT F F F F F F F F F F F

D. Write complete truth tables for the following sentences and mark the columnthat represents the possible truth values for the whole sentence.

1. (D & ¬D) → G

(D & ¬ D) → GT F F T T TT F F T T FF F T F T TF F T F T F

2. (¬P ∨ ¬M ) ↔ M

(¬ P ∨ ¬ M) ↔ MF T F F T T TF T T T F F FT F T F T T TT F T T F T F

3. ¬¬(¬A & ¬B)

¬ ¬ (¬ A & ¬ B)F T F T F F TF T F T F T FF T T F F F TT F T F T T F


4. [(D & R) → I ] → ¬(D ∨R)

[(D & R) → I] → ¬ (D ∨ R)T T T T T F F T T TT T T F F T F T T TT F F T T F F T T FT F F T F F F T T FF F T T T F F F T TF F T T F F F F T TF F F T T T T F F FF F F T F T T F F F

5. ¬[(D ↔ O ) ↔ A] → (¬D & O )

¬ [(D ↔ O) ↔ A] → (¬ D & O)F T T T T T T F T F TT T T T F F F F T F TT T F F F T F F T F FF T F F T F T F T F FT F F T F T T T F T TF F F T T F T T F T TF F T F T T T T F F FT F T F F F T T F F F

If you want additional practice, you can construct truth tables for any of thesentences and arguments in the exercises for the previous chapter.

Chapter 11

Entailment & validity

A. Revisit your answers to §10A. Determine which sentences were tautologies,which were contradictions, and which were neither tautologies nor contradictions.

1. A → A Tautology2. C → ¬C Neither3. (A ↔ B) ↔ ¬(A ↔ ¬B) Tautology4. (A → B) ∨ (B → A) Tautology5. (A & B) → (B ∨ A) Tautology6. ¬(A ∨ B) ↔ (¬A & ¬B) Tautology7.

[(A & B) & ¬(A & B)

]& C Contradiction

8. [(A & B) & C ] → B Tautology9. ¬

[(C ∨ A) ∨ B

]Neither

B. Use truth tables to determine whether these sentences are jointly consistent, orjointly inconsistent:

1. A → A, ¬A → ¬A, A & A, A ∨ A Jointly consistent (see line 1)

A A→A ¬A→¬A A & A A ∨AT TT T FTT FT T T T TTTF F T F TF T TF F F F F F F

2. A ∨ B , A → C , B → C Jointly consistent (see line 1)

22

CHAPTER 11. ENTAILMENT & VALIDITY 23

A B C A ∨B A→C B→CT T T TTT TT T TT TT T F TTT T F F T F FT F T TTT TT T F T TT F F TTF T F F F T FF T T FTF F T T TT TF T F FTT F T F T F FF F T F F F F T T F T TF F F F F F F T F F T F

3. B & (C ∨ A), A → B , ¬(B ∨C ) Jointly inconsistent

A B C B & (C ∨A) A→B ¬ (B ∨C )T T T T T T T T TT T F T T TT T F T T F T T TT T F T T FT F T F F T T T T F F F F T TT F F F F F T T T F F T F F FF T T T T T T F F T T F T T TF T F T F F F F F T T F T T FF F T F F T T F F T F F F T TF F F F F F F F F T F T F F F

4. A ↔ (B ∨C ), C → ¬A, A → ¬B Jointly consistent (see line 8)

A B C A↔(B ∨C ) C→¬A A→¬BT T T TT T T T T F FT T F FTT T F TT T T F F T FT T F FTT F T TT F T T T F FT TT TFT F F T F F F F F T FT TT TFF T T F F T T T TT TF F T FTF T F F F T T F F T TF F T FTF F T F F F T T TT TF F T TFF F F F T F F F F T TF F T TF

C. Use truth tables to determine whether each argument is valid or invalid.

1. A → A .Û. A Invalid (see line 2)

A A→A AT TT T TF F T F F

2. A → (A & ¬A) .Û. ¬A Valid


A A→(A & ¬A) ¬AT T F T F F T FTF F T F F T F TF

3. A ∨ (B → A) .Û. ¬A → ¬B Valid

A B A ∨ (B→A) ¬A→¬BT T TT T T T FTT FTT F TT F T T FTT TFF T F F T F F TF F FTF F FT F T F TF T TF

4. A ∨ B,B ∨C,¬A .Û. B & C Invalid (see line 6)

A B C A ∨B B ∨C ¬A B & CT T T TTT TTT FT T T TT T F TTT TT F FT T F FT F T TTF FTT FT F F TT F F TTF F F F FT F F FT T T FTT TTT TF T T TT T F FTT TT F TF T F FT F T F F F FTT TF F F TT F F F F F F F F TF F F F

5. (B & A) → C, (C & A) → B .Û. (C & B) → A Invalid (see line 5)

A B C (B & A)→C (C & A)→B (C & B)→AT T T T T T T T T T T T T T T T T TT T F T T T F F F F T T T F F T T TT F T F F T T T T T T F F T F F T TT F F F F T T F F F T T F F F F T TF T T T F F T T T F F T T T T T F FF T F T F F T F F F F T T F F T T FF F T F F F T T T F F T F T F F T FF F F F F F T F F F F T F F F F T F

D. Determine whether each sentence is a tautology, a contradiction, or a contin-gent sentence, using a complete truth table.

