15th Slovenian Competition in Logic
2000
1st and 2nd Year of Secondary School
1. Odds’ and Evens’ Island
Somewhere there is a small island where people of two kinds live, the Odds and the Evens. When the inhabitants of the island talk, the following rule applies: if a person says a sentence to another person, then the sentence is true if and only if the persons are of the same kind.
In the task there are inhabitants A, B, C, D and E. Precisely one of them had a gold coin. They were talking:
A to B: “D is an Odd if and only if C has the gold coin.”
B to C: “I am an Even or D is an Even.”
C to D: “A or B has the gold coin.”
D to E: “If C is an Odd, then I have the gold coin.”
E to A: “B is an Odd.”
Then B said to A either “C is an Odd or D is an Odd” or “C is an Odd and D is an Odd” or “C is an Odd” or “D is an Odd.”
The logician, who had heard what B had said, then deduced what the natives were and who had the gold coin.
What did B say to A? What are the inhabitants of the island and who had the gold coin?
2. Talking
a) Alternatives’ Island
On this island every inhabitant alternatively states the truth and the untruth (or vice versa). Inhabitants A, B and C have said:
A: “If there are gold and silver on the island, then there is platinum too.”
A: “If there is gold or silver on the island, then there is platinum too.”
B: “If there is gold on the island, then there is silver too.”
B: “There is gold or silver on the island.”
C: “There is gold on the island if and only if there is silver.”
C: “There is gold on the island or there is no platinum.”
Which metals are there on the island?
b) Lefties and Righties
Somewhere there is a small island where people of two kinds live, the Lefties and the Righties. When the inhabitants of the island communicate, the following rule applies: if a person says a sentence to another person, then the sentence is true if and only if the persons are of the same kind.
Inhabitants A, B and C once talked like this:
A to B: “None of us is a Lefty, but there is gold on the island.”
B to C: “Precisely one of us is a Lefty if and only if there is gold on the island.”
C to A: “At least two of us are Lefties or there is gold on the island.”
What are the natives? Is there gold on the island?
c) Knights, Knaves and Normals
We have three persons, A, B and C. One of them is a Knight, who always tells the truth, another one is a Knave, who always lies. The Normal may or may not tell the truth.
They once said:
A: “B is a Knave or C is a Knight.”
B: “I am not a Normal if and only if C is a Knight.”
What are persons A, B and C?
3. Six Mathematical Periodicals
Six mathematicians (Ara, Handelman, Berberian, von Neumann, Hafner, Roos) published their results about the construction of a regular ring in six mathematical periodicals (these periodicals’ abbreviations are: Ann. of Math., J. Algebra, Michigan Math. J., Trans. American math., C. R. Acad. Sci. Pari., Arch. der Math.) in different years (1939, 1967, 1972, 1974, 1978, 1984). Find out in which periodical a particular mathematician published his results and when they were published if you know:
1. The article of the year 1978 was not published in J. Algebra, C. R. Acad. Sci. Pari. or Arch. der Math.
2. The article in Arch. der Math. was not published by Berberian or Roos and was not written in 1967.
3. Von Neumann did not publish his article in 1972, 1984 or 1967.
4. The article in Michigan Math. J. was not published by Ara, Berberian or Roos.
5. The article of the year 1972 was not published by Ara, Hafner or Roos.
6. It was not Roos who published in J. Algebra, and Handelman published his article in 1978.
7. It was von Neumann who published his essay in Ann. of Math., and it was in 1974 that the essay was published in Michigan Math. J.
Fill in the chart:
Periodical
Author
Year
Ann. of Math.
J. Algebra
Michigan Math. J.
Trans. American math.
C. R. Acad. Sci. Pari.
Arch. der Math.
4. Tarski’s World
We are given two worlds of objects (A and B). The first one has three triangles of various sizes, two small squares and two small pentagons. An object is to the right of another object if the column in which the first object is placed is to the right of the column in which the other object is placed. An object can likewise be placed under (above) another object. Some of the objects are marked with letters.
Put a T in the table beside each of the following sentences if the sentence is true, and a U if it is untrue.
PICTURES
A
B
1.
Every triangle has a large square to its left.
2.
There are at most two pentagons that are to the left of a.
