Silogism once the investigator had to simultaneously interrogate three witnesses: Claude, Jacques and Dick. Their testimony contradicted each other, and each of. Why did the inspector come to this conclusion? Once the investigator had to interrogate three witnesses

. 18 years.

Decision

First method . By the condition of the problem, you can make an equation. Let the age of Dima - X years, then the age of sisters x / 3, and brother - x / 2; (x + x / 3 + x / 2): 3 \u003d 11. After solving this equation, we obtain that x \u003d 18. Dima turned 18 years old. It will be useful to bring a slightly different solution, "in parts".

Second way . If the ages of Dima, his brother and sisters to portray the segments, then the "Dimit Segment" consists of two "brother segments" or three "sister sections". Then, if the age of Dima is divided into 6 parts, then the age of the sister is two parts, and the age of the brother is three parts. Then the sum of their ages is 11 such parts. On the other hand, if average age Equal 11 years, then the sum of the ages is 33 years old. From where it follows that in one part - three years. So, Dima turned 18 years old.

Check criteria .

Complete correct solution - 7 Points.

The equation is true, but errors are allowed when solving 3 Point .

The correct answer is given and check is performed - 2 Point .

0 Points .

Answer . Sam Gray.

Decision .

From the conditions of the problem, it is clear that the statements of each of the witnesses are pronounced about the statements of the remaining two witnesses. Consider the statement of Bob Black. If what he says is true, then Sam Gray and John White Lgut. But from the fact that John White lies follows that not all the testimony of Sam Graya is a solid lie. And this is contrary to the words of Bob Black, to whom we decided to believe and that claims that Sam Gray is lying. So, the words Bob Black can not be true. So he lied, and we must recognize the words of Sam Gray True, and, consequently, John White's allegations are a lie. Answer: Not lied Sam Gray.

Check criteria .

A complete true analysis of the situation of the task is given and a correct answer is given - 7 Points .

A complete correct analysis of the situation is given, but for any reason it is an incorrect answer (for example, instead of someone who Litled, those who lied) are in response - 6 Points .

The correct analysis of the situation is given, but for some reason it is not given a correct answer (for example, it is proved that Bob Black lied, but further conclusions are not made) - 4 Point .

The correct answer is shown and it is shown that it satisfies the condition of the task (check), but it has not been proven that the answer is the only thing - 3 Point .

1 score .

0 Points .

Answer . One number 175.

Decision . First method . In the number of numbers that the number is written, no digit 0, otherwise the condition of the problem cannot be performed. This three digit number It was obtained by multiplying on 5 works of its digits, therefore, it is divided by 5. So, its entry ends with a digit 5. We obtain that the product of the numbers multiplied by 5 must be divided by 25. Note that there can be no even numbers in the recording of the number Otherwise, the product of the numbers would be zero. Thus, the three-digit number must be divided into 25 and do not contain even numbers. There are only five such numbers: 175, 375, 575, 775 and 975. The product of the figures of the desired number should be less than 200, otherwise, multiplied by 5, will give a four-digit number. Therefore, numbers 775 and 975 are not obviously suitable. Among the remaining three numbers, only 175 satisfies the condition of the task. Second way. Note (similar to the first method of solving) that the last figure of the desired number is 5. Leta. , b. , 5 - sequential figures of the desired number. By the condition of the task, we have: 100a. + 10 b. + 5 = a. · b. · 5 · 5. Objects both parts of the equation to 5, we get: 20a. + 2 b. + 1 = 5 aB . After subtracting from both parts of the equality 20a and mapping a common factor in the right-hand side, we get: 2b. + 1 = 5 a. (b. – 4 a.) (1 ). Considering that a. and b. can take natural values \u200b\u200bfrom 1 to 9, we obtain that possible values \u200b\u200bof A - only 1 or 2. But A \u003d 2 does not satisfy the equality (1 ), in the left part of which an odd number, and in the right when substituting A \u003d 2, it turns out even. So, the only possibility is a \u003d 1. Substituting this value in (1 ), we get: 2 b. + 1 = 5 b. - 20, from where b. \u003d 7. Answer: The only number is 175.

Check criteria .