1. ¬B & B Contradiction

2. ¬D ∨D Tautology

3. (A & B) ∨ (B & A) Contingent

4. ¬[A → (B → A)] Contradiction

5. A ↔ [A → (B & ¬B)] Contradiction


6. [(A & B) ↔ B] → (A → B) Contingent

E. Determine whether each the following sentences are logically equivalent usingcomplete truth tables. If the two sentences really are logically equivalent, write“equivalent.” Otherwise write, “Not equivalent.”

1. A and ¬A2. A & ¬A and ¬B ↔ B3. [(A ∨ B) ∨C ] and [A ∨ (B ∨C )]4. A ∨ (B & C ) and (A ∨ B) & (A ∨C )5. [A & (A ∨ B)] → B and A → B

F. Determine whether each the following sentences are logically equivalent usingcomplete truth tables. If the two sentences really are equivalent, write “equivalent.”Otherwise write, “not equivalent.”

1. A → A and A ↔ A2. ¬(A → B) and ¬A → ¬B3. A ∨ B and ¬A → B4. (A → B) → C and A → (B → C )5. A ↔ (B ↔ C ) and A & (B & C )

G. Determine whether each collection of sentences is jointly consistent or jointlyinconsistent using a complete truth table.

1. A & ¬B , ¬(A → B), B → A

A & ¬ B ¬ (A → B) B → A ConsistentT F F T F T T T T T TT T T F T T F F F T TF F F T F F T T T F FF F T F F F T F F T F

2. A ∨ B , A → ¬A, B → ¬B

A ∨ B A → ¬ A B → ¬ B InconsistentT T T T F F T T F F TT T F T F F T F T T FF T T F T T F T F F TF F F F T T F F T T F


3. ¬(¬A ∨ B), A → ¬C , A → (B → C ) Consistent

¬ (¬ A ∨ B) A → ¬ C A → (B → C)F F T T T T F F T T T T T TF F T T T T T T F T F T F FT F T F F T F F T T T F T TT F T F F T T T F T T F T FF T F T T F T F T F F T T TF T F T T F T T F F T T F FF T F T F F T F T F T F T TF T F T F F T T F F T F T F

4. A → B , A & ¬B Inconsistent5. A → (B → C ), (A → B) → C , A → C Consistent

H. Determine whether each collection of sentences is jointly consistent or jointlyinconsistent, using a complete truth table.

1. ¬B , A → B , A Inconsistent

2. ¬(A ∨ B), A ↔ B , B → A Consistent

3. A ∨ B , ¬B , ¬B → ¬A Inconsistent

4. A ↔ B , ¬B ∨ ¬A, A → B Consistent

5. (A ∨ B) ∨C , ¬A ∨ ¬B , ¬C ∨ ¬B Consistent

I. Determine whether each argument is valid or invalid, using a complete truthtable.

1. A → B , B .Û. A Invalid2. A ↔ B , B ↔ C .Û. A ↔ C Valid3. A → B , A → C .Û. B → C Invalid.4. A → B , B → A .Û. A ↔ B Valid

J. Determine whether each argument is valid or invalid, using a complete truthtable.

1. A ∨[A → (A ↔ A)

].Û. A Invalid

2. A ∨ B , B ∨C , ¬B .Û. A & C Valid

3. A → B , ¬A .Û. ¬B Invalid

4. A, B .Û. ¬(A → ¬B) Valid

5. ¬(A & B), A ∨ B , A ↔ B .Û. C Valid

K. Answer each of the questions below and justify your answer.


1. Suppose that A and B are logically equivalent. What can you say aboutA ↔ B?A and B have the same truth value on every line of a complete truth table,so A ↔ B is true on every line. It is a tautology.

2. Suppose that (A & B) → C is neither a tautology nor a contradiction. Whatcan you say about whether A,B .Û. C is valid?Since the sentence (A & B) → C is not a tautology, there is some line onwhich it is false. Since it is a conditional, on that line, A and B are true andC is false. So the argument is invalid.

3. Suppose that A, B and C are jointly inconsistent. What can you say about(A & B & C)?Since the sentences are jointly inconsistent, there is no valuation on whichthey are all true. So their conjunction is false on every valuation. It is acontradiction

4. Suppose that A is a contradiction. What can you say about whether A,B � C?Since A is false on every line of a complete truth table, there is no line onwhich A and B are true and C is false. So the entailment holds.

5. Suppose that C is a tautology. What can you say about whether A,B � C?Since C is true on every line of a complete truth table, there is no line onwhich A and B are true and C is false. So the entailment holds.

6. Suppose that A and B are logically equivalent. What can you say about(A ∨ B)?Not much. Since A and B are true on exactly the same lines of the truthtable, their disjunction is true on exactly the same lines. So, their disjunctionis logically equivalent to them.