3.
Everything that is large is to the right of every small pentagon.
4.
If and only if there is a large triangle, then there is a small square to the right of a large square.
5.
If there is a large square, then there is also a small square.
6.
If a is a square, then b is a triangle.
7.
If there is a square, then b is a triangle.
8.
Every triangle is to the left of every square.
9.
Every triangle is to the left of a square.
10.
There is a triangle that is larger than any square.
3rd and 4th Year of Secondary School and Students
1. Odds’ and Evens’ Island
Somewhere there is a small island where people of two kinds live, the Odds and the Evens. When the inhabitants of the island talk, the following rule applies: if a person says a sentence to another person, then the sentence is true if and only if the persons are of the same kind.
In the task there are inhabitants A, B, C, D and E. Precisely one of them had a gold coin. They were talking:
A to B: “D is an Odd if and only if C has the gold coin.”
B to C: “I am an Even or D is an Even.”
C to D: “A and B have the gold coin.”
D to E: “If C is an Odd, then I have the gold coin.”
E to A: “B is an Odd.”
Then B said to A either “C is an Odd or D is an Odd” or “C is an Odd and D is an Odd” or “C is an Odd” or “D is an Odd.”
The logician, who had heard what B had said, then deduced what the natives were, but he could not know who had the gold coin.
What did B say to A? What are the inhabitants of the island?
2. Talking
a) Alternatives’ Island
On this island every inhabitant alternatively states the truth and the untruth (or vice versa). Inhabitants A, B and C have said:
A: “If there are gold and silver on the island, then there is platinum too.”
A: “There is gold or silver on the island.”
B: “If there is gold on the island, then there is silver or there is no platinum.”
B: “There is gold or silver on the island.”
C: “There is gold on the island if and only if there is silver.”
C: “There is gold on the island or there is no platinum.”
Which metals are there on the island?
b) Lefties and Righties
Somewhere there is a small island where people of two kinds live, the Lefties and the Righties. When the inhabitants of the island communicate, the following rule applies: if a person says a sentence to another person, then the sentence is true if and only if the persons are of the same kind.
Inhabitants A, B and C once talked like this:
A to B: “If there is gold on the island, then we are all Lefties.”
B to C: “Precisely one of us is a Lefty if and only if I am a Righty.”
C to A: “Precisely two of us are Lefties if and only if you are a Righty.”
What are the natives? Is there gold on the island?
c) Knights, Knaves and Normals
We have three persons, A, B and C. One of them is a Knight, who always tells the truth, another one is a Knave, who always lies. The Normal may or may not tell the truth.
They once said:
A: “B is not a Knave and C is not a Knight.”
C: “I am not a Normal if and only if A is a Knight.”
What are persons A, B and C?
3. Equivalential Calculus
Seven logicians, Mihailescu, Tanaka, Lukasiewicz, Surma, Lesniewski, Wajsberg and Hafner (not necessarily in this order), published their results about equivalential calculus in seven scientific periodicals, Glasnik matematički, Fundamenta Mathematicae, Collectanea Logica, Monatshefte für Mathematik, Annales scientifiques, Ruch Filozoficzny and Proceedings of the Japan Academy. The essays were published in the following years: 1929, 1932, 1937, 1939, 1966, 1971, 1980. Find out in which periodical a particular logician published his essay and in which year if you know:
1. The essays of Mihailescu, Tanaka and Surma were not published in Fundamenta Mathematicae.
2. Wajsberg did not publish his essay in the years 1937, 1966 and 1980.
3. The article in Annales scientifiques was not published in the years 1932 and 1939 and was not written by Surma.
4. The article of the year 1932 was not published in Glasnik matematički, Collectanea Logica or Proceedings of the Japan Academy.
5. Mihailescu, Tanaka and Wajsberg did not publish in the year 1929.
6. Articles of the years 1937, 1939 and 1980 were not published in Proceedings of the Japan Academy.
7. The article in Fundamenta Mathematicae was not written by Wajsberg and was not published in 1939.
8. Surma and Lesniewski did not write the article of the year 1932.
9. Hafner wrote an article for Glasnik matematički, but not in the year 1929.
10. Tanaka did not publish his article in 1937, but it was in 1939 that Lukasiewicz published his.
11. It was not Surma who wrote for the Proceedings of the Japan Academy. Ruch Filozoficzny had an article published in 1971. The article in Annales scientifiques was not published in 1980.
