Once the investigator had to interrogate three witnesses of the robbery. Syllogisms Once an investigator had to interrogate three witnesses at the same time: Claude, Jacques and Dick

Once the investigator had to simultaneously interrogate three witnesses: Claude, Jacques and Dick. Their testimonies contradicted each other, and each of them accused someone of lying. Claude claimed that Jacques was lying, Jacques accused Dick of lying, and Dick persuaded the investigator not to believe either Claude or Jacques. But the investigator quickly brought them to clean water without asking them a single question. Which of the witnesses spoke the truth

Ilya Muromets, Dobrynya Nikitich and Alyosha Popovich were given 6 coins for their faithful service: 3 gold and 3 silver. Each received two coins. Ilya Muromets does not know which coins Dobrynya got and which Alyosha got, but he knows which coins he got himself. Think of a question to which Ilya Muromets will answer "yes", "no" or "I don't know", and by the answer to which you can understand what coins he got

Rules of syllogisms 1. In a syllogism there should be only three statements and only three terms. ZhG All the sightseers fled in different directions, Petrov is a sightseer, which means he fled in different directions. 3. If both premises are private statements, then it is impossible to draw a conclusion. 2. If one of the premises is a private statement, then the conclusion must be private. 4. If one of the premises is a negative statement, then the conclusion is also a negative statement. 5. If both premises are negative statements, then the conclusion cannot be made. 6. The middle term must be distributed in at least one of the premises. 7. A term cannot be distributed in a conclusion if it is not distributed in a premise.

All cats have four legs. All dogs have four legs. All dogs are cats. All people are mortal. All dogs are not human. Dogs are immortal (not mortal). Ukraine occupies a vast territory. Crimea is part of Ukraine. Crimea occupies a vast territory

. 18 years.

Decision

First way . According to the condition of the problem, you can write an equation. Let Dima's age be x years, then the sister's age is x/3, and the brother's age is x/2; (x + x / 3 + x / 2): 3 \u003d 11. After solving this equation, we get that x=18. Dima is 18 years old. It will be useful to give a slightly different solution, "in parts".

Second way . If the ages of Dima, his brother and sister are represented by segments, then "Dimin's segment" consists of two "brother segments" or three "sister segments". Then, if Dima's age is divided into 6 parts, then the sister's age is two such parts, and the brother's age is three such parts. Then the sum of their ages is 11 such parts. On the other hand, if the median age is 11 years, then the sum of ages is 33 years. Whence it follows that in one part - three years. So Dima is 18 years old.

Verification Criteria .

Complete correct solution 7 points.

The equation is correct, but errors were made in the solution - 3 points .

Correct answer given and verification done - 2 points .

0 points .

Answer . Sam Grey.

Decision .

It is clear from the condition of the problem that the statements of each of the witnesses are uttered in relation to the statements of the other two witnesses. Consider Bob Black's statement. If what he says is true, then Sam Gray and John White are lying. But from the fact that John White is lying, it follows that not all of Sam Gray's testimony is a complete lie. And this contradicts the words of Bob Black, whom we decided to believe and who claims that Sam Gray is lying. So Bob Black's words can't be true. So he lied, and we must admit that Sam Gray's words are true, and therefore John White's statements are false. Answer: Sam Gray did not lie.

Verification Criteria .

A complete correct analysis of the situation of the problem is given and the correct answer is given - 7 points .

A complete correct analysis of the situation is given, but for some reason an incorrect answer is given (for example, instead of the one who did NOT lie, the answer indicates those who lied) - 6 points .

A correct analysis of the situation is given, but for some reason the correct answer is not given (for example, it is proved that Bob Black lied, but no further conclusions are drawn) - 4 points .

The correct answer is given and it is shown that it satisfies the condition of the problem (the test was carried out), but it was not proved that the answer is the only one - 3 points .

1 score .

0 points .

Answer . One number 175.

