Syllogisms One day, the investigator had to simultaneously interrogate three witnesses: Claude, Jacques and Dick. Their testimony contradicted each other, and each

Problem 35

One person got a job with a salary of $1,000 a year. During the discussion of the conditions upon admission, he was promised that if he performed well, his salary would be increased. Moreover, you can choose the amount of the increase from two options at your discretion: in one case, an increase of $50 was offered every six months, starting from the second half, in the other - $200 every year, starting from the second. By providing freedom of choice, employers wanted to not only try to save on wages, but also test how quickly the new employee thinks. After thinking for a minute, he confidently named the terms of the increase.

Which option was preferred?

Problem 36

One day, the investigator had to simultaneously interrogate three witnesses: Claude, Jacques and Dick. Their testimony 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 out into the open without asking them a single question.

Which witness was telling the truth?

Problem 37

A terrible accident, inspector, said the museum employee. - 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 financial affairs in order. I was just sitting at this desk and looking through the accounts when I suddenly saw a shadow on the right side. The window was open.

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

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

You believe that I really; Will I believe your fabrications?

The inspector remarked irritably. “No one has ever managed to mislead me, and you won’t be the first.”

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

Problem 38

The missing person's body was found wrapped in a sheet that had a laundry 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 did not know each other and did not have any contact with the deceased and his relatives. No other evidence of their involvement in the murder was established.

During the verification process, were there any errors in the completeness and correctness of the information received?

Problem 39

Potapov, Shchedrin, and Semenov 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 the controls of the aircraft;

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

Semyon, while in a rest home, met Shchedrin and the weather forecaster's sister: Potapov and Shchedrin, in their free time from work, play chess with the flight mechanic and the pilot; Konovalov, Semenov and the weather forecaster are fond of boxing; The radio operator is not into 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; she snatched my purse with money and immediately disappeared.

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

And in fact, the aunt immediately saw her bag, which was standing on the bench between the two women. It was revealed. When the inspector took a careful look at the bag, both women, noticing this, stood up and walked to the other end of the room. The purse remained on the bench.

But I don't know which one stole my bag. “Yana didn’t have time to see her,” said her aunt.

“Well, it’s nothing,” the nephew answered. - We’ll interrogate both of them, but I think that the one who stole your bag was the one from...

Which?

Problem 41

Having received a message that a gray Chevrolet with a license plate starting with six had hit a woman and fled, the inspector and his assistant went to the villa of a gentleman whose car seemed to match the description. Less than half an hour had passed before they were there.

A gray Chevrolet was parked in front of the house. Seeing the police, the owner came down to them in his pajamas.

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

The assistant, having meanwhile managed to inspect the car, 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 doesn’t 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 lying?

Problem 42

The president of the company informs the investigator about a theft committed at his home.

Arriving at work, I remembered that I had forgotten the necessary documents at home. I gave the key to the home safe to my assistant and sent him to get a folder of documents. We have been working together for a long time, I have trusted 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, upon entering the room, he saw that the door of the wall safe was open and papers were scattered throughout the office. I arrived home and discovered 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, I found papers scattered on the floor and an open safe door. I immediately called my boss and reported what I had seen. After that, I jumped out onto the landing and called the butler. In response to my cry, a maid appeared from the living room on the lower floor and asked what was the matter. I told her what I saw. At her call, the butler came running from the yard. When I asked, 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 was doing my usual work on the ground floor and did not see anyone or hear anything unusual. The maid did not leave the kitchen in front of me. When an employee of our owner, who had known me 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 me about the theft from the owner’s office.”

The maid said that after breakfast she was in the kitchen, did not go anywhere, and only when she heard the assistant’s cry did she go out into the living room. The assistant reported a theft in the house and asked to know the butler.

In response to the investigator's question, the assistant replied that he did not touch anything in the office except the telephone, and did not rearrange it. The butler and maid said that they did not go into the office at all.

When examining the office, the investigator did not find any traces of fingers on the office door, safe door, objects or telephone on the table. Having examined the lock of the safe door, the specialist did not find any traces of any object or foreign key on its parts.

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

1. Select elementary (simple) statements from the problem statement and label them with letters.

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

3. Create a single logical expression for the requirements of the task.

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

5. Choose a solution – set of values simple statements in which the constructed logical expression is true.

6. Check whether the resulting solution satisfies the conditions of the problem.

Example:

Task 1:“Trying to remember the winners of last year's tournament, five former spectators of the tournament stated that:

1. Anton was second, and Boris was fifth.

2. Victor was second, and Denis was third.

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

4. Anton was third, and Evgeniy was sixth.

5. Victor was third, and Evgeniy 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) Let us denote by the first letter in the name of the tournament participant, a, 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 Let's carry out equivalent transformations, we get: .

