forall x: Calgary Remix. Solutions to Selected Exercisespeople.ucalgary.ca/~rzach/static/forallx/solutions/forallxsol.pdf · forall x Calgary Remix Solutions to Selected Exercises

forall xCalgary Remix

Solutions to Selected Exercises

By P. D. MagnusTim Buttonwith additions byJ. Robert Loftisremixed and revised byAaron Thomas-BolducRichard Zach

Summer 2017 bis

ii

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

Tim ButtonUniversity of Cambridge

used under a CC BY-SA 3.0 license, which is based in turn on forallx, by

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

used under a CC BY-SA 3.0 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-SA 3.0 license, and from forallx: Lorain County Remix, by CathalWoods and J. Robert Loftis, used under a CC BY-SA 4.0 license.

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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 notin any way that suggests the licensor endorses you or your use.

. If you remix, transform, or build upon the material, you must distributeyour contributions under the same license as the original.

The LATEX source for this book is available on GitHub. This version is revision9e6553e (2017-07-01).

The preparation of this textbook was made possible by a grant from the TaylorInstitute for Teaching and Learning.

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Contents1 Arguments 12 Valid arguments 23 Other logical notions 45 Connectives 86 Sentences of TFL 1410 Complete truth tables 1511 Semantic concepts 2112 Truth table shortcuts 2713 Partial truth tables 3015 Basic rules for TFL 3716 Additional rules for TFL 4517 Proof-theoretic concepts 4819 Derived rules 5720 Soundness and Completeness 5922 Sentences with one quantifier 6123 Multiple generality 6524 Identity 7025 Definite descriptions 7126 Sentences of FOL 7528 Truth in FOL 7630 Using Interpretations 7832 Basic rules for FOL 8033 Conversion of quantifiers 9434 Rules for identity 9935 Derived rules 10737 Normal Forms and Expressive Completeness 109

iii

CHAPTER 1

ArgumentsHighlight 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 mur-

der. And Reverend Green had the candlestick in the ballroom, and weknow that there is no blood on his hands. Hence Colonel Mustard did itin the kitchen with the lead-piping. Recall, after all, that the gun had notbeen fired.

1

CHAPTER 2

Valid argumentsA. 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..Û. Bob is now 20 years old. Valid

An 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:

2

CHAPTER 2. VALID ARGUMENTS 3

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, thereforeSocrates is 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 into thepremises. For an argument is valid if and only if it is impossible for all thepremises to be true and the conclusion false. If the premises are contra-dictory, 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 newpremise? No.An argument is valid if and only if it is impossible for all the premises tobe true and the conclusion false. Adding another premise will only makeit harder for the premises all to be true together.

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

CHAPTER 3

Other logicalnotionsA. For each of the following: Is it necessarily true, necessarily false, or contin-gent?

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

CHAPTER 3. OTHER LOGICAL NOTIONS 5

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

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 M32. Sentences M2, M3, and M4


3. 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,it will 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 false-hoodNo. 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 whose con-clusion is contingent

2. A valid argument with true premises and a false conclusion3. A jointly possible collection of sentences that contains two sentences that

are not necessarily equivalent4. A jointly possible collection of sentences, all of which are contingent5. A false necessary truth


6. 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

ConnectivesA. 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.

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

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

8

CHAPTER 5. CONNECTIVES 9

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 satisfy-ing.((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 he

is 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)


F. 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 embassywill be in an uproar regardless.((C ∨ ¬C ) ∧G )

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

G. 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 foodto 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 )

H. For each argument, write a symbolization key and symbolize all of the sen-tences of 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 it

snows, 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 )).

I. 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 makea dirty 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 theymust solve the water shortage. But the only way to solve the water shortageis to divert almost all the water from the Yangzi river northward. Thereforethe Chinese government will go with the project to divert water from thesouth to the north.

J. 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 ofTFLA. For each of the following: (a) Is it a sentence of TFL, strictly speaking? (b)Is it 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 10

Complete truthtablesA. Complete truth tables for each of the following:

1. A → A

A A→AT T TTF F TF

2. C → ¬CC C →¬CT T F FTF F TTF

3. (A ↔ B) ↔ ¬(A ↔ ¬B)

A B (A↔B)↔¬(A↔¬B)T T T T T TT T F F TT F T F F TF T T T FF T F F T TF F T F TF F F T F TT 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 10. 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 TF F TT F F T T F T F TFT FF T F F T T T T F F F TF F T F F F T T FTT 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 FTT T F T T T F F T T T F FT F T T F F FT T F F FTT F F T F F FT T F F F FF T T F F T FT F F F FTF T F F F T FT F F T F FF F T F F F FT F F T FTF F F F F F FT F F F F F

8. [(A ∧ B) ∧C ] → B

A B C [(A∧B)∧C ]→BT T T T T T T T TTT T F T T T F F TTT F T T F F F T TFT F F T F F F F TFF T T F F T F T TTF T F F F T F F TTF F T F F F F T TFF F F F F F F F TF

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§10.3, 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 TT T T T T TT T F T T T F F T F T F FT F T T F F FT T F F F TT F F T F F F F T F F F FF T T F F T FT F F T T TF T F F F T F F F F T F FF F T F F F FT 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 T T T T TT T F T T T T F T T T T FT F T T T F TT T T F T TT F F T T F T F T T F F FF T T F T T TT F T T T TF T F F T T T F F T T T FF F T F F F TT F T 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 TT T T T T TT T F T T T F F T T T F FT F T T T F TT T T F F TT F F T T F F F T T F F FF T T F T T TT F T T T TF T F F T T F F F F T F FF F T F F F FT 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 TT T T T T TT T F T T T F F T F T F FT F T T F F TT T T F T TT F F T F F T F T T F T FF T T F T T TT F T T T TF T F F T T F F F T T F FF F T F T F TT 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 TT T T T T TT T F T T T F F T F T F FT F T T F F FT T F F F TT F F T F F T F T T F T FF T T F F T FT F F T T TF T F F F T T F F T T F FF F T F T F TT 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

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

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,or jointly inconsistent:

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

A A→A ¬A→¬A A ∧ A A ∨ AT T TT FT TFT T TT T TTF F TF TF TTF F F F F F F

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

21

CHAPTER 11. SEMANTIC CONCEPTS 22

A B C A ∨ B A→C B→CT T T T TT T TT T TTT T F T TT T F F T F FT F T T TT T TT F TTT F F T TF T F F F T FF T T F TF F TT T TTF T F F TT F T F T F FF F T F F F F TT F TTF 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 T TT F T T TT T F T T F T T T TT 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 TT F T T TF T F T F F F F F TT F T T FF F T F F T T F F TF F F T TF F F F F F F F F TF 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 T T T T T T F FT T F FTT T F T T T T F F TFT T F FTT F T T T F T T T F FT T TTFT F F T F F F F F TFT T TTFF T T F F T T T T TTF F TFTF T F F F T T F F TTF F TFTF F T F F F T T T TTF F TTFF F F F T F F F F TTF F TTF

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 T TT TF F TF F

2. A → (A ∧ ¬A) .Û. ¬A Valid

A A→(A∧¬A) ¬ AT T F T F F T FTF F T F FT F TF

3. A ∨ (B → A) .Û. ¬A → ¬B Valid


A B A ∨ (B→A) ¬A→¬BT T T T T T T FT TFTT F T T F T T FT TTFF T F F T F F TF F FTF F F T F T F TF TTF

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 T TT T TT FT T TTT T F T TT T T F FT T F FT F T T TF F TT FT F FTT F F T TF F F F FT F F FT T T F TT T TT TF T TTT T F F TT T T F TF T F FT F T F F F F TT TF F FTT 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 TT T T T TT T T T TTT T F T T T F F F F T TT F F T TTT F T F F T TT T T T F F T F F TTT F F F F T T F F F T TF F F F TTF T T T F F TT T F F TT T T T F FF T F T F F T F F F F TT F F T TFF F T F F F TT T F F TF T F F TFF F F F F F T F F F F TF F F F TF

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

1. ¬B ∧ B Contradiction2. ¬D ∨D Tautology3. (A ∧ B) ∨ (B ∧ A) Contingent4. ¬[A → (B → A)] Contradiction5. A ↔ [A → (B ∧ ¬B)] Contradiction6. [(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 “equiva-lent.” 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 Inconsistent2. ¬(A ∨ B), A ↔ B , B → A Consistent3. A ∨ B , ¬B , ¬B → ¬A Inconsistent4. A ↔ B , ¬B ∨ ¬A, A → B Consistent5. (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 Valid3. A → B , ¬A .Û. ¬B Invalid4. A, B .Û. ¬(A → ¬B) Valid5. ¬(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?Aand 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 trueand C 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 Aand B are true on exactly the same lines of the truth ta-ble, their disjunction is true on exactly the same lines. So, their disjunctionis logically equivalent to them.

7. Suppose that Aand B are not logically equivalent. What can you say about(A∨ B)?Aand 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 containsA (either as a premise, or as the conclusion). The validity of the argumentwould be una�ected, if we replaced Awith 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 Awas among the premises orthe conclusion, then no valuation makes all the premises true and the conclusionfalse, when we replace Awith B.

CHAPTER 12

Truth tableshortcutsA. 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

27

CHAPTER 12. TRUTH TABLE SHORTCUTS 28

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 FF TT T F FF TT F T FF TT F F FF 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 29

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 truthtablesA. 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 FF F FT F FF T FF T F F FF F F T F

30

CHAPTER 13. PARTIAL TRUTH TABLES 31

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 TF

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 TT

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 TF 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) Contradiction2. (A → B) ∨ (B → A) Tautology3. [(A → B) → A] → A Tautology4. ¬[(A → B) ∨ (B → A)] Contradiction5. (A ∧ B) ∨ (A ∨ B) Contingent6. ¬(A ∧ B) ↔ A Contingent7. A → (B ∨C ) Contingent8. (A ∧ ¬A) → (B ∨C ) Tautology9. (B ∧D) ↔ [A ↔ (A ∨C )] Contingent10. ¬[(A → B) ∨ (C → D)] Contingent

F. Determine whether each the following pairs of sentences are logically equiva-lent using complete truth tables. If the two sentences really are logically equiva-lent, 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 Consistent2. A → ¬A, ¬A → A Inconsistent3. A ∨ B , A → C , B → C Consistent4. A ∨ B , A → C , B → C , ¬C Inconsistent5. B ∧ (C ∨ A), A → B , ¬(B ∨C ) Inconsistent6. (A ↔ B) → B , B → ¬(A ↔ B), A ∨ B Consistent7. A ↔ (B ∨C ), C → ¬A, A → ¬B Consistent8. A ↔ B , ¬B ∨ ¬A, A → B Consistent9. A ↔ B , A → C , B → D , ¬(C ∨D) Consistent10. ¬(A ∧ ¬B), B → ¬A, ¬B Consistent

H. Determine whether each argument is valid or invalid. Justify your answerwith a 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 forTFLA. The following two ‘proofs’ are incorrect. Explain the mistakes they make.