Decision . First way . The composition of the digits with which the number is written does not contain the digit 0, otherwise the condition of the problem cannot be fulfilled. This three-digit number is obtained by multiplying by 5 the product of its digits, therefore, it is divisible by 5. Hence, its entry ends with the number 5. We get that the product of digits multiplied by 5 must be divisible by 25. Note that there are even digits in the number entry cannot, otherwise the product of the digits would be zero. Thus, a three-digit number must be divisible by 25 and contain no even digits. There are only five such numbers: 175, 375, 575, 775 and 975. The product of the digits of the desired number must be less than 200, otherwise, multiplied by 5, it will give a four-digit number. Therefore, the numbers 775 and 975 are obviously not suitable. Among the remaining three numbers, only 175 satisfies the condition of the problem. Second way. Note (similarly to the first solution method) that the last digit of the desired number is 5. Leta , b , 5 - consecutive digits of the desired number. According to the condition of the problem, we have: 100a + 10 b + 5 = a · b 5 5. Dividing both sides of the equation by 5, we get: 20a + 2 b + 1 = 5 ab . After subtracting the equality 20a from both sides and bracketing the common factor on the right side, we get: 2b + 1 = 5 a (b – 4 a) (1 ). Given that a and b can take natural values from 1 to 9, we get that the possible values of a are only 1 or 2. But a=2 does not satisfy the equality (1 ), on the left side of which there is an odd number, and on the right side, when a = 2 is substituted, an even number is obtained. So the only possibility is a=1. Substituting this value into (1 ), we get: 2 b + 1 = 5 b- 20, from where b =7. Answer: the only desired number is 175.

Verification Criteria .

Complete correct solution 7 points .

The correct answer is received and there are arguments that significantly reduce the enumeration of options, but there is no complete solution - 4 points .

The equation is correctly composed and transformations and reasoning are given that allow solving the problem, but the solution is not brought to the end - 4 points .

The enumeration of options is reduced, but there is no explanation why, and the correct answer is indicated - 3 points .

The equation is correct, but the problem is not solved - 2 points .

There are arguments in the solution that allow excluding any numbers from consideration or considering numbers with certain properties (for example, ending with the number 5), but there is no further significant progress in the solution - 1 score .

Only the correct answer or the answer with verification is given - 1 score .

Answer . 75° .

Decision . Consider triangle AOC, where O is the center of the circle. This triangle is isosceles, since OS and OA are radii. So, by the property of an isosceles triangle, angles A and C are equal. Let's draw a perpendicular SM to the side AO and consider a right triangle OMC. According to the condition of the problem, the leg of the SM is half of the hypotenuse of the OS. Hence, the value of the angle COM is 30°. Then, according to the theorem on the sum of the angles of a triangle, we obtain that the angle CAO (or CAB) is 75 °.

Verification Criteria .

Correct substantiated solution of the problem - 7 points.

Correct reasoning is given, which is a solution to the problem, but for some reason an incorrect answer is given (for example, the angle COA is indicated instead of the angle CAO) - 6 points.

In general, correct reasoning is given, in which errors are made that do not have a fundamental nature for the essence of the decision, and the correct answer is given - 5 points.

The correct solution of the problem is given in the absence of justifications: all intermediate conclusions are indicated without indicating the links between them (references to theorems or definitions) - 4 points.

Additional constructions and designations were made on the drawing, from which the course of the solution is clear, the correct answer is given, but the reasoning itself is not given - 3 points.

The correct answer is given with incorrect reasoning - 0 points.

Only correct answer given 0 points.

Answer . See drawing.

Decision . We transform this equation by highlighting the full square under the root sign: . The expression on the right side only makes sense when x = 9. Substituting this value into the equation, we get: 9 2 – y 4 = 0. We factorize the left side: (3 –y)(3 + y)(9 + y 2 ) = 0. Whence y= 3 or y = -3. This means that the coordinates of only two points (9; 3) or (9; -3) satisfy this equation. The graph of the equation is shown in the figure.

Verification criteria.

The correct transformations and reasoning have been carried out and the graph has been correctly built - 7 points.

Correct transformations performed, but meaning is lost y = -3; one point is indicated as a graph -3 points.

One or two suitable points are indicated, possibly with verification, but without other explanations or after incorrect transformations -1 score.

Correct transformations are carried out, but it is declared that the expression under the root (or on the right side after squaring) is negative and the graph is an empty set of points - 1 score.

Reasoning has been carried out that led to the indication of two points, but these points are somehow connected (for example, by a segment) - 1 score.

Two points are indicated without explanation, which are somehow connected - 0 points.

In other cases - 0 points.

Answers to the tasks of the second stage of the Olympiad

Answer . They can.

Decision . If a \u003d, b \u003d -, then a \u003d b + 1 and a 2 \u003d b 2

You can also solve the system of equations:

Verification criteria.