Fill in the chart:
Periodical
Author
Year
Glasnik matematički
Fundamenta Mathematicae
Collectanea Logica
Monatshefte für Mathematik
Annales scientifiques
Ruch Filozoficzny
Proc. of the Japan Academy
4. Tarski’s World
We are given two worlds of objects (A and B). The first one has three triangles of various sizes and another small one, a small square, two medium squares, three small pentagons and a large pentagon. An object is to the right of another object if the column in which the first object is placed is to the right of the column in which the other object is placed. An object can likewise be placed under (above) another object. Some of the objects are marked with letters.
Put a T in the table beside each of the following sentences if the sentence is true, and a U if it is untrue.
PICTURES
A
B
1.
Every triangle has a large square to its left.
2.
There are at most two pentagons that are to the left of a.
3.
Everything that is large is to the right of every small pentagon.
4.
If and only if there is a large triangle, then there is a small square to the right of a large square.
5.
If there is a large triangle, then there is also a small square.
6.
If a is a square, then b is a triangle.
7.
If there is a square, then b is a triangle.
8.
Every triangle is to the left of every square.
9.
Every triangle is to the left of a square.
10.
There is a triangle that is larger than any square.
SOLUTIONS TO THE TASKS
1st and 2nd Year of Secondary School
1. Odds’ and Evens’ Island
After the first five sentences three possibilities exist:
(a, b, c, ce, e, -ca, -cb, -cc, -cd, -d(,
(a, b, cc, d, e, -c, -ca, -cb, -cd, -ce(,
(cc, d, e, -a, -b, -c, -ca, -cb, -cd, -ce(.
With a we denote that A is an Odd, and with -a that A is an Even (similarly for b, c, d and e). Likewise ca means that A has the gold coin, and -ca that A does not have the gold coin (similarly for the other persons).
A and B are of the same kind, so the sixth sentence must be true. If B had said “C is an Odd or D is an Odd”, the logician would still not know the solution, since the sentence is true in all three cases. If B had said “C is an Odd and D is an Odd”, this would be a paradox.
If B had said “C is an Odd”, we would get (a, b, c, ce, e, -ca, -cb, -cc, -cd, -d(. This is a unique solution.
If B had said “D is Odd”, we would have two possibilities: (a, b, cc, d, e, -c, -ca, -cb, -cd, -ce( and (cc, d, e, -a, -b, -c, -ca, -cb, -cd, -ce(.
The answer: A, B, C and E are Odds, D is an Even. E has the gold coin.
2. Talking
a) There is only gold on the island.
b) A is a Lefty, B is a Righty, C is a Lefty. There is gold on the island.
c) A is a Knight, B is a Knave, C is a Normal.
3. Six Mathematical Periodicals
Periodical
Author
Year
Ann. of Math.
von Neumann
1939
J. Algebra
Berberian
1972
Michigan Math. J.
Hafner
1974
Trans. American math.
Handelman
1978
C. R. Acad. Sci. Pari.
Roos
1967
Arch. der Math.
Ara
1984
4. Tarski’s World
World A: U, T, T, U, T, T, T, U, U, T.
World B: U, T, U, T, T, T, T, U, T, U.
3rd and 4th Year of Secondary School and Students
1. Odds’ and Evens’ Island
After the first five sentences five possibilities exist:
(a, b, c, ca, e, -cb, -cc, -cd, -ce, -d(,
(a, b, c, cb, e, -ca, -cc, -cd, -ce, -d(,
(a, b, c, ce, e, -ca, -cb, -cc, -cd, -d(,
(a, b, cc, d, e, -c, -ca, -cb, -cd, -ce(,
(cc, d, e, -a, -b, -c, -ca, -cb, -cd, -ce(.
With a we denote that A is an Odd, and with -a that A is an Even (similarly for b, c, d and e). Likewise ca means that A had the gold coin, and -ca that A did not have the gold coin (similarly for the other persons).
A and B are of the same kind, so the sixth sentence must be true. If B had said “C is an Odd or D is an Odd”, the logician would still not know the solution, since the sentence proves true in all five cases.