1 ¬L → (A ∧ L)

2 ¬L

3 A →E 1, 2

4 L

5 ⊥ ⊥I 4, 2

6 A ⊥E 5

7 A TND 2–3, 4–6

→E on line 3 should yield ‘A ∧L’. ‘A’could then be obtained by ∧E.⊥I on line 5 illicitly refers to a linefrom a closed subproof (line 2).

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 the argu-ment that corresponds to each proof.

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

37

CHAPTER 15. BASIC RULES FOR TFL 38

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 ⊥ ⊥I 7, 2

9 J ⊥E 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. ¬ J1 J → ¬ J

2 J

3 ¬ J →E 1, 2

4 ⊥ ⊥I 2, 3

5 ¬ J ¬I 2–42. Q → (Q ∧ ¬Q ) .Û. ¬Q

1 Q → (Q ∧ ¬Q )

2 Q

3 Q ∧ ¬Q →E 1, 2

4 ¬Q ∧E 3

5 ⊥ ⊥I 2, 4

6 ¬Q ¬I 2–53. 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 F

4 H →E 2, 3

5 G ∨H ∨I 4

6 ¬F

7 G →E 1, 6

8 G ∨H ∨I 7

9 G ∨H TND 3–5, 6–8


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 ∨ E1 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 ⊥ ⊥I 8, 4

10 Q ∨ E ⊥E 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. ¬Q1 ¬(P → Q )

2 Q

3 P

4 Q ∧Q ∧I 2, 2

5 Q ∧E 4

6 P → Q →I 3–5

7 ⊥ ⊥I 6, 1

8 ¬Q ¬I 2–712. ¬(P → Q ) .Û. P

1 ¬(P → Q )

2 P

3 P ∧ P ∧I 2, 2

4 P ∧E 3

5 ¬P

6 P

7 ⊥ ⊥I 6, 5

8 Q ⊥E 7

9 P → Q →I 6–8

10 ⊥ ⊥I 9, 1

11 P ⊥E 10

12 P TND 2–4, 5–11

CHAPTER 16

Additional rulesfor TFLA. 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 ⊥ ⊥I 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 ⊥ ⊥I 9, 6

11 ¬Z ¬I 7–10

12 N ∧ ¬C →E 2, 11

13 N ∧E 12

14 ⊥ ⊥I 13, 5

15 ¬¬(N ∨C ) ¬I 3–14

16 N ∨C DNE 15

B. Give a proof for each of these arguments:

45

CHAPTER 16. ADDITIONAL RULES FOR TFL 46

1. E ∨ F , F ∨G , ¬F .Û. E ∧G1 E ∨ F

2 F ∨G

3 ¬F

4 E DS 1, 3

5 G DS 2, 3

6 E ∧G ∧I 4, 5

CHAPTER 16. ADDITIONAL RULES FOR TFL 47

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, 84. (X ∧Y ) ∨ (X ∧ Z ), ¬(X ∧D), D ∨M .Û.M

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 ⊥ ⊥I 10, 2

12 ¬D ¬I 9–11

13 M DS 3, 12

CHAPTER 17

Proof-theoreticconceptsA. 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

2 N ∨ ¬N ∨I 1

3 ¬N

4 N ∨ ¬N ∨I 3

5 N ∨ ¬N TND 1–2, 3–4

3. J ↔ [ J ∨ (L ∧ ¬L)]

48

CHAPTER 17. PROOF-THEORETIC CONCEPTS 49

1 J

2 J ∨ (L ∧ ¬L) ∨I 1

3 J ∨ (L ∧ ¬L)

4 L ∧ ¬L

5 L ∧E 4

6 ¬L ∧E 4

7 ⊥ ⊥I 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 ⊥ ⊥I 4, 2

6 B ⊥E 5

7 A → B →I 4–6

8 ⊥ ⊥I 7, 3

9 ¬¬A ¬I 2

10 A DNE 9

11 ((A → B) → A) → A →I 1–10

B. Provide proofs to show each of the following:

1. C → (E ∧G ),¬C → G ` G1 C → (E ∧G )

2 ¬C → G

3 C

4 E ∧G →E 1, 3

5 G ∧E 4

6 ¬C

7 G →E 2, 6

8 G TND 3–5, 6–7

2. M ∧ (¬N → ¬M ) ` (N ∧M ) ∨ ¬M


1 M ∧ (¬N → ¬M )

2 M ∧E 1

3 ¬N → ¬M ∧E 1

4 ¬N

5 ¬M →E 3, 4

6 ⊥ ⊥I 2, 5

7 ¬¬N ¬I 4–6

8 N DNE 7

9 N ∧M ∧I 8, 2

10 (N ∧M ) ∨ ¬M ∨I 9


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 ⊥ ⊥I 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 ⊥ ⊥I 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 ⊥ ⊥I 5, 4

7 ¬(U ∧ ¬I ) ¬I 2–6

1 ¬(U ∧ ¬I )

2 U

3 ¬I

4 U ∧ ¬I ∧I 2, 3

5 ⊥ ⊥I 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 ⊥ ⊥I 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 ⊥ ⊥I 5, 1

7 ¬D ¬I 2–6

8 ¬C

9 C

10 ⊥ ⊥I 9, 8

11 D ⊥E 10

12 C → D →I 9–11

13 ⊥ ⊥I 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 ⊥ ⊥I 3, 5

7 ¬G

8 H ↔E 1, 7

9 G ↔E 2, 8

10 ⊥ ⊥I 9, 7

11 ⊥ TND 3–6, 7–10

12 ¬(G ↔ H ) ¬I 2–11


1 ¬(G ↔ H )

2 ¬G

3 ¬H

4 G

5 ⊥ ⊥I 4, 2

6 H ⊥E 5

7 H

8 ⊥ ⊥I 7, 3

9 G ⊥E 8

10 G ↔ H ↔I 4–6, 7–9

11 ⊥ ⊥I 10, 1

12 ¬¬H ¬I 3–11

13 H DNE 12

14 H

15 G

16 G

17 H R 14

18 H

19 G R 15

20 G ↔ H ↔I 16–17, 18–19

21 ⊥ ⊥I 20, 1

22 ¬G ¬I 15–21

23 ¬G ↔ H ↔I 2–13, 14–22

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 proofwith assumption A that ends with B, and no undischarged assumptions otherthan A. Now, if we start a proof with assumption (A∧ C), we can obtain A by∧E. We can now copy and paste the original proof of B from A, adding 1 toevery line number and line number citation. The result will be a proof of B fromassumption A.

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.Why did we claim this? (Hint: think of a sentence that would be a theorem i� A

and B were 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 havea deduction 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. Soif we can show that A↔ B is a theorem, we can show that Aand B are provablyequivalent.

Conversely, suppose we can show thatAandBare provably equivalent. Thenwe can prove B from the assumption of A in m lines, say, and prove A from theassumption 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

Derived rulesA. Provide proof schemes that justify the addition of the third and fourth DeMorgan rules as derived rules.

Third rule:m ¬A∧ ¬B

k ¬A ∧E m

k + 1 ¬B ∧E m

k + 2 A∨ B

k + 3 A

k + 4 ⊥ ⊥I k + 3, k

k + 5 B

k + 6 ⊥ ⊥I k + 5, k + 1

k + 7 ⊥ ∨E k + 2, k + 3–k + 4, k + 5–k + 6

k + 8 ¬(A∨ B) ¬I k + 2–k + 7

Fourth rule:

57

CHAPTER 19. DERIVED RULES 58

m ¬(A∨ B)

k A

k + 1 A∨ B ∨I k

k + 2 ⊥ ⊥I k + 1, m

k + 3 ¬A ¬I k–k + 2

k + 4 B

k + 5 A∨ B ∨I k + 4

k + 6 ⊥ ⊥I k + 5, m

k + 7 ¬B ¬I k + 4–k + 6

k + 8 ¬A∧ ¬B ∧I k + 3, k + 7B. The proofs you o�ered in response to the practice exercises of §§16–17 usedderived rules. Replace the use of derived rules, in such proofs, with only basicrules. You will find some ‘repetition’ in the resulting proofs; in such cases, o�era streamlined proof using only basic rules. (This will give you a sense, both ofthe power of derived rules, and of how all the rules interact.)

CHAPTER 20

Soundness andCompletenessPractice 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

59

CHAPTER 20. SOUNDNESS AND COMPLETENESS 60

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.

CHAPTER 22

Sentences withone quanti�erA.Here are the syllogistic figures identified by Aristotle and his successors, alongwith their medieval names:

• Barbara. All G are F. All H are G. So: All H are F∀x(Gx → Fx),∀x(H x → Gx) .Û. ∀x(H x → Fx)

• Celarent. No G are F. All H are G. So: No H are F∀x(Gx → ¬Fx),∀x(H x → Gx) .Û. ∀x(H x → ¬Fx)

• Ferio. No G are F. Some H is G. So: Some H is not F∀x(Gx → ¬Fx),∃x(H x ∧Gx) .Û. ∃x(H x ∧ ¬Fx)

• Darii. All G are H. Some H is G. So: Some H is F.∀x(Gx → Fx),∃x(H x ∧Gx) .Û. ∃x(H x ∧ Fx)

• Camestres. All F are G. No H are G. So: No H are F.∀x(Fx → Gx),∀x(H x → ¬Gx) .Û. ∀x(H x → ¬Fx)

• Cesare. No F are G. All H are G. So: No H are F.∀x(Fx → ¬Gx),∀x(H x → Gx) .Û. ∀x(H x → ¬Fx)

• Baroko. All F are G. Some H is not G. So: Some H is not F.∀x(Fx → Gx),∃x(H x ∧ ¬Gx) .Û. ∃x(H x ∧ ¬Fx)

• Festino. No F are G. Some H are G. So: Some H is not F.∀x(Fx → ¬Gx),∃x(H x ∧Gx) .Û. ∃x(H x ∧ ¬Fx)

• Datisi. All G are F. Some G is H. So: Some H is F.∀x(Gx → Fx),∃x(Gx ∧H x) .Û. ∃x(H x ∧ Fx)

• Disamis. Some G is F. All G are H. So: Some H is F.∃x(Gx ∧ Fx),∀x(Gx → H x) .Û. ∃x(H x ∧ Fx)

• Ferison. No G are F. Some G is H. So: Some H is not F.∀x(Gx → ¬Fx),∃x(Gx ∧H x) .Û. ∃x(H x ∧ ¬Fx)

• Bokardo. Some G is not F. All G are H. So: Some H is not F.∃x(Gx ∧ ¬Fx),∀x(Gx → H x) .Û. ∃x(H x ∧ ¬Fx)

• Camenes. All F are G. No G are H So: No H is F.∀x(Fx → Gx),∀x(Gx → ¬H x) .Û. ∀x(H x → ¬Fx)

61

CHAPTER 22. SENTENCES WITH ONE QUANTIFIER 62

• Dimaris. Some F is G. All G are H. So: Some H is F.∃x(Fx ∧Gx),∀x(Gx → H x) .Û. ∃x(H x ∧ Fx)

• Fresison. No F are G. Some G is H. So: Some H is not F.∀x(Fx → ¬Gx),∃x(Gx ∧H x) .Û. ∃(H x ∧ ¬Fx)

Symbolize each argument in FOL.