Complete correct solution - 7 Points .

The right answer is obtained and there are reasoning, substantially cutting over the options, but there is no complete solution - 4 Point .

The equation is true and transformations and reasoning are presented, allowing to solve the problem, but the solution is not brought to the end - 4 Point .

Browse options reduced, but there is no explanation, why, and the right answer is indicated - 3 Point .

The equation is true, but the task is not solved - 2 Point .

The solution has reasoning, allowing you to exclude any numbers from consideration or consider numbers with certain properties (for example, an ending digit 5), but there is no significant advancement in the solution - there is no 1 score .

Only the right answer or a response to check is given - 1 score .

Answer . 75 ° .

Decision . Consider the triangle AOC, where about the center of the circle. This triangle is a preceded, as OS and OA - radii. So, by property equal triangle, Angles A and C are equal. We will carry out perpendicular see to the side of the JSC and consider right triangle OMS. Under the condition of the task, katat cm - half of the OS hypotenuse. So, the value of the angle of som is 30 °. Then, by the theorem on the sum of the corners of the triangle, we obtain that the angle of CAO (or Sav) is 75 °.

Check criteria .

The faithful reasonable solution of the problem - 7 Points.

The right arguments are given, which are the solution of the problem, but for some reason it is an incorrect answer (for example, the CAO angle instead of the CAO angle) - 6 Points.

In general, loyal reasoning, in which errors that do not have a decision of a fundamental nature are made, and given a correct answer - 5 Points.

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

Additional construction and designations are made in the drawing, of which the decision is clear, the correct answer is given, but the arguments themselves are not given - 3 Point.

A true answer is given with incorrect reasoning - 0 Points.

Only the right answer is given - 0 Points.

Answer . See drawing.

Decision . We transform this equation, allocation of the root of the root full of the root :. The expression in the right part makes sense only at x \u003d 9. Substituting this value to the equation, we get: 9 2 – y. 4 \u003d 0. Spread the left part on factors: (3 -y.)(3 + y.)(9 + y. 2 ) \u003d 0. From where y. \u003d 3 or y. \u003d -3. So, the coordinates of only two points (9; 3) or (9; -3) satisfy this equation. The graph of the equation is shown in the figure.

Check criteria.

The right transformations and reasoning were performed and the schedule was correctly built - 7 Points.

Conducted correct transformations, but lost value y. \u003d -3; As a graph, one point is indicated -3 Point.

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

The correct transformations were carried out, but it was announced that the expression under the root (or on the right side after the construction of the square) negatively and the schedule is a blank multiple points - 1 score

Conducted arguments that led to the direction of two points, but these points are somehow connected (for example, a segment) - 1 score

Are indicated without explanation two points that are somehow connected - 0 Points.

In other cases - 0 Points.

Answers of the tasks of the second stage of the Olympics

Answer . May.

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:

Check criteria.

The right answer indicating numbers a. and b. – 7 Points .

A system of equations is compiled, but an arithmetic error is allowed when it is solved - 3 Point .

Only the answer - 1 score .

Answer . For 12 seconds .

Decision . Between the first and fourth floors of 3 span, and between the fifth and first - 4. According to the condition, Petya 4 span runs for 2 seconds longer than the mother rides on the elevator, and three spans - for 2 seconds faster mom. So, in 4 seconds, Petya runs one span. Then from the fourth floor to the first (i.e. on 3 span) Peter runs off for 4 * 3 \u003d 12 seconds.

Check criteria.

The right answer fulfilled - 7 Points .

It is explained that one flight requires 4 seconds, 4 seconds are in response - 5 Points .

The right substantiation suggests that the path from the fifth floor for the first 1.25 times the path from the fourth floor to the first and answer is 16 seconds - 3 Point .

Only the answer - 0 Points .

Answer . See drawing.

Decision . Because H. 2 =| H. | 2 , then u =| H. |, Moreover, x ≠ 0.

Can also, using the module definition, obtain that (at x = 0 function is not defined).

Check criteria.

Faithful schedule with an explanation - 7 Points .

Faithful schedule without any explanation - 5 Points .

Function schedule U. \u003d | x | without a drop point -3 Point .

Answer . Yes .