7. Suppose that A and B are not logically equivalent. What can you say about(A ∨ B)?A and B have di�erent truth values on at least one line of a complete truthtable, and (A ∨ B) will be true on that line. On other lines, it might betrue or false. So (A ∨ B) is either a tautology or it is contingent; it is not acontradiction.

L. Consider the following principle:

• Suppose A and B are logically equivalent. Suppose an argument contains A(either as a premise, or as the conclusion). The validity of the argumentwould be una�ected, if we replaced A with B.

Is this principle correct? Explain your answer.The principle is correct. Since A and B are logically equivalent, they have thesame truth table. So every valuation that makes A true also makes B true, andevery valuation that makes A false also makes B false. So if no valuation makesall the premises true and the conclusion false, when A was among the premises or


the conclusion, then no valuation makes all the premises true and the conclusionfalse, when we replace A with B.

Chapter 12

Truth table shortcuts

A. Using shortcuts, determine whether each sentence is a tautology, a contradic-tion, or neither.

1. ¬B & B Contradiction

B ¬B & BT F FF F

2. ¬D ∨D Tautology

D ¬D ∨DT TF T T

3. (A & B) ∨ (B & A) Neither

A B (A & B) ∨ (B & A)T T T TT F F F FF T F F FF F F F F

4. ¬[A → (B → A)] Contradiction

A B ¬[A→(B→A)]T T F T TT F F T TF T F TF F F T

5. A ↔ [A → (B & ¬B)] Contradiction

29

CHAPTER 12. TRUTH TABLE SHORTCUTS 30

A B A↔[A→(B & ¬B)]T T F F F FT F F F FF T F TF F F T

6. ¬(A & B) ↔ A Neither

A B ¬(A & B)↔AT T F T FT F T F TF T T F FF F T F F

7. A → (B ∨C ) Neither

A B C A→(B ∨C )T T T T TT T F T TT F T T TT F F F FF T T TF T F TF F T TF F F T

8. (A & ¬A) → (B ∨C ) Tautology

A B C (A & ¬A)→(B ∨C )T T T F F TT T F F F TT F T F F TT F F F F TF T T F TF T F F TF F T F TF F F F T

9. (B & D) ↔ [A ↔ (A ∨C )] Neither

CHAPTER 12. TRUTH TABLE SHORTCUTS 31

A B C D (B & D)↔[A↔(A∨C )]T T T T T T T TT T T F F F T TT T F T T T T TT T F F F F T TT F T T F F T TT F T F F F T TT F F T F F T TT F F F F F T TF T T T T F F TF T T F F T F TF T F T T T T FF T F F F F T FF F T T F T F TF F T F F T F TF F F T F F T FF F F F F F T F

Chapter 13

Partial truth tables

A. Use complete or partial truth tables (as appropriate) to determine whetherthese pairs of sentences are logically equivalent:

1. A, ¬A Not logically equivalent

A A ¬AT T F

2. A, A ∨ A Logically equivalent

A A A ∨ AT T TT T T

3. A → A, A ↔ A Logically equivalent

A A → A A ↔ AT T TF T T

4. A ∨ ¬B , A → B Not logically equivalent

A B A ∨ ¬B A → BT F T F

5. A & ¬A, ¬B ↔ B Logically equivalent

A B A & ¬A ¬B↔BT T F F F FT F F F T FF T F F FF F F T F

32

CHAPTER 13. PARTIAL TRUTH TABLES 33

6. ¬(A & B), ¬A ∨ ¬B Logically equivalent

A B ¬ (A & B) ¬A ∨ ¬BT T F T F F FT F T F F TTF T T F T TFF F T F T TT

7. ¬(A → B), ¬A → ¬B Not logically equivalent

A B ¬(A→B) ¬A→¬BT T F T F T F

8. (A → B), (¬B → ¬A) Logically equivalent

A B (A → B) (¬B→¬A)T T T F TT F F T F FF T T F TF F T T T T

B. Use complete or partial truth tables (as appropriate) to determine whetherthese sentences are jointly consistent, or jointly inconsistent:

1. A & B , C → ¬B , C Jointly inconsistent

A B C A & B C→¬B CT T T T F F TT T F T T FT F T F T T TT F F F T FF T T F F F TF T F F T FF F T F T T TF F F F T F

2. A → B , B → C , A, ¬C Jointly inconsistent

A B C A → B B → C A ¬CT T T T T T FT T F T F T TT F T F T T FT F F F T T TF T T T T F FF T F T F F TF F T T T F FF F F T T F T


3. A ∨ B , B ∨C , C → ¬A Jointly consistent

A B C A ∨ B B ∨C C→¬AF T T T T T T

4. A, B , C , ¬D , ¬E, F Jointly consistent

A B C D E F A B C ¬D ¬E FT T T F F T T T T T T T

C. Use complete or partial truth tables (as appropriate) to determine whethereach argument is valid or invalid:

1. A ∨[A → (A ↔ A)

].Û. A Invalid

A A ∨[A→(A↔A)

]A

F T T F

2. A ↔ ¬(B ↔ A) .Û. A Invalid

A B A↔¬(B ↔ A) AF F T F T F

3. A → B,B .Û. A Invalid

A B A → B B AF T T T F

4. A ∨ B,B ∨C,¬B .Û. A & C Valid

A B C A ∨ B B ∨C ¬B A & CT T T TT T F F FT F T TT F F T F T FF T T F FF T F F FF F T F T FF F F F T F

5. A ↔ B,B ↔ C .Û. A ↔ C Valid


A B C A ↔ B B ↔ C A ↔ CT T T TT T F T F FT F T TT F F F FF T T F FF T F TF F T T F FF F F T

D. Determine whether each sentence is a tautology, a contradiction, or a contin-gent sentence. Justify your answer with a complete or partial truth table whereappropriate.

