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(1) Formalize the following compound propositions from natural language into propositional logic:
[1] It is raining and it is cold.
p ∧ q
[2] It is not raining and it is cold.
¬p ∧ q
[3] It is raining or it is cold.
p ∨ q
[4] Either it is raining, or it is cold.
(p ∨ q) ∧ ¬(p ∧ q)
[5] It is raining and cold, or it is snowing.
(p ∧ q) ∨ r
[6] It is not true that it is raining and cold.
¬(p ∧ q)
[7] It is not true that it is not raining and not cold.
¬(¬p ∧ ¬q)
[8] If it rains, it is cold.
p → q
[9] If it rains, then if it is cold, it will snow.
p → (q → r)
[10] It is not true that if it rains and snows or it is cold, then it must be cold.
¬[(p ∧ r) ∨ q → q]
(2) Formalize the following compound propositions from natural language into propositional logic:
[11] He is blond and has blue eyes or is tall.
(p ∧ q) ∨ r
[12] It is not true that he is blond and has blue eyes.
¬(p ∧ q)
[13] He is not blond and does not have blue eyes.
¬p ∧ ¬q
[14] When it rains, the ground gets wet.
p → q
[15] If the dog barks again, I will bite it.
p → q
[16] Keep shaking the glass and the water will spill.
p → q
[17] Eat and be quiet.
p ∧ q
[18] There were pieces of metal, newspapers, and food scraps.
p ∧ q ∧ r
[19] The moon is indifferent to our verses; if it were not, it would have left long ago.
p ∧ (¬p → q)
[20] Rain does not imply that the ground gets wet.
¬(p → q)
[21] When it rains the ground gets wet, but it is not true that it only gets wet when it rains.
(p → q) ∧ ¬(q → p)
[22] If I eat a lot, I gain weight. And if I gain weight, I feel bad.
p → q, q → r
[23] If I study, I get sleepy. If I get sleepy, I sleep. If I sleep, I wake up nervous for not having studied. If I wake up nervous for not having studied, then I study. Therefore, I get sleepy.
Variables:
p = I study
q = I get sleepy
r = I sleep
s = I wake up nervous
Formalization:
p → q, q → r, r → (¬p ∧ s), (¬p ∧ s) → p ⊦ q
[24] If the King of Argentina is bald, then there is a King of Argentina. If the King of Argentina is not bald, then there is a King of Argentina. There is no King of Argentina. Therefore, the King of Argentina is bald if and only if the King of Argentina is not bald.
p → q, ¬p → q, ¬q ⊦ p ↔ ¬p
[25] If evil exists in the world and does not originate from human actions, then God cannot or does not want to prevent it. Evil exists in the world. If God cannot prevent evil in this world, then He is not omnipotent. If God does not want to prevent the existence of evil, then He is not good. But God is omnipotent and good. Therefore, the evil that exists in this world originates from human actions.
p ∧ ¬q → (¬r ∨ ¬s), p, ¬r → ¬t, ¬s → ¬w, t ∧ w ⊦ q
(3) Formalize the following hypothetical sentences correctly according to propositional logic:
[26] If you want a dog, then you must have time.
p → q
[27] Only if you are not wearing sneakers can you enter the club.
q → ¬p
"Only if" indicates a necessary condition, not a sufficient one.
[28] You can come to my house whenever you want.
p → q
"Whenever" is equivalent to "if".
[29] It is sufficient to get a 5 to enter university.
p → q
"It is sufficient" indicates a sufficient condition.
[30] It is necessary to take the exam to enter university.
q → p
"It is necessary" indicates a necessary condition.
[31] To have friends, you must have health, money, fame, and charisma.
p → (q ∧ r ∧ s ∧ t)
[32] Unless you prevent the murder, the agent will succeed.
¬p → q
"Unless" is equivalent to "if not".
[33] Unless you buy a lot of food, it will run out soon.
¬p → q
"Unless" is equivalent to "if not".