Correct answer with numbers a and b – 7 points .

A system of equations was compiled, but an arithmetic error was made in its solution - 3 points .

Only answer is 1 score .

Answer . In 12 seconds .

Decision . There are 3 spans between the first and fourth floors, and 4 spans between the fifth and first. According to the condition, Petya runs 4 spans 2 seconds longer than mom rides the elevator, and three spans are 2 seconds faster than mom. So, in 4 seconds Petya runs through one span. Then Petya runs from the fourth floor to the first (that is, 3 flights) in 4*3=12 seconds.

Verification criteria.

Correct answer with complete solution - 7 points .

It is explained that it takes 4 seconds for one span, the answer says 4 seconds − 5 points .

Correct justification assuming that the path from the fifth floor to the first is 1.25 times the distance from the fourth floor to the first and the answer is 16 seconds - 3 points .

Only answer is 0 points .

Answer . See drawing.

Decision . Because X 2 =| X | 2 , then =| X |, with x≠ 0.

It is also possible, using the definition of the module, to obtain that (for x = 0 function not defined).

Verification criteria.

Correct graph with explanation - 7 points .

Correct graph without any explanation - 5 points .

function graph y =|x| without punched point3 points .

Answer . Yes .

Decision . Let us divide the given square with side 5 by straight lines parallel to its sides into 25 squares with side 1 (see the figure). If there were no more than 4 marked points in each such square, then no more than 25 * 4 = 100 points would be marked, which contradicts the condition. Therefore, at least one of the resulting squares must contain 5 of the marked points.

Verification criteria.

The right decision - 7 points .

Only answer is 0 points .

Answer . Eight ways.

Decision . From point a) it follows that the coloring of all points with integer coordinates is uniquely determined by the coloring of the points corresponding to the numbers 0, 1, 2, 3, 4, 5 and 6. The point 0=14-2*7 must be colored in the same way as 14, those. red. Similarly, point 1=71-107 should be colored blue, point 3=143-20*7 blue, and 6=20-2*7 red. Therefore, it remains only to calculate how many different ways you can color the points corresponding to the numbers 2, 4 and 5. Since each point can be colored in two ways - red or blue - there are only 2 * 2 * 2 = 8 ways. Note. When counting the number of ways to color points 2, 4 and 5, you can simply list all the ways, for example, in the form of a table:

Verification Criteria .

Correct answer with correct reasoning 7 points .

The problem is reduced to counting the number of ways to color 3 points, but the answer is 6 or 7 - 4 points .

The problem is reduced to counting the number of ways to color 3 points, but there is no count of the number of ways or the answer is different from the ones indicated above - 3 points .

Answer (including the correct one) without justification - 0 points .

Answer . 4 times.

Decision .

Let's draw segments MK and AC . The MVKE quadrilateral consists of

triangles MVK and MKE , and quadrilateral AECD- from triangles

1 way . Triangles MVK and ACD- rectangular and the legs of the first are 2 times smaller than the legs of the second, therefore they are similar and the area of triangle ACD 4 times the area of triangle MBK. Because M and K – the midpoints of AB and BC, respectively, then MK– , so MK || AS and MK = 0.5AC . From the parallelism of straight lines MK and AS follows the similarity

triangles MKE and AEC, and since similarity coefficient is 0.5, then the area of triangle AEC is 4 times the area of triangle MKE. Now: S AES D=SAEC+SACD= 4 SMKE+ 4 SMBK= 4 (SMKE+SMBK)= 4 SMBKE.

2 way . Let the area of the rectangle ABCD is equal to S. Then the area of triangle ACD is equal to ( the diagonal of a rectangle divides it into two equal triangles), and the area of \u200b\u200bthe triangle MVK is equal to MV × VK \u003d T.k. M and K – midpoints of segments AB and BC, then AK and SM – medians of triangle ABC, so E – point of intersection of medians of triangle ABC, those. the distance from E to AC ish, where h- altitude of triangle ABC, drawn from vertex B. Then the area of triangle AEC is. Then for the area of the quadrilateral AECD, equal to the sum of the areas of triangles AEC and ACD, we get: Next, because MK– midline of triangle ABC, then the area of the triangle MKE is equal to* h -* h ) = h )=(AC * h )== S . Therefore, for the area of the quadrilateral MVKE, equal to the sum of the areas of triangles MVK and MKE, we get: . Thus, the ratio of the areas of quadrilaterals AECD and MVKE is the same.