1. A → ¬A Contingent

A A → ¬ AT T F F TF F T T F

2. A → (A & (A ∨ B)) Tautology

A B A → (A & (A ∨ B ))T T T T T T T T TT F T T T T T T FF T F T F F F T TF F F T F F F F F

3. (A → B) ↔ (B → A) Contingent

A B (A → B ) ↔ (B → A)T T T T T T T T TT F T F F F F T TF T F T T F T F FF F F T F T F T F

4. A → ¬(A & (A ∨ B)) Contingent

A B A → ¬ (A & (A ∨ B ))T T T F F T T T T TT F T F F T T T T FF T F T T F F F T TF F F T T F F F F F


5. ¬B → [(¬A & A) ∨ B] Contingent

A B ¬ B → ((¬ A & A) ∨ B )T T F T T F T F T T TT F T F F F T F T F FF T F T T T F F F T TF F T F F T F F F F F

6. ¬(A ∨ B) ↔ (¬A & ¬B) Tautology

A B ¬ (A ∨ B ) ↔ (¬ A & ¬ B )T T F T T T T F T F F TT F F T T F T F T F T FF T F F T T T T F F F TF F T F F F T T F T T F

7. [(A & B) & C ] → B Tautology

A B C ((A & B ) & C ) → BT T T T T T T T T TT T F T T T F F T TT F T T F F F T T FT F F T F F F F T FF T T F F T F T T TF T F F F T F F T TF F T F F F F T T FF F F F F F F F T F

8. ¬[(C ∨ A) ∨ B

]Contingent

A B C ¬ ((C ∨ A) ∨ B )T T T F T T T T TT T F F F T T T TT F T F T T T T FT F F F F T T T FF T T F T T F T TF T F F F F F T TF F T F T T F T FF F F T F F F F F


9.[(A & B) & ¬(A & B)

]& C Contradiction

A B C ((A & B ) & ¬ (A & B )) & CT T T T T T F F T T T F TT T F T T T F F T T T F FT F T T F F F T T F F F TT F F T F F F T T F F F FF T T F F T F T F F T F TF T F F F T F T F F T F FF F T F F F F T F F F F TF F F F F F F T F F F F F

10. (A & B)] → [(A & C ) ∨ (B & D)] Contingent

A B C D ((A & B )) → ((A & C ) ∨ (B & D))

T T T T T T T T T T T T T T TT T F F T T T F T F F F T F F

E. Determine whether each sentence is a tautology, a contradiction, or a contin-gent sentence. Justify your answer with a complete or partial truth table whereappropriate.

1. ¬(A ∨ A) Contradiction

2. (A → B) ∨ (B → A) Tautology

3. [(A → B) → A] → A Tautology

4. ¬[(A → B) ∨ (B → A)] Contradiction

5. (A & B) ∨ (A ∨ B) Contingent

6. ¬(A & B) ↔ A Contingent

7. A → (B ∨C ) Contingent

8. (A & ¬A) → (B ∨C ) Tautology

9. (B & D) ↔ [A ↔ (A ∨C )] Contingent

10. ¬[(A → B) ∨ (C → D)] Contingent

F. Determine whether each the following pairs of sentences are logically equivalentusing complete truth tables. If the two sentences really are logically equivalent,write “equivalent.” Otherwise write, “not equivalent.”

1. A and A ∨ A2. A and A & A3. A ∨ ¬B and A → B4. (A → B) and (¬B → ¬A)


5. ¬(A & B) and ¬A ∨ ¬B6. ((U → (X ∨ X )) ∨U ) and ¬(X & (X & U ))

7. ((C & (N ↔ C )) ↔ C ) and (¬¬¬N → C )8. [(A ∨ B) & C ] and [A ∨ (B & C )]9. ((L & C ) & I ) and L ∨C

G. Determine whether each collection of sentences is jointly consistent or jointlyinconsistent. Justify your answer with a complete or partial truth table whereappropriate.

1. A → A, ¬A → ¬A, A & A, A ∨ A Consistent

2. A → ¬A, ¬A → A Inconsistent

3. A ∨ B , A → C , B → C Consistent

4. A ∨ B , A → C , B → C , ¬C Inconsistent

5. B & (C ∨ A), A → B , ¬(B ∨C ) Inconsistent

6. (A ↔ B) → B , B → ¬(A ↔ B), A ∨ B Consistent

7. A ↔ (B ∨C ), C → ¬A, A → ¬B Consistent

8. A ↔ B , ¬B ∨ ¬A, A → B Consistent

9. A ↔ B , A → C , B → D , ¬(C ∨D) Consistent

10. ¬(A & ¬B), B → ¬A, ¬B Consistent

H. Determine whether each argument is valid or invalid. Justify your answer witha complete or partial truth table where appropriate.