[34] You will only achieve success if you have the best grades.
p → q
[35] When you pass, everything will be wonderful.
p → q
[36] Adrian is a logic professor if and only if he has a degree.
p ↔ q
"If and only if" indicates a biconditional.
[37] In case you like the exercises, you will hire the teacher.
p → q
[38] It is enough for John to arrive to ruin the party.
p → q
"It is enough" indicates a sufficient condition.
[39] Pablo's company is essential for the night to be complete.
q → p
"Essential" indicates a necessary condition.
[40] When one does what one can, one is not obligated to do more.
p → ¬q
(4) Formalize the following disjunction arguments using propositional logic:
[41] You either buy a PlayStation or a Nintendo.
p ∨ q
Inclusive disjunction.
[42] You either pass or fail.
(p ∨ q) ∧ ¬(p ∧ q)
Exclusive disjunction: you cannot pass and fail at the same time.
[43] Either he is single, or he is not.
p ∨ ¬p
Tautology: law of excluded middle.
[44] We are looking for someone who speaks English or German.
p ∨ q
Inclusive disjunction: they may know both languages.
[45] Either it is an analytic proposition or it is a synthetic proposition.
(p ∨ q) ∧ ¬(p ∧ q)
Exclusive disjunction: a proposition cannot be both.
(5) Formalize the following arguments using predicate logic with simple quantification:
[46] Everyone is mortal. Therefore, no one is immortal.
∀xMx ⊦ ¬∃x¬Mx
[47] Third world countries are not industrialized. Some third world countries possess great wealth. Therefore, there are third world countries that possess great wealth and are not industrialized.
∀x(Px ∧ Tx → ¬Ix), ∃x(Px ∧ Tx ∧ Gx) ⊦ ∃x(Px ∧ Tx ∧ Gx ∧ ¬Ix)
[48] No athlete who aspires to participate in the Olympics drinks alcoholic beverages. There are athletes who drink alcoholic beverages. Therefore, some athletes do not participate in the Olympics.
∀x(Dx ∧ Ax → ¬Ix), ∃x(Dx ∧ Ix) ⊦ ∃x(Dx ∧ ¬Ax)
[49] Doctors and engineers are professionals. Professionals and managers are respected. Therefore, doctors are respected.
∀x(Mx ∨ Ix → Px), ∀x(Px ∨ Dx → Rx) ⊦ ∀x(Mx → Rx)
[50] There are intelligent men. Therefore, it is not the case that no man is intelligent.
∃x(Hx ∧ Ix) ⊦ ¬∀x(Hx → ¬Ix)
[51] Philosophers and only philosophers are intelligent. Therefore, those who are not philosophers are not intelligent.
¬∃x(¬Fx ∧ Ix) ⊦ ∀x(¬Fx → ¬Ix)
[52] No perfect being is immoral. Any individual who does not value intellectual honesty is imperfect. No moral individual who values intellectual honesty can condemn agnosticism. It follows that if God is perfect, He cannot condemn agnosticism.
∀x(Px → Mx), ∀x(¬Hx → ¬Px), ∀x(Mx ∧ Hx → ¬Cx) ⊦ Pd → ¬Cd
[53] Mathematical propositions are necessary. Only synthetic propositions have content. There are no synthetic a priori propositions. Every proposition is either a priori or a posteriori. Therefore, mathematical propositions are synthetic a posteriori.
∀x(Mx → Nx), ¬∃x(¬Sx ∧ Cx), ¬∃x(Sx ∧ Ax), ∀x[Ax ∨ Bx ∧ ¬(Ax ∧ Bx)] ⊦ ∀x(Mx → Sx ∧ Bx)
[54] Mathematical propositions are necessary. A posteriori propositions are not necessary. Mathematical propositions have content. Only propositions with content are synthetic. Therefore, mathematical propositions are synthetic a priori.
∀x(Mx → Nx), ∀x(Bx → ¬Nx), ∀x(Mx → Cx), ¬∃x(¬Cx ∧ Sx) ⊦ ∀x(Mx → Sx ∧ Ax)
(6) Formalize the following arguments using predicate logic with multiple quantification:
[55] If Watson can catch Moriarty, Holmes can. Holmes cannot. Therefore, Watson cannot either.