B. Using the following symbolization key:

domain: peopleKx : x knows the combination to the safeSx : x is a spyV x : x is a vegetarianh: Hofthori : Ingmar

symbolize the following sentences in FOL:

1. Neither Hofthor nor Ingmar is a vegetarian.¬V h ∧ ¬V i

2. No spy knows the combination to the safe.∀x(Sx → ¬Kx)

3. No one knows the combination to the safe unless Ingmar does.∀x¬Kx ∨ Ki

4. Hofthor is a spy, but no vegetarian is a spy.Sh ∧ ∀x(V x → ¬Sx)

C. Using this symbolization key:

domain: all animalsAx : x is an alligator.Mx : x is a monkey.Rx : x is a reptile.Zx : x lives at the zoo.a: Amosb : Bouncerc : Cleo

symbolize each of the following sentences in FOL:

1. Amos, Bouncer, and Cleo all live at the zoo.Za ∧ Zb ∧ Zc

2. Bouncer is a reptile, but not an alligator.Rb ∧ ¬Ab

3. Some reptile lives at the zoo.∃x(Rx ∧ Zx)

4. Every alligator is a reptile.∀x(Ax → Rx)

5. Any animal that lives at the zoo is either a monkey or an alligator.


∀x(Zx → (Mx ∨ Ax))6. There are reptiles which are not alligators.

∃x(Rx ∧ ¬Ax)7. If any animal is an reptile, then Amos is.

∃xRx → Ra8. If any animal is an alligator, then it is a reptile.

∀x(Ax → Rx)

D. For each argument, write a symbolization key and symbolize the argument inFOL.

1. Willard is a logician. All logicians wear funny hats. So Willard wears afunny hat

domain: peopleLx : x is a logicianH x : x wears a funny hati : Willard

Li,∀x(Lx → H x) .Û. H i2. Nothing on my desk escapes my attention. There is a computer on my

desk. As such, there is a computer that does not escape my attention.

domain: physical thingsDx : x is on my deskEx : x escapes my attentionCx : x is a computer

∀x(Dx → ¬Ex),∃x(Dx ∧Cx) .Û. ∃x(Cx ∧ ¬Ex)3. All my dreams are black and white. Old TV shows are in black and white.

Therefore, some of my dreams are old TV shows.

domain: episodes (psychological and televised)Dx : x is one of my dreamsBx : x is in black and whiteOx : x is an old TV show

∀x(Dx → Bx),∀x(Ox → Bx) .Û. ∃x(Dx ∧Ox).Comment: generic statements are tricky to deal with. Does the secondsentence mean that all old TV shows are in black and white; or that mostof them are; or that most of the things which are in black and white are oldTV shows? I have gone with the former, but it is not clear that FOL dealswith these well.

4. Neither Holmes nor Watson has been to Australia. A person could see akangaroo only if they had been to Australia or to a zoo. Although Watsonhas not seen a kangaroo, Holmes has. Therefore, Holmes has been to azoo.

domain: peopleAx : x has been to AustraliaKx : x has seen a kangaroo


Zx : x has been to a zooh: Holmesa: Watson

¬Ah ∧ ¬Aa,∀x(Kx → (Ax ∨ Zx)),¬Ka ∧ Kh .Û. Zh5. No one expects the Spanish Inquisition. No one knows the troubles I’ve

seen. Therefore, anyone who expects the Spanish Inquisition knows thetroubles I’ve seen.

domain: peopleSx : x expects the Spanish InquisitionT x : x knows the troubles I’ve seenh: Holmesa: Watson

∀x¬Sx,∀x¬T x .Û. ∀x(Sx → T x)6. All babies are illogical. Nobody who is illogical can manage a crocodile.

Berthold is a baby. Therefore, Berthold is unable to manage a crocodile.

domain: peopleBx : x is a babyI x : x is illogicalCx : x can manage a crocodileb : Berthold

∀x(Bx → I x),∀x(I x → ¬Cx),Bb .Û. ¬Cb

CHAPTER 23

MultiplegeneralityA. Using this symbolization key:

domain: all animalsAx : x is an alligatorMx : x is a monkeyRx : x is a reptileZx : x lives at the zooLxy : x loves y

a: Amosb : Bouncerc : Cleo

symbolize each of the following sentences in FOL:

1. If Cleo loves Bouncer, then Bouncer is a monkey.Lcb → Mb

2. If both Bouncer and Cleo are alligators, then Amos loves them both.(Ab ∧ Ac ) → (Lab ∧ Lac )

3. Cleo loves a reptile.∃x(Rx ∧ Lcx)Comment: this English expression is ambiguous; in some contexts, it canbe read as a generic, along the lines of ‘Cleo loves reptiles’. (Compare ‘Ido love a good pint’.)

4. Bouncer loves all the monkeys that live at the zoo.∀x((Mx ∧ Zx) → Lbx)

5. All the monkeys that Amos loves love him back.∀x((Mx ∧ Lax) → Lxa)

6. Every monkey that Cleo loves is also loved by Amos.∀x((Mx ∧ Lcx) → Lax)

65

CHAPTER 23. MULTIPLE GENERALITY 66

7. There is a monkey that loves Bouncer, but sadly Bouncer does not recipro-cate this love.∃x(Mx ∧ Lxb ∧ ¬Lbx)


domain: all animalsDx : x is a dogSx : x likes samurai moviesLxy : x is larger than y

r : Raveh: Shaned : Daisy


1. Rave is a dog who likes samurai movies.Dr ∧ Sr

2. Rave, Shane, and Daisy are all dogs.Dr ∧Dh ∧Dd

3. Shane is larger than Rave, and Daisy is larger than Shane.Lhr ∧ Ldh

4. All dogs like samurai movies.∀x(Dx → Sx)

5. Only dogs like samurai movies.∀x(Sx → Dx)Comment: the FOL sentence just written does not require that anyone likessamurai movies. The English sentence might suggest that at least somedogs do like samurai movies?

6. There is a dog that is larger than Shane.∃x(Dx ∧ Lxh)

7. If there is a dog larger than Daisy, then there is a dog larger than Shane.∃x(Dx ∧ Lxd) → ∃x(Dx ∧ Lxh)

8. No animal that likes samurai movies is larger than Shane.∀x(Sx → ¬Lxh)

9. No dog is larger than Daisy.∀x(Dx → ¬Lxd)

10. Any animal that dislikes samurai movies is larger than Rave.∀x(¬Sx → Lxr )Comment: this is very poor, though! For ‘dislikes’ does not mean the sameas ‘does not like’.

11. There is an animal that is between Rave and Shane in size.∃x((Lbx ∧ Lxh) ∨ (Lhx ∧ Lxr ))

12. There is no dog that is between Rave and Shane in size.∀x (Dx → ¬

[(Lbx ∧ Lxh) ∨ (Lhx ∧ Lxr )

] )13. No dog is larger than itself.

∀x(Dx → ¬Lxx)


14. Every dog is larger than some dog.∀x(Dx → ∃y(Dy ∧ Lxy))Comment: the English sentence is potentially ambiguous here. I have re-solved the ambiguity by assuming it should be paraphrased by ‘for everydog, there is a dog smaller than it’.

15. There is an animal that is smaller than every dog.∃x∀y(Dy → Lyx)

16. If there is an animal that is larger than any dog, then that animal does notlike samurai movies.∀x(∀y(Dy → Lxy) → ¬Sx)Comment: I have assumed that ‘larger than any dog’ here means ‘largerthan every dog’.

C. Using the symbolization key given, translate each English-language sentenceinto FOL.

domain: candiesCx : x has chocolate in it.Mx : x has marzipan in it.Sx : x has sugar in it.T x : Boris has tried x .Bxy : x is better than y .

1. Boris has never tried any candy.2. Marzipan is always made with sugar.3. Some candy is sugar-free.4. The very best candy is chocolate.5. No candy is better than itself.6. Boris has never tried sugar-free chocolate.7. Boris has tried marzipan and chocolate, but never together.8. Any candy with chocolate is better than any candy without it.9. Any candy with chocolate and marzipan is better than any candy that lacks

both.

D. Using the following symbolization key:

domain: people and dishes at a potluckRx : x has run out.T x : x is on the table.Fx : x is food.P x : x is a person.Lxy : x likes y .e : Elif : Francescag : the guacamole

symbolize the following English sentences in FOL:


1. All the food is on the table.∀x(Fx → T x)

2. If the guacamole has not run out, then it is on the table.¬Rg → T g

3. Everyone likes the guacamole.∀x(P x → Lxg )

4. If anyone likes the guacamole, then Eli does.∃x(P x ∧ Lxg ) → Le g

5. Francesca only likes the dishes that have run out.∀x [(Lfx ∧ Fx) → Rx

]6. Francesca likes no one, and no one likes Francesca.

∀x [P x → (¬Lfx ∧ ¬Lxf)]

7. Eli likes anyone who likes the guacamole.∀x((P x ∧ Lxg ) → Lex)

8. Eli likes anyone who likes the people that he likes.∀x [ (P x ∧ ∀y[(P y ∧ Ley) → Lxy]

)→ Lex

]9. If there is a person on the table already, then all of the food must have run

out.∃x(P x ∧T x) → ∀x(Fx → Rx)

E. Using the following symbolization key:

domain: peopleDx : x dances ballet.Fx : x is female.Mx : x is male.Cxy : x is a child of y .Sxy : x is a sibling of y .e : Elmerj : Janep : Patrick


1. All of Patrick’s children are ballet dancers.∀x(Cxp → Dx)

2. Jane is Patrick’s daughter.C j p ∧ F j

3. Patrick has a daughter.∃x(Cxp ∧ Fx)

4. Jane is an only child.¬∃xSx j

5. All of Patrick’s sons dance ballet.∀x [(Cxp ∧Mx) → Dx

]6. Patrick has no sons.

¬∃x(Cxp ∧Mx)7. Jane is Elmer’s niece.


