Two equivalent chess players play chess. Equivalent transformations. Simplification of formulas. How can you play this game
Section 2. Logical equivalence of formulas. Normal Forms for Propositional Algebra Formulas
Equivalence ratio
Using truth tables, it is possible to establish for which sets of truth values of incoming variables a formula will take a true or false value (as well as a statement that has an appropriate logical structure), which formulas will be tautologies or contradictions, and also establish whether two given formulas are equivalent.
In logic, they say that two sentences are equivalent if they are both true or false at the same time. The word "simultaneously" in this phrase is ambiguous. So, for the sentences "Tomorrow will be Tuesday" and "Yesterday was Sunday" this word has a literal meaning: on Monday they are both true, and on the rest of the week - both are false. For equations “ x = 2"And" 2x = 4"" At the same time "means" at the same values of the variable ". The forecasts "It will rain tomorrow" and "It is not true that it will not rain tomorrow" will simultaneously be confirmed (turn out to be true) or not confirmed (turn out to be false). In essence, this is one and the same forecast, expressed in two different forms, which can be represented by the formulas X and. These formulas simultaneously take the value "true" or the value "false". To check it is enough to compile a truth table:
| X | ||
| 1 | 0 | 1 |
| 0 | 1 | 0 |
We see that the truth values in the first and last columns are the same. It is natural to consider such formulas, as well as the sentences corresponding to them, to be equivalent.
Formulas F 1 and F 2 are called equivalent if their equivalent is a tautology.
The equivalence of two formulas is written as follows: (read: formula F 1 is equivalent to the formula F 2).
There are three ways to check whether the formulas are equivalent: 1) compose their equivalence and use the truth table to check whether it is not a tautology; 2) compile a truth table for each formula and compare the final results; if in the final columns with the same sets of variable values the truth values of both formulas will be equal, then the formulas are equivalent; 3) using equivalent transformations.
Example 2.1: Find out if the formulas are equivalent: 1),; 2),.
1) We will use the first method to determine equivalence, that is, find out whether the equivalence of formulas is also a tautology.
Let's compose the equivalent of the formulas:. The resulting formula contains two different variables ( BUT and IN) and 6 operations: 1); 2); 3); four) ; five) ; 6). This means that the corresponding truth table will have 5 rows and 8 columns:
| BUT | IN | ||||||
| 1 | 1 | 0 | 0 | 0 | 1 | 0 | 1 |
| 1 | 0 | 0 | 1 | 1 | 0 | 1 | 1 |
| 0 | 1 | 1 | 0 | 1 | 0 | 1 | 1 |
| 0 | 0 | 1 | 1 | 1 | 0 | 1 | 1 |
From the final column of the truth table, it can be seen that the compiled equivalence is a tautology and, therefore,.
2) In order to find out whether the formulas are and are equivalent, we use the second method, that is, we compose a truth table for each of the formulas and compare the final columns. ( Comment... In order to effectively use the second method, it is necessary that all compiled truth tables begin in the same way, that is, the sets of variable values were the same in the corresponding lines .)
The formula has two different variables and 2 operations, which means that the corresponding truth table has 5 rows and 4 columns:
| BUT | IN | ||
| 1 | 1 | 1 | 0 |
| 1 | 0 | 0 | 1 |
| 0 | 1 | 1 | 0 |
| 0 | 0 | 1 | 0 |
The formula has two different variables and 3 operations, which means that the corresponding truth table has 5 rows and 5 columns:
| BUT | IN | |||
| 1 | 1 | 0 | 0 | 1 |
| 1 | 0 | 0 | 1 | 1 |
| 0 | 1 | 1 | 0 | 0 |
| 0 | 0 | 1 | 1 | 1 |
Comparing the final columns of the compiled truth tables (since the tables begin the same way, we can ignore the sets of variable values), we see that they do not coincide and, therefore, the formulas are not equivalent ().
An expression is not a formula (since the "" symbol does not refer to any logical operation). It expresses attitude between formulas (as well as equality between numbers, parallelism between lines, etc.).
The theorem on the properties of the equivalence relation is valid:
Theorem 2.1. Equivalence relation between propositional algebra formulas:
1) reflexively:;
2) symmetrically: if, then;
3) transitively: if and, then.
The laws of logic
The equivalences of formulas in propositional logic are often called laws of logic... Let's list the most important of them:
1. - the law of identity.
