Question

Составить таблицу истинности для логической функции: F=(A v ¬B) v (¬A&B) ПОМОГИТЕ ПЖ!!!!

Answer 1

Ответ:

ниже

Объяснение:

вот таблица, дай коронку.

Answer 2

Truth Table for Logical Function F=(A+v+¬B)+v+(¬A&B)

To create a truth table for the logical function F=(A+v+¬B)+v+(¬A&B), we need to consider all possible combinations of the variables A and B and evaluate the function for each combination.

Let's break down the function and evaluate it step by step:

1. The expression A+v+¬B represents the logical OR operation between A and ¬B. - A is one of the variables, and ¬B represents the negation of B. - The logical OR operation returns true if at least one of the operands is true.

2. The expression (A+v+¬B)+v+(¬A&B) represents the logical OR operation between the previous result and (¬A&B). - ¬A represents the negation of A, and ¬A&B represents the logical AND operation between ¬A and B. - The logical OR operation returns true if at least one of the operands is true.

Now, let's create the truth table by considering all possible combinations of A and B:

| A | B | ¬B | A+v+¬B | ¬A | ¬A&B | (A+v+¬B)+v+(¬A&B) | |---|---|----|--------|----|------|-------------------| | 0 | 0 | 1 | 1 | 1 | 0 | 1 | | 0 | 1 | 0 | 0 | 1 | 0 | 0 | | 1 | 0 | 1 | 1 | 0 | 0 | 1 | | 1 | 1 | 0 | 1 | 0 | 0 | 1 |

In the truth table above, the columns represent the variables A, B, ¬B, A+v+¬B, ¬A, ¬A&B, and the final result (A+v+¬B)+v+(¬A&B). The rows represent all possible combinations of the variables.

To evaluate the function for each combination, we substitute the values of A, B, ¬B, ¬A, and ¬A&B into the corresponding expressions and calculate the result.

For example, when A=0 and B=0: - ¬B = 1 - A+v+¬B = 0+1+1 = 1 - ¬A = 1 - ¬A&B = 1*0 = 0 - (A+v+¬B)+v+(¬A&B) = 1+1+0 = 1

We repeat this process for all possible combinations of A and B to fill in the truth table.

Therefore, the truth table for the logical function F=(A+v+¬B)+v+(¬A&B) is as follows:

| A | B | F | |---|---|---| | 0 | 0 | 1 | | 0 | 1 | 0 | | 1 | 0 | 1 | | 1 | 1 | 1 |

Note: The truth table shows that the function F evaluates to 1 (true) for three out of the four possible combinations of A and B.