If B had said “C is an Odd and D is an Odd”, this would be a paradox.
If B had said “C is an Odd”, we would get three possibilities:
(a, b, c, ca, e, -cb, -cc, -cd, -ce, -d(,
(a, b, c, cb, e, -ca, -cc, -cd, -ce, -d(,
(a, b, c, ce, e, -ca, -cb, -cc, -cd, -d(.
This is a unique solution for the kinds.
If B had said “D is an Odd”, two possibilities would exist:
(a, b, cc, d, e, -c, -ca, -cb, -cd, -ce(,
(cc, d, e, -a, -b, -c, -ca, -cb, -cd, -ce(.
But the kind of the natives could not be deduced.
Therefore B said, “C is an Odd.”
A, B, C and E are Odds, D is an Even.
2. Talking
a) There is only platinum on the island.
b) Everyone is a Lefty, we cannot know about the gold.
c) A is a Normal, B is a Knight, C is a Knave.
3. Equivalential Calculus
Periodical
Author
Year
Glasnik matematički
Hafner
1980
Fundamenta Mathematicae
Lesniewski
1929
Collectanea Logica
Lukasiewicz
1939
Monatshefte für Mathematik
Wajsberg
1932
Annales scientifiques
Mihailescu
1937
Ruch Filozoficzny
Surma
1971
Proc. of the Japan Academy
Tanaka
1966
4. Tarski’s World
World A: U, T, U, U, T, T, T, U, T, T.
World B: U, U, U, U, T, T, T, U, U, T.
14th Slovenian Competition in Logic
1999
1st and 2nd Year of Secondary School
1. Odds’ and Evens’ Island
Somewhere there is a small island where people of two kinds live, the Odds and the Evens. When the inhabitants of the island talk, the following rules apply:
1. If a person says a sentence to another person, then the sentence is true if and only if the persons are of the same kind.
2. If the sentence, however, is not said to another person (let us suppose the person is talking to themselves), the Odd’s sentence will be true and the Even’s untrue (so they act like Knights and Knaves).
In the tasks the inhabitants will be marked A, B, C... Generally they will not represent the same inhabitants in all the tasks. That is to say, A is the first person appearing in a task, B is the second one etc. When A says to B a sentence X, this will be written as A to B: “X”. If A speaks in general, it will be A: “X”.
a) This time we have two inhabitants, A and B.
A: “It is not true that at least one of us is an Odd.”
B to A: “You are an Odd.”
What are they?
In each of the following tasks three inhabitants marked A, B and C appear.
b)
A: “B and me, we are Odds, and C is an Even.”
B to C: “A is an Odd and I am an Even.”
C to A: “B is an Even.”
What can be deduced?
c)
A: “We are all Odds.”
B to C: “None of us are Odds.”
C to A: “At least two of us are Odds.”
What can be deduced?
d)
A to B: “We are all Odds.”
B: “It is not true that none of us are Odds.”
C to A: “At least two of us are Odds.”
What can be deduced?
Now the question is also whether there is gold on the island.
e)
A to B: “If there is gold on the island, then we are all Odds.”
B: “Precisely one of us is an Odd if and only if I am an Even.”
C to A: “If at least two are Odds, then you are an Even.”
What can be deduced?
2. Logical Fallacies
Logical fallacies were discussed already by the Greeks. Here we have seven fallacies with their Latin names (one of them is quaternio terminorum). Find out their English names and the persons who committed them (one of these is Fiona). We know:
1. Non distributus medius is the fallacy of undistributed middle. But it was not committed by Dora or Helen.
2. Ann did not commit any of these three fallacies: non distributus medius; inductio per enumerationem simplicem; post hoc, ergo propter hoc.
3. Dora did not commit the fallacy of simple enumeration, the fallacy of four terms or the fallacy of hasty generalization.
4. The fallacy of simple enumeration was not committed by Barbara, Gloria or Helen.
5. The fallacy of four terms was not committed by Barbara or Helen.
6. The fallacy of hasty generalization was not committed by Barbara or Ann.
7. Fallacia (fallacy) post hoc, ergo propter hoc is the ‘after this therefore because of this’ fallacy, but it was not committed by Dora.