2.- the law of the excluded third
3. - the law of contradiction
4. - disjunction with zero
5. - conjunction with zero
6. - disjunction with unit
7. - conjunction with one
8. - the law of double negation
9. - commutativity of a conjunction
10. - commutativity of disjunction
11. - associativity of conjunction
12. - associativity of disjunction
13. - distributiveness of a conjunction
14. - disjunction distributiveness
15.- laws of idempotency
16. 
; 
- laws of absorption
17. 
; 
- de Morgan's laws
18. 
- the law expressing implication through disjunction
19. 
- the law of counterposition
20.- laws expressing equivalence through other logical operations
The laws of logic are used to simplify complex formulas and to prove that formulas are equally true or false.
Equivalent transformations. Simplification of formulas
If the same formula is substituted everywhere in equivalent formulas instead of some variable, then the newly obtained formulas will also turn out to be equivalent in accordance with the substitution rule. In this way, from each equivalence, as many new equivalences can be obtained.
Example 1: If in de Morgan's law instead of X substitute, and instead of Y substitute, we get a new equivalence. The validity of the obtained equivalence can be easily checked using the truth table.
If any formula that is part of the formula F, replaced by a formula equivalent to the formula, then the resulting formula will be equivalent to the formula F.
Then, for the formula from Example 2, the following changes can be made:
- the law of double negation;
- de Morgan's law;
- the law of double negation;
- the law of associativity;
- the law of idempotency.
By the transitivity property of the equivalence relation, we can assert that 
.
Replacing one formula with another, equivalent to it, is called equivalent transformation formulas.
Under simplification formulas that do not contain signs of implication and equivalence understand an equivalent transformation that leads to a formula that does not contain negations of non-elementary formulas (in particular, double negations) or contains in aggregate fewer signs of conjunction and disjunction than the original one.
Example 2.2: Let's simplify the formula 
.
In the first step, we applied the law that transforms implication into disjunction. In the second step, we applied the commutative law. At the third step, we applied the law of idempotency. The fourth is de Morgan's law. And on the fifth - the law of double negation.
Remark 1... If a certain formula is a tautology, then any formula equivalent to it is also a tautology.
Thus, equivalent transformations can also be used to prove the identical truth of certain formulas. To do this, this formula must be reduced by equivalent transformations to one of the formulas that are tautologies.
Remark 2... Some tautologies and equivalences are combined in pairs (the law of contradiction and the law of alternative, commutative, associative laws, etc.). In these correspondences, the so-called duality principle .
Two formulas that do not contain signs of implication and equivalence are called ambivalent if each of them can be obtained from the other by replacing the signs, respectively, with.
The duality principle states the following:
Theorem 2.2: If two formulas that do not contain signs of implication and equivalence are equivalent, then their dual formulas are also equivalent.
Normal forms
Normal form Is a syntactically unambiguous way of writing a formula that implements a given function.
Using the well-known laws of logic, any formula can be transformed into an equivalent formula of the form 
, where and each is either a variable, or a negation of a variable, or a conjunction of variables or their negations. In other words, any formula can be reduced to an equivalent formula of a simple standard form, which will be a disjunction of elements, each of which is a conjunction of separate different logical variables, either with a negative sign or without it.
Example 2.3: In large formulas or for multiple transformations, it is customary to omit the conjunction sign (by analogy with the multiplication sign):. We see that after the performed transformations the formula is a disjunction of three conjunctions.
This form is called disjunctive normal form (DNF). A separate element of DNF is called elementary conjunction or constituent unit.
Similarly, any formula can be reduced to an equivalent formula, which will be a conjunction of elements, each of which will be a disjunction of logical variables with or without a negative sign. That is, each formula can be reduced to an equivalent formula of the form 
, where and each is either a variable, or a negation of a variable, or a disjunction of variables or their negations. This form is called conjunctive normal form
(CNF).
Example 2.4:
A separate element of the CNF is called elementary disjunction or a constituent of zero.
Obviously, each formula has infinitely many DNFs and CNFs.
Example 2.5: Let us find several DNFs for the formula 
.
Perfect normal forms
SDNF (perfect DNF) is a DNF in which each elementary conjunction contains all elementary statements, or their negations one time, elementary conjunctions are not repeated.
SKNF (perfect CNF) is a CNF in which each elementary disjunction contains all elementary statements, or their negations one time, elementary disjunctions are not repeated.
Example 2.6: 1) - SDNF
2) 1 - SKNF
Let us formulate the characteristic features of SDNF (SKNF).