Decision . We divide this square with a side of 5 straight, parallel to its parties, by 25 squares with a side 1 (see Fig.). If there were no more than 4 marked points in each such square, then no more than 25 * 4 \u003d 100 points would be noted, which contradicts the condition. Therefore, at least in one of the coming squares should be 5 of the noted points.

Check criteria.

The right decision - 7 Points .

Only the answer - 0 Points .

Answer . Eight ways.

Decision . From paragraph A) it follows that the coloring of all points with integer coordinates is uniquely determined by the coloring points corresponding to the numbers 0, 1, 2, 3, 4, 5 and 6. The point 0 \u003d 14-2 * 7 should be painted as much as 14, those. red. Similarly, the point 1 \u003d 71-107 should be painted blue, point 3 \u003d 143-20 * 7 - blue, and 6 \u003d 20-2 * 7 - red. Therefore, it remains only to calculate, how many different ways you can paint the points corresponding to the numbers 2, 4 and 5. Since each point can be painted in two ways - red or blue - then all methods 2 * 2 * 2 \u003d 8. Note. When calculating the number of ways to color dots 2, 4 and 5, you can simply list all the methods, for example, in the form of a table:

Check criteria .

The right answer with the right substantiation - 7 Points .

The task is reduced to the calculation of the number of ways to color 3 points, but a response is received 6 or 7 - 4 Point .

The task is reduced to the calculation of the number of ways to color 3 points, but the counting of the number of methods is missing or received a response other than those of the previously 3 Point .

The answer (including the right) without justification - 0 Points .

Answer . 4 times.

Decision .

We carry out the segments of MK and AS . Quadrangle MVKE consists of

triangles MVK and MKA , a quadrangle AES.D - from triangles

1 method . Triangles MVK and ASD - rectangular and kartets of the first 2 times less waters of the second, so they are similar and the area of \u200b\u200bthe AC triangleD. 4 times larger than the area of \u200b\u200bthe triangle MVK. Because M and K. – middle AB and Sun respectively, then MK– , Therefore, MK || AC and MK = 0.5As. . From the parallelism of direct MK and AU follows

triangles MKA and AES, and because The likeness ratio is 0.5, then the AES triangle area is 4 times larger than the area of \u200b\u200bthe triangle MKA. Now: S. AES. D \u003d SAEC + SACD \u003d 4 SMKE +. 4 SMBK \u003d. 4 (SMKE + SMBK) \u003d 4 Smbke.

2 method . Let the area of \u200b\u200bthe ABC rectangleD. equal S. Then the area of \u200b\u200bthe triangleD. equal ( the diagonal of the rectangle divides it into two equal triangles), and the area of \u200b\u200bthe triangle of the MVK is equal to MV × VK \u003d because. M and K. – mid Segments AB and Sun, then ak and cm – medians of the triangle ABS, Therefore, E. – point of intersection Median Triangle ABC, those. Distance from E to AC is equalh, Where h - the height of the triangle AVS., spent from the top in. Then the area of \u200b\u200bthe AES triangle is equal. Then for the square of the quadrangle AESD, equal to the sum of the area of \u200b\u200bthe triangles AES and ASD, we get: Next, because MK– the middle line of the triangle ABC, then the triangle area of \u200b\u200bthe MKA is equal* H - * H) \u003d H) \u003d (AC * H) \u003d\u003d S . Therefore, for the square of the quadrangle MVK, equal to the amount of the triangles of the MVK and the IC, we get :. Thus, the ratio of the square of quadrangles AESD. And the MVKE is equal.

Check criteria.

The right decision and the right answer -7 points .

The right decision, but the answer is incorrect because of an arithmetic error -5 points .

5. Summing up and rewarding winners

The final indicators of the jury's competitive tasks performed inaccordance with the developed assessment criteria;

For the winners of the Olympics defined by the highest quantity points,there are three prizes;

The results of the competition are made up by the report of the organizer of the Olympiad.

Winners are awarded with diplomas and valuable gifts.

In case of disagreement with an assessment of the jury, the participant may submitwritten appeal within an hour after declaring results.

The publicity of the competition is ensured - according to the results of the competition announced Winners winners.