1. A → (A & ¬A) .Û. ¬A Valid

2. A ∨ B , A → B , B → A .Û. A ↔ B Valid

3. A ∨ (B → A) .Û. ¬A → ¬B Valid

4. A ∨ B , A → B , B → A .Û. A & B Valid

5. (B & A) → C , (C & A) → B .Û. (C & B) → A Invalid

6. ¬(¬A ∨ ¬B), A → ¬C .Û. A → (B → C ) Invalid

7. A & (B → C ), ¬C & (¬B → ¬A) .Û. C & ¬C Valid

8. A & B , ¬A → ¬C , B → ¬D .Û. A ∨ B Invalid

9. A → B .Û. (A & B) ∨ (¬A & ¬B) Invalid

10. ¬A → B ,¬B → C ,¬C → A .Û. ¬A → (¬B ∨ ¬C ) Invalid

I. Determine whether each argument is valid or invalid. Justify your answer witha complete or partial truth table where appropriate.


1. A ↔ ¬(B ↔ A) .Û. A Invalid

2. A ∨ B , B ∨C , ¬A .Û. B & C Invalid

3. A → C , E → (D ∨ B), B → ¬D .Û. (A ∨C ) ∨ (B → (E & D)) Invalid

4. A ∨ B , C → A, C → B .Û. A → (B → C ) Invalid

5. A → B , ¬B ∨ A .Û. A ↔ B Valid

Chapter 15

Basic rules for TFL

A. The following two ‘proofs’ are incorrect. Explain the mistakes they make.

1 (¬L ∧ A) ∨ L

2 ¬L & A

3 ¬L & E 3

4 A & E 1

5 L

6 ⊥ ¬E 5, 3

7 A X 6

8 A ∨E 1, 2–4, 5–7

& E on line 4 can’t be applied to line 1,since it is not of the form A & B. ‘A’could be obtained by & E, but fromline 2.

⊥I on line 5 illicitly refers to a linefrom a closed subproof (line 3).

1 A & (B & C )

2 (B ∨C ) → D

3 B & E 1

4 B ∨C ∨I 3

5 D →E 4, 2

& E on line 3 should yield ‘B & C ’. ‘B ’could then be obtained by & E again.

The citation for line 5 is the wrongway round: it should be ‘→E 2, 4’.

B. The following three proofs are missing their citations (rule and line numbers).Add them, to turn them into bona �de proofs. Additionally, write down theargument that corresponds to each proof.

40

CHAPTER 15. BASIC RULES FOR TFL 41

1 P & S

2 S → R

3 P & E 1

4 S & E 1

5 R →E 2, 4

6 R ∨ E ∨I 5

Corresponding argument:P & S,S → R .Û. R ∨ E

1 A → D

2 A & B

3 A & E 2

4 D →E 1, 3

5 D ∨ E ∨I 4

6 (A & B) → (D ∨ E) →I 2–5

Corresponding argument:A → D .Û. (A & B) → (D ∨ E)

1 ¬L → ( J ∨ L)

2 ¬L

3 J ∨ L →E 1, 2

4 J

5 J & J & I 4, 4

6 J & E 5

7 L

8 ⊥ ¬E 7, 2

9 J X 8

10 J ∨E 3, 4–6, 7–9

Corresponding argument:¬L → ( J ∨ L),¬L .Û. J

C. Give a proof for each of the following arguments:

1. J → ¬ J .Û. ¬ J

1 J → ¬ J

2 J

3 ¬ J →E 1, 2

4 ⊥ ¬E 2, 3

5 ¬ J ¬I 2–4

2. Q → (Q & ¬Q ) .Û. ¬Q

1 Q → (Q & ¬Q )

2 Q

3 Q & ¬Q →E 1, 2

4 ¬Q & E 3

5 ⊥ ¬E 2, 4

6 ¬Q ¬I 2–5

3. A → (B → C ) .Û. (A & B) → C


1 A → (B → C )

2 A & B

3 A & E 2

4 B → C →E 1, 3

5 B & E 2

6 C →E 4, 5

7 (A & B) → C →I 2–6

4. K & L .Û. K ↔ L1 K & L

2 K

3 L & E 1

4 L

5 K & E 1

6 K ↔ L ↔I 2–3, 4–5

5. (C & D) ∨ E .Û. E ∨D1 (C & D) ∨ E

2 C & D

3 D & E 2

4 E ∨D ∨I 3

5 E

6 E ∨D ∨I 5

7 E ∨D ∨E 1, 2–4, 5–6

6. A ↔ B,B ↔ C .Û. A ↔ C1 A ↔ B

2 B ↔ C

3 A

4 B ↔E 1, 3

5 C ↔E 2, 4

6 C

7 B ↔E 2, 6

8 A ↔E 1, 7

9 A ↔ C ↔I 3–5, 6–8

7. ¬F → G ,F → H .Û. G ∨H


1 ¬F → G

2 F → H

3 ¬(G ∨H )