1) All members of the disjunction (conjunction) are different;
2) All members of each conjunction (disjunction) are different;
3) Not a single conjunction (disjunction) contains both a variable and its negation;
4) Each conjunction (disjunction) contains all the variables included in the original formula.
As we can see, characteristic features (but not forms!) Satisfy the definition of duality, therefore, it is enough to deal with one form in order to learn how to receive both.
From DNF (CNF), using equivalent transformations, one can easily obtain SDNF (SKNF). Since the rules for obtaining perfect normal forms are also dual, we will analyze in detail the rule for obtaining the SDNF, and formulate the rule for obtaining the SKNF yourself using the definition of duality.
General rule reducing the formula to SDNF using equivalent transformations:
In order to give the formula F, which is not identically false, to SDNF, it is enough:
1) bring her to some DNF;
2) remove the members of the disjunction containing the variable together with its negation (if any);
3) remove all but one of the identical members of the disjunction (if any);
4) remove all but one of the identical members of each conjunction (if any);
5) if any conjunction does not contain a variable from the number of variables included in the original formula, add a term to this conjunction and apply the corresponding distributive law;
6) if the resulting disjunction contains the same terms, use Prescription 3.
The resulting formula is the SDNF of this formula.
Example 2.7: Find SDNF and SKNF for the formula 
.
Since the DNF for this formula has already been found (see Example 2.5), we will start by obtaining the SDNF:
2) in the resulting disjunction there are no variables along with their negations;
3) there are no identical members in the disjunction;
4) there are no identical variables in any conjunction;
5) the first elementary conjunction contains all the variables included in the original formula, and the second elementary conjunction lacks a variable z, so let's add a member to it and apply the distributive law:;
6) it is easy to see that identical terms have appeared in the disjunction, so we remove one (prescription 3);
3) remove one of the same disjunctions: 
;
4) there are no identical members in the remaining clauses;
5) none of the elementary disjunctions contains all the variables included in the original formula, so we supplement each of them with the conjunction:;
6) in the resulting conjunction there are no identical disjunctions, therefore, the found conjunctive form is perfect.
Since in the aggregate SKNF and SDNF formulas F 8 members, they are most likely found correctly.
Each executable (refutable) formula has one single SDNF and one unique SKNF. A tautology has no SKNF, and a contradiction has no SDNF.
1. Two equal players play a game in which there is no draw. What is the probability for the first player to win: a) one game out of two? b) two out of four? c) three out of six?
Answer: but) ; b); in)
3. Segment AB divided by a dot FROM in a ratio of 2: 1. Four points are thrown at random on this segment. Find the probability that two of them will be to the left of point C, and two to the right.
Answer:
4. Find the probability that event A will occur exactly 70 times in 243 trials if the probability of this event occurring in each trial is 0.25.
Answer: .
5. The probability of having a boy is 0.515. Find the probability that among 100 newborns, boys and girls will be equally.
Answer: 0,0782
6. The store received 500 glass bottles. The probability that any of the bottles will be broken during transportation is 0.003. Find the probability that the store will receive broken bottles: a) exactly two; b) less than two; c) at least two; d) at least one.
Answer: a) 0.22; b) 0.20; c) 0.80; d) 0.95
7. The automobile plant produces 80% of cars without significant defects. What is the likelihood that there will be at least 500 cars without significant defects among the 600 cars delivered from the factory to the car exchange?
Answer: 0,02.
8. How many times should a coin be thrown so that with a probability of 0.95 one can expect that the relative frequency of the appearance of the coat of arms will deviate from the probability R= 0.5 appearance of the coat of arms with one toss of a coin by no more than 0.02?
Answer: n ≥ 2401.
9. The probability of occurrence of an event in each of 100 independent events is constant and equal to p= 0.8. Find the probability that the event will appear: a) at least 75 times and no more than 90 times; b) at least 75 times; c) no more than 74 times.
Answer: a B C) .
10. The probability of occurrence of an event in each of the independent tests is 0.2. Find what deviation of the relative frequency of occurrence of an event from its probability can be expected with a probability of 0.9128 for 5000 tests.
Answer:
11. How many times should a coin be thrown so that with a probability of 0.6 one could expect that the deviation of the relative frequency of appearance of the coat of arms from the probability p= 0.5 turns out to be no more than 0.01 in absolute value.
Answer: n = 1764.