∃x(Sxe ∧C jx ∧ F j )8. Patrick is Elmer’s brother.S pe ∧Mp

9. Patrick’s brothers have no children.∀x [(S px ∧Mx) → ¬∃yC yx ]

10. Jane is an aunt.F j ∧ ∃x(Sx j ∧ ∃yC yx)

11. Everyone who dances ballet has a brother who also dances ballet.∀x [Dx → ∃y(M y ∧ S yx ∧Dy)]

12. Every woman who dances ballet is the child of someone who dances ballet.∀x [(Fx ∧Dx) → ∃y(Cxy ∧Dy)]

CHAPTER 24

IdentityA. Explain why:

• ‘∃x∀y(Ay ↔ x = y)’ is a good symbolization of ‘there is exactly one apple’.We might naturally read this in English thus:

• There is something, x, such that, if you choose any object at all, if youchose an apple then you chose x itself, and if you chose x itself thenyou chose an apple.

The x in question must therefore be the one and only thing which is anapple.

• ‘∃x∃y [¬x = y ∧∀z (Az ↔ (x = z ∨ y = z )]’ is a good symbolization of ‘there

are exactly two apples’.Similarly to the above, we might naturally read this in English thus:

• There are two distinct things, x and y, such that if you choose anyobject at all, if you chose an apple then you either chose x or y, andif you chose either x or y then you chose an apple.

The x and y in question must therefore be the only things which are apples,and since they are distinct, there are two of them.

70

CHAPTER 25

De�nitedescriptionsA. Using the following symbolization key:

domain: peopleKx : x knows the combination to the safe.Sx : x is a spy.V x : x is a vegetarian.T xy : x trusts y .h: Hofthori : Ingmar


1. Hofthor trusts a vegetarian.∃x(V x ∧T hx)

2. Everyone who trusts Ingmar trusts a vegetarian.∀x [T xi → ∃y(T xy ∧V y)]

3. Everyone who trusts Ingmar trusts someone who trusts a vegetarian.∀x [T xi → ∃y (T xy ∧ ∃z (T yz ∧V z )) ]

4. Only Ingmar knows the combination to the safe.∀x(Ki → x = i )Comment: does the English claim entail that Ingmar does know the combi-nation to the safe? If so, then we should formalise this with a ‘↔’.

5. Ingmar trusts Hofthor, but no one else.∀x(T ix ↔ x = h)

6. The person who knows the combination to the safe is a vegetarian.∃x [Kx ∧ ∀y(Ky → x = y) ∧V x

]7. The person who knows the combination to the safe is not a spy.

∃x [Kx ∧ ∀y(Ky → x = y) ∧ ¬Sx]

Comment: the scope of negation is potentially ambiguous here; I have readit as inner negation.

71

CHAPTER 25. DEFINITE DESCRIPTIONS 72


domain: cards in a standard deckBx : x is black.Cx : x is a club.Dx : x is a deuce.J x : x is a jack.Mx : x is a man with an axe.Ox : x is one-eyed.Wx : x is wild.

symbolize each sentence in FOL:

1. All clubs are black cards.∀x(Cx → Bx)

2. There are no wild cards.¬∃xW x

3. There are at least two clubs.∃x∃y(¬x = y ∧Cx ∧C y)

4. There is more than one one-eyed jack.∃x∃y(¬x = y ∧ J x ∧Ox ∧ J y ∧Oy)

5. There are at most two one-eyed jacks.∀x∀y∀z [( J x ∧Ox ∧ J y ∧Oy ∧ J z ∧Oz ) → (x = y ∨ x = z ∨ y = z )

]6. There are two black jacks.

∃x∃y(¬x = y ∧ Bx ∧ J x ∧ By ∧ J y)Comment: I am reading this as ‘there are at least two. . . ’. If the suggestionwas that there are exactly two, then a di�erent FOL sentence would berequired, namely:∃x∃y (¬x = y ∧ Bx ∧ J x ∧ By ∧ J y ∧ ∀z [(Bz ∧ J z ) → (x = z ∨ y = z )]

)7. There are four deuces.

∃w∃x∃y∃z (¬w = x ∧ ¬w = y ∧ ¬w = z ∧ ¬x = y ∧ ¬x = z ∧ ¬y = z ∧Dw ∧Dx ∧Dy ∧Dz )Comment: I am reading this as ‘there are at least four. . . ’. If the suggestionis that there are exactly four, then we should o�er instead:∃w∃x∃y∃z (¬w = x ∧ ¬w = y ∧ ¬w = z ∧ ¬x = y ∧ ¬x = z ∧ ¬y = z ∧Dw ∧Dx ∧Dy ∧Dz ∧ ∀v [Dv → (v = w ∨ v = x ∨ v = y ∨ v = z )]

)8. The deuce of clubs is a black card.

∃x [Dx ∧Cx ∧ ∀y ((Dy ∧C y) → x = y)∧ Bx

]9. One-eyed jacks and the man with the axe are wild.

∀x [( J x ∧Ox) →Wx]∧ ∃x [Mx ∧ ∀y(M y → x = y) ∧Wx

]10. If the deuce of clubs is wild, then there is exactly one wild card.

∃x (Dx ∧Cx ∧∀y [(Dy ∧C y) → x = y]∧Wx

)→ ∃x (Wx ∧∀y(W y → x = y)

)Comment: if there is not exactly one deuce of clubs, then the above sen-tence is true. Maybe that’s the wrong verdict. Perhaps the sentence shoulddefinitely be taken to imply that there is one and only one deuce of clubs,and then express a conditional about wildness. If so, then we might sym-


bolize it thus:∃x (Dx ∧Cx ∧ ∀y [(Dy ∧C y) → x = y

]∧[Wx → ∀y(W y → x = y)

] )11. The man with the axe is not a jack.

∃x [Mx ∧ ∀y(M y → x = y) ∧ ¬ J x]

12. The deuce of clubs is not the man with the axe.∃x∃y (Dx ∧Cx ∧∀z [(Dz ∧C z ) → x = z ] ∧M y ∧∀z (Mz → y = x) ∧ ¬x = y

)C. Using the following symbolization key:

domain: animals in the worldBx : x is in Farmer Brown’s field.H x : x is a horse.P x : x is a Pegasus.Wx : x has wings.


1. There are at least three horses in the world.∃x∃y∃z (¬x = y ∧ ¬x = z ∧ ¬y = z ∧H x ∧H y ∧H z )

2. There are at least three animals in the world.∃x∃y∃z (¬x = y ∧ ¬x = z ∧ ¬y = z )

3. There is more than one horse in Farmer Brown’s field.∃x∃y(¬x = y ∧H x ∧H y ∧ Bx ∧ By)

4. There are three horses in Farmer Brown’s field.∃x∃y∃z (¬x = y ∧ ¬x = z ∧ ¬y = z ∧H x ∧H y ∧H z ∧ Bx ∧ By ∧ Bz )Comment: I have read this as ‘there are at least three. . . ’. If the suggestionwas that there are exactly three, then a di�erent FOL sentence would berequired.

5. There is a single winged creature in Farmer Brown’s field; any other crea-tures in the field must be wingless.∃x [Wx ∧ Bx ∧ ∀y ((W y ∧ By) → x = y)

]6. The Pegasus is a winged horse.

∃x [P x ∧ ∀y(P y → x = y) ∧Wx ∧H x]

7. The animal in Farmer Brown’s field is not a horse.∃x [Bx ∧ ∀y(By → x = y) ∧ ¬H x

]Comment: the scope of negation might be ambiguous here; I have read itas inner negation.

8. The horse in Farmer Brown’s field does not have wings.∃x [H x ∧ Bx ∧ ∀y ((H y ∧ By) → x = y

)∧ ¬Wx

]Comment: the scope of negation might be ambiguous here; I have read itas inner negation.

D. In this chapter, we symbolized ‘Nick is the traitor’ by ‘∃x(T x ∧ ∀y(T y → x =y) ∧ x = n)’. Two equally good symbolizations would be:

• T n ∧ ∀y(T y → n = y)


This sentence requires that Nick is a traitor, and that Nick alone is a traitor.Otherwise put, there is one and only one traitor, namely, Nick. Otherwiseput: Nick is the traitor.

• ∀y(T y ↔ y = n)This sentence can be understood thus: Take anything you like; now, if youchose a traitor, you chose Nick, and if you chose Nick, you chose a traitor.So there is one and only one traitor, namely, Nick, as required.

Explain why these would be equally good symbolizations.

CHAPTER 26

Sentences ofFOLA. Identify which variables are bound and which are free. We underline thebound variables, and overline the free variables.

1. ∃xLxy ∧ ∀yLyx2. ∀xAx ∧ Bx3. ∀x(Ax ∧ Bx) ∧ ∀y(Cx ∧Dy)4. ∀x∃y[Rxy → ( J z ∧ Kx)] ∨Ryx5. ∀x1(Mx2 ↔ Lx2x1) ∧ ∃x2Lx3x2

75

CHAPTER 28

Truth in FOLA. Consider the following interpretation:

• The domain comprises only Corwin and Benedict• ‘Ax ’ is to be true of both Corwin and Benedict• ‘Bx ’ is to be true of Benedict only• ‘N x ’ is to be true of no one• ‘c ’ is to refer to Corwin

Determine whether each of the following sentences is true or false in that inter-pretation:

1. Bc False2. Ac ↔ ¬N c True3. N c → (Ac ∨ Bc ) True4. ∀xAx True5. ∀x¬Bx False6. ∃x(Ax ∧ Bx) True7. ∃x(Ax → N x) False8. ∀x(N x ∨ ¬N x) True9. ∃xBx → ∀xAx True

B. Consider the following interpretation:

• The domain comprises only Lemmy, Courtney and Eddy• ‘Gx ’ is to be true of Lemmy, Courtney and Eddy.• ‘H x ’ is to be true of and only of Courtney• ‘Mx ’ is to be true of and only of Lemmy and Eddy• ‘c ’ is to refer to Courtney• ‘e ’ is to refer to Eddy


1. H c True2. H e False

76

CHAPTER 28. TRUTH IN FOL 77

3. Mc ∨Me True4. Gc ∨ ¬Gc True5. Mc → Gc True6. ∃xH x True7. ∀xH x False8. ∃x¬Mx True9. ∃x(H x ∧Gx) True10. ∃x(Mx ∧Gx) True11. ∀x(H x ∨Mx) True12. ∃xH x ∧ ∃xM x True13. ∀x(H x ↔ ¬Mx) True14. ∃xGx ∧ ∃x¬Gx False15. ∀x∃y(Gx ∧H y) True