8. Charlotte committed the fallacy of illicit major, which is not the same as inductio per enumerationem simplicem.
9. Ann did not commit the fallacy of simple enumeration nor the illicitus processus fallacy.
10. Fallacia disjunctionis is the fallacy of affirming a disjunct. Gloria committed the fictae universalitatis fallacy.
Fill in the chart:
Latin Expression
English Expression
Committed by
quaternio terminorum
non distributus medius
illicitus processus
disjunctionis
inductio per enumerationem simplicem
fictae universalitatis
post hoc, ergo propter hoc
3. Children’s Ages
Peter once asked Paul, who is a good logician, about the ages of his four children (in years). The conversation went like this:
Peter: “The product of their ages is 79, 80 or 81.”
Paul: “That does not help me much with deducing the ages.”
Peter: “The sum of their ages matches your age.”
Paul (after thinking for a few minutes): “I still do not have enough information.”
Peter: “None of them were born last year.”
Paul: “Now I know their ages.”
What age are Peter’s children?
3rd and 4th Year of Secondary School
1. Odds’ and Evens’ Island
Somewhere there is a small island where people of two kinds live, the Odds and the Evens. When the inhabitants of the island talk, the following rules apply:
1. If a person says a sentence to another person, then the sentence is true if and only if the persons are of the same kind.
2. If the sentence, however, is not said to another person (let us suppose the person is talking to themselves), the Odd’s sentence will be true and the Even’s untrue (so they act like Knights and Knaves).
In the tasks the inhabitants will be marked by the letters A, B, C... Generally they will not represent the same inhabitants in all the tasks. That is to say, A is the first person appearing in a task, B is the second one etc. When A says to B a sentence X, this will be written as A to B: “X”. If A speaks in general, it will be A: “X”.
a) This time we have two inhabitants, A and B.
A: “We are both Odds.”
B to A: “At least one of us is an Odd.”
What are they?
In each of the following tasks three inhabitants marked A, B and C appear.
b)
A: “B and C are Odds.”
B to C: “A is an Odd or I am an Odd.”
C to A: “If B is an Odd, then you are too.”
What can be deduced?
c)
A to B: “We are all Odds.”
B: “Precisely one of us is an Odd.”
C to A: “At least two of us are Odds.”
What can be deduced?
Now the question is also whether there is gold on the island.
d)
A: “At least one of us is an Odd and there is no gold on the island.”
B to C: “Precisely one of us is an Odd and there is gold on the island.”
C to A: “If at least two of us are Odds, then there is no gold on the island.”
What can be deduced?
e)
A: “None of us are Odds and there is gold on the island.”
B to C: “Precisely one of us is an Odd if and only if I am an Even.”
C to A: “If at least two are Odds, then you are an Even.”
What can be deduced?
2. Fallacies of Argument
Eight people committed eight different fallacies (or used various »tricks« during their argumentation). The fallacies are mentioned with their Latin names (there is also the ad verecundiam fallacy) and their approximate English translations. For each fallacy you should find its translation and the person who committed it.
1. The ad baculum fallacy is not the argument to the person and was not committed by Charles or Henry.
2. Compassion was not appealed to by Felix, Ian or Andrew.
3. Henry did not commit the ad misericordiam, ad hominem or tu quoque fallacy.
4. The appeal to force was not used by Felix or Ian. The latter did not need the you too argument.
5. The ad misericordiam argument was not used by Felix or Dan. The latter did not need the tu quoque argument either.
6. Ad populum means appeal to the people and was used by Barry.
7. Ad baculum is not the you too argument and was not used by Ian.
8. Ad ignorantiam means appeal to ignorance; it was Gabriel that used this argument.
9. Dan used the argument to the person. Henry appealed to authority. Non sequitur is the fallacy of false cause.
Fill in the chart:
Latin Expression
English expression
Committed by
ad baculum
ad populum
ad misericordiam
ad hominem
tu quoque
ad ignorantiam
ad verecundiam
non sequitor
3. Children’s Ages
Peter once asked Paul, who is a good logician, about the ages of his four children (in years). The conversation went like this:
Peter: “The product of their ages is 47, 48, 49 or 50.”