Awm → Ahm, ¬Ahm ⊦ ¬Awm
[56] Only Holmes can catch Moriarty. Holmes cannot. Therefore, no one can.
∀x(Axm → Ahm), ¬Ahm ⊦ ¬∃x(Axm)
[57] If anyone can catch Moriarty, then Holmes can. Holmes cannot. Therefore, no one can catch him.
∃x(Axm) → Ahm, ¬Ahm ⊦ ¬∃x(Axm)
[58] Everyone is related to everyone. Therefore, everyone is related to themselves.
∀x∀y(Rxy) ⊦ ∀x(Rxx)
[59] Every boy is younger than his father. Charles is a boy who is not younger than Louis. Whoever is married to Mary is Charles's father. Therefore, Louis is not married to Mary.
∀x(Cx → Jxf(x)), Cc ∧ ¬Jcl, ∀x(Mxm → x = f(c)) ⊦ ¬Mlm
[60] Every empiricist admires Hume. Some idealists do not esteem anyone who admires Hume. Consequently, some idealists do not esteem any empiricist.
∀x(Ex → Axh), ∃x(Ix ∧ ∀y(Ayh → ¬Exy)) ⊦ ∃x(Ix ∧ ∀y(Ey → ¬Exy))
[61] There is a man whom all men admire. Therefore, there is at least one man who admires himself.
∃x(Hx ∧ ∀y(Hy → Ayx)) ⊦ ∃x(Hx ∧ Axx)
[62] Colonels command sergeants and sergeants command soldiers. Anyone who is commanded by another receives orders from them. Anyone who commands someone who in turn commands a third person, commands that third person. P is a Colonel, H is a Sergeant, and B is a soldier. Therefore, B receives orders from P.
∀x∀y(Cx ∧ Sy → Mxy), ∀x∀y(Sx ∧ Dy → Mxy), ∀x∀y(Mxy → Ryx), ∀x∀y∀z(Mxy ∧ Myz → Mxz), Cp, Sh, Db ⊦ Rbp
[63] It is a criminal who sells an unregistered pistol. All weapons that John owns were bought by him from Luis's store or Jose's store. Thus, if one of John's weapons is an unregistered pistol, then if John has never bought anything from Jose's store, Luis is a criminal.
∀x∀y(Vxy ∧ Py ∧ ¬Ry → Dx), ∀x(Ajx → Cxl ∨ Cxj) ⊦ ∃x(Ajx ∧ Px ∧ ¬Rx) → (¬∃x(Cxj) → Dl)
[64] Anyone who reads Freud will misinterpret him unless they have psychiatric training. Anyone who reads Freud and misinterprets him contributes to their own mental illness. An immature person is incapable of correctly interpreting Freud. Not everyone who reads Freud and has psychiatric training is a mature person. Therefore, there are people with psychiatric training who contribute to their own mental illness.
∀x(Lxf ∧ ¬Px → Ix), ∀x(Lxf ∧ Ix → Cx), ∀x(¬Mx → Ix), ¬∀x(Lxf ∧ Px → Mx) ⊦ ∃x(Px ∧ Cx)
(7) Formalize the following compound propositions using predicate logic with functors and identity:
[65] Peter's father is Louis.
f(p) = l
Where f(x) = "the father of x"
[66] Peter's father is a football referee.
Af(p)
Where f(x) = "the father of x", A = "is a football referee"
[67] The sum of two and three is a prime number.
Ps(2,3)
Where s(x,y) = "the sum of x and y", P = "is a prime number"
[68] There are at least two natural numbers whose sum equals six.
∃x∃y(Nx ∧ Ny ∧ x ≠ y ∧ s(x,y) = 6)
[69] There are at least two natural numbers such that their sum with themselves equals their product with themselves.
∃x∃y(Nx ∧ Ny ∧ x ≠ y ∧ s(x,x) = p(x,x) ∧ s(y,y) = p(y,y))
[70] For every natural number, the product of that number with itself equals the square of that number.