Verification criteria.

Right decision and right answer7 points .

Correct solution, but the answer is incorrect due to an arithmetic error -5 points .

5. SUMMING UP AND AWARDING THE WINNERS

The final indicators of the completed competitive tasks are determined by the jury inin accordance with the developed assessment criteria;

For the winners of the Olympiad, determined by the highest number of points,three prizes are established;

The results of the competition are documented by the report of the organizer of the Olympiad.

The winners are awarded with diplomas and valuable gifts.

In case of disagreement with the score given by the jury, the participant may submitwritten appeal within an hour after the announcement of the results.

The publicity of the competition is ensured - the results of the competition are announcedprize-winners.

We can single out the following sequence of steps in solving logical problems.

1. Select elementary (simple) statements from the condition of the problem and designate them with letters.

2. Write down the condition of the problem in the language of the algebra of logic, combine simple statements into complex ones using logical operations.

3. Compose a single logical expression for the requirements of the problem.

4. Using the laws of the algebra of logic, try to simplify the resulting expression and calculate all its values, or build a truth table for the expression in question.

5. Choose a solution - value set simple propositions, in which the constructed logical expression is true.

6. Check whether the obtained solution satisfies the condition of the problem.

Example:

Task 1:“In an attempt to recall the winners of last year’s tournament, five former tournament viewers stated that:

1. Anton was second, and Boris was fifth.

2. Viktor was second, and Denis was third.

3. Gregory was the first, and Boris was the third.

4. Anton was third, and Evgeny was sixth.

5. Viktor was third and Evgeny was fourth.

Subsequently, it turned out that each viewer was mistaken in one of his two statements. What was the true distribution of places in the tournament.

1) Denote by the first letter in the name of the participant of the tournament, and - the number of the place that he has, i.e. we have.

2) 1. ; 3. ; 5. .

3) A single logical expression for all the requirements of the task: .

4) In the formula L we carry out equivalent transformations, we get: .

5) From paragraph 4 it follows:,.

6) Distribution of places in the tournament: Anton was third, Boris was fifth, Viktor was second, Grigory was first, and Evgeny was fourth.

Task 2:“Ivanov, Petrov, Sidorov appeared before the court on charges of robbery. The investigation found:

1. if Ivanov is not guilty or Petrov is guilty, then Sidorov is guilty;

2. if Ivanov is not guilty, then Sidorov is not guilty.

Is Ivanov guilty?

1) Consider the statements:

BUT: "Ivanov is guilty", AT: "Petrov is guilty", With: "Sidorov is guilty."

2) Facts established by the investigation:,.

3) A single logical expression: . It is true.

Let's make a truth table for it.

BUT	AT	With	L

To solve a problem means to indicate for what values of A the resulting complex statement L is true. If, but, then the investigation does not have enough facts to accuse Ivanov of a crime. Analysis of the table shows and, i.e. Ivanov is guilty of robbery.

Questions and tasks.

1. Compile RCS for the formulas:

2. Simplify RCS:

3. Based on this switching circuit, construct a logical formula corresponding to it.

4. Check the equivalence of the RCS:

5. Construct a circuit of three switches and a light bulb so that the light comes on only when exactly two switches are in the “on” position.

6. Using this conductivity table, build a circuit of functional elements with three inputs and one output that implements the formula.

x	y	z	F

7. Analyze the diagram shown in the figure and write out the formula for the function F.

8. Task: “Once the investigator had to interrogate three witnesses at the same time: Claude, Jacques, Dick. Their testimonies contradicted each other, and each of them accused someone of lying.

1) Claude claimed that Jacques was lying.

2) Jacques accused Dick of lying.

3) Dick persuaded the investigator not to believe either Claude or Jacques.

But the investigator quickly brought them to clean water without asking them a single question. Which witness was telling the truth?

9. Determine which of the four students passed the exam, if it is known that:

1) If the first passed, then the second passed.

2) If the second passed, then the third passed or the first did not pass.

3) If the fourth did not pass, then the first passed, and the third did not pass.

4) If the fourth passed, then the first passed.

10. When asked which of the three students studied logic, the answer was received: if he studied the first, then he studied the third, but it is not true that if he studied the second, then he studied the third. Who studied logic?