C. Following the diagram conventions introduced at the end of §23, consider thefollowing interpretation:

1 2

3 4 5


1. ∃xRxx True2. ∀xRxx False3. ∃x∀yRxy True4. ∃x∀yRyx False5. ∀x∀y∀z ((Rxy ∧Ryz ) → Rxz ) False6. ∀x∀y∀z ((Rxy ∧Rxz ) → Ryz ) False7. ∃x∀y¬Rxy True8. ∀x(∃yRxy → ∃yRyx) True9. ∃x∃y(¬x = y ∧Rxy ∧Ryx) True10. ∃x∀y(Rxy ↔ x = y) True11. ∃x∀y(Ryx ↔ x = y) False12. ∃x∃y(¬x = y ∧Rxy ∧ ∀z (Rzx ↔ y = z )) True

CHAPTER 30

UsingInterpretationsA. Show that each of the following is neither a logical truth nor a contradiction:

1. Da ∧Db2. ∃xT xh3. Pm ∧ ¬∀xP x4. ∀z J z ↔ ∃y J y5. ∀x(Wxmn ∨ ∃yLxy)6. ∃x(Gx → ∀yM y)7. ∃x(x = h ∧ x = i )

B. Show that the following pairs of sentences are not logically equivalent.

1. J a, Ka2. ∃x J x , J m3. ∀xRxx , ∃xRxx4. ∃xP x → Qc , ∃x(P x → Qc )5. ∀x(P x → ¬Qx), ∃x(P x ∧ ¬Qx)6. ∃x(P x ∧Qx), ∃x(P x → Qx)7. ∀x(P x → Qx), ∀x(P x ∧Qx)8. ∀x∃yRxy , ∃x∀yRxy9. ∀x∃yRxy , ∀x∃yRyx

C. Show that the following sentences are jointly consistent:

1. Ma,¬N a,P a,¬Qa2. Lee,Le g ,¬Lge,¬Lg g3. ¬(Ma ∧ ∃xAx),Ma ∨ Fa,∀x(Fx → Ax)4. Ma ∨Mb,Ma → ∀x¬Mx5. ∀yG y,∀x(Gx → H x),∃y¬I y6. ∃x(Bx ∨ Ax),∀x¬Cx,∀x [(Ax ∧ Bx) → Cx

]78

CHAPTER 30. USING INTERPRETATIONS 79

7. ∃xX x,∃xY x,∀x(X x ↔ ¬Y x)8. ∀x(P x ∨Qx),∃x¬(Qx ∧ P x)9. ∃z (N z ∧Ozz ),∀x∀y(Oxy → Oyx)10. ¬∃x∀yRxy,∀x∃yRxy11. ¬Raa, ∀x(x = a ∨Rxa)12. ∀x∀y∀z [(x = y ∨ y = z ) ∨ x = z ], ∃x∃y ¬x = y13. ∃x∃y((Zx ∧ Zy) ∧ x = y), ¬Zd , d = e

D. Show that the following arguments are invalid:

1. ∀x(Ax → Bx) .Û. ∃xBx2. ∀x(Rx → Dx),∀x(Rx → Fx) .Û. ∃x(Dx ∧ Fx)3. ∃x(P x → Qx) .Û. ∃xP x4. N a ∧ N b ∧ N c .Û. ∀xN x5. Rde,∃xRxd .Û. Red6. ∃x(Ex ∧ Fx),∃xF x → ∃xGx .Û. ∃x(Ex ∧Gx)7. ∀xOxc,∀xOcx .Û. ∀xOxx8. ∃x( J x ∧ Kx),∃x¬Kx,∃x¬ J x .Û. ∃x(¬ J x ∧ ¬Kx)9. Lab → ∀xLxb,∃xLxb .Û. Lbb10. ∀x(Dx → ∃yT yx) .Û. ∃y∃z ¬y = z

CHAPTER 32

Basic rules forFOLA. The following two ‘proofs’ are incorrect. Explain why both are incorrect. Also,provide interpretations which would invalidate the fallacious argument forms the‘proofs’ enshrine:

1 ∀xRxx2 Raa ∀E 1

3 ∀yRay ∀I 24 ∀x∀yRxy ∀I 3

When using ∀I, you must replace allnames with the new variable. So line3 is bogus. As a counterinterpretation,consider the following:

1 2

1 ∀x∃yRxy2 ∃yRay ∀E 1

3 Raa

4 ∃xRxx ∃I 35 ∃xRxx ∃E 2, 3–4

The instantiating constant, ‘a’, occursin the line (line 2) to which ∃E is to beapplied on line 5. So the use of ∃E online 5 is bogus. As a counterinterpre-tation, consider the following:

1 2

B. The following three proofs are missing their citations (rule and line numbers).Add them, to turn them into bona fide proofs.

1 ∀x∃y(Rxy ∨Ryx)2 ∀x¬Rmx3 ∃y(Rmy ∨Rym) ∀E 1

4 Rma ∨Ram

5 ¬Rma ∀E 2

6 Ram DS 4, 5

7 ∃xRxm ∃I 68 ∃xRxm ∃E 3, 4–7

80

CHAPTER 32. BASIC RULES FOR FOL 81

1 ∀x(∃yLxy → ∀zLzx)2 Lab

3 ∃yLay → ∀zLza ∀E 1

4 ∃yLay ∃I 25 ∀zLza →E 3, 4

6 Lca ∀E 5

7 ∃yLcy → ∀zLzc ∀E 1

8 ∃yLcy ∃I 69 ∀zLzc →E 7, 8

10 Lcc ∀E 9

11 ∀xLxx ∀I 10

1 ∀x( J x → Kx)

2 ∃x∀yLxy3 ∀x J x4 ∀yLay5 Laa ∀E 4

6 J a ∀E 3

7 J a → Ka ∀E 1

8 Ka →E 7, 6

9 Ka ∧ Laa ∧I 8, 5

10 ∃x(Kx ∧ Lxx) ∃I 911 ∃x(Kx ∧ Lxx) ∃E 2, 4–10

C. In §22 problem A, we considered fifteen syllogistic figures of Aristotelian logic.Provide proofs for each of the argument forms. NB: You will find it much easierif you symbolize (for example) ‘No F is G’ as ‘∀x(Fx → ¬Gx)’.I shall prove the four Figure I syllogisms; the rest are extremely similar.

Barbara1 ∀x(Gx → Fx)

2 ∀x(H x → Gx)

3 Ga → Fa ∀E 1

4 H a → Ga ∀E 2

5 H a

6 Ga →E 4, 5

7 Fa →E 3, 6

8 H a → Fa →I 5–7

9 ∀x(H x → Fx) ∀I 8

Celerant is exactly as Barbara, re-placing ‘F ’ with ‘¬F ’ throughout.

Ferio1 ∀x(Gx → ¬Fx)

2 ∃x(H x ∧Gx)3 H a ∧Ga

4 H a ∧E 3

5 Ga ∧E 3

6 Ga → ¬Fa ∀E 1

7 ¬Fa →E 6, 5

8 H a ∧ ¬Fa ∧I 4, 7

9 ∃x(H x ∧ ¬Fx) ∃I 810 ∃x(H x ∧ ¬Fx) ∃E 2, 3–9

Darii is exactly as Ferio, replacing‘¬F ’ with ‘F ’ throughout.

D. Aristotle and his successors identified other syllogistic forms which dependedupon ‘existential import’. Symbolize each of the following argument forms inFOL and o�er proofs.


• Barbari. Something is H. All G are F. All H are G. So: Some H is F∃xH x,∀x(Gx → Fx),∀x(H x → Gx) .Û. ∃x(H x ∧ Fx)1 ∃xH x2 ∀x(Gx → Fx)

3 ∀x(H x → Gx)

4 H a

5 H a → Ga ∀E 3

6 Ga →E 5, 4

7 Ga → Fa ∀E 2

8 Fa →E 7, 6

9 H a ∧ Fa ∧I 4, 8

10 ∃x(H x ∧ Fx) ∃I 911 ∃x(H x ∧ Fx) ∃E 1, 4–10

• Celaront. Something is H. No G are F. All H are G. So: Some H is not F∃xH x,∀x(Gx → ¬Fx),∀x(H x → Gx) .Û. ∃x(H x ∧ ¬Fx)Proof is exactly as for Barbari, replacing ‘F ’ with ‘¬F ’ throughout.

• Cesaro. Something is H. No F are G. All H are G. So: Some H is not F.∃xH x,∀x(Fx → ¬Gx),∀x(H x → Gx) .Û. ∃x(H x ∧ ¬Fx)1 ∃xH x2 ∀x(Fx → ¬Gx)

3 ∀x(H x → Gx)

4 H a

5 H a → Ga ∀E 3

6 Ga →E 5, 4

7 Fa → ¬Ga ∀E 2

8 Fa

9 ¬Ga →E 7, 8

10 ⊥ ⊥I 6, 9

11 ¬Fa ¬I 8–10

12 H a ∧ ¬Fa ∧I 4, 11

13 ∃x(H x ∧ ¬Fx) ∃I 1214 ∃x(H x ∧ ¬Fx) ∃E 1, 4–13


• Camestros. Something is H. All F are G. No H are G. So: Some H is notF.∃xH x,∀x(Fx → Gx),∀x(H x → ¬Gx) .Û. ∃x(H x ∧ ¬Fx)1 ∃xH x2 ∀x(Fx → Gx)

3 ∀x(H x → ¬Gx)

4 H a

5 H a → ¬Ga ∀E 3

6 ¬Ga →E 5, 4

7 Fa → Ga ∀E 2

8 ¬Fa MT 7, 6

9 H a ∧ ¬Fa ∧I 4, 8

10 ∃x(H x ∧ ¬Fx) ∃I 911 ∃x(H x ∧ ¬Fx) ∃E 1, 4–10

• Felapton. Something is G. No G are F. All G are H. So: Some H is not F.∃xGx,∀x(Gx → ¬Fx),∀x(Gx → H x) .Û. ∃x(H x ∧ ¬Fx)1 ∃xGx2 ∀x(Gx → ¬Fx)

3 ∀x(Gx → H x)

4 Ga

5 Ga → H a ∀E 3

6 H a →E 5, 4

7 Ga → ¬Fa ∀E 2

8 ¬Fa →E 7, 4

9 H a ∧ ¬Fa ∧I 6, 8

10 ∃x(H x ∧ ¬Fx) ∃I 911 ∃x(H x ∧ Fx) ∃E 1, 4–10

• Darapti. Something is G. All G are F. All G are H. So: Some H is F.∃xGx,∀x(Gx → Fx),∀x(Gx → H x) .Û. ∃x(H x ∧ Fx)Proof is exactly as for Felapton, replacing ‘¬F ’ with ‘F ’ throughout.