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

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

Task 2:“Ivanov, Petrov, Sidorov appeared in court on charges of robbery. The investigation established:

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:

A: “Ivanov is guilty”, IN: “Petrov is guilty,” WITH: “Sidorov is guilty.”

2) Facts established by the investigation: , .

3) Single logical expression: . It is true.

Let's create a truth table for it.

A	IN	WITH	L

To solve a problem means to indicate at what values of A the resulting complex statement L is true. If , a , 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 assignments.

1. Compose the RKS for the formulas:

2. Simplify the RKS:

3. Using this switching circuit, construct a corresponding logical formula.

4. Check the equivalence of the RKS:

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

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

x	y	z	F

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

8. Problem: “Once the investigator had to simultaneously interrogate three witnesses: Claude, Jacques, Dick. Their testimony 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 tried to persuade the investigator not to believe either Claude or Jacques.

But the investigator quickly brought them to light 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 one passes, then the second one passes.

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 one passes, then the first one passes.

10. To the question 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?

. 18 years old.

Solution

First way . Based on the conditions of the problem, you can create an equation. Let Dima’s age be x years, then the age of the sister is x/3, and the age of the brother is x/2; (x + x/3 + x/2):3=11. After solving this equation we find that x=18. Dima turned 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 depicted by segments, then “Dima’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 middle age is 11 years, then the sum of the ages is 33 years. It follows that in one part there are three years. This means Dima is 18 years old.

Verification criteria .

Completely correct solution - 7 points.

The equation was drawn up correctly, but errors were made in the solution - 3 points .

The correct answer is given and the check is completed - 2 points .

0 points .

Answer . Sam Gray.

Solution .

From the conditions of the problem it is clear that the statements of each of the witnesses are made regarding 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 the fact that John White lies means that not all of Sam Gray's testimony is a complete lie. And this contradicts the words of Bob Black, whom we choose to believe and who claims that Sam Gray is lying. So what Bob Black said can't be true. This means that he lied, and we must accept Sam Gray's words as true, and, therefore, John White's statements as lies. Answer: Sam Gray did not lie.

Verification criteria .

A complete correct analysis of the problem situation 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 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 proven 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 conditions of the problem (verification has been carried out), but it has not been proven that the answer is the only one - 3 points .

1 point .

0 points .

Answer . One number 175.

Solution . First way . The digits used to write a number do not contain the number 0, otherwise the condition of the task cannot be fulfilled. Given three digit number obtained by multiplying the product of its digits by 5, therefore, it is divisible by 5. This means that its notation ends with the number 5. We obtain that the product of digits multiplied by 5 must be divisible by 25. Note that there cannot be even digits in the notation of a number, otherwise the product of the digits would be equal to 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 conditions of the problem. Second way. Note (similar 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 conditions 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 20a from both sides of the equation and taking out the common factor on the right side, we get: 2b + 1 = 5 a (b – 4 a) (1 ). Considering that a And b can take natural values from 1 to 9, we get that possible values a – 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 substituting a=2, 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 number required is 175.

Verification criteria .

Completely correct solution - 7 points .

The correct answer is received and there are arguments that significantly reduce the search for options, but complete solution No - 4 points .

The equation is drawn up correctly and transformations and reasoning are given to solve the problem, but the solution is not completed - 4 points .

The list 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 .

The solution contains reasoning that makes it possible to exclude any numbers from consideration or consider numbers with certain properties (for example, ending with the number 5), but there is no further significant progress in the solution - 1 point .

Only the correct answer or a verified answer is given - 1 point .

Answer . 75° .

Solution . Consider the triangle AOC, where O is the center of the circle. This triangle is isosceles, since OC and OA are radii. So, by property isosceles triangle, angles A and C are equal. Let us draw a perpendicular CM to the side AO and consider right triangle Compulsory medical insurance. According to the conditions of the problem, leg CM is half of the hypotenuse OS. This means that the angle COM is 30°. Then, according to the theorem on the sum of the angles of a triangle, we find that the angle CAO (or CAB) is equal to 75°.

Verification criteria .

A correct, well-founded solution to the problem – 7 points.

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

Generally correct reasoning is given, in which errors were made that are not fundamental to the essence of the decision, and the correct answer is given - 5 points.

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

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

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

Only the correct answer is given - 0 points.

Answer . See picture.