∀x(Nx → p(x,x) = c(x))
Where p(x,y) = "product of x and y", c(x) = "square of x"
[71] The square of three is even.
Pc(3)
Where c(x) = "square of x", P = "is even"
[72] The product of three times four is a multiple of two.
Mp(3,4)2
Where M = "is a multiple of"
[73] The product of two times any natural number is even.
∀x(Nx → Pp(2,x))
[74] It is not true that three is a divisor of the cube of every odd natural number.
¬∀x(Nx ∧ Ix → D3cb(x))
Where cb(x) = "cube of x", D = "is a divisor of", I = "is odd"
(8) Formalize the following compound propositions using predicate logic with functors, descriptors, and identity:
[75] The cube of two equals the product of two times four.
cb(2) = p(2,4)
[76] There exist two natural numbers such that their product equals five.
∃x∃y(Nx ∧ Ny ∧ p(x,y) = 5)
[77] There do not exist two natural numbers such that their product equals five.
¬∃x∃y(Nx ∧ Ny ∧ x ≠ y ∧ p(x,y) = 5)
[78] There exists a natural number whose square is the cube of four.
∃x(Nx ∧ c(x) = cb(4))
[79] There exists a natural number whose cube is a divisor of any odd natural number.
∃x(Nx ∧ ∀y(Ny ∧ Iy → Dcb(x)y))
[80] The author of Capital is Marx.
ιx(Axc) = m
Where ι is the definite descriptor
[81] The writer of The Republic is Plato.
ιx(Exr) = p
[82] The author of The Alchemist is a writer.
Eιx(Axa)
[83] The logic professor at the Zaragoza science faculty studied philosophy.
Fιx(Pxlz)
Where P = "is a logic professor at", F = "studied philosophy"
[84] Adrian's director is an excellent researcher.
Iιx(Dxa)
Where D = "is director of", I = "is an excellent researcher"
(9) Formalize the following compound propositions using predicate logic with numerical quantification:
[85] There is at least one God.
∃xDx
[86] There are at least two Gods.
∃x∃y(Dx ∧ Dy ∧ x ≠ y)
[87] There is at most one God.
∀x∀y(Dx ∧ Dy → x = y)
Or equivalently: ¬∃x∃y(Dx ∧ Dy ∧ x ≠ y)
[88] There are at most two Gods.
∀x∀y∀z(Dx ∧ Dy ∧ Dz → x = y ∨ x = z ∨ y = z)
[89] There is exactly one God.
∃x(Dx ∧ ∀y(Dy → x = y))
At least one and at most one.
[90] There are exactly two Gods.
∃x∃y(Dx ∧ Dy ∧ x ≠ y ∧ ∀z(Dz → z = x ∨ z = y))
[91] Some philosopher tames exactly one tiger.
∃x(Fx ∧ ∃y(Ty ∧ Dxy ∧ ∀z(Tz ∧ Dxz → z = y)))
[92] A tiger is tamed by exactly one philosopher.
∃x(Tx ∧ ∃y(Fy ∧ Dyx ∧ ∀z(Fz ∧ Dzx → z = y)))
[93] At least two linesmen who use at most one flag.
∃x∃y(Jx ∧ Jy ∧ x ≠ y ∧ ∀z∀w(Bz ∧ Bw ∧ Uxz ∧ Uxw → z = w) ∧ ∀z∀w(Bz ∧ Bw ∧ Uyz ∧ Uyw → z = w))
[94] Exactly two linesmen use exactly the same flag.
∃x∃y∃z(Jx ∧ Jy ∧ Bz ∧ x ≠ y ∧ Uxz ∧ Uyz ∧ ∀w(Jw ∧ Uwz → w = x ∨ w = y))
(10) Formalize the following arguments using predicate logic:
[95] An eccentric university professor set his office hours from 6 to 7 in the morning, with the following reasoning: "Students who need to talk to me will come to my office even at that hour, but those who don't need to will not come. Therefore, a student will come to my office if and only if they need to talk to me."