• Calemos. Something is H. All F are G. No G are H. So: Some H is not F.∃xH x,∀x(Fx → Gx),∀x(Gx → ¬H x) .Û. ∃x(H x ∧ ¬Fx)1 ∃xH x2 ∀x(Fx → Gx)

3 ∀x(Gx → ¬H x)

4 H a

5 Ga → ¬H a ∀E 3

6 Ga

7 ¬H a →E 5, 6

8 ⊥ ⊥I 4, 7

9 ¬Ga ¬I 6–8

10 Fa → Ga ∀E 2

11 ¬Fa MT 10, 9

12 H a ∧ ¬Fa ∧I 4, 11

13 ∃x(H x ∧ Fx) ∃I 1214 ∃x(H x ∧ Fx) ∃E 1, 4–13

• Fesapo. Something is G. No F is G. All G are H. So: Some H is not F.∃xGx,∀x(Fx → ¬Gx),∀x(Gx → H x) .Û. ∃x(H x ∧ ¬Fx)1 ∃xGx2 ∀x(Fx → ¬Gx)

3 ∀x(Gx → H x)

4 Ga

5 Ga → H a ∀E 3

6 H a →E 5, 4

7 Fa → ¬Ga ∀E 2

8 Fa

9 ¬Ga →E 7, 8

10 ⊥ ⊥I 4, 9

11 ¬Fa ¬I 8–10

12 H a ∧ ¬Fa ∧I 6, 11

13 ∃x(H x ∧ Fx) ∃I 1214 ∃x(H x ∧ Fx) ∃E 1, 4–13


• Bamalip. Something is F. All F are G. All G are H. So: Some H are F.∃xF x,∀x(Fx → Gx),∀x(Gx → H x) .Û. ∃x(H x ∧ Fx)1 ∃xF x2 ∀x(Fx → Gx)

3 ∀x(Gx → H x)

4 Fa

5 Fa → Ga ∀E 2

6 Ga →E 5, 4

7 Ga → H a ∀E 3

8 H a →E 7, 6

9 H a ∧ Fa ∧I 8, 4

10 ∃x(H x ∧ Fx) ∃I 911 ∃x(H x ∧ Fx) ∃E 1, 4–10

E. Provide a proof of each claim.

1. ` ∀xF x ∨ ¬∀xF x1 ∀xF x2 ∀xF x ∨ ¬∀xF x ∨I 1

3 ¬∀xF x4 ∀xF x ∨ ¬∀xF x ∨I 3

5 ∀xF x ∨ ¬∀xF x TND 1–2, 3–4

2. ` ∀z (P z ∨ ¬P z )1 P a

2 P a ∨ ¬P a ∨I 1

3 ¬P a

4 P a ∨ ¬P a ∨I 3

5 P a ∨ ¬P a TND 1–2, 3–4

6 ∀x(P x ∨ ¬P x) ∀I 53. ∀x(Ax → Bx),∃xAx ` ∃xBx


1 ∀x(Ax → Bx)

2 ∃xAx3 Aa

4 Aa → Ba ∀E 1

5 Ba →E 4, 3

6 ∃xBx ∃I 57 ∃xBx ∃E 2, 3–6


4. ∀x(Mx ↔ N x),Ma ∧ ∃xRxa ` ∃xN x1 ∀x(Mx ↔ N x)

2 Ma ∧ ∃xRxa3 Ma ∧E 2

4 Ma ↔ N a ∀E 1

5 N a ↔E 4, 3

6 ∃xN x ∃I 55. ∀x∀yGxy ` ∃xGxx

1 ∀x∀yGxy2 ∀yGay ∀E 1

3 Gaa ∀E 2

4 ∃xGxx ∃I 36. ` ∀xRxx → ∃x∃yRxy

1 ∀xRxx2 Raa ∀E 1

3 ∃yRay ∃I 24 ∃x∃yRxy ∃I 35 ∀xRxx → ∃x∃yRxy →I 1–4

7. ` ∀y∃x(Qy → Qx)1 Qa

2 Qa R 1

3 Qa → Qa →I 1–2

4 ∃x(Qa → Qx) ∃I 35 ∀y∃x(Qy → Qx) ∀I 4

8. N a → ∀x(Mx ↔ Ma),Ma,¬Mb ` ¬N a


1 N a → ∀x(Mx ↔ Ma)

2 Ma

3 ¬Mb

4 N a

5 ∀x(Mx ↔ Ma) →E 1, 4

6 Mb ↔ Ma ∀E 5

7 Mb ↔E 6, 2

8 ⊥ ⊥I 7, 3

9 ¬N a ¬I 4–8


9. ∀x∀y(Gxy → Gyx) ` ∀x∀y(Gxy ↔ Gyx)1 ∀x∀y(Gxy → Gyx)

2 Gab

3 ∀y(Gay → Gya) ∀E 1

4 Gab → Gba ∀E 3

5 Gba →E 4, 2

6 Gba

7 ∀y(Gby → Gyb) ∀E 1

8 Gba → Gab ∀E 7

9 Gab →E 8, 6

10 Gab ↔ Gba ↔I 2–5, 6–9

11 ∀y(Gay ↔ Gya) ∀I 1012 ∀x∀y(Gxy ↔ Gyx) ∀I 11

10. ∀x(¬Mx ∨ L jx),∀x(Bx → L jx),∀x(Mx ∨ Bx) ` ∀xL j x1 ∀x(¬Mx ∨ L jx)2 ∀x(Bx → L jx)

3 ∀x(Mx ∨ Bx)4 ¬Ma ∨ L jx ∀E 1

5 Ba → L ja ∀E 2

6 Ma ∨ Ba ∀E 3

7 ¬Ma

8 Ba DS 6, 7

9 L ja →E 5, 8

10 L ja

11 L ja R 10

12 L ja ∨E 4, 7–9, 10–11

13 ∀xL j x ∀I 12

F. Write a symbolization key for the following argument, symbolize it, and proveit:

There is someone who likes everyone who likes everyone that shelikes. Therefore, there is someone who likes herself.

Symbolization key:


domain: all peopleLxy : x likes y

∃x∀y(∀z (Lxz → Lyz ) → Lxy) .Û. ∃xLxx1 ∃x∀y(∀z (Lxz → Lyz ) → Lxy)

2 ∀y(∀z (Laz → Lyz ) → Lay)

3 ∀z (Laz → Laz ) → Laa ∀E 2

4 Lac

5 Lac R 4

6 Lac → Lac →I 4–5

7 ∀z (Laz → Laz ) ∀I 68 Laa →E 3, 7

9 ∃xLxx ∃I 810 ∃xLxx ∃E 1, 2––9

G. Show that each pair of sentences is provably equivalent.

1. ∀x(Ax → ¬Bx), ¬∃x(Ax ∧ Bx)2. ∀x(¬Ax → Bd ), ∀xAx ∨ Bd3. ∃xP x → Qc , ∀x(P x → Qc )

H. For each of the following pairs of sentences: If they are provably equivalent,give proofs to show this. If they are not, construct an interpretation to show thatthey are not logically equivalent.

1. ∀xP x → Qc,∀x(P x → Qc ) Not logically equivalentCounter-interpretation: let the domain be the numbers 1 and 2. Let ‘c ’name 1. Let ‘P x ’ be true of and only of 1. Let ‘Qx ’ be true of, and only of,2.

2. ∀x∀y∀zBxyz,∀xBxxx Not logically equivalentCounter-interpretation: let the domain be the numbers 1 and 2. Let ‘Bxyz ’be true of, and only of, 〈1,1,1〉 and 〈2,2,2〉.

3. ∀x∀yDxy,∀y∀xDxy Provably equivalent

1 ∀x∀yDxy2 ∀yDay ∀E 1

3 Dab ∀E 2

4 ∀xDxb ∀I 35 ∀y∀xDxy ∀I 4

1 ∀y∀xDxy2 ∀xDxa ∀E 1

3 Dba ∀E 2

4 ∀yDby ∀I 35 ∀x∀yDxy ∀I 4


4. ∃x∀yDxy,∀y∃xDxy Not logically equivalentCounter-interpretation: let the domain be the numbers 1 and 2. Let ‘Dxy ’hold of and only of 〈1,2〉 and 〈2,1〉. This is depicted thus:

1 2

5. ∀x(Rca ↔ Rxa),Rca ↔ ∀xRxa Not logically equivalentCounter-interpretation, consider the following diagram, allowing ‘a’ toname 1 and ‘c ’ to name 2:

1 2

I. For each of the following arguments: If it is valid in FOL, give a proof. If it isinvalid, construct an interpretation to show that it is invalid.

1. ∃y∀xRxy .Û. ∀x∃yRxy Valid1 ∃y∀xRxy2 ∀xRxa3 Rba ∀E 2

4 ∃yRby ∃I 35 ∃yRby ∃E 1, 2–4

6 ∀x∃yRxy ∀I 52. ∀x∃yRxy .Û. ∃y∀xRxy Not valid

Counter interpretation: let the domain be the numbers 1 and 2. Let ‘Rxy ’be true of 1 and 2, and of 2 and 1 (but not 1 and itself or 2 and itself).

3. ∃x(P x ∧ ¬Qx) .Û. ∀x(P x → ¬Qx) Not validCounter interpretation: let the domain be the numbers 1 and 2. Let ‘P x ’be true of everything in the domain. Let ‘Qx ’ be true of, and only of, 2.

4. ∀x(Sx → T a),Sd .Û. T a Valid1 ∀x(Sx → T a)

2 Sd

3 Sd → T a ∀E 1

4 T a →E 3, 2

5. ∀x(Ax → Bx),∀x(Bx → Cx) .Û. ∀x(Ax → Cx) Valid


1 ∀x(Ax → Bx)

2 ∀x(Bx → Cx)

3 Aa → Ba ∀E 1

4 Ba → Ca ∀E 2

5 Aa

6 Ba →E 3, 5

7 Ca →E 4, 6

8 Aa → Ca →I 5–7

9 ∀x(Ax → Cx) ∀I 86. ∃x(Dx ∨ Ex),∀x(Dx → Fx) .Û. ∃x(Dx ∧ Fx) Invalid

Counter-interpretation: let the domain be the number 1 . Let ‘Dx ’ hold ofnothing. Let both ‘Ex ’ and ‘Fx ’ hold of everything.

7. ∀x∀y(Rxy ∨Ryx) .Û. R j j Valid1 ∀x∀y(Rxy ∨Ryx)2 ∀y(R j y ∨Ry j ) ∀E 1

3 R j j ∨R j j ∀E 2

4 R j j

5 R j j R 4

6 R j j

7 R j j R 6

8 R j j ∨E 3, 4–5, 6–7

8. ∃x∃y(Rxy ∨Ryx) .Û. R j j InvalidCounter-interpretation: consider the following diagram, allowing ‘ j ’ toname 2.

1 2

9. ∀xP x → ∀xQx,∃x¬P x .Û. ∃x¬Qx InvalidCounter-interpretation: let the domain be the number 1. Let ‘P x ’ be trueof nothing. Let ‘Qx ’ be true of everything.