Solution . Let's transform this equation by selecting a complete square under the root sign: . The expression on the right side makes sense only when x = 9. Substituting this value into the equation, we get: 9 2 – y 4 = 0. Let’s factorize the left side: (3 –y)(3 + y)(9 + y 2 ) = 0. Where 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 constructed - 7 points.

The correct conversions were carried out, but the meaning was 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 point.

The correct transformations were carried out, but it was 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 point.

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

Two points are indicated without explanation that are somehow connected - 0 points.

In other cases - 0 points.

Answers to tasks of the second stage of the Olympiad

Answer . They can.

Solution . If a = , b = - , then a = b+1 and a 2 = 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 when solving it - 3 points .

The only answer is 1 point .

Answer . In 12 seconds .

Solution . There are 3 flights between the first and fourth floors, and 4 between the fifth and first floors. According to the condition, Petya runs 4 flights 2 seconds longer than his mother takes the elevator, and three flights is 2 seconds faster than his mother. This means that Petya runs one flight in 4 seconds. Then Petya runs from the fourth floor to the first (i.e., 3 flights) in 4*3=12 seconds.

Verification criteria.

Correct answer with complete solution - 7 points .

It is explained that one flight takes 4 seconds, the answer indicates 4 seconds - 5 points .

The correct justification is based on the assumption that the path from the fifth floor to the first is 1.25 times longer than the path from the fourth floor to the first and the answer is 16 seconds - 3 points .

The only answer is 0 points .

Answer . See picture.

Solution . Because X 2 =| X | 2 , then =| X |, and x≠ 0.

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

Verification criteria.

Correct graph with explanation - 7 points .

True graph without any explanation - 5 points .

Graph of function =|x| without a punctured point -3 points .

Answer . Yes .

Solution . Let's divide given square with a side of 5 straight lines parallel to its sides, into 25 squares with a side of 1 (see figure). If each such square had no more than 4 marked points, then in total 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 is 7 points .

The only answer is 0 points .

Answer . Eight ways.

Solution . 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. Point 0=14-2*7 should 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, all that remains is to count how many in various 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 a total of 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 justification - 7 points .

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

The task 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 received is different from those indicated earlier - 3 points .

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

Answer . 4 times.

Solution .

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

triangles MVK and MKE , and the quadrilateral AESD – 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, so they are similar and the area of the triangle is ACD 4 times more area triangle MVK. Because M and K – the middles of AB and BC respectively, then MK– , therefore MK || AS and MK = 0.5AC . From the parallelism of straight lines MK and AC, the similarity follows

triangles MKE and AEC, and because similarity coefficient is 0.5, then the area of triangle AEC is 4 times greater than 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 rectangle ABCD equal to S. Then the area of triangle ACD equal to ( the diagonal of a rectangle divides it into two equal triangle), and the area of the triangle MVK is equal to MV×VK=T.k. M and K – midpoints of segments AB and BC, then AK and SM – median of triangle ABC, therefore E – point of intersection of the 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 AESD, 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 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 the quadrilaterals AESD and MVKE is equal.

Verification criteria.

The right solution and the right answer -7 points .

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

5. SUMMING UP THE RESULTS 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 in a report from the organizer of the Olympiad.

The winners are awarded certificates and valuable gifts.

In case of disagreement with the assessment 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 - based on the results of the competition,winners.

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

1. Select elementary (simple) statements from the problem statement and label them with letters.

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

3. Create a single logical expression for the requirements of the task.

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

5. Choose a solution – set of values simple statements in which the constructed logical expression is true.

6. Check whether the resulting solution satisfies the conditions of the problem.

Example:

Task 1:“Trying to remember the winners of last year's tournament, five former spectators of the tournament stated that:

1. Anton was second, and Boris was fifth.

2. Victor was second, and Denis was third.

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

4. Anton was third, and Evgeniy was sixth.

5. Victor was third, and Evgeniy 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) Let us denote by the first letter in the name of the tournament participant, a, 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 Let's carry out equivalent transformations, we get: .

5) From point 4 it follows: , .

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

Task 2:“Ivanov, Petrov, Sidorov appeared in court on charges of robbery. The investigation established:

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:

A: “Ivanov is guilty”, IN: “Petrov is guilty,” WITH: “Sidorov is guilty.”

2) Facts established by the investigation: , .

3) Single logical expression: . It is true.

Let's create a truth table for it.

A	IN	WITH	L

To solve a problem means to indicate at what values of A the resulting complex statement L is true. If so, 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 assignments.