You can select the following sequence of steps in solving logical tasks.

1. Allocate the problem of the problem of elementary (simple) statements and designate them with letters.

2. Write the task condition in the language of the Logic algebra, to combine simple statements in difficult using logical operations.

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

4. Using the laws of logic algebra, try to simplify the resulting expression and calculate all its values \u200b\u200bor build a truth table for the expression under consideration.

5. Select a solution - set of values Simple statements in which the constructed logical expression is true.

6. Check whether the solution obtained is satisfying the task condition.

Example:

Task 1: "Trying to remember the winners of last year's tournament, five former viewers of the tournament said that:

1. Anton was the second, and Boris - the fifth.

2. Victor was the second, and Denis is the third.

3. Grigory was the first, and Boris - Third.

4. Anton was the third, and Evgeny - sixth.

5. Victor was the third, and Evgeny is the 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 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 will carry out equivalent transformations, we get :.

5) From paragraph 4 follows: ,.

6) The distribution of places in the tournament: Anton was the third, Boris - the fifth, Victor - the Second, Grigory - the first, and Evgeny - the fourth.

Task 2: "On charges of robbery, Ivanov, Petrov, Sidorov appeared before the court. The consequence of:

1. If Ivanov is not guilty or Petrov guilty, then the sidors are guilty;

2. If Ivanov is not guilty, then the sidors are not guilty.

Is Ivanov guilty? "

1) Consider sayings:

BUT: "Ivanov is guilty," IN: "Petrov is guilty", FROM: "Sidorov is guilty."

2) Facts established by the consequence: ,.

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

We will make a table of truth for it.

BUT	IN	FROM	L.

Solve the task - it means to specify at what values \u200b\u200band the obtained complex statement l is true. If, and then the investigation is not enough facts in order to accuse Ivanov in a crime. The analysis of the table shows and, i.e. Ivanov in robbery is guilty.

Questions and tasks.

1. Make a RCC for formulas:

2. Simplify RKS:

3. For this switching scheme, construct the proper logical formula.

4. Check the balance of RCC:

5. Construct a diagram of three switches and light bulbs so that the light is lit only when exactly two switches are in the "Included" position.

6. According to this table of conductivity, construct a diagram from functional elements with three inputs and one output that implements the formula.

x.	y.	z.	F.

7. Analyze the scheme shown in the figure and write down the formula for the function F..

8. Task: "Once again, the investigator had to simultaneously interrogate three witnesses: Claude, Jacques, Dick. Their testimony was contrary to each other, and each of them accused someone in lies.

1) Claude argued that Jacques Lzhet.

2) Jacques accused of lies Dick.

3) Dick persuaded the investigator not to believe neither Claude or Jacques.

But the investigator quickly brought them to clean waterwithout asking them any question. Which of the Witnesses spoke the truth?

9. Determine who 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. To the question of which of the three students has studied logic, the answer was received: if I studied the first, I studied the third, but it is not true that if I studied the second, I studied the third. Who studied logic?

1. a) ( switching disjunction );

b)

(communction commutativity );

2. a) ( associativity for disjunction );

b) ( associative conjunction );

3. a) ( distribution of disjunction relative to conjunction );

b) ( distribution of conjunction relative to disjunction );

4.

and

–laws de Morgana .

5.

;

;

;

6.

(or

) (the law of an excluded third );

(or

(law of contradiction );

7.

(or

);

(or

);

(or

);

(or

).

These properties are usually used to convert and simplify logical formulas. Here are the properties of only three logical operations (disjunction, conjunction and denial), but then it will be shown that all other operations can be expressed through them.

Using logical ligaments, you can make logical equations, and solve logical tasks is similar to how arithmetic tasks are solved using systems of conventional equations.

Example.Once the investigator had to simultaneously interrogate three witnesses: Claude, Jacques and Dick. Their testimony was contrary to each other, and each of them accused someone in lies. Claude argued that Jacques Lzhet, Jacques accused of lies Dick, and Dick persuaded the investigator not to believe in neither 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?

Decision. Consider sayings:

(Claude speaks the truth);

(Jacques tells the truth);

(Dick speaks the truth).