Paul: “That does not help me much with deducing the ages.”
Peter: “The sum of their ages matches your age.”
Paul (after thinking for a few minutes): “I still do not have enough information.”
Peter: “Sammy was born last year.”
Paul: “Now I know their ages.”
What age are Peter’s children?
SOLUTIONS TO THE TASKS
1st and 2nd Year of Secondary School
1. Odds’ and Evens’ Island
a) A is an Even, B is an Odd.
b) A and B are Odds, C is an Even.
c) A and B are Evens, C is an Odd.
d) Everyone is an Odd.
e) C is an Even, there is no gold on the island. A and B are of the same kind.
2. Logical Fallacies
Latin Expression
English Expression
Committed by
quaternio terminorum
four terms
Ann
non distributus medius
undistributed middle
Barbara
illicitus processus
illicit major
Charlotte
disjunctionis
affirming a disjunct
Dora
inductio per enumeracionem simplicem
simple enumeration
Fiona
fictae universalitatis
hasty generalization
Gloria
post hoc, ergo propter hoc
after this therefore because of this
Helen
3. Children’s Ages
First we divide the numbers 79, 80 and 81 into their respective factors (including 1):
{{79, 1, 1, 1}}
{{80, 1, 1, 1,}, {40, 2, 1, 1}, {20, 4, 1, 1}, {20, 2, 2, 1}, {16, 5, 1, 1}, {10, 8, 1, 1}, {10, 4, 2, 1}, {10, 2, 2, 2}, {8, 5, 2, 1}, {5, 4, 4, 1}, {5, 4, 2, 2}}
{{81, 1, 1, 1}, {27, 3, 1, 1}, {9, 9, 1, 1}, {9, 3, 3, 1}, {3, 3, 3, 3}}
Then we write down the sums of the divisions: {82},
{83, 44, 26, 25, 23, 20, 17, 16, 16, 14, 13},
{84, 32, 20, 16, 12}.
Number 79 cannot be the product. The only case when Paul cannot ascertain the ages is when the sum of the ages is 20 or 16. The information that none of the children are one year old gives the result 10, 2, 2, 2.
3rd and 4th Year of Secondary School and Students
1. Odds’ and Evens’ Island
a) They are both Odds.
b) They are all Odds.
c) A is an Odd, B is an Even, C is an Odd.
d) A is an Odd, B is an Even, C is an Odd, there is no gold.
e) A is an Even, B is an Odd, C is an Even. We do not know about the gold.
2. Fallacies of Argument
Latin Expression
English expression
Committed by
ad baculum
to force
Andrew
ad populum
to the people
Barry
ad misericordiam
to compassion
Charles
ad hominem
to the person
Dan
tu quoque
you too
Felix
ad ignorantiam
to ignorance
Gabriel
ad verecundiam
to authority
Henry
non sequitor
false cause
Ian
3. Children’s Ages
We first divide the numbers 47, 48, 49 and 50 into their respective factors (including 1):
{{47, 1, 1, 1}}
{{48, 1, 1, 1}, {24, 2, 1, 1}, {16, 3, 1, 1}, {12, 4, 1, 1}, {12, 2, 2, 1}, {8, 6, 1, 1}, {8, 3, 2, 1}, {6, 4, 2, 1}, {6, 2, 2, 2}, {4, 4, 3, 1}, {4, 3, 2, 2}}
{{49, 1, 1, 1}, {7, 7, 1, 1}}
{{50, 1, 1, 1}, {25, 2, 1, 1}, {10, 5, 1, 1}, {5, 5, 2, 1}}
Then we write down the sums of the divisions: {50},
{51, 28, 21, 18, 17, 16, 14, 13, 12,12,11},
{52, 16},
{53, 29, 17, 13}.
Number 47 cannot be the product. The cases when Paul cannot ascertain the ages are when the addition is 12, 13, 16 or 17:
12
6,
2,
2,
2
4,
4,
3,
1
13
6,
4,
2,
1
5,
5,
2,
1
16
7,
7,
1,
1
8,
6,
1,
1
17
10,
5,
1,
1
12,
2,
2,
1
When Paul finds out that at least one child is one year old, he can ascertain the children’s ages, so the solution is 4, 4, 3, 1.