10. ∃xM x → ∃xN x , ¬∃xN x .Û. ∀x¬Mx Valid


1 ∃xM x → ∃xN x2 ¬∃xN x3 Ma

4 ∃xM x ∃I 35 ∃xN x →E 1, 4

6 ⊥ ⊥I 5, 2

7 ¬Ma ¬I 3–6

8 ∀x¬Mx ∀I 7

CHAPTER 33

Conversion ofquanti�ersA. Show in each case that the sentences are provably inconsistent:

1. Sa → T m,T m → Sa,T m ∧ ¬Sa1 Sa → T m

2 T m → Sa

3 T m ∧ ¬Sa

4 T m ∧E 3

5 ¬Sa ∧E 3

6 Sa →E 2, 4

7 ⊥ ⊥I 5, 62. ¬∃xRxa,∀x∀yRyx

1 ¬∃xRxa2 ∀x∀yRyx3 ∀x¬Rxa CQ 1

4 ¬Rba ∀E 3

5 ∀yRya ∀E 2

6 Rba ∀E 5

7 ⊥ ⊥I 6, 43. ¬∃x∃yLxy,Laa

94

CHAPTER 33. CONVERSION OF QUANTIFIERS 95

1 ¬∃x∃yLxy2 Laa

3 ∀x¬∃yLxy CQ 1

4 ¬∃yLay ∀E 3

5 ∀y¬Lay CQ 4

6 ¬Laa ∀E 5

7 ⊥ ⊥I 2, 64. ∀x(P x → Qx),∀z (P z → Rz ),∀yP y,¬Qa ∧ ¬Rb

1 ∀x(P x → Qx)

2 ∀z (P z → Rz )

3 ∀yP y4 ¬Qa ∧ ¬Rb

5 ¬Qa ∧E 4

6 P a → Qa ∀E 1

7 ¬P a MT 6, 5

8 P a ∀E 3

9 ⊥ ⊥I 8, 7

B. Show that each pair of sentences is provably equivalent:

1. ∀x(Ax → ¬Bx),¬∃x(Ax ∧ Bx)

1 ∀x(Ax → ¬Bx)

2 ∃x(Ax ∧ Bx)3 Aa ∧ Ba

4 Aa ∧E 3

5 Ba ∧E 3

6 Aa → ¬Ba ∀E 1

7 ¬Ba →E 6, 4

8 ⊥ ⊥I 5, 7

9 ⊥ ∃E 2, 3–8

10 ¬∃x(Ax ∧ Bx) ¬I 2–9

1 ¬∃x(Ax ∧ Bx)2 ∀x¬(Ax ∧ Bx) CQ 1

3 ¬(Aa ∧ Ba) ∀E 2

4 Aa

5 Ba

6 Aa ∧ Ba ∧I 4, 5

7 ⊥ ⊥I 6, 3

8 ¬Ba ¬I 5–7

9 Aa → ¬Ba →I 4–8

10 ∀x(Ax → ¬Bx) ∀I 9

2. ∀x(¬Ax → Bd ),∀xAx ∨ Bd


1 ∀x(¬Ax → Bd )

2 ¬Aa → Bd ∀E 1

3 Bd

4 ∀xAx ∨ Bd ∨I 6

5 ¬Bd

6 ¬¬Aa MT 2, 5

7 Aa DNE 6

8 ∀xAx ∀E 7

9 ∀xAx ∨ Bd ∨I 8

10 ∀xAx ∨ Bd TND 3–4, 5–9

1 ∀xAx ∨ Bd2 ¬Aa

3 ∀xAx4 Aa ∀E 3

5 ⊥ ⊥I 4, 2

6 ¬∀xAx ¬I 3–5

7 Bd DS 1, 6

8 ¬Aa → Bd →I 2–7

9 ∀x(Ax → Bd ) ∀I 8

C. In §22, we considered what happens when we move quantifiers ‘across’ variouslogical operators. Show that each pair of sentences is provably equivalent:

1. ∀x(Fx ∧Ga),∀xF x ∧Ga

1 ∀x(Fx ∧Ga)2 Fb ∧Ga ∀E 1

3 Fb ∧E 2

4 Ga ∧E 6

5 ∀xF x ∀I 36 ∀xF x ∧Ga ∧I 5, 4

1 ∀xF x ∧Ga2 ∀xF x ∧E 1

3 Ga ∧E 1

4 Fb ∀E 2

5 Fb ∧Ga ∧I 4, 3

6 ∀x(Fx ∧Ga) ∀I 5

2. ∃x(Fx ∨Ga),∃xF x ∨Ga

1 ∃x(Fx ∨Ga)2 Fb ∨Ga

3 Fb

4 ∃xF x ∃I 35 ∃xF x ∨Ga ∨I 4

6 Ga

7 ∃xF x ∨Ga ∨I 6

8 ∃xF x ∨Ga ∨E 2, 3–5, 6–7

9 ∃xF x ∨Ga ∃E 1, 2–8

1 ∃xF x ∨Ga2 ∃xF x3 Fb

4 Fb ∨Ga ∨I 3

5 ∃x(Fx ∨Ga) ∃I 46 ∃x(Fx ∨Ga) ∃E 2, 3–5

7 Ga

8 Fb ∨Ga ∨I 7

9 ∃x(Fx ∨Ga) ∃I 810 ∃x(Fx ∨Ga) ∨E 1, 2–6, 7–9


3. ∀x(Ga → Fx),Ga → ∀xF x

1 ∀x(Ga → Fx)

2 Ga → Fb ∀E 1

3 Ga

4 Fb →E 2, 3

5 ∀xF x ∀I 46 Ga → ∀xF x →I 3–5

1 Ga → ∀xF x2 Ga

3 ∀xF x →E 1, 2

4 Fb ∀E 3

5 Ga → Fb →I 2–4

6 ∀x(Ga → Fx) ∀I 5

4. ∀x(Fx → Ga),∃xF x → Ga

1 ∀x(Fx → Ga)

2 ∃xF x3 Fb

4 Fb → Ga ∀E 1

5 Ga →E 4, 3

6 Ga ∃E 2, 3–5

7 ∃xF x → Ga →I 2–6

1 ∃xF x → Ga

2 Fb

3 ∃xF x ∃I 24 Ga →E 1, 3

5 Fb → Ga →I 2–4

6 ∀x(Fx → Ga) ∀I 5

5. ∃x(Ga → Fx),Ga → ∃xF x

1 ∃x(Ga → Fx)

2 Ga

3 Ga → Fb

4 Fb →E 3, 2

5 ∃xF x ∃I 46 ∃xF x ∃E 1, 3–5

7 Ga → ∃xF x →I 2–6

1 Ga → ∃xF x2 Ga

3 ∃xF x4 Fb

5 Ga

6 Fb R 4

7 Ga → Fb →I 5–6

8 ∃x(Ga → Fx) ∃I 79 ∃x(Ga → Fx) ∃E 3, 4–8

10 ¬Ga

11 Ga

12 ⊥ ⊥I 11, 10

13 Fb ⊥E 12

14 Ga → Fb →E 11–13

15 ∃x(Ga → Fx) ∃I 1416 ∃x(Ga → Fx) TND 2–9, 10–15


6. ∃x(Fx → Ga),∀xF x → Ga

1 ∃x(Fx → Ga)

2 ∀xF x3 Fb → Ga

4 Fb ∀E 2

5 Ga →E 3, 4

6 Ga ∃E 1, 3–5

7 ∀xF x → Ga →I 2–6

1 ∀xF x → Ga

2 ∀xF x3 Ga →E 1, 2

4 Fb

5 Ga R 3

6 Fb → Ga →I 4–5

7 ∃x(Fx → Ga) ∃I 68 ¬∀xF x9 ∃x¬Fx CQ 8

10 ¬Fb

11 Fb

12 ⊥ ⊥I 11, 10

13 Ga ⊥E 12

14 Fb → Ga →I 11–13

15 ∃x(Fx → Ga) ∃I 1416 ∃x(Fx → Ga) ∃E 9, 10–15

17 ∃x(Fx → Ga) TND 2–7, 8–16

NB: the variable ‘x ’ does not occur in ‘Ga’.When all the quantifiers occur at the beginning of a sentence, that sentence is

said to be in prenex normal form. Together with the CQ rules, these equivalencesare sometimes called prenexing rules, since they give us a means for putting anysentence into prenex normal form.

CHAPTER 34

Rules foridentityA. Provide a proof of each claim.

1. P a ∨Qb,Qb → b = c,¬P a ` Qc1 P a ∨Qb

2 Qb → b = c

3 ¬P a

4 Qb DS 1, 3

5 b = c →E 2, 4

6 Qc =E 5, 4

2. m = n ∨ n = o,An ` Am ∨ Ao1 m = n ∨ n = o

2 An

3 m = n

4 Am =E 3, 2

5 Am ∨ Ao ∨I 4

6 n = o

7 Ao =E 6, 7

8 Am ∨ Ao ∨I 7

9 Am ∨ Ao ∨E 1, 3–5, 6–8

3. ∀x x = m,Rma ` ∃xRxx

99

CHAPTER 34. RULES FOR IDENTITY 100

1 ∀x x = m2 Rma

3 a = m ∀E 1

4 Raa =E 3, 2

5 ∃xRxx ∃I 44. ∀x∀y(Rxy → x = y) ` Rab → Rba

1 ∀x∀y(Rxy → x = y)

2 Rab

3 ∀y(Ray → a = y) ∀E 1

4 Rab → a = b ∀E 3

5 a = b →E 4, 2

6 Raa =E 5, 2

7 Rba =E 5, 6

8 Rab → Rba →I 2–7

5. ¬∃x¬x = m ` ∀x∀y(P x → P y)1 ¬∃x¬x = m2 ∀x¬¬x = m CQ 1

3 ¬¬a = m ∀E 2

4 a = m DNE 3

5 ¬¬b = m ∀E 2

6 b = m DNE 5

7 P a

8 Pm =E 3, 7

9 Pb =E 5, 8

10 P a → Pb →I 7–9

11 ∀y(P a → P y) ∀I 1012 ∀x∀y(P x → P y) ∀I 11

6. ∃x J x,∃x¬ J x ` ∃x∃y ¬x = y


1 ∃x J x2 ∃x¬ J x3 J a

4 ¬ J b

5 a = b

6 J b =E 5, 3

7 ⊥ ⊥I 6, 4

8 ¬a = b ¬I 5–7

9 ∃y¬a = y ∃I 810 ∃x∃y¬x = y ∃I 911 ∃x∃y¬x = y ∃E 2, 4–10

12 ∃x∃y¬x = y ∃E 1, 3–11

7. ∀x(x = n ↔ Mx),∀x(Ox ∨ ¬Mx) ` On1 ∀x(x = n ↔ Mx)