4 F

5 H →E 2, 4

6 G ∨H ∨I 5

7 ⊥ ¬E 3, 6

8 ¬F ¬I 4–7

9 G →E 1, 8

10 G ∨H ∨I 9

11 ⊥ ¬E 3, 10

12 G ∨H IP 3–11

8. (Z & K ) ∨ (K & M ),K → D .Û. D1 (Z & K ) ∨ (K & M )

2 K → D

3 Z & K

4 K & E 3

5 K & M

6 K & E 5

7 K ∨E 1, 3–4, 5–6

8 D →E 2, 7

9. P & (Q ∨R),P → ¬R .Û. Q ∨ E


1 P & (Q ∨R)

2 P → ¬R

3 P & E 1

4 ¬R →E 2, 3

5 Q ∨R & E 1

6 Q

7 Q ∨ E ∨I 6

8 R

9 ⊥ ¬E 8, 4

10 Q ∨ E X 9

11 Q ∨ E ∨E 5, 6–7, 8–10


10. S ↔ T .Û. S ↔ (T ∨ S )1 S ↔ T

2 S

3 T ↔E 1, 2

4 T ∨ S ∨I 3

5 T ∨ S

6 T

7 S ↔E 1, 6

8 S

9 S & S & I 8, 8

10 S & E 9

11 S ∨E 5, 6–7, 8–10

12 S ↔ (T ∨ S ) ↔I 2–4, 5–11

11. ¬(P → Q ) .Û. ¬Q

1 ¬(P → Q )

2 Q

3 P

4 Q & Q & I 2, 2

5 Q & E 4

6 P → Q →I 3–5

7 ⊥ ¬E 6, 1

8 ¬Q ¬I 2–7

12. ¬(P → Q ) .Û. P

1 ¬(P → Q )

2 ¬P

3 P

4 ⊥ ¬E 3, 2

5 Q X 4

6 P → Q →I 3–5

7 ⊥ ¬E 6, 1

8 P IP 2–7

Chapter 16

Proof-theoretic concepts

A. Show that each of the following sentences is a theorem:

1. O → O1 O

2 O R 1

3 O → O →I 1–2

2. N ∨ ¬N1 ¬(N ∨ ¬N )

2 N

3 N ∨ ¬N ∨I 2

4 ⊥ ¬E 1, 3

5 ¬N ¬I 2–4

6 N ∨ ¬N ∨I 5

7 ⊥ ¬E 1, 6

8 N ∨ ¬N IP 1–7

46

CHAPTER 16. PROOF-THEORETIC CONCEPTS 47

3. J ↔ [ J ∨ (L & ¬L)]

1 J

2 J ∨ (L & ¬L) ∨I 1

3 J ∨ (L & ¬L)

4 L & ¬L

5 L & E 4

6 ¬L & E 4

7 ⊥ ¬E 5, 6

8 ¬(L & ¬L) ¬I 4–7

9 J DS 3, 8

10 J ↔ [ J ∨ (L & ¬L)] ↔I 1–2, 3–9

4. ((A → B) → A) → A1 (A → B) → A

2 ¬A

3 ¬(A → B) MT 1, 2

4 A

5 ⊥ ¬E 4, 2

6 B X 5

7 A → B →I 4–6

8 ⊥ ¬E 7, 3

9 A IP 1–8

10 ((A → B) → A) → A →I 1–9

B. Provide proofs to show each of the following:

1. C → (E & G ),¬C → G ` G


1 C → (E & G )

2 ¬C → G

3 ¬G

4 C

5 E & G →E 1, 4

6 G & E 5

7 ⊥ ¬E 3, 6

8 ¬C ¬I 4–7

9 G →E 2, 8

10 ⊥ ¬E 3, 9

11 G IP 3–10

2. M & (¬N → ¬M ) ` (N & M ) ∨ ¬M1 M & (¬N → ¬M )

2 M & E 1

3 ¬N → ¬M & E 1

4 ¬N

5 ¬M →E 3, 4

6 ⊥ ¬E 2, 5

7 N IP 4–6

8 N & M & I 7, 2

9 (N & M ) ∨ ¬M ∨I 8

3. (Z & K ) ↔ (Y & M ),D & (D → M ) `Y → Z1 (Z & K ) ↔ (Y & M )

2 D & (D → M )

3 D & E 2

4 D → M & E 2

5 M →E 4, 3

6 Y

7 Y & M & I 6, 5

8 Z & K ↔E 1, 7

9 Z & E 8

10 Y → Z →I 6–9


4. (W ∨ X ) ∨ (Y ∨ Z ),X →Y,¬Z `W ∨Y1 (W ∨ X ) ∨ (Y ∨ Z )