1. a) ( commutativity of disjunction );

b)

(conjunction commutativity );

2. a) ( disjunction associativity );

b) ( conjunction associativity );

3. a) ( distributivity of disjunction with respect to conjunction );

b) ( distributivity of conjunction with respect to disjunction );

4.

and

–de Morgan's laws .

5.

;

;

;

6.

(or

) (law of the excluded middle );

(or

(law of contradiction );

7.

(or

);

(or

);

(or

);

(or

).

The listed properties are commonly used to transform and simplify logical formulas. Here the properties of only three logical operations (disjunction, conjunction and negation) are given, but it will be shown later that all other operations can be expressed through them.

With the help of logical connectives, you can compose logical equations, and solve logical problems in the same way as arithmetic problems are solved using systems of ordinary equations.

Example. Once the investigator had to simultaneously interrogate three witnesses: Claude, Jacques and Dick. Their testimonies contradicted each other, and each of them accused someone of lying. Claude claimed that Jacques was lying, Jacques accused Dick of lying, and Dick persuaded the investigator not to believe either Claude or Jacques. But the investigator quickly brought them to clean water without asking them a single question. Which witness was telling the truth?

Decision. Consider the statements:

(Claude tells the truth);

(Jacques tells the truth);

(Dick is telling the truth).

We do not know which of them are correct, but we do know the following:

1) either Claude told the truth, and then Jacques lied, or Claude lied, and then Jacques told the truth;

2) either Jacques told the truth, and then Dick lied, or Jacques lied, and then Dick told the truth;

3) either Dick told the truth, and then Claude and Jacques lied, or Dick lied, and then it is not true that both other witnesses lied (i.e. at least one of these witnesses told the truth).

We express these statements in the form of a system of equations:

The condition of the problem will be fulfilled if these three statements are true at the same time, which means that their conjunction is true. We multiply these equalities (i.e., take their conjunction)

But

if and only if

, a

. Therefore, Jacques is telling the truth, while Claude and Dick are lying.

Any -term operation, denoted, for example,

, will be completely determined if it is established for what values of statements

the result will be true or false. One way to specify such an operation is to fill in a table of values:

In the table of meanings of the statement formed from the simplest statements

, available lines. The value column also has positions. Therefore, there is

various options for filling it, and, accordingly, the number of all -term operations is equal to

. At

the number of one-term operations is 4, with

the number of binomial - 16, with

the number of three-membered ones is 256, etc.

Consider some special types of formulas.

The formula is called elementary conjunction if it is a conjunction of variables and negations of variables. For example, formulas ,

,

,

are elementary conjunctions.

A formula that is a disjunction (possibly one-term) of elementary conjunctions is called disjunctive normal form (D.Sc.). For example, formulas ,

,

.

Theorem 1(on reduction to D.Sc.). For any formula , who is d. f. .

This theorem and Theorem 2 following it will be proved in the next subsection. By applying these theorems, one can standardize the form of logical formulas.

The formula is called elementary disjunction if it is a disjunction of variables and negations of variables. For example, formulas

,

,

etc.

A formula that is a conjunction (possibly one-term) of elementary disjunctions is called conjunctive normal form (PhD). For example, formulas

,

.

Theorem 2(on reduction to Ph.D.). For any formula one can find an equivalent formula , which is Ph.D. f.

Problem 35

One man went to work with a salary of $1,000 a year. During the discussion of conditions upon admission, he was promised that in the case of good work, an increase would be made to the salary. Moreover, the amount of the increase can be chosen from two options at your discretion: in one case, an increase of $ 50 every six months, starting from the second half, was offered, in the other - $ 200 every year, starting from the second. Given the freedom of choice, employers wanted not only to try to save on wages, but also to check how fast the new employee thinks. After thinking for a moment, he confidently named the conditions for the increase.

Which option was preferred?

Problem 36

Once the investigator had to simultaneously interrogate three witnesses: Claude, Jacques and Dick. Their testimonies contradicted each other, and each of them accused someone of lying. Claude claimed that Jacques was lying. Jacques accused Dick of lying, and Dick persuaded the investigator not to believe either Claude or Jacques. But the investigator quickly led them to clean water without asking them a single question.

Which witness was telling the truth?