2 ∀x(Ox ∨ ¬Mx)

3 n = n ↔ Mn ∀E 1

4 n = n =I

5 Mn ↔E 3, 4

6 On ∨ ¬Mn ∀E 2

7 ¬On

8 ¬Mn DS 6, 7

9 ⊥ ⊥I 5, 8

10 ¬¬On ¬I 7–9

11 On DNE 10

8. ∃xDx,∀x(x = p ↔ Dx) ` Dp1 ∃xDx2 ∀x(x = p ↔ Dx)

3 Dc

4 c = p ↔ Dc ∀E 2

5 c = p ↔E 4, 3

6 Dp =E 5, 3

7 Dp ∃E 1, 3–6


9. ∃x [(Kx ∧ ∀y(Ky → x = y)) ∧ Bx],Kd ` Bd

1 ∃x [(Kx ∧ ∀y(Ky → x = y) ∧ Bx]

2 Kd

3 (Ka ∧ ∀y(Ky → a = y)) ∧ Ba

4 Ka ∧ ∀y(Ky → a = y) ∧E 3

5 Ka ∧E 4

6 ∀y(Ky → a = y) ∧E 4

7 Kd → a = d ∀E 6

8 a = d →E 7, 2

9 Ba ∧E 3

10 Bd =E 8, 9

11 Bd ∃E 1, 3–10

10. ` P a → ∀x(P x ∨ ¬x = a)1 P a

2 b = a

3 Pb =E 2, 1

4 Pb ∨ ¬b = a ∨I 3

5 ¬b = a

6 Pb ∨ ¬b = a ∨I 5

7 Pb ∨ ¬b = a TND 2–4, 5–6

8 ∀x(P x ∨ ¬x = a) ∀I 79 P a → ∀x(P x ∨ ¬x = a) →I 1–8

B. Show that the following are provably equivalent:

• ∃x ([Fx ∧ ∀y(F y → x = y)] ∧ x = n)

• Fn ∧ ∀y(F y → n = y)

And hence that both have a decent claim to symbolize the English sentence ‘Nickis the F’.In one direction:


1 ∃x ([Fx ∧ ∀y(F y → x = y)] ∧ x = n)

2 [Fa ∧ ∀y(F y → a = y)] ∧ a = n

3 a = n ∧E 2

4 Fa ∧ ∀y(F y → a = y) ∧E 2

5 Fa ∧E 4

6 Fn =E 3, 5

7 ∀y(F y → a = y) ∧E 4

8 ∀y(F y → n = y) =E 3, 7

9 Fn ∧ ∀y(F y → n = y) ∧I 6, 8

10 Fn ∧ ∀y(F y → n = y) ∃E 1, 2–9

And now in the other:1 Fn ∧ ∀y(F y → n = y)

2 n = n =I

3 [Fn ∧ ∀y(F y → n = y)] ∧ n = n ∧I 1, 2

4 ∃x ([Fx ∧ ∀y(F y → x = y)] ∧ x = n) ∃I 3

C. In §24, we claimed that the following are logically equivalent symbolizationsof the English sentence ‘there is exactly one F’:

• ∃xF x ∧ ∀x∀y [(Fx ∧ F y) → x = y]

• ∃x [Fx ∧ ∀y(F y → x = y)]

• ∃x∀y(F y ↔ x = y)

Show that they are all provably equivalent. (Hint: to show that three claims areprovably equivalent, it su�ces to show that the first proves the second, the secondproves the third and the third proves the first; think about why.)It su�ces to show that the first proves the second, the second proves the thirdand the third proves the first, for we can then show that any of them prove anyothers, just by chaining the proofs together (numbering lines, where necessary.Armed with this, we start on the first proof:


1 ∃xF x ∧ ∀x∀y [(Fx ∧ F y) → x = y]

2 ∃xF x ∧E 1

3 ∀x∀y [(Fx ∧ F y) → x = y]

∧E 1

4 Fa

5 ∀y [(Fa ∧ F y) → a = y] ∀E 3

6 (Fa ∧ Fb) → a = b ∀E 5

7 Fb

8 Fa ∧ Fb ∧I 4, 7

9 a = b →E 6, 8

10 Fb → a = b →I 7–9

11 ∀y(F y → a = y) ∀I 1012 Fa ∧ ∀y(F y → a = y)) ∧I 4, 11

13 ∃x [Fx ∧ ∀y(F y → x = y)] ∃I 12

14 ∃x [Fx ∧ ∀y(F y → x = y)] ∃E 2, 4–13

Now for the second proof:1 ∃x [Fx ∧ ∀y(F y → x = y)

]2 Fa ∧ ∀y(F y → a = y)

3 Fa ∧E 2

4 ∀y(F y → a = y) ∧E 2

5 Fb

6 Fb → a = b ∀E 4

7 a = b →E 6, 5

8 a = b

9 Fb =E 8, 3

10 Fb ↔ a = b ↔I 5–7, 8–9

11 ∀y(F y ↔ a = y) ∀I 1012 ∃x∀y(F y ↔ x = y) ∃I 1113 ∃x∀y(F y ↔ x = y) ∃E 1, 2–12

And finally, the third proof:


1 ∃x∀y(F y ↔ x = y)

2 ∀y(F y ↔ a = y)

3 Fa ↔ a = a ∀E 2

4 a = a =I

5 Fa ↔E 3, 4

6 ∃xF x ∃I 57 Fb ∧ F c

8 Fb ∧E 7

9 Fb ↔ a = b ∀E 2

10 a = b ↔E 9, 8

11 F c ∧E 7

12 F c ↔ a = c ∀E 2

13 a = c ↔E 12, 11

14 b = c =E 10, 13

15 (Fb ∧ F c ) → b = c →I 8–14

16 ∀y [(Fb ∧ F y) → b = y] ∀I 15

17 ∀x∀y [(Fx ∧ F y) → x = y] ∀I 16

18 ∃xF x ∧ ∀x∀y [(Fx ∧ F y) → x = y]

∧I 6, 17

19 ∃xF x ∧ ∀x∀y [(Fx ∧ F y) → x = y] ∃E 1, 2–18

D. Symbolize the following argument

There is exactly one F. There is exactly one G. Nothing is both F andG. So: there are exactly two things that are either F or G.

And o�er a proof of it.

1. ∃x [Fx ∧ ∀y(F y → x = y)]

2. ∃x [Gx ∧ ∀y(Gy → x = y)]

3. ∀x(¬Fx ∨ ¬Gx) .Û..Û. ∃x∃y [¬x = y ∧ ∀z ((F z ∨Gz ) → (x = z ∨ y = z ))

]


1 ∃x [Fx ∧ ∀y(F y → x = y)]

2 ∃x [Gx ∧ ∀y(Gy → x = y)]

3 ∀x(¬Fx ∨ ¬Gx)

4 Fa ∧ ∀y(F y → a = y)

5 Fa ∧E 4

6 ∀y(F y → a = y) ∧E 4

7 ¬Fa ∨ ¬Ga ∀E 3

8 ¬Ga DS 7, 5

9 Gb ∧ ∀y(Gy → b = y)

10 Gb ∧E 9

11 ∀y(Gy → b = y) ∧E 9

12 a = b

13 Ga =E 12, 10

14 ⊥ ⊥I 13, 8

15 ¬a = b ¬I 12–14

16 F c ∨Gc

17 F c

18 F c → a = c ∀E 6

19 a = c →E 18, 17

20 a = c ∨ b = c ∨I 19

21 Gc

22 Gc → b = c ∀E 11

23 b = c →E 22, 21

24 a = c ∨ b = c ∨I 23

25 a = c ∨ b = c ∨E 16, 17–20, 21–24

26 (F c ∨Gc ) → (a = c ∨ b = c ) →I 16–25

27 ∀z ((F z ∨Gz ) → (a = z ∨ b = z )) ∀I 2628 ¬a = b ∧ ∀z ((F z ∨Gz ) → (a = z ∨ b = z )) ∧I 15, 27

29 ∃y [¬a = y ∧ ∀z ((F z ∨Gz ) → (a = z ∨ y = z ))] ∃I 28

30 ∃x∃y [¬x = y ∧ ∀z ((F z ∨Gz ) → (x = z ∨ y = z ))] ∃I 29

31 ∃x∃y [¬x = y ∧ ∀z ((F z ∨Gz ) → (x = z ∨ y = z ))] ∃E 2, 9–30

32 ∃x∃y [¬x = y ∧ ∀z ((F z ∨Gz ) → (x = z ∨ y = z ))] ∃E 1, 4–31

CHAPTER 35

Derived rulesA. O�er proofs which justify the addition of the second and fourth CQ rules asderived rules.Justification for the second rule:1 ¬∃xAx2 Aa

3 ∃xAx ∃I 24 ⊥ ⊥I 3, 1

5 ¬Aa ¬I 2–4

6 ∀x¬Ax ∀I 5The fourth rule is harder to justify. Here is a proof that is relatively straightfor-ward, but uses the derived rule DNE:1 ¬∀xAx2 ¬∃x¬Ax3 ¬Aa

4 ∃x¬Ax ∃I 35 ⊥ ⊥I 4, 2

6 ¬¬Aa ¬I 3–5

7 Aa DNE 6

8 ∀xAx ∀I 79 ⊥ ⊥I 8, 1

10 ¬¬∃x¬Ax ¬I 2–9

11 ∃x¬Ax DNE 10

And here is a proof that does not use any derived rules:

107

CHAPTER 35. DERIVED RULES 108

1 ¬∀xAx2 ∃x¬Ax3 ∃x¬Ax ∧ ∃x¬Ax ∧I 2

4 ∃x¬Ax ∧E 3

5 ¬∃x¬Ax6 Ab

7 Ab ∧ Ab ∧I 6

8 Ab ∧E 7

9 ¬Ab

10 ∃x¬Ax ∃I 911 ⊥ ⊥I 10, 5

12 Ab ⊥E 11

13 Ab TND 6–8, 9–12

14 ∀xAx ∀I 1315 ⊥ ⊥I 14, 1

16 ∃x¬Ax ⊥E 15

17 ∃x¬Ax TND 2–4, 5–16

CHAPTER 37

Normal Formsand ExpressiveCompletenessPractice exercises

A. Consider the following sentences:

1. (A → ¬B)2. ¬(A ↔ B)3. (¬A ∨ ¬(A ∧ B))4. (¬(A → B) ∧ (A → C ))5. (¬(A ∨ B) ↔ ((¬C ∧ ¬A) → ¬B))6. ((¬(A ∧ ¬B) → C ) ∧ ¬(A ∧D))

For each sentence, find a tautologically equivalent sentence in DNF and one inCNF.

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