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

1) Either Claude said the truth, and then Jacques lied, or Claude licked, and then Jacques told the truth;

2) either Jacques told the truth, and then Dick Lital Malgar, or Jacques Lital, 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 wrong that both other witnesses lied (ie, at least one of these witnesses told the truth).

Express these statements in the form of a system of equations:

The task condition will be executed if these three statements are simultaneously true, which means that their conjunction is true. Move these equalities (i.e., take their conjunction)

But

in that and only the case if

, but

. Consequently, Jacques tells the truth, and Claude and Dick Lgut.

Any -nual operation, designated, for example,

will be fully defined if installed under what values \u200b\u200bof statements

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

In the table of values \u200b\u200bof the statement formed from simplest statements

, there is lines. The values \u200b\u200bcolumn also has positions. Therefore, there is

different options for filling, and, accordingly, the number of all -Chentable operations equal to

. For

the number of single operations is 4, with

the number of bounced - 16, with

the number of three-mended - 256, etc.

Consider some special types of formulas.

Formula called elementary conjunction if it is a conjunction of variables and denial of variables. For example, formulas ,

,

,

- Elementary conjunctions.

A formula for disjunction (possibly unicked) elementary conjunctions, called disjunctive normal form (d. n. f.). For example, formulas ,

,

.

Theorem 1.(On bringing to d. n. f.). For any formula , D.N. f. .

This theorem and the following theorem 2 will be proved in the next paragraph. Applying these theorems, it is possible to standardize the type of logical formulas.

Formula called elementary disjunction If it is a disjunction of variable variables and denials. For example, formulas

,

,

etc.

The formula that is conjunction (possibly universal) elementary disjunction is called conjunctive normal form (k. n. f.). For example, formulas

,

.

Theorem 2.(On bringing to. n. f.). For any formula can be found equivalent to her formula , which is to. n. f.

Once the investigator had to simultaneously interrogate three witnesses: Claude, Jacques and Dick. Their testimony was contrary to each other, and each of them accused someone in lies. Claude argued that Jacques Lzhet, Jacques accused of lies Dick, and Dick persuaded the investigator not to believe in neither Claude or Jacques. But the investigator quickly brought them to clean water without asking them a single question. Who of the witnesses spoke the truth

Ilya Muromtsu, Dobryne Nikitich and Alyosha Popovichu for the faithful service given 6 coins: 3 gold and 3 silver. Everyone got two coins. Ilya Muromets does not know which coins got good, and what aleas, but knows which coins got himself. Come up with the question for which Ilya Murometh will answer "yes", "no" or "I don't know," and, after answering which you can understand what coins he got

The rules of Sillogisoms 1. In terms of Slogism, there should be only three statements and only three terms. Lady all the excursions fled in different directions, Petrov's excursant, then he was ruined in different directions. 3. If both parcels are private statements, then the conclusion is impossible. 2. If one of the parcels is a private statement, then the conclusion should be private. 4. If one of the parcels is a negative statement, then the conclusion is a negative statement. 5. If both parcels negative statementsThe conclusion is impossible to do 6. The average term must be distributed at least in one of the parcels. 7. The term cannot be distributed in conclusion if it is not distributed in the parcel.

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

Task 35.

One person entered the work with salary in, $ 1000 per year. During the discussion of the conditions during the reception, he was promised that in case of good work, an increase will be made to the salary. Moreover, the amount of the increase can be chosen from two options at its discretion: in one case, an increase of 50 dollars was offered every six months, starting from the second half, in another - $ 200 each year, starting from the second. By providing freedom of choice, the employers wanted not only to try to save on a salary, but also to check how quickly the new employee cleans. Thinking for a minute, he confidently called the conditions of addition.

What option was preferred?

Task 36.

Once the investigator had to simultaneously interrogate three witnesses: Claude, Jacques and Dick. Their testimony was contrary to each other, and each of them accused someone in lies. Claude claimed that Jacques Lgez. Jacques accused of lies Dick, and Dick persuaded the investigator not to believe her Claude or Jacques. But the investigator quickly brought them to pure water, without setting them a single question.

Which of the Witnesses spoke the truth?

Task 37.