Problem 37

Terrible misfortune, Inspector, the museum official said. - You can't imagine how excited I am. I'll tell you everything in order. I stayed at the museum today to work and get our finances in order. I was just sitting at this desk and looking through the accounts, when suddenly I saw a shadow on the right side. The window was open.

And you didn't hear any rustle? asked the inspector.

Absolutely none. The radio was playing music, and besides, I was too engrossed in what I was doing. Taking my eyes off the heat, I saw that a man had jumped out of the window. I immediately turned on the overhead light and discovered that two boxes with the most valuable collection of coins, which I had taken to my office for work, had disappeared. I am in a terrible state: after all, this collection is valued at 10,000 marks.

Do you believe that I really do; do I believe your thoughts?

The Inspector was irritated. - No one has yet managed to mislead me, and you will not be the first.

How did the inspector guess that they were trying to deceive him?

Problem 38

The corpse of the missing person was found wrapped in a sheet that had a laundry number tag on it. A family was identified that used such tags, however, during the verification process, it turned out that the members of this family were not familiar with and had no contact with the deceased and his relatives. No other evidence of their involvement in the murder was found.

Were there any errors in the completeness and correctness of obtaining information during the check?

Problem 39

Potapov, Shchedrin, Semyonov serve in the aviation unit. Konovalov and Samoilov. Their specialties are: pilot, navigator, flight mechanic, radio operator and weather forecaster.

Determine what specialty each of them has if the following facts are known.

Shchedrin and Konovalov are not familiar with aircraft controls;

Potapov and Konovalov are preparing to become navigators; Shchedrin's and Samoilov's apartments are next to the radio operator's apartment;

Semyon, while in a rest home, met Shchedrin and the forecaster's sister: Potapov and Shchedrin play chess with the flight engineer and pilot in their free time; Konovalov, Semyonov and the weatherman are fond of boxing; The radio operator is not fond of boxing.

Problem 40

The aunt, who was waiting for her nephew, the inspector, rushed to meet him, not hiding her impatience.

Some woman just now; snatched my purse with money and immediately disappeared.

Most likely she disappeared into the very savings bank where you were, - the inspector noted. - Let's try to find it.

And in fact, the aunt immediately saw her bag, which was standing on a bench between two women. She was exposed. When the inspector took a careful look at the bag, both women, noticing this, got up and went to the other end of the room. The handbag remained on the bench.

But I don't know which one of them stole my bag. Yana managed to see her, - said the aunt.

Well, it's nothing, - answered the nephew. - We'll interrogate both, but I think that the bag was stolen from you by the one who ...

Which?

Problem 41

Having received a message that a gray Chevrolet with a number starting with six hit a woman and disappeared, the inspector and his assistant drove to the villa of the gentleman, whose car seemed to match the description. Less than half an hour later they were there.

A gray Chevrolet stood in front of the house. Seeing the police, the owner went down to them right in his pajamas.

I didn't go anywhere today,” he said after listening to the inspector. - Yes, and I could not: yesterday I lost the ignition key, and the new one will be ready only on Friday.

The assistant, having managed to inspect the car in the meantime, whispered to the inspector:

Apparently he is telling the truth. There are no signs of a collision on the car.

The inspector, leaning on the hood of the car, answered:

This does not mean anything, the blow was not strong, because the victim is alive. And your alibi, sir, seems extremely suspicious to me. Why are you trying to hide from me that you just arrived here in this very car?

What gave the inspector a reason to suspect the gentleman of a lie?

Problem 42

The president of the firm informs the investigator about the theft at his home.

Arriving at work, I remembered that I forgot the necessary documents at home. I gave the key to the home safe to my assistant and sent him for the file folder. We have been working together for a long time, I trust him for a long time, and often sent him home to take something from the safe. This time, shortly after leaving, he called me on the phone and said that when he entered the room, he saw that the door of the wall safe was open, and papers were scattered all over the office. I arrived home and found that, in addition to scattered documents, jewelry and money had disappeared from the safe.

Assistant's testimony: “When I arrived, the butler let me in and I went up to the second floor of the apartment. Entering the office, he found papers scattered on the floor and an open safe door. I immediately phoned my boss and reported what I had seen. After that, I jumped out to the landing of the stairs and called the butler. At my cry, a maid appeared from the living room downstairs and asked what was the matter. I told her what I saw. At her call, the butler came running from the courtyard. To my question, they said that no one came to the apartment after the owner left, and they did not hear any noise in the house.