∀x(Ex ∧ Nxa → Dxa), ∀x(Ex ∧ ¬Nxa → ¬Dxa) ⊦ ∀x(Ex → (Dxa ↔ Nxa))
[96] When, in her journey through Wonderland, Alice addresses the cat, which has suddenly appeared at the top of a tree, and asks him for directions, the cat says: "We're all mad here: you're mad, I'm mad." "How do you know you're mad?" replied Alice. "To begin with," said the cat, "a dog is not mad. Do you agree?... Well then," continued the cat, "a dog growls when it's angry, and wags its tail when it's happy. Now, I growl when I'm happy and wag my tail when I'm angry. Therefore, I am mad."
Variables:
Px = x is a dog, Lx = x is mad, Gx = x growls, Mx = x wags its tail, Ex = x is angry, Cx = x is happy, g = the cat
Formalization:
∀x(Px → ¬Lx), ∀x(Px ∧ Ex → Gx), ∀x(Px ∧ Cx → Mx), Cg → Gg, Eg → Mg ⊦ Lg
[97] There is an island populated exclusively by "knights" and "knaves." The only thing that differentiates them is that the former always tell the truth and the latter always lie. On one occasion when three inhabitants -A, B, C- met in the garden, a passing stranger asked A "Are you a knight or a knave?" A replied but so confusingly that he could not hear what was said. Then the stranger asked B "What did he say?" and B replied that "A said he is a knave." But the third man C immediately replied: "Don't believe B, he's lying."
Analysis:
- If A is a knight, he tells the truth, so he would say "I am a knight".
- If A is a knave, he lies, so he would also say "I am a knight".
- Therefore, A said "I am a knight".
- B says that A said "I am a knave", which is false. Therefore B is a knave.
- C says that B is lying, which is true. Therefore C is a knight.
Conclusion: B is a knave, C is a knight. We cannot determine what A is.
[98] During the presentation of the summary regarding an important robbery that occurred in London, Inspector Craig asked his assistant, Sergeant McPherson: "What would you do with these facts?": (1) If A is guilty and B innocent, then C is guilty. (2) C never works alone. (3) A never works with C. (4) No one other than A, B, C was involved, and at least one of them is guilty.
Formalization:
1. (Ca ∧ ¬Cb) → Cc
2. Cc → (Ca ∨ Cb)
3. ¬(Ca ∧ Cc)
4. Ca ∨ Cb ∨ Cc
Analysis:
From (3): If C is guilty, A is innocent.
From (2): If C is guilty, A or B is guilty. With (3), B must be guilty.
From (1): If A guilty and B innocent → C guilty. With (3), this forces: if A guilty, B guilty.
Conclusion: B is guilty. A may or may not be guilty. If A is guilty, B is also guilty.
[99] Mr. McGregor, a London merchant, phoned Scotland Yard to say that his store had been robbed. Three suspects -A, B, C- were arrested for questioning and the following facts were established: (1) Each of the three men had been in the store on the day of the robbery, and no one else had been in the store. (2) If A was guilty, then he had an accomplice, and only one. (3) If B was innocent, so was C. (4) If two and only two are guilty, then A is one of them. (5) If C is innocent, so is B. Who did Inspector Craig indict?
Formalization:
1. Everyone was in the store, no one else involved.
2. Ca → [(Cb ∧ ¬Cc) ∨ (Cc ∧ ¬Cb)]
3. ¬Cb → ¬Cc
4. [(Ca ∧ Cb ∧ ¬Cc) ∨ (Ca ∧ ¬Cb ∧ Cc) ∨ (¬Ca ∧ Cb ∧ Cc)] → Ca
5. ¬Cc → ¬Cb
Analysis:
From (3) and (5): B is innocent ↔ C is innocent. They are both guilty or both innocent.
From (2): If A is guilty, he has exactly one accomplice. But B and C go together.
Therefore: A is innocent. B and C are guilty.
Conclusion: Inspector Craig indicted B and C.