A terrible misfortune, inspector, said an employee of the museum. - You can not imagine how excited me. I'll tell you everything in order. I stayed today at the museum in order to work and put our financial affairs in order. I just sat at this writing desk and looked through the bills, as suddenly saw the shadow on the right side. The window was open.

And you did not hear any rustle? - asked the inspector.

Absolutely no. The radio has played music, besides, I was too passionate about my occupation. Tinging the eye from the game and, I saw that some person jumped out of the window. ITHOT turned on the top light and found that two boxes were disappeared with the most valuable collection of coins, which I took to my office for work. Jaw a terrible state: after all, this collection is estimated at 10 thousand marks.

You think I'm really; I believe your fabrication?

Izynously noticed the inspector. - No one has yet managed to mislead me, and you will not be the first.

How did the inspector guessed that he was trying to deceive?

Task 38.

The corpse of the missing facial was found wrapped in a sheet, which had a license plate laundry tag. Installed a family that used such tags, however, in the verification process it turned out that the members of this family were not familiar and had no contacts with the dead and his loved ones. There were no other evidence of involvement in their murder.

Is it not allowed when checking the error in the completeness and correctness of obtaining information?

Task 39.

In the aviation unit serve Potapov, Shchedrin, Semenov. Konovalov and Samoilov. Their specialties are: pilot, navigator, bornemaker, radist and synoptic.

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

Shchedrin and Konovalov are not familiar with the aircraft management;

Potapov and Konovalov are preparing to become assaults; Apartments Shchedrin and Samoilov are near the apartment of a radar;

Semenom, being in the holiday home, met Shchedrin and sister Sinsinstika: Potapov and Shchedrin in his free time playing chess with a bornemaker and pilot; Konovalov, Semenov and weather forecasters are fond of boxing; Radine box is not fond of.

Task 40.

Aunt, who was waiting for his nephew - inspector, rushed towards him, without hiding his impatience.

Some woman just that; I rang my handbag with money and immediately disappeared.

Most likely she disappeared in the savings cassation itself, where you were, - I noticed the inspector. - Let's try to find it.

And in fact, Aunt immediately saw her bag that stood on a bench between two women. It was revealed. When the inspector threw a careful look at the bag, both women, noticing this, got up and went to the other end of the room. The handbag remained a bench.

But I do not know which of them stole my bag. Yane managed to see her, "Tiet said.

Well, these are trifles, "the nephew answered. - Interrogate both, but I think that the bag has stolen the bag that ...

What?

Task 41.

Having received a message that the gray "Chevrolet" with the number starting to the sixth, hit the woman and disappeared, the inspector and his assistant left for the Mr. Villa, whose car seems to be: complied with the description. It was not half an hour as they were in place.

Gray "Chevrolet" stood in front of the house. Seeing the police, the owner went down to them right in pajamas.

Nicatuda did not leave today, "he said, after hearing the inspector. - Yes, and could not: Yesterday I lost the key from the ignition, and the new will be ready only on Friday.

Assistant, having time to inspect the car, whispered inspector:

Apparently, he tells the truth. There are no traces of collision on the car.

Inspector, borrowing a car hood, answered:

It does not mean anything, the blow was mesmer, because the victim is alive. And your Alibi, Mr., seems to me extremely suspicious. Why are trying to hide from me that you just came here on this machine itself?

What did the inspector give a reason to suspect a lord in lies?

Task 42.

The president of the company informs the investigator about the stealing at him.

Arriving to work, I remembered that I forgot at home required documents. I gave the key to my home safe my assistant and sent him to the folder with the documents. We have been working together with him 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 the departure, he called me on the phone and said that by entering the room, I saw that the door of the wall safe was open, and the paper was scattered across the cabinet. I came home and found that, in addition to scattered documents, jewels and money disappeared from the safe.