2 X →Y

3 ¬Z

4 W ∨ X

5 W

6 W ∨Y ∨I 5

7 X

8 Y →E 2, 7

9 W ∨Y ∨I 8

10 W ∨Y ∨E 4, 5–6, 7–9

11 Y ∨ Z

12 Y DS 11, 3

13 W ∨Y ∨I 12

14 W ∨Y ∨E 1, 4–10, 11–13

C. Show that each of the following pairs of sentences are provably equivalent:

1. R ↔ E, E ↔ R

1 R ↔ E

2 E

3 R ↔E 1, 2

4 R

5 E ↔E 1, 4

6 E ↔ R ↔I 2–3, 4–5

1 E ↔ R

2 E

3 R ↔E 1, 2

4 R

5 E ↔E 1, 4

6 R ↔ E ↔I 4–5, 2–3

2. G , ¬¬¬¬G

1 G

2 ¬¬¬G

3 ¬G DNE 2

4 ⊥ ¬E 1, 3

5 ¬¬¬¬G ¬I 2–4

1 ¬¬¬¬G

2 ¬¬G DNE 1

3 G DNE 2


3. T → S , ¬S → ¬T

1 T → S

2 ¬S

3 ¬T MT 1, 2

4 ¬S → ¬T →I 2–3

1 ¬S → ¬T

2 T

3 ¬S

4 ¬T →E 1, 3

5 ⊥ ¬E 2, 4

6 ¬¬S ¬I 3–5

7 S DNE 6

8 T → S →I 2–7

4. U → I , ¬(U & ¬I )

1 U → I

2 U & ¬I

3 U & E 2

4 ¬I & E 2

5 I →E 1, 3

6 ⊥ ¬E 5, 4

7 ¬(U & ¬I ) ¬I 2–6

1 ¬(U & ¬I )

2 U

3 ¬I

4 U & ¬I & I 2, 3

5 ⊥ ¬E 4, 1

6 ¬¬I ¬I 3–5

7 I DNE 6

8 U → I →I 2–7


5. ¬(C → D),C & ¬D

1 C & ¬D

2 C & E 1

3 ¬D & E 1

4 C → D

5 D →E 4, 2

6 ⊥ ¬E 5, 3

7 ¬(C → D) ¬I 4–6

1 ¬(C → D)

2 D

3 C

4 D R 2

5 C → D →I 3–4

6 ⊥ ¬E 5, 1

7 ¬D ¬I 2–6

8 ¬C

9 C

10 ⊥ ¬E 9, 8

11 D X 10

12 C → D →I 9–11

13 ⊥ ¬E 12, 1

14 ¬¬C ¬I 8–13

15 C DNE 14

16 C & ¬D & I 15, 7

6. ¬G ↔ H , ¬(G ↔ H )

1 ¬G ↔ H

2 G ↔ H

3 G

4 H ↔E 2, 3

5 ¬G ↔E 1, 4

6 ⊥ ¬E 3, 5

7 ¬G

8 H ↔E 1, 7

9 G ↔E 2, 8

10 ⊥ ¬E 9, 7

11 ⊥ LEM 3–6, 7–10

12 ¬(G ↔ H ) ¬I 2–11


1 ¬(G ↔ H )

2 ¬G

3 ¬H

4 G

5 ⊥ ¬E 4, 2

6 H X 5

7 H

8 ⊥ ¬E 7, 3

9 G X 8

10 G ↔ H ↔I 4–6, 7–9

11 ⊥ ¬E 10, 1

12 H IP 3–11

13 H

14 G

15 G

16 H R 13

17 H

18 G R 14

19 G ↔ H ↔I 15–16, 17–18

20 ⊥ ¬E 19, 1

21 ¬G ¬I 14–20

22 ¬G ↔ H ↔I 2–12, 13–21

D. If you know that A ` B, what can you say about (A & C) ` B? What about(A ∨ C) ` B? Explain your answers.If A ` B, then (A & C) ` B. After all, if A ` B, then there is some proof withassumption A that ends with B, and no undischarged assumptions other than A.Now, if we start a proof with assumption (A & C), we can obtain A by & E. Wecan now copy and paste the original proof of B from A, adding 1 to every linenumber and line number citation. The result will be a proof of B from assumptionA.

However, we cannot prove much from (A∨C). After all, it might be impossibleto prove B from C.


E. In this chapter, we claimed that it is just as hard to show that two sentencesare not provably equivalent, as it is to show that a sentence is not a theorem. Whydid we claim this? (Hint: think of a sentence that would be a theorem i� A and Bwere provably equivalent.)Consider the sentence A ↔ B. Suppose we can show that this is a theorem. Sowe can prove it, with no assumptions, in m lines, say. Then if we assume A andcopy and paste the proof of A ↔ B (changing the line numbering), we will have adeduction of this shape:1 A

m + 1 A ↔ B

m + 2 B ↔E m + 1, 1

This will show that A ` B. In exactly the same way, we can show that B ` A. So ifwe can show that A ↔ B is a theorem, we can show that A and B are provablyequivalent.

Conversely, suppose we can show that A and B are provably equivalent. Thenwe can prove B from the assumption of A in m lines, say, and prove A fromthe assumption of B in n lines, say. Copying and pasting these proofs together(changing the line numbering where appropriate), we obtain:1 A

m B

m + 1 B

m + n A

m + n + 1 A ↔ B ↔I 1–m, m + 1–m + n

Thus showing that A ↔ B is a theorem.There was nothing special about A and B in this. So what this shows is that

the problem of showing that two sentences are provably equivalent is, essentially,the same problem as showing that a certain kind of sentence (a biconditional) isa theorem.

