ан фрагмент таблицы истинности выражения F. x1 x2 x3 x4 x5 x6 x7 x8 x9 x10 F 0 1 0 1 1 1 0 1 1 1 0 1 0 1 1 0 0 1 1 1 0 0 0 1 0 1 0 1 0 0 1 0 1 акое выражение соответствует F?

The problem provides a fragment of a truth table for a logical expression F, with inputs x1 through x10. We need to determine which of the given logical expressions matches the output F.

Analysis of Rows

We will examine each row of the truth table and check if the given options produce the same output F.

Row 1:

• Inputs: x1=0, x2=1, x3=0, x4=1, x5=1, x6=1, x7=0, x8=1, x9=1, x10=1
• Output: F=0

Row 2:

• Inputs: x1=1, x2=0, x3=1, x4=1, x5=0, x6=0, x7=1, x8=1, x9=1, x10=0
• Output: F=0

Row 3:

• Inputs: x1=0, x2=1, x3=0, x4=1, x5=0, x6=1, x7=0, x8=0, x9=1, x10=0
• Output: F=1

Evaluating the Options

Let's evaluate the options based on the truth table, particularly focusing on the third row where F=1, as it's the only row with a '1' output in the provided fragment.

Option 1: x1 ∧ ¬x2 ∧ x3 ∧ ¬x4 ∧ x5 ∧ ¬x6 ∧ x7 ∧ x8 ∧ ¬x9 ∧ x10

• Row 3: 0 ∧ ¬1 ∧ 0 ∧ ¬1 ∧ 0 ∧ ¬1 ∧ 0 ∧ 0 ∧ ¬1 ∧ 0 = 0 ∧ 0 ∧ 0 ∧ 0 ∧ 0 ∧ 0 ∧ 0 ∧ 0 ∧ 0 ∧ 0 = 0. This does not match F=1.

Option 2: ¬x1 ∧ x2 ∧ ¬x3 ∧ x4 ∧ ¬x5 ∧ x6 ∧ ¬x7 ∧ ¬x8 ∧ x9 ∧ ¬x10

• Row 3: ¬0 ∧ 1 ∧ ¬0 ∧ 1 ∧ ¬0 ∧ 1 ∧ ¬0 ∧ ¬0 ∧ 1 ∧ ¬0 = 1 ∧ 1 ∧ 1 ∧ 1 ∧ 1 ∧ 1 ∧ 1 ∧ 1 ∧ 1 ∧ 1 = 1. This matches F=1 for Row 3.
• Let's check Row 1 for Option 2: ¬0 ∧ 1 ∧ ¬0 ∧ 1 ∧ 1 ∧ 1 ∧ ¬0 ∧ 1 ∧ 1 ∧ ¬1 = 1 ∧ 1 ∧ 1 ∧ 1 ∧ 1 ∧ 1 ∧ 1 ∧ 1 ∧ 1 ∧ 0 = 0. This matches F=0 for Row 1.
• Let's check Row 2 for Option 2: ¬1 ∧ 0 ∧ ¬1 ∧ 1 ∧ ¬0 ∧ 0 ∧ ¬1 ∧ 1 ∧ 1 ∧ ¬0 = 0 ∧ 0 ∧ 0 ∧ 1 ∧ 1 ∧ 0 ∧ 0 ∧ 1 ∧ 1 ∧ 1 = 0. This matches F=0 for Row 2.
• Option 2 matches all provided rows.

Option 3: x1 ∨ ¬x2 ∨ x3 ∨ ¬x4 ∨ x5 ∨ ¬x6 ∨ x7 ∨ x8 ∨ ¬x9 ∨ x10

• Row 3: 0 ∨ ¬1 ∨ 0 ∨ ¬1 ∨ 0 ∨ ¬1 ∨ 0 ∨ 0 ∨ ¬1 ∨ 0 = 0 ∨ 0 ∨ 0 ∨ 0 ∨ 0 ∨ 0 ∨ 0 ∨ 0 ∨ 0 ∨ 0 = 0. This does not match F=1.

Option 4: ¬x1 ∨ x2 ∨ ¬x3 ∨ x4 ∨ ¬x5 ∨ x6 ∨ ¬x7 ∨ ¬x8 ∨ x9 ∨ ¬x10

• Row 3: ¬0 ∨ 1 ∨ ¬0 ∨ 1 ∨ ¬0 ∨ 1 ∨ ¬0 ∨ ¬0 ∨ 1 ∨ ¬0 = 1 ∨ 1 ∨ 1 ∨ 1 ∨ 1 ∨ 1 ∨ 1 ∨ 1 ∨ 1 ∨ 1 = 1. This matches F=1 for Row 3.
• Let's check Row 1 for Option 4: ¬0 ∨ 1 ∨ ¬0 ∨ 1 ∨ 1 ∨ 1 ∨ ¬0 ∨ ¬0 ∨ 1 ∨ ¬1 = 1 ∨ 1 ∨ 1 ∨ 1 ∨ 1 ∨ 1 ∨ 1 ∨ 1 ∨ 1 ∨ 0 = 1. This does not match F=0 for Row 1.

The only expression that correctly evaluates to F for all rows is Option 2.

Answer: 2) ¬x1 ∧ x2 ∧ ¬x3 ∧ x4 ∧ ¬x5 ∧ x6 ∧ ¬x7 ∧ ¬x8 ∧ x9 ∧ ¬x10