12. The probability of occurrence of an event in each of 10,000 independent tests is 0.75. Find the probability that the relative frequency of occurrence of an event will deviate from its probability in absolute value by no more than 0.01.
Answer: .
13. The probability of occurrence of an event in each of the independent tests is 0.5. Find the number of trials n, in which, with a probability of 0.7698, it can be expected that the relative frequency of occurrence of an event will deviate from its probability in absolute value by no more than 0.02.
Open lesson in mathematics "Bernouli scheme. Solving problems according to Bernouli and Laplace scheme"
Didactic: the acquisition of skills and abilities to work with the Bernoulli scheme for calculating probabilities.
Developing: the development of skills in applying knowledge in practice, the formation and development of functional thinking of students, the development of comparison, analysis and synthesis skills, skills of working in pairs, the expansion of professional vocabulary.
How can you play this game:
Educational: fostering interest in the subject through the practical application of theory, achieving the conscious assimilation of the educational material of students, the formation of the ability to work in a team, the correct use computer terms, interest in science, respect for the future profession.
Scientific character of knowledge: B
Lesson type: combined lesson:
- consolidation of the material passed in previous lessons;
- thematic, information-problem technology;
- generalization and consolidation of the material studied in this lesson.
Teaching method: explanatory - illustrative, problematic.
Knowledge control: frontal survey, problem solving, presentation.
Material and technical equipment of the lesson. computer, multimedia projector.
Methodological support: reference materials, presentation on the topic of the lesson, crossword puzzle.
During the classes
1. Organizational moment: 5 min.
(greeting, readiness of the group for the lesson).
2. Knowledge check:
Check the questions frontally on the slides: 10 min.
- definitions of the section "Probability Theory"
- the main concept of the section "Probability Theory"
- what events does "Probability Theory" study
- characteristic of a random event
- classical definition of probabilities
Summarizing. 5 minutes.
3. Solving problems in rows: 5 min.
Problem 1. A dice is thrown. What is the probability that an even and less than 5 points will be rolled?
Problem 2. There are nine identical radio tubes in the box, three of which were in use. During the working day, the foreman had to take two radio tubes to repair the equipment. What is the likelihood that both lamps taken were in use?
Problem 3. Three different films are shown in three halls of the cinema. The probability that there are tickets for a certain hour at the box office of the 1st hall is 0.3, at the box office of the 2nd hall - 0.2, and at the box office of the 3rd hall - 0.4. What is the probability that at a given hour it is possible to buy a ticket for at least one movie?
4. Checking the ways of solving problems at the board. Appendix 1.5 min.
5th Conclusion on solving problems:
The probability of occurrence of an event is the same for each task: m and n - const
6. Goal-setting through a task: 5 min.
A task. Two equivalent chess players play chess. What is the probability of winning two games out of four?
What is the probability of winning three games out of six (draws are not taken into account)?
Question. Think and name how the questions of this task differ from the questions of the previous tasks?
Reasoning, comparing to achieve an answer: in questions m and n are different.
7. Lesson topic:
Calculation of the probability of occurrence of an event k times out of n experiments with p-const.
If tests are carried out in which the probability of occurrence of event A in each test does not depend on the outcomes of other tests, then such tests are called independent of event A. Tests in each of which the probability of occurrence of the event is the same.
Bernoulli's formula. The probability that in n independent trials, in each of which the probability of the occurrence of an event is p (0 or Appendix 2 Bernoulli formula, where k, n are small numbers where q = 1-p Solution: Equivalent chess players are playing, so the probability of winning is p = 1/2; therefore, the probability of losing q is also 1/2. Since in all games the probability of winning is constant and it does not matter in what sequence the games will be won, the Bernoulli formula is applicable. 5 minutes Let's find the probability that two games out of four will be won: Let us find the probability that three games out of six will be won: Since P4 (2)> P6 (3), it is more likely to win two games out of four than three out of six. Find the probability that event A will occur exactly 70 times in 243 trials if the probability of this event occurring in each trial is 0.25. k = 70, n = 243 Hence k and n are large numbers. This means that it is difficult to count using the Bernoulli formula. For such cases, the local Laplace formula is applied: Appendix 3 for positive values of x is given in Appendix 4; for negative values of x use the same table and =.8. Task.
9. We compose an algorithm for solving the problem: 5 min.
10. Solving the problem with analysis at the blackboard. Appendix 5.10 min.
11. Summarizing lesson information through presentations