Chapter 19

Additional rules for TFL

A. The following proofs are missing their citations (rule and line numbers). Addthem wherever they are required:

1 W → ¬B

2 A &W

3 B ∨ ( J & K )

4 W & E 2

5 ¬B →E 1, 4

6 J & K DS 3, 5

7 K & E 6

1 L ↔ ¬O

2 L ∨ ¬O

3 ¬L

4 ¬O DS 2, 3

5 L ↔E 1, 4

6 ⊥ ¬E 5, 3

7 ¬¬L ¬I 3–6

8 L DNE 7

1 Z → (C & ¬N )

2 ¬Z → (N & ¬C )

3 ¬(N ∨C )

4 ¬N & ¬C DeM 3

5 ¬N & E 4

6 ¬C & E 4

7 Z

8 C & ¬N →E 1, 7

9 C & E 8

10 ⊥ ¬E 9, 6

11 ¬Z ¬I 7–10

12 N & ¬C →E 2, 11

13 N & E 12

14 ⊥ ¬E 13, 5

15 ¬¬(N ∨C ) ¬I 3–14

16 N ∨C DNE 15

B. Give a proof for each of these arguments:

1. E ∨ F , F ∨G , ¬F .Û. E & G

54

CHAPTER 19. ADDITIONAL RULES FOR TFL 55

1 E ∨ F

2 F ∨G

3 ¬F

4 E DS 1, 3

5 G DS 2, 3

6 E & G & I 4, 5

2. M ∨ (N → M ) .Û. ¬M → ¬N1 M ∨ (N → M )

2 ¬M

3 N → M DS 1, 2

4 ¬N MT 3, 2

5 ¬M → ¬N →I 2–4

3. (M ∨ N ) & (O ∨ P ), N → P , ¬P .Û. M & O1 (M ∨ N ) & (O ∨ P )

2 N → P

3 ¬P

4 ¬N MT 2, 3

5 M ∨ N & E 1

6 M DS 5, 4

7 O ∨ P & E 1

8 O DS 7, 3

9 M & O & I 6, 8

4. (X &Y ) ∨ (X & Z ), ¬(X & D), D ∨M .Û.M

CHAPTER 19. ADDITIONAL RULES FOR TFL 56

1 (X &Y ) ∨ (X & Z )

2 ¬(X & D)

3 D ∨M

4 X &Y

5 X & E 4

6 X & Z

7 X & E 6

8 X ∨E 1, 4–5, 6–7

9 D

10 X & D & I 8, 9

11 ⊥ ¬E 10, 2

12 ¬D ¬I 9–11

13 M DS 3, 12

Chapter 20

Soundness and Completeness

Practice exercises

A. Use either a derivation or a truth table for each of the following.

1. Show that A → [((B & C ) ∨D) → A] is a tautology.

2. Show that A → (A → B) is not a tautology

3. Show that the sentence A → ¬A is not a contradiction.

4. Show that the sentence A ↔ ¬A is a contradiction.

5. Show that the sentence ¬(W → ( J ∨ J )) is contingent

6. Show that the sentence ¬(X ∨ (Y ∨ Z )) ∨ (X ∨ (Y ∨ Z )) is not contingent

7. Show that the sentence B → ¬S is equivalent to the sentence ¬¬B → ¬S

8. Show that the sentence ¬(X ∨O ) is not equivalent to the sentence X & O

9. Show that the sentences ¬(A ∨ B), C , C → A are jointly inconsistent.

10. Show that the sentences ¬(A ∨ B), ¬B , B → A are jointly consistent

11. Show that ¬(A ∨ (B ∨C )) .Û.¬C is valid.

12. Show that ¬(A & (B ∨C )) .Û.¬C is invalid.

B. Use either a derivation or a truth table for each of the following.

1. Show that A → (B → A) is a tautology

2. Show that ¬(((N ↔ Q ) ∨Q ) ∨ N ) is not a tautology

57

CHAPTER 20. SOUNDNESS AND COMPLETENESS 58

3. Show that Z ∨ (¬Z ↔ Z ) is contingent

4. show that (L ↔ ((N → N ) → L)) ∨H is not contingent

5. Show that (A ↔ A) & (B & ¬B) is a contradiction

6. Show that (B ↔ (C ∨ B)) is not a contradiction.

7. Show that ((¬X ↔ X ) ∨ X ) is equivalent to X

8. Show that F & (K & R) is not equivalent to (F ↔ (K ↔ R))

9. Show that the sentences ¬(W →W ), (W ↔W ) &W , E∨(W → ¬(E &W ))

are inconsistent.

10. Show that the sentences ¬R ∨ C , (C & R) → ¬R, (¬(R ∨ R) → R) areconsistent.

11. Show that ¬¬(C ↔ ¬C ), ((G ∨C ) ∨G ) .Û. ((G → C ) & G ) is valid.

12. Show that ¬¬L, (C → ¬L) → C ) .Û. ¬C is invalid.

forall x: msu edition. solutions to selected exercisesloighic.net/logic/forallx-solutions.pdf · ii...

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