1. Compose the RKS for the formulas:

2. Simplify the RKS:

3. Using this switching circuit, construct a corresponding logical formula.

4. Check the equivalence of the RKS:

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

6. Using this conductivity table, construct 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 down the formula for the function F.

8. Problem: “Once the investigator had to simultaneously interrogate three witnesses: Claude, Jacques, Dick. Their testimony 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 tried to persuade the investigator not to believe either Claude or Jacques.

But the investigator quickly brought them to light 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 one passes, then the second one passes.

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 one passes, then the first one passes.

1. a) ( commutativity of disjunction );

b)

(commutativity of conjunction );

2. a) ( associativity of disjunction );

b) ( conjunction associativity );

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

b) ( distributivity of conjunction relative 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 given properties are usually used to transform and simplify logical formulas. Here the properties of only three logical operations (disjunction, conjunction and negation) are given, but further it will be shown 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. One day, the investigator had to simultaneously interrogate three witnesses: Claude, Jacques and Dick. Their testimony 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 light without asking them a single question. Which witness was telling the truth?

Solution. Let's look at the statements:

(Claude is telling the truth);

(Jacques is telling the truth);

(Dick is telling the truth).

We don't know which ones are true, 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).

Let us express these statements in the form of a system of equations:

The condition of the problem will be fulfilled if these three statements are simultaneously true, which means their conjunction is true. Let's multiply these equalities (i.e. take their conjunction)

But

if and only if

, A

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

Any -member operation, denoted, for example,

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

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

In the table of meanings of the statement formed from simple sayings

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

different options for filling it, and, accordingly, the number of all -member operations are equal to

. At

the number of one-term operations is 4, with

the number of binomial ones is 16, with

number of three-term ones – 256, etc.

Let's look at 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 ,

,

,

– elementary conjunctions.

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

,

.

Theorem 1(about reduction to D.N.F.). For any formula , who is a doctor of science. f. .

This theorem and the following Theorem 2 will be proved in the next section. By applying these theorems, it is possible to 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(about reduction to Ph.D.). For any formula one can find an equivalent formula , who is a Ph.D. f.

One day, the investigator had to simultaneously interrogate three witnesses: Claude, Jacques and Dick. Their testimony 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 light without asking them a single question. Which of the witnesses was telling the truth?

Ilya Muromets, Dobrynya Nikitich and Alyosha Popovich were given 6 coins for their faithful service: 3 gold and 3 silver. Everyone got two coins. Ilya Muromets does not know which coins went to Dobrynya and which to Alyosha, but he knows which coins he got. Come up with 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. A syllogism must have only three statements and only three terms. ZhG All the excursionists fled to different sides, Petrov is an excursionist, which means he ran away in different directions. 3. If both premises are private statements, then the conclusion cannot be drawn. 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 a negative statement. 5. If both premises are negative statements, then the conclusion cannot be drawn. 6. The middle term must be distributed in at least one of the premises. 7. A term cannot be distributed in the conclusion if it is not distributed in the premise.

All cats have four legs. All dogs have four legs. All dogs are 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

Problem 35

Which option was preferred?

Problem 36

Which witness was telling the truth?

Problem 37

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

You believe that I really; Will I believe your fabrications?

The inspector remarked irritably. “No one has ever managed to mislead me, and you won’t be the first.”

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

Problem 38

During the verification process, were there any errors in the completeness and correctness of the information received?

Problem 39

Potapov, Shchedrin, and Semenov 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 the controls of the aircraft;

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

Problem 40

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

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

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

But I don't know which one stole my bag. “Yana didn’t have time to see her,” said her aunt.

“Well, it’s nothing,” the nephew answered. - We’ll interrogate both of them, but I think that the one who stole your bag was the one from...

Which?

Problem 41

A gray Chevrolet was parked in front of the house. Seeing the police, the owner came down to them in his pajamas.

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

The assistant, having meanwhile managed to inspect the car, 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:

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

Problem 42

The president of the company informs the investigator about a theft committed at his home.

One day, the investigator had to simultaneously interrogate three witnesses: Claude, Jacques and Dick. Their testimony 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 light without asking them a single question. Which of the witnesses was telling the truth?

Rules of syllogisms 1. A syllogism must have only three statements and only three terms. ZhG All the excursionists fled in different directions, Petrov is a excursionist, which means he fled in different directions. 3. If both premises are private statements, then the conclusion cannot be drawn. 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 a negative statement. 5. If both premises are negative statements, then the conclusion cannot be drawn. 6. The middle term must be distributed in at least one of the premises. 7. A term cannot be distributed in the conclusion if it is not distributed in the premise.