Assistant testimony: "When I arrived, I was let the butler and I rose to the second floor of the apartment. Entering the office, discovered papers scattered across semi and an open safe door. I immediately called my chief on the phone and reported on what he saw. After that, I jumped over the ladder platform and called the butler. The servant appeared on my cry from the living room of the lower floor and asked what's the matter. I informed her about what he saw. By her call from the courtyard came rangly. They said to my question that no one in the apartment after the departure of the host came and no noise in the house they did not hear. "

The butler explained: "After in the morning the owner went, I do the usual work on the lower floor and did not see anyone and I did not hear anything unusual. The servant with me from the kitchen did not come out. When the employee of our owner arrived a long-friendly person, he went to the stairs to the second floor, and went to the courtyard. A few minutes later the cook I called me and I entered the house where the assistant said about the steal from the host's office. "

The maid said that after breakfast was Pa Kitchen, did not go anywhere and only, having heard the crock of the assistant, went into the living room. The assistant said about the stealing in the house and asked to know the butler.

To the question of the investigator, the assistant replied that he did not touch anything in the office, except for the phone, and did not rearrange. The butler and the maid said that they did not go to the office at all.

When inspecting in the office, the investigator did not find the cabinet door, the door of the safe, items and the phone on the table there are no trace of fingers. After examining the lock of the safe door, the specialist did not find a part of its details of the traces of any object or foreign key.

Task 35.

What option was preferred?

Task 36.

Once the investigator had to simultaneously interrogate three witnesses: Claude, Jacques and Dick. Their testimony was contrary to each other, and each of them accused someone in lies. Claude claimed that Jacques Lgez. Jacques accused of lies Dick, and Dick persuaded the investigator not to believe her Claude or Jacques. But the investigator quickly brought them to pure water, without setting them a single question.

Which of the Witnesses spoke the truth?

Task 37.

And you did not hear any rustle? - asked the inspector.

You think I'm really; I believe your fabrication?

Izynously noticed the inspector. - No one has yet managed to mislead me, and you will not be the first.

How did the inspector guessed that he was trying to deceive?

Task 38.

Is it not allowed when checking the error in the completeness and correctness of obtaining information?

Task 39.

In the aviation unit serve Potapov, Shchedrin, Semenov. Konovalov and Samoilov. Their specialties are: pilot, navigator, bornemaker, radist and synoptic.

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

Shchedrin and Konovalov are not familiar with the aircraft management;

Potapov and Konovalov are preparing to become assaults; Apartments Shchedrin and Samoilov are near the apartment of a radar;

Task 40.

Aunt, who was waiting for his nephew - inspector, rushed towards him, without hiding his impatience.

Some woman just that; I rang my handbag with money and immediately disappeared.

Most likely she disappeared in the savings cassation itself, where you were, - I noticed the inspector. - Let's try to find it.

But I do not know which of them stole my bag. Yane managed to see her, "Tiet said.

Well, these are trifles, "the nephew answered. - Interrogate both, but I think that the bag has stolen the bag that ...

What?

Task 41.

Gray "Chevrolet" stood in front of the house. Seeing the police, the owner went down to them right in pajamas.

Nicatuda did not leave today, "he said, after hearing the inspector. - Yes, and could not: Yesterday I lost the key from the ignition, and the new will be ready only on Friday.

Assistant, having time to inspect the car, whispered inspector:

Apparently, he tells the truth. There are no traces of collision on the car.

Inspector, borrowing a car hood, answered:

What did the inspector give a reason to suspect a lord in lies?

Task 42.

The president of the company informs the investigator about the stealing at him.

Having arrived at work, I remembered that I forgot the necessary documents. I gave the key to my home safe my assistant and sent him to the folder with the documents. We have been working together with him 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 the departure, he called me on the phone and said that by entering the room, I saw that the door of the wall safe was open, and the paper was scattered across the cabinet. I came home and found that, in addition to scattered documents, jewels and money disappeared from the safe.

The rules of Sillogisoms 1. In terms of Slogism, there should be only three statements and only three terms. Lady all the excursions fled in different directions, Petrov's excursant, then he was ruined in different directions. 3. If both parcels are private statements, then the conclusion is impossible. 2. If one of the parcels is a private statement, then the conclusion should be private. 4. If one of the parcels is a negative statement, then the conclusion is a negative statement. 5. If both parcels are negative statements, then the conclusion is not possible 6. The average term must be distributed at least in one of the parcels. 7. The term cannot be distributed in conclusion if it is not distributed in the parcel.