The butler explained: “After the owner left in the morning, I did my usual work on the lower floor and did not see anyone or hear anything unusual. The maid never left the kitchen with me. When an employee of our host, whom I had known for a long time, arrived, he went to the stairs to the second floor and went out into the courtyard. A few minutes later the cook called me and I entered the house, where the assistant told about the theft from the owner's office.

The maid said that after breakfast she was in the kitchen, did not go out anywhere, and only, having heard the cry of the assistant, went out into the living room. The assistant told about the theft in the house and asked to know the butler.

When asked by the investigator, the assistant replied that he had not touched anything in the office, except for the telephone, and had not rearranged it. The butler and the maid said they didn't go to the office at all.

During the inspection in the office, the investigator did not find any traces of fingers on the office door, the safe door, objects and the telephone on the table. Having examined the lock of the safe door, the specialist did not find traces of any object or foreign key on its details.

We can single out the following sequence of steps in solving logical problems.

1. Select elementary (simple) statements from the condition of the problem and designate them with letters.

2. Write down the condition of the problem in the language of the algebra of logic, combine simple statements into complex ones using logical operations.

3. Compose a single logical expression for the requirements of the problem.

4. Using the laws of the algebra of logic, try to simplify the resulting expression and calculate all its values, or build a truth table for the expression in question.

5. Choose a solution - value set simple propositions, in which the constructed logical expression is true.

6. Check whether the obtained solution satisfies the condition of the problem.

Example:

Task 1:“In an attempt to recall the winners of last year’s tournament, five former tournament viewers stated that:

1. Anton was second, and Boris was fifth.

2. Viktor was second, and Denis was third.

3. Gregory was the first, and Boris was the third.

4. Anton was third, and Evgeny was sixth.

5. Viktor was third and Evgeny was fourth.

Subsequently, it turned out that each viewer was mistaken in one of his two statements. What was the true distribution of places in the tournament.

1) Denote by the first letter in the name of the participant of the tournament, and - the number of the place that he has, i.e. we have .

2) 1. ; 3. ; 5. .

3) A single logical expression for all the requirements of the task: .

4) In the formula L we carry out equivalent transformations, we get: .

5) From paragraph 4 it follows: , , , , .

6) Distribution of places in the tournament: Anton was third, Boris was fifth, Viktor was second, Grigory was first, and Evgeny was fourth.

Task 2:“Ivanov, Petrov, Sidorov appeared before the court on charges of robbery. The investigation found:

1. if Ivanov is not guilty or Petrov is guilty, then Sidorov is guilty;

2. if Ivanov is not guilty, then Sidorov is not guilty.

Is Ivanov guilty?

1) Consider the statements:

BUT: "Ivanov is guilty", AT: "Petrov is guilty", With: "Sidorov is guilty."

2) Facts established by the investigation:,.

3) A single logical expression: . It is true.

Let's make a truth table for it.

BUT	AT	With	L

To solve a problem means to indicate for what values of A the resulting complex statement L is true. If , and , then the investigation does not have enough facts to accuse Ivanov of a crime. Analysis of the table shows and , i.e. Ivanov is guilty of robbery.

Questions and tasks.

1. Compile RCS for the formulas:

2. Simplify RCS:

3. Based on this switching circuit, construct a logical formula corresponding to it.

4. Check the equivalence of the RCS:

5. Construct a circuit of three switches and a light bulb so that the light comes on only when exactly two switches are in the “on” position.

6. Based on this conductivity table, build a circuit of functional elements with three inputs and one output that implements the formula.

x	y	z	F

7. Analyze the diagram shown in the figure and write out the formula for the function F.

8. Task: “Once the investigator had to interrogate three witnesses at the same time: Claude, Jacques, Dick. Their testimonies contradicted each other, and each of them accused someone of lying.

1) Claude claimed that Jacques was lying.

2) Jacques accused Dick of lying.

3) Dick persuaded the investigator not to believe either Claude or Jacques.

But the investigator quickly brought them to clean water without asking them a single question. Which witness was telling the truth?

9. Determine which of the four students passed the exam, if it is known that:

1) If the first passed, then the second passed.

2) If the second passed, then the third passed or the first did not pass.

3) If the fourth did not pass, then the first passed, and the third did not pass.

4) If the fourth passed, then the first passed.

Portal for the student. Self-training

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